Gecikmeli Seçim Kuantum Silgisi — Etkileşimli Simülasyon

The user is asking in Turkish for an interactive simulation of the delayed-choice quantum eraser experiment in physics, with realistic optical element layout. Let me translate: "Make and run an interactive simulation of the delayed-choice quantum eraser experiment in physics. The layout of the optical elements in the experiment should be realistic." So I need to create an interactive HTML page (single file) that simulates the delayed-choice quantum eraser experiment (the famous Scully-Druhl/Walborn or the Kim et al. 2000 version). The user wants: 1. Interactive simulation 2. Realistic optical element layout 3. In Turkish language (since they asked in Turkish, I should respond in Turkish) Let me think about the delayed-choice quantum eraser experiment (Kim et al., 2000): Setup: - A laser pumps a BBO (beta barium borate) crystal via double slit - SPDC (spontaneous parametric down-conversion) creates entangled photon pairs (signal and idler) - Signal photon goes through the double slit and then to detector D0 (movable along x-axis, connected to coincidence counter) - Idler photon goes through a prism/beam-splitter setup: - After the slits, there are Glan-Thompson prisms - The idler paths encounter beam splitters BSa, BSb, BSc, and mirrors Ma, Mb - If idler goes through BSa or BSb (transmits), it reaches BSc and then detectors D1 or D2 — these provide "which-path" information (erased in a way... actually let me recall correctly) Actually let me recall the Kim et al. 2000 setup precisely: - Pump laser hits double slit, behind which is a BBO crystal. Signal and idler photons are created at the slits. - Signal photon goes up to D0 (via a lens and fiber), scanned along x to build up pattern. - Idler photon goes down. After the slits, the idler passes through: - GL (Glan-Thompson prism) to separate signal and idler? Actually the signal and idler are separated by Glan-Thompson prisms. - The idler from slit A encounters beam splitter BSa; idler from slit B encounters BSb. - If the idler is reflected by BSa or BSb, it goes to mirror Ma or Mb, then to detector D3 or D4. These detections reveal which-path information (D3 → slit A, D4 → slit B). The coincidence counts D0&D3 and D0&D4 show no interference (single-slit diffraction blobs). - If the idler transmits through BSa or BSb, it goes to BSc (50/50 beam splitter) where the paths combine, then to D1 or D2. The which-path information is "erased". Coincidence counts D0&D1 show interference fringes, D0&D2 show anti-fringes. - The "delayed choice" aspect: the idler's path (which-path or erasure) is decided AFTER the signal photon is already detected at D0. The distance to D0 is shorter than the idler path, so D0 detection happens first. Key physics: - D0 alone (all coincidences summed): no interference — sum of fringes and anti-fringes. - D0 & D1: interference fringes. - D0 & D2: anti-fringes. - D0 & D3: no interference, centered pattern corresponding to slit A. - D0 & D4: no interference, centered pattern corresponding to slit B. Now I want to make this an interactive, realistic simulation. Design considerations: 1. **Realistic optical layout**: Draw the optical table top-down view like a real optics paper diagram: - Laser at top left - Double slit + BBO crystal - Signal path going up/right to D0 on a movable translation stage - Idler path going down through BSa, BSb, Ma, Mb, BSc, to D1, D2, D3, D4 - Use proper optical symbols: beam splitters as squares with diagonal, mirrors as hatched rectangles/lines, detectors as semicircles or labeled blocks 2. **Interactive elements**: - Start/pause photon emission - Photon animation: show pump photon going to slits, splitting into signal (red) and idler (blue), traveling through the setup - Click a detector or select coincidence channel (D0&D1, D0&D2, D0&D3, D0&D4, D0 alone) - Move D0's position (slider for x position) — or auto-scan mode - Counter for accumulated coincidence counts at each D0 position - Toggle "insert BSc" (delayed choice): you can decide after the signal is detected whether to erase or not — this is the essence of delayed choice. Actually in Kim's setup, BSc is always there, but the choice happens at the quantum level per photon. But for pedagogy, we could allow toggling beam splitters BSa/BSb (remove them → all idlers go to erasure path; insert → 50/50 chance). - Show graphs: D0 coincidence histograms as a function of x position, compared with theoretical curves. 3. **Physics simulation**: - Signal photon detection at D0 at position x: probability distribution. - For entangled photons from SPDC at two slits, the state is (|signal_A, idler_A⟩ + |signal_B, idler_B⟩)/√2. - D0 detection amplitude at x: ψ_A(x) + ψ_B(x) where ψ_A and ψ_B are single-slit diffraction envelopes with phase difference φ = 2πdx/λL or similar. - When idler is detected at D3 (which-path A): probability = |ψ_A(x)|². At D4: |ψ_B(x)|². - When idler detected at D1: erasure with a particular relative phase, P ∝ |ψ_A + ψ_B|²/... hmm, actually for D1 and D2 with proper erasure: P(D1|x) ∝ |ψ_A(x) + ψ_B(x)|²/2? Let me think. In Kim's experiment, D1 shows fringes, D2 anti-fringes. The exact form: P(x, D1) = ½|ψ_A + ψ_B|² · something, P(x, D2) = ½|ψ_A − ψ_B|². Actually more carefully: The joint state |Ψ⟩ = (|A⟩_s|A⟩_i + |B⟩_s|B⟩_i)/√2. D1 receives idler amplitude from A and B with equal phase (say), so idler measurement at D1 projects onto (|A⟩_i + |B⟩_i)/√2. Then the signal collapses to (ψ_A(x) + ψ_B(x))/√2 normalized. So coincidence rate ∝ |ψ_A + ψ_B|². D2 projects onto (|A⟩_i − |B⟩_i)/√2 → |ψ_A − ψ_B|². In reality, Kim et al. had extra phases from the geometry, and D1/D2 aren't perfectly pure fringes/antifringes in the actual data — they show shifted fringes. But for the simulation, clean fringes and anti-fringes are fine and pedagogically correct. Single-slit diffraction: ψ_A(x) = sinc(β) e^{iφ/2}, ψ_B(x) = sinc(β) e^{−iφ/2}, where β = π a x/(λ L) (a = slit width), φ = 2π d x/(λ L) (d = slit separation). Then: - |ψ_A|² = |ψ_B|² = sinc²(β) — single-slit envelope - |ψ_A + ψ_B|² = 4 sinc²(β) cos²(φ/2) — fringes - |ψ_A − ψ_B|² = 4 sinc²(β) sin²(φ/2) — anti-fringes - Sum (D0 alone) = 2 sinc²(β) — no interference ✓ 4. **Animation of photons**: Show individual photon events. Each pump pulse: - Green pump photon travels from laser to slit/BBO - At BBO, splits into signal (one color, e.g., red/orange) and idler (another, e.g., blue) - Signal travels to D0, gets detected (flash at D0, count registered with x position) - Idler travels through its path (randomly chosen: reflected at BSa → Ma → D3, etc.), detected at D1–D4 - Since idler path is longer, idler detection happens AFTER signal detection → "delayed choice" - Coincidence counter pairs the events: plot the D0 x-position into the histogram of the corresponding Dk channel. 5. **Controls**: - Play/pause, speed - D0 position: manual slider or auto-scan - Select which coincidence channel to display - Show/hide theoretical curves - Toggle beam splitters (BSa, BSb, BSc) — the delayed choice: decide whether to erase which-path info. Hmm, realistic: BSa and BSb are removable. If removed, all idlers go to erasure branch (D1/D2). If inserted, 50% get which-path (D3/D4), 50% erased (D1/D2). BSc removable: if removed, then idlers that transmitted BSa/BSb go to... hmm, in Kim's setup, if BSc removed, those photons go to D2? Let me think about geometry. Kim et al. 2000 layout (from the PRL figure): - Pump laser (UV, 351.1 nm) at top. - Double slit with BBO behind it. - Signal photons (going up-left direction? Actually they're separated) — Glan Thompson prism directs signal up to lens, fiber optic, to D0 which is on a scanning stage. - Idler goes down: - From slit A: idler hits BSa (beam splitter). Transmitted → continues to BSc. Reflected → to Ma (mirror) → to D3? Hmm, actually I recall: reflected by BSa goes to Ma, then to D3? No wait. Let me recall the figure more carefully. In Kim et al. (PRL 84, 1 (2000), "Delayed 'Choice' Quantum Eraser"): The figure shows: - Iris, lens, argon ion laser - Double slit - BBO - Signal and idler separated via GT prisms - The signal goes up to D0 (movable, with lens L and fiber) - The idler goes down: - The idler from slit A: hits BSa. If reflected → Ma → D3? Or transmitted? From the paper's description: "If the idler photon is detected at D3, it is known that the signal photon passed through slit A (or B for D4). If detected at D1 or D2, which-path info is erased." Geometry: Idler from slit A hits BSa; transmitted photons continue to BSc; reflected photons go to Ma, then reflected to D3. Hmm, no — I think reflected by BSa goes to Ma, and Ma directs to D3. And idler from slit B hits BSb; transmitted continues to BSc; reflected goes to Mb → D4. Then at BSc: 50% transmitted → D1, 50% reflected → D2 (after coming from both paths). The two paths (from A via BSa and from B via BSb) overlap at BSc, erasing path info. D1 and D2 each can receive from either slit, so no which-path info. Wait, but I should double check: D3 receives only from slit A path (via Ma), D4 only from slit B (via Mb). Yes, that's right. Actually in the actual paper: "the idler photon... after reflecting from mirrors Ma/Mb..." Hmm. Let me just be careful about which is reflected/transmitted — it doesn't matter much for physics; the key topology: - Idler-A: BSa (50/50): (T) → BSc; (R) → Ma → D3. - Idler-B: BSb (50/50): (T) → BSc; (R) → Mb → D4. - BSc (50/50): receives combined idler from both A and B: (T) → D1; (R) → D2. That's the standard topology. Good. For "delayed choice" interactivity, I can add the ability to insert/remove BSa and BSb — but the real delayed choice is that the signal detection at D0 happens before the idler reaches BSa/BSb/BSc. In the simulation, we can literally show this timing: the signal photon arrives at D0 first (shorter path), THEN the idler hits the beam splitters. The user can toggle BSc (or BSa/BSb) while idler is in flight? That would be a fun interactive "delayed choice" feature: a button "silgiyi tak / çıkar" (insert/remove eraser) that can be pressed while the photon pair is in flight. Hmm, to keep it manageable but rich, I'll include: - Toggle for each beam splitter (BSa, BSb, BSc) with realistic behavior: - BSa/BSb removed → idler goes straight to BSc → erasure channel always - BSc removed → idler goes straight through to... hmm, geometry: if BSc is removed, the transmitted-through-BSc path would continue straight. In Kim's setup, D1 and D2 are positioned such that BSc sends one to D1 (reflected) and other to D2 (transmitted). If BSc removed, all go to D2 (straight). Hmm, but that changes probabilities. Actually, maybe simpler and more pedagogical: keep the topology fixed but let the user choose whether which-path info is available by toggling "quantum eraser" mode... Actually no — the delayed choice aspect is inherently about the random 50/50 at beam splitters. Each photon pair randomly "chooses" whether to reveal or erase path info. The user's "choice" is which coincidence channel to look at (D1–D4). That's actually the deep point: the choice of whether to see interference or not is made when you choose which coincidence data set to look at — after D0 detection. But an interactive element that really sells "delayed choice": allow the user to insert/remove BSa and BSb AFTER the signal photon is already registered (photon in flight). Since in the sim we control time, we can literally let the user click "BSc'yi çıkar" while idler is mid-flight, and the outcome probabilities change for that in-flight photon. That dramatizes the delayed choice. Let me design the interactivity: **Main controls:** 1. Başlat/Durdur (start/stop) photon pairs 2. Hız (speed slider) 3. D0 position mode: manual (slider x) or auto-scan 4. Coincidence channel display selector: buttons for D0 (toplam), D0∩D1, D0∩D2, D0∩D3, D0∩D4 5. Beam splitter toggles: BSa, BSb, BSc — clickable even mid-flight 6. Reset data 7. Show theoretical curve toggle **Views:** - Left: optical table diagram (SVG or canvas) with animated photons - Right: coincidence histogram(s) building up + counts + explanations **The physics detail:** Parameters: - Slit separation d, slit width a, signal wavelength λ_s, distance from slits to D0 = L. - The coincidence rate as function of D0 position x: - P(x|D3) ∝ sinc²(π a x/(λ L)) — single slit envelope (path A only) - P(x|D4) ∝ same envelope (path B only) — in symmetric geometry, identical - P(x|D1) ∝ envelope × cos²(π d x/(λ L) + θ1) — fringes - P(x|D2) ∝ envelope × sin²(π d x/(λ L)) — anti-fringes (with possible phase shift; Kim's real data had D1 and D2 both showing shifted fringes due to geometry, but sin²/cos² is the idealized clean version) - P(x) total ∝ envelope — no fringes Hmm, actually, to be more faithful to Kim's experiment: In the actual experiment, because of the geometry (the idler paths to D1 and D2 have different lengths from A and B), D1 shows fringes with phase φ and D2 shows fringes with phase φ + π (anti-fringes) — actually the paper's data: D1 coincidence shows fringes, D2 shows anti-fringes (π-shifted). Yes, the published figures show "unfolded" patterns where D1 has fringes and D2 has anti-fringes. Good, sin²/cos² is fine. Let me define parameters to make a nice visible pattern: - λ_s = 700 nm (signal photon, degenerate SPDC from 351 nm pump → 702 nm pairs; realistic!) - d (slit separation) = 0.25 mm? L (distance to D0) = 1 m? Then fringe spacing = λL/d = 700e-9 × 1 / 0.25e-3 = 2.8 mm. Scan range ±5 mm → ~3-4 fringes. Reasonable. - a (slit width) = 0.1 mm: envelope first zero at λL/a = 7 mm. So over ±5mm scan, envelope gradually falls. Fine. In Kim's experiment: d = 0.7 mm? Actually I recall slit separation 0.7 mm... not sure. Doesn't matter; I'll pick values that render nicely and state them. Let me use: λ_s = 702 nm (degenerate), a = 0.1 mm, d = 0.3 mm, L = 1.5 m. Fringe spacing = 702e-9*1.5/3e-4 = 3.51 mm. Scan range ±7 mm? Envelope zero at 702e-9*1.5/1e-4 = 10.5 mm. OK, scan ±8 mm, ~4-5 fringes visible. **Simulation mechanics:** Photon pair event: 1. Pump photon (green/violet) travels laser → slits/BBO. 2. At BBO: SPDC (show a subtle sparkle); signal photon (red) and idler photon (dark red/blue... in Kim's experiment both are 702 nm; but for visualization, color-code: signal = orange, idler = cyan. Realistically they'd both be same wavelength but different polarization... I'll color-code and note it's for visualization). 3. Signal: travels to D0 (short path). On arrival: flash + count at position x_D0. But we don't yet know the coincidence channel! The x-position gets recorded in a "pending" list. 4. Idler: travels long path through BSa/BSb → maybe Ma/Mb → D3/D4, or → BSc → D1/D2. Random outcomes with proper probabilities. 5. On idler detection: coincidence established → add count to the appropriate channel histogram at the recorded x position. Flash the coincidence counter. 6. The delay between D0 hit and idler hit is visible — that's the "delayed choice" (governed by optical path length difference; in Kim's exp, idler path ~2.3 m longer... something like that). Probabilities per pair (with all BS inserted): - P(idler → D3) = 1/4 (reflect at BSa) - P(idler → D4) = 1/4 (reflect at BSb) - P(idler → D1) = 1/4 (transmit both, then transmit/reflect at BSc... wait) Careful: idler from A transmits BSa (1/2), reaches BSc: transmits→D1 or reflects→D2? Each with 1/2. Similarly from B. So: - P(D1) = P(transmit BSa or BSb) × 1/2 = 1/2 × 1/2 = 1/4 - P(D2) = 1/4 - P(D3) = 1/4 - P(D4) = 1/4 Total 1 ✓. If BSa and BSb removed: idlers go straight to BSc: P(D1) = 1/2, P(D2) = 1/2, P(D3) = P(D4) = 0. If BSc removed: photons that would hit BSc continue straight to D2's direction? Hmm. In my layout, I'll position BSc such that: idler-A transmitted goes right toward BSc, idler-B transmitted goes up (or left) toward BSc. At BSc: one output goes to D1, other to D2. If BSc removed, the beam passes through straight → to whichever detector is straight ahead. This gets complicated. Alternative approach for interactivity: Make the toggles physical and the geometry consistent: - Layout (top-down): - BBO at center-top. Signal goes up to D0. - Idler-A goes down-left at 45°, hits BSa at 45° orientation. Transmitted (continues down-left) → to BSc location. Reflected (90° turn, now going down-right... or up-right?) → Ma → D3. - Idler-B goes down-right at 45°, hits BSb. Transmitted (continues down-right) → BSc. Reflected → Mb → D4. - BSc at bottom-center, oriented 45°. Receives beam from up-left (A) and beam from up-right (B) — these two must be collinear/overlapping for erasure! Hmm wait, that's a problem: for path erasure at BSc, the two input beams must be mode-matched (same direction). In Kim's actual setup, the geometry has the two transmitted beams converging onto BSc from... let me think. Actually in the real experiment, mirrors are used to make both paths meet at BSc collinearly. The figure shows: idler from A transmits BSa, goes to... there are extra mirrors? Let me recall the figure... Kim et al. figure: It shows the pump laser at top, double slit, BBO. Signal goes up-left to D0? And idler goes down. BSa and BSb are on the left and right. After BSa/BSb transmission, beams go down to BSc which is at bottom center. The two beams converge at BSc at an angle from left and right. Then BSc outputs go to D1 (left) and D2 (right)? Hmm, but then the two paths hitting BSc from different angles aren't collinear... Actually for the erasure to work, what matters is that each output port of BSc receives amplitudes from both slits, and the detector (with some aperture) can't distinguish. Perfect mode matching isn't essential for a schematic simulation. In the Kim figure, the two transmitted beams meet at BSc from symmetric angles (left beam going down-right, right beam going down-left), and BSc is oriented to split them: one output toward D1 (down-left?), other toward D2. Hmm, but geometrically, a single 45° beamsplitter with two inputs at ±45°... outputs would be at... Let me not overthink; the schematic in the paper: the beams from BSa and BSb travel down and inward, meet at BSc. BSc splits: one branch continues down to D2, the other reflects to the side... Let me look at it from the actual figure description (from memory of the famous figure): - Top: laser - Double slit + BBO below - Signal path: from BBO up... hmm, actually the signal and idler are emitted in slightly different directions (noncollinear SPDC), separated by Glan-Thompson prism. The signal (going up) goes to lens, fiber, D0 on translational stage. - Idler (going down): splits at the slits into two beams. Left beam → BSa, right beam → BSb. - BSa reflects down-left... The figure shows: BSa at top-left of the idler section; transmitted goes straight down to... Ma? Hmm. I recall the figure like this: The idler beams go diagonally outward-down. BSa (left) and BSb (right). At BSa: reflected beam goes down to Ma (mirror at bottom left), which reflects it to D3 (pointing up? detectors drawn as semicircles). Hmm, honestly the exact figure: Let me reconstruct from the physics: From the paper text: "The idler photon... is detected by D1, D2, D3, or D4... When the idler is detected at D3 (D4), it is known that the signal passed through slit A (B). Detections at D1 and D2 provide no which-path info." And the figure caption: "Delayed choice quantum eraser... P12 is a prism (Glan-Thompson)..." The figure: after the slits, both signal and idler from each slit... The signal and idler photons are emitted at slightly different angles. P1 (prism) directs signal up. The idler continues down. OK here's my plan for a clean, realistic-looking schematic that preserves the essential topology (this matches typical textbook redrawings of Kim's setup): ``` [Laser] | v [Slits + BBO] / \ signal / \ idler (two beams: A left, B right) v \ [D0] (movable) \ (A) (B) v v D3 <-- [Ma] <-- [BSa] [BSb] --> [Mb] --> D4 \ / (transmit) \ / (transmit) v v [BSc] / \ v v [D1] [D2] ``` Hmm wait, that has the two transmitted beams converging at BSc — beam A goes down-right, beam B goes down-left, meeting at BSc from symmetric directions. Then BSc (45°) splits each: For a beamsplitter at 45° (diagonal from lower-left to upper-right): a beam coming from up-left going down-right... hmm, I need to work out consistent geometry. Simpler approach: make the idler paths go straight down in parallel then use mirrors to bring them together. Like: - Idler A: goes down-left at 45° from BBO, hits BSa (oriented 45°): transmitted continues down-left... that runs off to the left. Not great. Alternative cleaner layout (common in redrawings): - Idler A travels down, slightly left. Idler B travels down, slightly right. - BSa on the left path, oriented 45° (diagonal "\" or "/"): - Transmit: continues down → toward BSc at bottom center. - Reflect: turns left → Ma mirror → down to D3? Hmm, wait. Let me define with coordinates. Actually, let me design a symmetric "bowtie" layout: Positions (in SVG coords, y down): - Laser: (600, 30) pointing down. - Slits+BBO: (600, 110). - From BBO: signal goes straight up? No wait, laser comes from top... Laser at top, beam goes down through slits into BBO, SPDC happens at BBO. Then signal and idler are emitted downward at small angles (non-collinear type-I... in Kim's experiment it was type-II with GT prism separation). Schematic: signal reflects off prism and goes up-right to D0; idler continues down. Actually in the Kim figure, I'm fairly confident: pump comes from top, goes through slits, BBO. Signal photons are deflected (by prism) upward back toward D0 which is up and to the side, with a lens. Idler photons continue downward. The "delayed" part: idler path is ~ meters longer. But visually, having the laser beam and signal beam overlap/parallel is messy. Many redrawings place D0 to the upper-left or upper-right of the BBO, with the signal deflected by a prism/dichroic at 90°. Let me design my layout concretely (canvas ~1000×760): - **Laser**: box at top-center-right, say at (620, 40), beam goes down. - **Iris/lens**: small circle at (620, 75) — optional detail. - **Double slit**: at (620, 110) — small icon with two slits. - **BBO crystal**: at (620, 145) — small green rectangle. - **Glan-Thompson prism (P)**: at (620, 185): deflects signal photon to the left (horizontally), idler continues down. - **Signal path**: goes left from prism at y=185 to x=250 area, then... D0 should be movable along a direction perpendicular to the signal beam — in the real experiment, D0 scans along x (perpendicular to the signal beam propagation? No...). Hmm, important physics point: D0 is at distance L from the slits, scanning transversely. The interference pattern appears in the far field of the slits. In the schematic, the signal beam after the slits spreads out (diffraction), and D0 sits on a translation stage at distance L, scanning laterally. Schematic simplification used by everyone: draw D0 on a horizontal rail, with the signal beam arriving from below/right... Honestly, the cleanest pedagogical schematic: signal photon goes up from the BBO region to D0 at top (D0 movable left-right on a horizontal rail at top of the diagram). The idler goes down through the eraser network at the bottom. This creates a nice symmetric layout, and the "delay" is visible: signal path short (reaches top quickly), idler path long (winding through bottom). But then the laser beam would coincide with the signal path... Put the laser off to the side: laser at top-left corner, beam goes right, then mirror redirects down into the slits from above? If D0 rail is at top, the pump beam crossing the rail area is ugly but manageable: laser at left (mid-height), beam goes right at y≈155, hits slits+BBO at (600,155)? Then signal continues right to D0 on a vertical rail at right? Hmm. Alternative: This is a schematic; many textbook figures do this: - Pump laser at left, beam travels right. - Double slit + BBO in the beam path. - Behind BBO, a dichroic/prism splits: signal continues straight (right) toward D0 (movable vertically on a rail at the right side — vertical scan direction = "x" axis transverse); idler is deflected down into the eraser network at the bottom. Wait, but the pump would then be blocked/need to be separated... In Kim's real setup: pump comes down, hits BBO. Down-converted light goes at angles. GT prism separates signal (goes up-left to D0) from idler (goes down). The pump continues straight down and is dumped/blocked (it's absorbed or passes through). Actually the figure shows the pump laser coming from the top, through the slits, and the BBO right after the slits; the signal and idler emerge downward; a prism redirects the signal up to D0. The transmitted pump... in reality the pump is focused and continues, but detectors have filters. For a clean schematic I'll do: Top-down optical table view: - **Laser** (box, top-left area at (140, 60)): beam goes right along y=60 to x=520. - **Mirror M0** at (520,60)? No — simpler: laser points down from top-center: **Laser box at (620, 25–55)**, beam down. - Actually, let me go with the classic layout that best matches the Kim figure: **Kim-style layout (my version):** Canvas 1100×820 (I'll make it responsive-ish, viewBox). 1. **Laser (Ar+ 351 nm)**: box at top, centered at x=560, y≈28. Emission aperture pointing down. Label "UV Lazer (351 nm)". 2. Pump beam: vertical line from (560, 48) down to slits at (560, 105). 3. **Double slit (S)**: at (560, 105): vertical bar icon with two gaps. Label "Çift yarık (d = 0.3 mm)". 4. **BBO crystal**: at (560, 140): small green rotated square/rect. Label "BBO kristali (SPDC)". 5. **Glan-Thompson prism P1**: at (560, 178): right triangle. Signal photon deflected to the LEFT horizontally; idler continues DOWN. Hmm wait — but if signal is deflected horizontally left and D0 is on a vertical rail at the left... then the signal beam after the slits should be spreading vertically (perpendicular to its propagation). D0 scans vertically. That's fine: D0 on a vertical translation stage on the left side of the canvas, at "distance L". The diffraction spreading of signal shown as a widening fan from the slits toward the left. But actually, hold on: physically, the signal photon must travel from the slit it passed through to D0. The two slits are vertically separated (since beam is vertical and slits bar is horizontal? or...). Hmm, orientation of slits: pump beam travels down; the double slit must be perpendicular to the beam, i.e., a horizontal plane... in top-down view, the slit plane is a horizontal line, slits separated horizontally! So slits A and B are side by side horizontally (left slit A, right slit B). The diffraction fan of the signal spreads horizontally. If the signal is deflected to travel horizontally left, its transverse spread direction (x) would be vertical in the drawing... no wait. Ugh, let me think about the geometry properly in top-down view: - Pump beam direction: down (in drawing) = +y. - Slit plane: horizontal line (perpendicular to pump). Slits at (560−dx, 105) and (560+dx, 105) — slit A left, slit B right, separated along x (drawing horizontal). - Signal photon after slit: travels to D0 at distance L "behind" the slits, i.e., below (further along y) — unless deflected. Its transverse position x at D0 varies along drawing-x. So D0 should scan along drawing-x (horizontal), positioned at y = 105 + L·scale below the slits. - Idler similarly travels down, with transverse x determined by which slit. So both signal and idler travel downward from the slits. To separate them, the GT prism deflects the signal sideways (90°) — say to the right — while idler continues down. Then D0 sits on a horizontal rail at the right side, scanning in drawing-x... but wait, after the 90° deflection, the signal travels in +x direction, and its transverse coordinate becomes drawing-y. Then D0 scans vertically on the right side. Hmm, that works too — and this is actually what the Kim figure does! In the Kim figure, the signal is deflected and D0 with its lens/fiber is on a stage. Actually in the real Kim figure, I believe the signal goes up-left via prism and D0 is up-left on a scanning stage, scanning diagonally/horizontally. You know what, since this is a 2D schematic, I have freedom. Cleanest option that keeps physics legible: **Option A (my choice):** - Pump laser at top-center, pointing down. - Slits (A left, B right) + BBO below it. - **Signal**: deflected by prism P1 to the RIGHT, traveling horizontally to the right, arriving at detector **D0** mounted on a **vertical translation rail** on the right side of the canvas. The "x" scan coordinate = vertical position on the rail. The diffraction fan from the two slits... hmm, but here's the subtlety: the signal photons from slit A and slit B are deflected at the prism — both get bent 90° — so slit-A signal and slit-B signal both travel right, parallel, separated vertically? No — separated by their emission angle. Hmm, this is getting confusing. Let me step back and think about what actually gets drawn for photon animation: For the animation, each photon pair event: - Pump pulse travels down from laser through slits/BBO (choose slit A or B randomly, 50/50). - SPDC at BBO: two photons born, both continue roughly along pump direction (non-collinear but schematically down). - Prism P separates: signal bends right (90°), idler continues down. - Signal photon: travels right toward D0. Its arrival point at D0 is sampled from the correct conditional distribution (given the eventual idler outcome!). Oh wait — that's the subtle point: the D0 x-position distribution depends on the idler outcome (that's the whole point: P(x|D3) is a blob, P(x|D1) is fringes). But causally, the D0 detection happens first in time. In a Monte Carlo simulation, we sample the joint distribution P(x, Dk) — sample Dk first (with its marginal probability), then sample x from P(x|Dk). But for the animation timing, we show the D0 flash first, then the idler detection, and only when the idler is detected do we add the x to that channel's histogram. This is exactly how the real experiment's coincidence logic works! The raw D0 position stream shows no interference; sorting by idler outcome reveals fringes. So animation-wise: 1. Sample idler outcome k ∈ {1,2,3,4} (or "no detection" if... no, keep it simple: always detected somewhere). 2. Sample x from P(x|k) (rejection sampling over the scan range). 3. Animate: pump → SPDC → signal to D0 (record x, flash) → idler through its path → detector Dk flash → coincidence counter clicks → add to histogram k at x. The D0 x position in the drawing: D0 is fixed on its rail at position x_current (user-controlled or auto-scan). Hmm wait — here's a design tension: **Design tension**: In the real experiment, D0 is at a fixed position x, and you count coincidences at that x for a long time, then move D0, repeat. The interference pattern is built from many D0 positions over time. For the simulation, two modes: a. **Realistic mode**: D0 at fixed x (user slider or auto-scan slowly); each photon pair either hits D0 (if sampled x ≈ current x... probability ∝ P(x_current|k)) or misses. Hmm, but then most photons "miss" and we need them to go somewhere (detector has finite size; photons that would hit elsewhere are just not detected — they're absorbed elsewhere or D0 blocks them? Actually D0 is a small detector/fiber tip; photons not hitting it are lost). Hmm, but if we do full realism, the count rate varies strongly with x (envelope), and most events are "no coincidence". That's realistic but slow/boring. b. **Fast pedagogical mode**: assume D0 magically detects every signal photon at its sampled x — but then x jumps around per event, which contradicts "D0 at fixed position". c. **Hybrid (common in sims)**: D0 has a finite aperture covering a range... no. d. **Auto-scan mode**: D0 steps through positions; at each position, accumulate N photon pairs; count coincidences at that x. Histogram builds across x. In manual mode, user drags D0 and each pair contributes to the bin at D0's current position with probability proportional to P(x|k) — so at envelope nodes, few counts. Realistic! I'll do: D0 is at position x₀ (user slider or auto-scan stepping x₀). For each photon pair: sample k, sample x_true from P(x|k) over the full range... no wait, more realistic: the signal photon's actual landing x is sampled; if |x_true − x₀| < aperture/2 (D0 has small aperture Δx), it's detected (coincidence with pending idler); otherwise "lost" (no D0 click). With aperture Δx = bin width, this works nicely: histogram bins = aperture positions. With auto-scan: sweep x₀ across range; dwell time per position ∝ enough events; histogram of coincidence counts vs x builds and converges to ∝ P(x|k)·Δx. Rate consideration: acceptance probability ≈ Δx·P(x₀|k)/∫P dx — with Δx = 0.5mm, range 16mm, envelope... average P ~ 1/16mm, so acceptance ~ 3%. That means ~30 pairs per count. With animation at, say, 10–50 pairs/sec visible... slow. Hmm. Compromise: make D0 aperture a bit generous (e.g., 0.8 mm) and allow high speed mode (many pairs per frame, animation shows only a sample). Or: in "fast" mode, skip the miss animation. Let me add a speed control from 1× to 200×; at high speeds, only animate every Nth pair (or animate a few representative ones and process the rest statistically in background). That's a good approach: "Turbo" mode computes counts without full animation. Actually simpler and quite elegant: **Dwell logic**: Each "trial" = one pump pulse → at most one pair (in reality SPDC is rare per pulse; here every pulse makes a pair for animation richness — fine). The signal photon is detected by D0 iff it lands within the aperture. Photons that miss are shown continuing past/being absorbed by a beam dump? D0 could be drawn as a fiber-lens assembly on the rail; missed photons just fly past beyond the detector into... let's just fade them out (absorbed by the enclosure). Realistic enough — detectors are inside a dark box; stray photons are absorbed. Hmm, but if most signal photons visibly miss the small D0, and D0 is on a rail, visually the photons would need to travel to various x positions — the signal beam is a fan (diffraction). We draw the fan lightly, and each photon as a dot traveling along its angle to its landing x. That's actually beautiful and educational: shows the diffraction fan from the slits, photons landing spread out, D0 only catching the ones near x₀. And with "Gözləmə"... ok let me also think about whether to make misses invisible at high speed. I'll implement speed scaling: pairsPerFrame = speed setting; animate only up to ~3 pairs' photons simultaneously (the rest contribute counts silently with correct statistics). At speed=1, animate every pair. **Idler path animation & delayed choice:** The idler from slit A (or B — wait, both signal AND idler pass through the same slit (they're entangled in path). The pair is born at slit A or slit B. So the idler also originates from slit A or B position! Hmm, right: in Kim's experiment, SPDC happens in thin BBO crystals placed right at each slit (actually two BBO crystals, one at each slit, with orthogonal optic axes for type-II... in Kim's it was a type-I... let me recall: Kim et al. used two thin BBO crystals behind the slits? The famous setup: "double slit" and BBO — the paper says a 0.1mm BBO... I believe it's placed after the double slit so both slits pump the same crystal region... but then the down-converted photons from either slit are collinear with pump... and the GT prism separates signal/idler by polarization. The path entanglement comes from both slits contributing. For the schematic, "BBO behind the double slit" is standard.) So: pair born at (560±dx_s, 140) where dx_s = slit offset. Both photons continue down, slightly diverging (non-collinear SPDC). Prism P1 at (560, 185): reflects signal to the right; idler transmits through, continues down. Wait — both signal and idler from slit A are offset to the left (x = 560−dx). The prism is an extended element (a plate), so it can intercept both. OK. But then the idler continues down along x = 560−dx (slit A) or 560+dx (slit B) — two parallel vertical beams, separated by 2dx. - Idler A (left, x=560−dx): hits **BSa** at (560−dx, y_BSa). BSa oriented 45° ("/" or "\"): - Reflected → turns right (horizontal) → travels right to **Ma** at (560+something, y_BSa) → reflects down → **D3** at bottom? Hmm, but then Ma's output goes down and would cross idler-B's path... Let me lay out carefully. Let me set up coordinates concretely (SVG viewBox 0 0 1160 860, drawing top-down): - Laser: centered x=580, y from 18 to 52 (box), aperture at bottom. - Pump beam: (580,52) → (580,100). - Double slit: horizontal bar at y=100, from x=545 to 615, with two gaps at x=566 and 594 (slit A left, slit B right). Slit centers: A=(573,100), B=(587,100)? That's too close to draw nicely. Exaggerate for drawing: slit separation drawn as, say, 60px: A at x=550, B at x=610, with the pump beam at 580 centered. Fine — schematic scale. Hmm wait, actually, I realize the pump must pass through BOTH slits (it illuminates both). So the pump beam should be wide enough (or the slit separation small): draw pump beam as a slightly diverging beam from laser covering both slits. Nice detail: laser beam drawn expanding to cover both slit gaps. - BBO: at y=130, centered x=580, a green plate ~90px wide covering both beam positions. - Pairs born at slit positions: idler-A continues down at x=550; idler-B at x=610. Signal photons: both get deflected at prism P1... but wait, the signal photons also travel down from slits through BBO to the prism. Prism P1 at y≈165, wide enough to cover x from 545 to 615. Signal from A reflects off P1 → heads right (horizontal) at y=165; signal from B → also right at y=165. But hold on — if signal from A and signal from B both end up traveling horizontally right at the same y, they'd overlap perfectly... that would itself be an eraser (they'd interfere!). That's wrong for the schematic: the signal photons from A and B should remain distinguishable in the drawing by their propagation direction until they land on D0's plane. Correct physics/drawing: after the slits, both signal photons propagate downward, spreading by diffraction. The signal from slit A and slit B overlap immediately (they're coherent!). The D0 detection position x is the transverse coordinate on the detection plane at distance L. In top-down view, the signal fan spreads along x (drawing horizontal) as it travels down. So actually the signal should travel DOWN to a D0 rail at the bottom... but that conflicts with the idler network at the bottom. OK here's the resolution used by the Kim figure itself: the signal is deflected sideways (to the left in the real figure, I think) AFTER the slits, and D0 is placed at the side at distance L with a lens (Fourier-transforming lens — actually in Kim's setup, the signal goes through lens L and fiber to D0; the lens makes D0 scan in the Fourier plane? Hmm, in Kim's experiment, D0 was scanned and measured the far-field? Let me not go down this hole — the schematic shows D0 on a translation stage at the side). For the drawing, the deflection by the prism maps the signal's transverse coordinate x to position along the D0 rail: photons emitted from the slits at different angles → after 90° deflection, they travel at different angles horizontally → they spread out vertically as they travel right → land on the vertical D0 rail at different heights. Wait no: deflection by a flat prism preserves angle. Signal traveling down-and-slightly-left (from slit A, spreading) → after reflection off a 45° surface → travels right-and-slightly-... hmm, reflection off a 45° mirror: a downward ray (0,1) → becomes (−1,0)? Let me compute: mirror surface along direction (1,1)/√2 (45°). Reflect direction d=(0,1): reflection formula d' = d − 2(d·n)n with n = (1,−1)/√2. d·n = −1/√2. d' = (0,1) − 2(−1/√2)(1,−1)/√2 = (0,1) + (1,−1) = (1,0). Yes: downward → rightward. A ray going down-left (−ε, 1) → d·n = (−ε−1)/√2·... let me just: d' = d + ( (ε+1)/√2 )·... ugh. Roughly: down-left → right-up-ish? The angular deviation flips sign in one axis: direction (−ε, 1) → (1, −ε)? Let me verify with formula: d=(−ε,1), n=(1,−1)/√2. d·n = (−ε − 1)/√2. d' = d − 2(d·n)n = (−ε,1) − 2(−ε−1)/√2 · (1,−1)/√2 = (−ε,1) + (ε+1)(1,−1) = (−ε+ε+1, 1−ε−1) = (1, −ε). So down-left → right-up. Down-right (ε,1) → (1, ε): right-down. So after the 45° prism, the angular spread maps directly: signal photons fan out vertically as they travel right. D0 on a vertical rail at the right side, at horizontal distance = L·(schematic). The landing height ∝ angle ∝ x. So: signal from slit A (going down with slight leftward angle if heading toward x<0 side...). Hmm wait, but which slit the photon came from doesn't determine its angle — the diffraction makes both slits contribute to all angles (that's interference). For the ANIMATION, we don't need to show "which slit" for the signal (quantum superposition — we can show the photon as a superposition... or just not commit visually: the signal photon is drawn as a dot starting from the BBO center region and traveling down-right per its sampled angle). Actually, nice pedagogical touch: draw the signal photon as TWO ghost dots (through slit A and slit B) that recombine — showing superposition — then a single detection at D0. Hmm, that might be overkill. Simpler: since we sampled k (idler outcome) BEFORE the animation... for D3/D4 outcomes (which-path known), we could animate the signal clearly through slit A or B; for D1/D2 outcomes, animate as a superposed wave packet. Ohh, that's actually a lovely pedagogical detail but it "reveals" the outcome too early (before the idler is detected!). Hmm, careful — pedagogically, the signal photon's behavior at D0 must be independent of the (future) idler outcome locally. If we animate differently based on k, we'd be showing "spooky action" misleadingly. Better: animate the signal always the same way (dot from BBO → prism → right → lands at sampled x on D0 rail... wait, if it lands at x within aperture, flash; else continue past and fade). The sampled x already encodes everything (fringes etc. emerge statistically). For the "through which slit" animation of the signal: just don't show it for signal (show pump through both slits, then the pair "born" — draw the SPDC as a sparkle at both slit positions? or at the sampled slit?). For the idler: which-path IS real when it goes to D3/D4 (it genuinely went through that slit). For D1/D2 the idler is also in superposition. But animating the idler: it must travel from the BBO region down through BSa/BSb. If idler outcome is D3, it physically went through slit A → travels along the A path (x=550) → BSa → reflect → Ma → D3. If outcome is D4: B path. If outcome is D1 or D2: which path did it take in the animation? Quantum mechanically it's superposed. But the drawing has it traveling along either the left or right beam... Physically, the idler photon's wavefunction splits: amplitude along path A and path B. For animation clarity, I could draw the idler as TWO ghost dots traveling both paths simultaneously (from both slits) whenever the outcome is D1/D2 (erasure: both paths contribute, that's why path info is erased), and a SINGLE dot along the known path when outcome is D3/D4. But again — this reveals the future outcome during animation... but only for the idler photon itself, which is fine! The idler's own destiny is its own; the "spookiness" is about the signal. The signal's animation must not depend on k. The idler's animation may depend on its own outcome (it's that photon's own path). Hmm, but even that is a bit questionable — when the idler is headed to BSc and then D1, before BSc it's still in superposition of A/B; drawing two ghost dots along both paths converging at BSc, then one dot to D1 — that's actually the correct quantum picture (two-path superposition). And for D3: single dot through A path. That's correct too (path A → D3 exclusively). And BSa reflection "collapses"... eh, whatever — for D3/D4, once it reflects at BSa toward Ma, path is known. So animation: - Outcome D3: single idler dot: BBO/slit A → down → BSa (reflect) → right → Ma (reflect down) → D3. Flash. - Outcome D4: single idler dot: slit B → down → BSb (reflect) → left → Mb (reflect down) → D4. - Outcome D1: two idler ghost dots: from slit A down → BSa (transmit) → down to BSc; from slit B down → BSb (transmit) → down to BSc; merge at BSc → single dot → transmitted → D1 (or reflected → D2). Wait, D1 vs D2 decided at BSc. For D1: after BSc, dot goes one way; D2: other way. - Signal: single dot from BBO region → P1 → right → D0 rail at sampled x. (Or two ghost dots from both slits converging at P1 then one dot — this shows the superposition nicely and is outcome-independent! I like this: pump dot goes through one of the slits? No — pump through both slits? Hmm, the pump photon is single and goes through both slits in superposition too! OK let me simplify the pump: draw the pump pulse as a dot that splits at the slit plane into two ghost dots (through A and B) — superposition — then at BBO each ghost creates a pair... this gets visually busy. Alternative accepted simplification (used in virtually every animation): pump dot travels to a randomly-chosen slit? That's wrong physics (which-path for pump would kill nothing though — actually pump through which slit doesn't matter for the entanglement... wait, it does: pair born at slit A or B is exactly the path entanglement source!). The pair's path entanglement |A,A⟩+|B,B⟩ comes from the pump diffracting through both slits. Pedagogical compromise: draw the pump pulse splitting into two faint ghost dots at the slit plane (superposition), and then: - If idler outcome is which-path (D3/D4): show one ghost becoming real (slit A or B), other fading (measurement via future idler detection)... but again this pre-empts the future outcome in the animation. Hmm. But honestly, this is unavoidable in any single-photon animation: you must pre-sample the outcome to animate consistently. The KEY teaching point to display prominently: "D0 tespiti idler tespitinden ÖNCE olur" (D0 detection happens BEFORE idler detection) — with a timeline. And the signal photon always looks the same (single dot, landing at x with the proper distribution); the fringes only appear when you SORT by the later idler outcome. So final animation decision: - Pump: single dot laser → slits (through the middle, it's a pulse... I'll draw it entering and at slit plane, it "splits" into two ghost dots (A and B) — labeled as superposition. Then at BBO, each ghost converts to a pair? Too busy. SIMPLER: pump dot hits slit plane; at BBO (below slits), sparkle; then signal dot and idler ghost(s) emerge. For the signal: single dot (superposition — draw it maybe as slightly translucent with a subtle double-trail? no—keep single solid dot). For the idler: - D3/D4 outcomes: single dot from the correct slit path. - D1/D2 outcomes: two translucent dots (A and B paths) merging at BSc, then single dot to D1/D2. And at the very start (pump→slits), just draw the pump dot going through the slit region centered — fine, nobody animates pump diffraction. Hmm, actually wait. There's a subtlety I should double check: when idler outcome is D1/D2 (transmitted through BSa/BSb), in the ANIMATION both ghost dots travel: ghost A goes down the LEFT path through BSa (transmitted = continues straight down), ghost B down the RIGHT path through BSb (transmitted = continues down). Then they must MEET at BSc. Geometry: left path continues down at x=550... BSc must be positioned where both beams cross. If left beam goes straight down at x=550 and right beam straight down at x=610, they never meet! Need mirrors or angled geometry. In Kim's real setup, the geometry: the idler beams from the two slits diverge (non-collinear SPDC: signal and idler emitted at ±3°, and the two slits offset...). The actual Kim figure: the idler beams from the slits travel down-and-outward (A to the left-down, B to the right-down)? Then BSa and BSb are placed out at the sides; transmitted beams continue down-inward?? A beam can't bend. Hmm — reflection at BSa: if BSa is 45°, transmitted keeps direction (down-left), reflected goes down-right (perpendicular-ish). Let me look at the actual Kim figure in my memory once more... The famous figure: laser at top; "double slit" horizontal; BBO below; then the signal goes DOWN-LEFT? and idler DOWN-RIGHT? no... I recall: the signal is deflected to the upper-left toward D0 (top-left of figure, on a stage labeled with x). The idler continues down; the figure's bottom half has a symmetric arrangement: BSa on the left, BSb on the right; Ma at bottom-left, Mb at bottom-right; D3 bottom-left corner, D4 bottom-right corner; BSc at bottom-center; D1 and D2 below-center. The idler beams: from BBO, the idler from slit A travels down-left to BSa; from slit B travels down-right to BSb. At BSa (45°, oriented like "/"): transmitted → continues down-left?? that heads to bottom-left where Ma is... Alternatively at BSa oriented "\": beam from up-right traveling down-left hits "\": transmitted continues down-left; reflected turns to down-right?? A "\" mirror (normal along (1,1)/√2... let me compute: mirror direction (−1,1)... I'll just trust: beam traveling down-left (−1,1)/√2 hitting mirror along direction (1,1)/√2 (a "/" ... no wait. Let me define: mirror line "/" has direction vector (1,1)/√2 (from bottom-left to top-right). Beam d=(−1,1)/√2 (down-left). n = perpendicular = (1,−1)/√2. d·n = (−1−1)/√2/... = (−1·1 + 1·(−1))/√2·√2 = (−2)/2 = −1. d' = d − 2(d·n)n = d + 2n = (−1,1)/√2 + √2(1,−1)/√2·... I'm messing up. d' = d − 2(d·n)n = (−1,1)/√2 − 2(−1)(1,−1)/√2 = (−1,1)/√2 + √2(1,−1)/√2·2/2... Cleaner: d·n = −1. d' = d − 2(−1)n = d + 2n = (−1,1)/√2 + 2(1,−1)/√2 = ((−1+2), (1−2))/√2 = (1,−1)/√2 = up-right. Hmm, that's reflection back up-right — a "/" mirror reflects a down-left beam to... yes physically: a beam moving down-left hitting a "/"-oriented mirror bounces to... imagine a "/" wall; a ball rolling toward down-left hits it and bounces to down-... Let me think physically: "/" mirror. Light coming from the upper-right heading to lower-left (direction down-left). It hits the mirror face (which faces up-right). It reflects to heading... The mirror "/" makes a 45° angle; incidence direction (−1,1)/√2; the mirror swaps the components along its normal. Components: along mirror (1,1)/√2: d·m = (−1+1)/2·... = ((−1)(1)+(1)(1))/2 = 0. Oh! The beam is perpendicular to the mirror surface — it reflects straight back up-right. Right — down-left is exactly anti-parallel to "/"... no wait, "/" direction is (1,1) (up-right). Beam (−1,1) is down-left. Is (−1,1) parallel to (1,1)? No: (−1,1) vs (1,1) — not parallel; (−1,1) is perpendicular to (1,1): dot = −1+1 = 0. Yes! So beam is along the normal → reflects straight back. Bad example. Let me redo: beam traveling straight DOWN: d=(0,1). "/" mirror, m=(1,1)/√2, n=(1,−1)/√2. d·n = −1/√2. d' = d − 2(−1/√2)(1,−1)/√2 = (0,1) + (2/2)(1,−1) = (0,1)+(1,−1) = (1,0) → travels RIGHT. So "/" mirror turns down → right. And beam traveling down-RIGHT d=(1,1)/√2 hits "/" → reflects to d' = d−2(d·n)n; d·n = (1−1)/√2=0 → passes unchanged?? No: d·n=0 means beam along mirror — grazing. Right, (1,1) IS the mirror direction. OK so: - "/" mirror: down→right, left→up? d=(−1,0): d·n = −1/√2; d' = (−1,0)+(1,−1) = (0,−1) → up. Yes: left→up. - "\" mirror (m=(1,−1)/√2, n=(1,1)/√2): down→left: d=(0,1), d·n=1/√2, d' = (0,1) − (1,1) = (−1,0) ✓. right→up: d=(1,0), d·n=1/√2, d'=(1,0)−(1,1)=(0,−1) ✓. Beam splitters: same as mirrors but 50% transmit (unchanged) + 50% reflect. Now let me design the idler network with clean geometry: Slits at top: A at x=550, B at x=610, y=100. BBO at y=130. Prism P1 at y≈170 deflects signal to the right. Idler continues down. To make the A and B idler beams CONVERGE toward BSc (needed for erasure), I can tilt their directions: idler A born at (550,130) traveling down-RIGHT at angle (i.e., direction (sin θ, cos θ) with θ>0 to the right); idler B born at (610,130) traveling down-LEFT. They cross at bottom-center! Where they cross, place BSc. And along the way, place BSa in path A and BSb in path B: - Idler A: from (550,130) going down-right. BSa placed at (585, 250) oriented "/"? Beam direction (≈ (0.25, 0.97) down-right-ish). Hmm, for a clean 90° reflection, BS should be at 45° to the beam. If the beam is nearly vertical going down with slight rightward tilt, a "/" splitter reflects it to nearly-horizontal right. So: - BSa at (590, 260), oriented "/" (surface facing up-left... the "/" line). Reflect: beam A → travels RIGHT (horizontal) to Ma at (740, 260) oriented... beam traveling right hits "\" → reflects UP? No, we want it to go DOWN to D3. Beam traveling right hits "/" mirror: d=(1,0), "/" n=(1,−1)/√2, d·n=1/√2, d'=(1,0)−(1,−1)=(0,1) → DOWN ✓. So Ma at (740,260) oriented "/" reflects down → D3 at (740, 380)? Hmm wait, but that region (x=740) is to the right — under the signal path region? Signal travels right along y≈170 to D0 rail at x≈1000. D3 at (740, 380+) is below that; OK no collision as long as y differs. But visually the signal beam line at y=170 crossing... the idler-A reflected segment at y=260 horizontal from x=590 to 740 — fine, no overlap. Hmm wait, but there's an issue: the reflected beam from BSa travels right and crosses idler-B's path (B goes down-left from x=610 toward bottom-center)! Idler B path: from (610,130) down-left toward, say, (560, 350) — crosses x=590..? yes around y≈180-ish at x≈600. And the horizontal reflected beam at y=260 from x=590→740: idler B at y=260 is at x ≈ 610 − (260−130)·(50/220) ≈ 610−30 = 580. So the horizontal beam from BSa starts at x=590 going right; idler B at y=260 is at x≈580 — just misses (left of BSa start). Close crossing though. In optics diagrams crossing beams are normal and fine (beams cross in free space all the time). Actually, hmm, wait: is it OK that the reflected beam from BSa passes through where BSb is? BSb is on path B: place BSb at (580, 260)?? That's exactly where idler B is at y=260 — but that's ON the reflected beam from BSa (which spans x=590→740 at y=260)! Collision of elements. Let me adjust geometry to be cleaner. Cleaner symmetric design: make the two idler beams travel STRAIGHT DOWN (parallel, at x=550 and x=610) to BSa/BSb, then use the beam splitters + mirrors to route: - BSa at (550, 300), oriented "/": - Reflect down→RIGHT: horizontal beam at y=300 going right, from x=550 to Ma at (760, 300), Ma oriented "/" reflects down → D3 at (760, 430). - Transmit: continues DOWN from (550,300) toward... BSc. - BSb at (610, 300), oriented "\": - Reflect down→LEFT: horizontal at y=300 from x=610 to Mb at (400, 300), Mb "\" reflects down → D4 at (400, 430). - Transmit: continues down at x=610 toward BSc. - BSc: needs to receive BOTH the transmitted A beam (x=550 going down) and transmitted B beam (x=610 going down) — but they're parallel, never meet! Problem again. Fix: angle the transmitted paths. Option: after BSa/BSb transmission, the beams continue down but we add the convergence via the INITIAL emission angle: idler A emitted down-right, idler B down-left (non-collinear-ish SPDC geometry). Then: - Idler A: from (550, 130), direction down-right: reaches BSa at (600, 280) [Δx=+50, Δy=150, angle ≈ 18°]. BSa oriented to reflect to horizontal-right: beam direction (0.316, 0.949); a mirror at 45° to this... The reflected direction should be roughly (0.949, −0.316) (perpendicular). Hmm, that's up-right, not horizontal. To send it exactly horizontal right, the splitter orientation must bisect. Fine — I'll just draw the splitter square rotated appropriately and route the reflected beam: from BSa at (600,280) going right, slight up-slope? Messy. You know what — let me use the actual Kim topology with straight-down parallel beams and a classic trick: place BSc in ONE of the paths and use a mirror to bring the other path into it. Like: - Idler A: down at x=550 → BSa (550, 280) "/" : reflect→right to Ma (720,280) "/" → down to D3 (720, 420); transmit→down to BSc? - Idler B: down at x=610 → BSb (610, 280) "\": reflect→left to Mb (440,280) "\" → down to D4 (440,420); transmit→down at x=610... - Now route B's transmitted beam into BSc which sits in A's transmitted path? They're parallel (x=550 and x=610)... Add a fold mirror Mc: B transmitted → down to Mc at (610, 360) oriented "/"? beam down→right: from (610,360) going right to... BSc at (700, 360)? And A transmitted → down at x=550 passes y=360... doesn't reach BSc at (700,360). Need A's transmitted beam also folded: another mirror? This is getting complicated but it's REALISTIC — real experiments have fold mirrors! But Kim's original doesn't have extra fold mirrors for this; instead the geometry works because the idler beams from the two slits are not parallel — they're actually converging or diverging from the slit pair, and BSc is placed where... Hmm hold on. Let me think about the actual Kim geometry once more. In the real experiment: the pump is focused onto the slits; SPDC happens at BBO right behind the slits. The down-converted pairs are emitted in a cone. The GT prism separates signal (one polarization) and idler (other) — actually type-II SPDC emits them at different cone angles. The signal from BOTH slits goes toward D0's collection lens (the lens + fiber collects a small angular range — and the fiber tip position/scan... in Kim's exp, D0 with lens is scanned along x). The idler from both slits goes to the lower half. The key: the idler beams from slit A and slit B both propagate into the lower region; since the slits are only 0.7mm... wait actually! In the REAL experiment the slit separation is small (sub-mm) and the idler optics are meters away — the beams from A and B heavily overlap everywhere downstream. The beam splitters BSa and BSb aren't hit by "A beam" and "B beam" separately — wait, no, they must be! D3 receives only slit-A light... OH WAIT. I remember now! In Kim's experiment, the two slits' idler beams are separated by... the figure clearly shows two distinct beams going down from the two slits to BSa and BSb separately. The slit separation in the figure is exaggerated (schematic). The real separation: the figure shows them diverging downward-outward: slit A's idler goes down-LEFT to BSa (left side), slit B's idler goes down-RIGHT to BSb (right side). Then: - BSa (left): reflected → down to Ma → to D3? And transmitted → continues down-right?? toward the center bottom... - BSb (right): reflected → down to Mb → D4; transmitted → down-left toward center bottom... - The two transmitted beams CONVERGE at BSc (bottom center)! YES — that's the geometry! The idler beams diverge from the slit pair (like a fan), BSa and BSb intercept them at the outer edges, and the transmitted beams continue diverging... wait no, if they diverge from the slits (A going down-left, B going down-right), transmitted beams continue down-left and down-right — they diverge further, never meet. Unless mirrors redirect. Hmm, alternatively the figure shows the idler beams going down and CONVERGING (crossing) — slit A's idler goes down-RIGHT, slit B's idler goes down-LEFT, crossing below the BBO. Then BSa is placed on the LEFT part... no wait, slit A's beam going down-right would be on the right side... Let me try to actually remember the figure... I picture it now: the figure has the laser at top center. Double slit. BBO. Then a prism (triangle) below-left deflects the SIGNAL up-left to D0 (D0 drawn at upper-left with a stage arrow). The IDLER continues down. The two idler beams (from A and B) go DOWN-DIVERGING (A to down-left, B to down-right). BSa is on the LEFT beam, BSb on the RIGHT beam. At BSa: REFLECTED beam turns to down-RIGHT?? or... The reflected beams must go to D3/D4. I believe D3 is bottom-left and D4 is bottom-right, with Ma and Mb redirecting. And the TRANSMITTED beams continue down-left and down-right... then there are additional mirrors? Hmm, no — I recall the figure showing BSa transmitted continuing DOWN (not down-left)... Honestly, I've spent enough on recalling the exact figure. Let me just design my own clean, physically-consistent, realistic topology (it will match the Kim topology in spirit and in labels): **My layout (final):** Drawing region: viewBox 0 0 1200 900. Top-down view. - **Laser**: box centered (600, 40), width 90, height 36. Label "UV Pompalama Lazeri (351 nm)". Beam exits bottom. - **Iris**: small double-line at (600, 88) — detail, optional. Skip or include tiny. - **Çift yarık (double slit)**: plate at y=105, x from 520 to 680: draw as a horizontal bar with two vertical gaps at x=560 (A) and x=640 (B). Labels "Yarık A", "Yarık B". Slit separation drawn 80px. - **BBO**: green plate at y=135, x 530–670, label "BBO (SPDC)". - **Prism P (GT)**: at (600, 178): draw a right-triangle prism spanning x 540–660 oriented to reflect the SIGNAL to the right (horizontal). Actually, both signal photons (from A and B paths) travel down through the BBO and hit the prism; the prism (large enough) reflects them all to the right along y=178. Idler transmits through the prism and continues down. Physically: GT prism separates by polarization (type-II SPDC: signal & idler have orthogonal polarizations). The prism is drawn as a wide 45° plate. Label "Glan-Thompson prizması (sinyal/idler ayırımı)". Wait, hmm — problem: the signal photons from slit A and slit B, after reflecting off the same 45° surface, both travel right along y=178 — they'd overlap exactly and lose their distinguishability... but as discussed, that's FINE and actually correct: signal photons from A and B are coherent and DO overlap (they must, to interfere at D0!). The distinguishability (which-path info) is carried by the IDLER, not the signal. The signal fan: photons leave the slits at various angles; the prism reflects each; a photon that left heading down-left reflects to right-up... wait I computed earlier: down-left → (1, −ε): right and UP slightly. Down-right → (1, +ε): right and down. So the signal beam after the prism is a fan spreading vertically as it travels right. The vertical spread ∝ tan(angle)·distance. D0 rail: vertical line at x=1060. The fan from the prism (y=178, x≈660 exit) to x=1060: spread over, say, ±150px vertically for the full scan range. D0 moves along this rail between y=40 and y=330 or so. Hmm, but wait: is it OK that the prism is in the near field (right after slits, before diffraction develops)? The far-field pattern forms at D0's plane... In Kim's experiment, the signal goes through a lens L (Fourier lens) then fiber to D0 — the lens makes the far-field (momentum) pattern appear at the fiber plane regardless of distance. I'll add a **lens L** in the signal path: a vertical ellipse at (860, 178), label "Fourier merceği L (f = 1.5 m)". Then D0 at x=1060. This is realistic (Kim had exactly this: lens + fiber-coupled D0 on scanning stage). The scanning direction: vertical in drawing (perpendicular to signal beam) ✓ matches physics (transverse x) ✓. Also add **beam dump for pump?** The pump passes through the BBO and prism... the pump (351nm) would also reflect off the GT prism? In reality, the pump is separated by dichroic mirrors/filters. Schematic: after the prism, the pump continues down with the idler... and gets absorbed by a beam block. Add small "pomp bloğu (beam dump)" below the prism at (600, 230)? But the idler beams pass through that region... Put the beam dump BETWEEN the two idler beams? The idler beams diverge from slit positions — pump goes straight down the center (x=600) between them. Beam dump at (600, 240): a small black hatched block. Nice realistic detail! The pump beam visually stops there. - **Idler beams**: from BBO: idler A from (560, 135) traveling down-LEFT? or down-RIGHT? Decide: I want the two transmitted beams to CONVERGE at BSc. So idler A (from x=560) should travel down-RIGHT, idler B (from x=640) down-LEFT. They cross at x=600 below. Place: - **BSa** on idler-A path: idler A direction: from (560,135) to (640, 330)? that's slope Δx=+80, Δy=+195, angle ≈ 22° from vertical. Place BSa at (620, 300) — hmm, wait, but then the reflected beam at BSa: we want it to go to Ma then D3. BSa must reflect the beam (which travels down-right at ~22°) into roughly horizontal or downward direction... Hmm, let me reconsider: maybe have idler beams travel down-DIVERGING (A: down-left, B: down-right) — more natural fan from slits (each slit emits a cone; schematic: A's idler goes down-left, B's down-right — wait, but actually each slit emits idler in ALL downward directions; the schematic picks the ray that reaches its BS). Then: - Idler A: from (560,135) down-left to BSa at (470, 320). - Idler B: from (640,135) down-right to BSb at (730, 320). - At BSa: transmit → continues down-left → goes off toward bottom-left... then place a mirror to fold it toward center? OR reflect → down-right toward center-bottom where BSc is! IDEA: Swap the roles! Let REFLECTED at BSa/BSb be the ERASURE path (routing to BSc at bottom center) and TRANSMITTED be the which-path path (going to D3/D4 directly below). Hmm, but physically it doesn't matter which port is which. Let's see the geometry: - Idler A arrives at BSa (470, 320) traveling down-left (direction ≈ (−0.38, 0.92)). BSa oriented 45° such that reflection sends it roughly horizontal-RIGHT: reflected direction should be ≈ (0.92, 0.38) (down-right-ish) — the mirror bisects. Place BSc at bottom-center (600, 430): beam from BSa (470,320) traveling right-down reaches (600, 430)? Δ=(130,110) — direction (0.76,0.64) — plausible with proper mirror angle. Similarly idler B at BSb (730,320) reflected left-down to BSc (600,430). The two beams CONVERGE at BSc ✓✓. Then BSc (50/50): one output down-left → D1 at (480, 560); other down-right → D2 at (720, 560). And the two inputs at BSc are symmetric ✓. - Transmitted at BSa: continues down-left from (470,320) → to Ma? We want D3 somewhere bottom-left. Transmit → down-left to Ma at (410, 430), Ma reflects → down to D3 at (410, 560)? Hmm wait, Ma reflecting a down-left beam to straight down: mirror at 22.5°-ish. OK fine — or simpler: transmitted continues down-left to D3 DIRECTLY at (390, 520)? But Kim's figure has Ma and Mb mirrors. Keep Ma/Mb for realism: - BSa transmit: (470,320) → down-left → Ma at (400, 425) → reflect down → D3 at (400, 545). - BSb transmit: (730,320) → down-right → Mb at (800, 425) → reflect down → D4 at (800, 545). - BSc outputs: down-left → D1 at (500, 560)? and down-right → D2 at (700, 560)? Let me place: BSc at (600, 440) oriented like "—"? The two inputs come from up-left and up-right at ±40°-ish; outputs go down-left and down-right at ∓40°-ish — a symmetric "X" crossing at BSc. That's a classic-looking beamsplitter junction! BSc drawn as a small square rotated 45°... The output directions: input A direction at BSc: from BSa to BSc: (130,110) normalized (0.76,0.64); input B: (−0.76,0.64). Outputs: D1 direction (−0.76,0.64)?? that equals input B reversed... Outputs should be, say, (−0.76, 0.64) → to D1 at (600−130, 440+110) = (470, 550); and (0.76,0.64) → D2 at (730,550). Wait but those directions are the same as the input directions — i.e., BSc just "passes through" both?! The junction: input A arrives going down-right; output D2 also goes down-right — so BSc transmits A's beam to D2. Input B arrives going down-left; output D1 goes down-left: BSc transmits B's beam to D1. Then what does BSc reflect? For a beam splitter with inputs from up-left and up-right (symmetric about vertical), a horizontal-ish splitter surface: input A (down-right, dir (0.76,0.64)) hitting a splitter whose surface is horizontal (—, normal vertical): transmit → continue down-right → D2 side ✓; reflect → flip vertical component → (0.76,−0.64) → up-right — no. Hmm, that's wrong. For the outputs to be down-left and down-right with BOTH inputs contributing to BOTH outputs, the splitter surface should be... Let me think of the classic Mach-Zehnder final beamsplitter: two input beams converge at ±θ from above onto a splitter; outputs leave at ±θ downward. Splitter surface: vertical line "|" (normal horizontal): input A arrives going down-right (0.76,0.64): transmit → continues down-right ✓ (output to lower-right = D2). Reflect → flip horizontal → (−0.76,0.64) → down-left ✓ (output to lower-left = D1). Input B arrives going down-left (−0.76,0.64): transmit → down-left → D1 ✓; reflect → (0.76,0.64) → down-right → D2 ✓. So BSc is a VERTICAL splitter surface (drawn as a vertical line/square, or rotated square) at (600,440), with inputs converging from up-left (BSa side) and up-right (BSb side), outputs to down-left (D1) and down-right (D2). Both slits' amplitudes reach both D1 and D2 ✓ erasure ✓. Wait, but hold on: BSc as a vertical surface with beams crossing at ±40°: input A transmits to D2 and reflects to D1; input B transmits to D1 and reflects to D2. Both outputs get both inputs ✓. PERFECT. And physically that's a plate beamsplitter standing vertical in the drawing — realistic (a plate BS at normal-ish incidence... at 40° incidence, fine). Hmm, however: for nice visuals, beamsplitter symbols are usually drawn as squares with a diagonal line, at 45° incidence. My BSc with vertical surface and ±40° beams — draw it as a small square rotated so its face is vertical. OK. But hmm, wait: now the WHICH-PATH detectors D3/D4 get the TRANSMITTED beams and the ERASURE happens on REFLECTED beams at BSa/BSb. In Kim's original, I believe reflected at BSa → Ma → D3 (which-path) and transmitted → BSc. Doesn't matter physically — I'll keep my geometry but let me double check the erasure condition: for erasure at D1/D2, the two paths (A via BSa-reflection → BSc-transmission → D1, and B via BSb-reflection → BSc-reflection → D1) must be indistinguishable — ✓ both reach D1. Good. Actually hmm, wait. Let me reconsider — there's an even cleaner classic layout that many textbook redrawings use, and I want maximum clarity. Alternative: idler beams travel straight down in parallel (A at x=550, B at x=650); BSa at (550,300) and BSb at (650,300): - BSa "/"-oriented (45°): reflect down→RIGHT: horizontal beam at y=300 going right... to Ma at (800,300)? then Ma "/" reflects down → D3 at (800, 450). But this horizontal beam CROSSES the B path (x=650) at y=300 — where BSb is! Collision again (BSb is at (650,300)). Argh. Place BSb at different y: BSa at (550, 270), BSb at (650, 340). Then A-reflected horizontal at y=270 crosses x=650 above BSb ✓ no collision; goes to Ma (820, 270) → down → D3 (820, 440). B-reflected horizontal at y=340 going LEFT crosses x=550 below BSa ✓; goes to Mb (380,340) → down → D4 (380, 440)... wait D4 on the left? I wanted D3 on A-side (left)... A is at x=550 (left slit) but its reflected beam goes RIGHT to D3 at x=820 (right side). And B (right slit, x=650) → D4 at x=380 (left). So D3 ends up on the right, D4 on the left — crossed. Could flip: BSa "\" reflect down→LEFT... BSa at (550,270) "\": reflect down-beam to LEFT: horizontal at y=270 going left to Ma (380,270) → down → D3 (380,440) ✓ D3 on left (same side as slit A ✓). BSb at (650,340) "/": reflect to RIGHT: to Mb (820,340) → down → D4 (820, 440) ✓ same side as B ✓. - Transmitted: A continues down at x=550; B continues down at x=650. They're parallel — need them to meet at BSc. Add fold mirrors? OR: tilt the transmitted beams using the splitter orientation! When a beam passes through a PLATE beamsplitter at 45° incidence, it refracts... a thin plate deflects slightly (and lateral shift) — negligible. Hmm. OK so parallel transmitted beams fundamentally can't meet without extra folds. The converging-fan layout (my earlier design) avoids extra mirrors. Let me also double check the fan layout's beam crossings: - Idler A: (560,135) → (470,320): passes region... crosses the pump beam (x=600 vertical)? A goes down-LEFT from x=560 — never crosses x=600. Idler B: (640,135)→(730,320): down-right, never crosses x=600 ✓. But both CROSS the signal beam region? Signal travels right at y≈178 from x≈680. Idler A at y=178 is at x ≈ 560−(43/185)·80·... ≈ 560 − (178−135)/185·80 ≈ 560−18.6 = 541 (left of prism start 540? borderline). Prism spans x 540–660 at y≈178. Idler A passes through the prism's left edge region?? The idler transmits through the prism — that's fine actually (GT prism transmits idler ✓ by design!). Idler B at y=178: x ≈ 640+18.6 = 659 — right edge of prism ✓ transmits. Good — actually both idlers DO pass through the prism (transmitted), consistent with the GT-prism story ✓. But wait — then the idler beams bend at the prism? No — they transmit straight through (schematically ignore refraction) ✓. And the pump beam: laser at x=600 goes down through slits (y=105), BBO (y=135), prism (y=178, transmitted — GT prism reflects signal polarization, transmits idler AND pump? whatever — schematically pump continues down), then beam dump at (600, 230). But the idler beams at y=230: A at x≈560−(95/185)·80 ≈ 560−41 = 519; B at x≈681. Beam dump at x=600, width ~30px — no collision ✓. Hmm wait, actually now the signal: signal photons reflect off the prism to the right. But which signal? Both A and B signal photons hit the prism and reflect right ✓. The signal FAN: from the slit pair, photons leave at angles ±θ; after prism reflection, travel right at angles ∓θ... The fan after the prism: spreads vertically around y=178. Lens at (880, 178). D0 rail at x=1080, vertical. The fan between prism and lens: ±θ·(880−660) px. With θ_max corresponding to scan range ±8mm... The mapping: x_mm ∝ tanθ. After lens (Fourier lens): all photons arriving at angle θ focus to position x_f = f·tanθ at the focal plane. D0 at x=1080 ≈ focal plane (f drawn as 200px). So D0's y position maps to tanθ·200px. Scan range ±8mm ↔ ±150px on the rail: y from 28 to 328. Rail at x=1080 from y=20 to y=340. Hmm, wait — but the photons must visibly travel along their angled path from the prism, through the lens, converging... With a Fourier lens, parallel rays at angle θ converge to the focal point. Since each photon has a specific θ, after the lens it heads to its focal point (1080, 178 + tanθ·200). Before the lens it travels at angle θ (from prism exit (660,178): at lens x=880: y = 178 ± tanθ·220 — for tanθ = 0.75: y = 178±165 — fan spread ±165px at the lens — lens must be tall (draw ellipse height ~340px? that's huge). Alternatively make the pre-lens propagation shorter or scale smaller. Let me scale: rail half-range 140px; lens at distance 220 from prism: pre-lens spread = 140·(220/200)... hmm if lens focal length (drawing) = 200px and lens-to-rail distance = 200px ✓, then tanθ_max = 140/200 = 0.7, pre-lens spread at lens = 0.7·220 = 154px → lens height ~340px. A Fourier lens collecting a big fan IS big — realistic (high NA lens). Draw the lens as a tall thin ellipse (like a biconvex seen from top = a circle-ish region? top-down view of a lens = a line segment (its aperture) — actually top-down, a lens is drawn as a thin line/rectangle with small width). In top-down optical schematics, lenses are drawn as short segments or small ellipses. A tall thin rounded rectangle works. Hmm, actually simpler: skip the lens and just put D0 far enough with the fan — "far field" L. Kim's setup had a lens though. I'll include the lens for realism — it also justifies D0 being close. Let me place: prism at x=660 exit; lens L at x=830 (draw as vertical rounded bar, height ~320, at y-center 178); D0 rail at x=1030, y range 20–340. Then tanθ mapping: focal length drawing = 1030−830 = 200 ✓. Pre-lens spread at x=830: tanθ·170. Wait, I need to double check the physical consistency of what x means: In Kim's exp, D0 position x (mm) vs coincidence rate — the fringe period ~0.4mm? Whatever — my sim: x in mm from −8 to +8, fringe spacing λL_eff/d... with the Fourier lens, the pattern at the focal plane: x_f = f·tanθ where θ is emission angle; far-field interference of double slit: intensity ∝ sinc²(πa sinθ/λ)·cos²(πd sinθ/λ). With x_f = f·tanθ ≈ f·sinθ: P(x) ∝ sinc²(πax/(λf))·cos²(πdx/(λf)). So effective L = f = 1.5m (label lens f=1.5m — wait, that gives fringe spacing λf/d = 702e-9·1.5/0.3e-3 = 3.5mm ✓ as computed before). So final physics params: - λ = 702 nm (both signal & idler, degenerate type-II SPDC from 351nm pump ✓ realistic) - slit width a = 0.1 mm - slit separation d = 0.3 mm — hmm wait, with d=0.3mm and a=0.1mm we get 2-3 fringes under the envelope: envelope first zero at λf/a = 10.5mm; fringe spacing 3.5mm → 3 fringes to first zero. Over ±8mm scan: ~4-5 fringe periods, envelope dropping to sinc²(8/10.5·π)... sinc(0.76π)≈... sin(0.76π)/(0.76π) = 0.68/2.39 = 0.286 → squared 0.08. Hmm, edges very dim. Maybe reduce scan to ±7mm and/or a=0.08mm (envelope zero at 13mm; at 7mm: sinc(0.54π)=sin(1.69)/1.69=0.993/1.69=0.59→0.35). Eh, a bit dim at edges but that's realistic diffraction. Alternatively d=0.25mm, a=0.07mm, f=1.2m: fringe = 702e-9·1.2/2.5e-4 = 3.37mm; envelope zero at 12mm. Scan ±7mm: sinc(7/12·π)= sinc(1.83 rad)= sin(1.83)/1.83 = 0.967/1.83 = 0.53 → 0.28 at edges. OK let me go: λ=702nm, d=0.25mm, a=0.07mm, f=1.2m, scan ±7mm, bin width 0.5mm → 28 bins. Fringe spacing 3.37mm ≈ 6.7 bins per fringe — decent resolution. Hmm, maybe bins of 0.35mm → 40 bins, ~10 bins per fringe — smoother curves. D0 aperture = bin width (0.35mm)... acceptance probability = 0.35·P(x)/∫P dx over 14mm ≈ 0.35/14 ≈ 2.5% average (for envelope~1). So ~40 pairs per coincidence count — need many pairs for a good histogram: with, say, 3000 detected pairs per channel... 3000·40 = 120k pairs?? At 60fps with speed... too slow if animated per-pair. Hmm, this suggests the "aperture + miss" realism might make buildup too slow. Options: 1. Bigger aperture (D0 collects a range of x — fiber with lens has some acceptance width; realistic ~0.4mm?). 2. Speed mode: process many pairs per frame statistically (only animate a few). At speed 500 pairs/frame·60fps = 30k pairs/s → 120k pairs in 4 seconds ✓. Fine! The speed slider goes 1× to ~1000×. At high speed, counts accumulate fast; animation shows a subsample of photon flights (e.g., up to 3 concurrent pairs in flight, throttled). Also, maybe reconsider: do we even show "missed" photons? If D0 aperture = 0.35mm and drawing maps ±7mm → ±140px, aperture = 7px — tiny. Photons landing outside → they fly past D0 and fade. At high speeds we won't animate misses (too many) — only animate detected ones occasionally. Actually here's a cleaner approach: **animate only events that end in detection at D0** (coincidence events) + optionally show the fan. But then the animation under-represents misses... For pedagogy, what matters: (1) D0 clicks happen FIRST at position x₀ (D0 is at a fixed position — clicked x is always within the aperture around x₀!); (2) later idler clicks at Dk; (3) histogram builds at x₀. Hmm wait — this is the REALISTIC mode and it's actually the most honest: D0 is at x₀; every animated pair: signal flies to D0 (lands within aperture — since we only animate detected pairs, draw the landing at x₀±small jitter within aperture) and clicks; idler flies its path and clicks at Dk; coincidence counter increments histogram[x₀ bin] for channel k. The relative RATES across channels come from P(x₀|k)·aperture — so as D0 scans, histogram heights follow the curves ✓. And "misses" just aren't animated (they're lost photons; mention in a note: "D0'ya gelmeyen sinyaller kaybolur (gösterilmiyor)" or show occasional ghost misses at low speed). I think at speed 1×, animate everything including misses (educational: shows D0 catching only a fraction); at higher speeds animate only detections. Hmm, but even at 1×, 97% misses means boring wait for a click. Alternative that keeps BOTH realism and watchability: make the aperture moderately large, e.g., 1.0mm?? Then bin width 1mm → 14 bins — too coarse for fringes (3.37mm spacing → 3.4 bins/fringe — marginal, actually OK-ish? A histogram with 3-4 bins per fringe period is enough to SEE fringes clearly. Hmm.) With 1mm bins over 14mm: 14 bins. Acceptance ~ 1/14 ≈ 7%. Still mostly misses. OK here's my decision: **decouple animation from statistics**: - The histogram/counters accumulate via fast statistical simulation (many pairs per tick, only counting those that land in the aperture, exactly P(x₀|k)·Δx rates). - The animation is a "visualization layer" that shows a representative subsample: at moderate rates, show every DETECTED pair's full flight (pump → SPDC → signal to D0 click → idler through network → Dk click → coincidence flash → histogram tick with a falling dot into the bin!). Misses: show them only at the slowest speed setting (and maybe 1 in N). Actually — simpler UI: two independent things: "Foton animasyonu" (on/off or density) and "İstatistik hızı" (pairs per second: 1× … 5000×). Default: animation on, speed ~50× (a few flights visible per second, histogram grows visibly). Hmm, wait — but if the histogram ticks from events that aren't animated, the connection "photon → click → bin" gets fuzzy. Compromise: every animated pair corresponds to a REAL statistical event (a detection); additional non-animated events also add counts (shown as small "hızlı sayaç" indicator). The falling-dot-into-histogram animation for animated events makes the causality clear. I think that's good. Also the **D0 scan modes**: - "Otomatik tarama" (auto-scan): x₀ steps bin by bin, dwelling T seconds each (e.g., enough for ~equal counts); histogram fills across all bins — the classic buildup. - "Manuel": slider sets x₀; counts accumulate only at that bin. And **channel selection**: which histogram to display — D0 total, or coincidence with D1/D2/D3/D4. Since data for ALL channels accumulates simultaneously regardless of selection (realistic! the coincidence counter sorts all), switching views is instant. ✓ (Important teaching point: all datasets exist simultaneously; the pattern appears or not depending on which you look at.) **The delayed-choice interactivity** — the star feature: - Toggle switches for BSa, BSb, BSc ("Takılı / Çıkarılmış"). Removing BSa: idler-A always transmits → no which-path via D3; D3 goes dark. Removing both BSa & BSb: ALL idlers go to BSc → all detections are erasure type (D1/D2 only). Removing BSc: erasure path idlers continue straight... where? In my geometry, BSc is where the two transmitted... wait in MY layout it's reflected beams that go to BSc. Ugh, let me re-fix my layout decision. Let me re-derive with my converging-fan geometry: My layout: idler A: (560,135) → down-left → BSa at (470,320). Idler B: (640,135) → down-right → BSb at (730,320). - At BSa (470,320): beam direction ≈ normalize(−90,185) ≈ (−0.44,0.90). Options: transmit → continue down-left; reflect → direction mirrored. To route to BSc at (600,440): from BSa, direction (130,120) normalized (0.73,0.67) — the reflection of (−0.44,0.90) about a surface normal... reflect d about mirror with normal n: d' = d − 2(d·n)n. We need d'=(0.73,0.67), d=(−0.44,0.90). d+d' = (0.29,1.57) ∝ normal? Reflection: d'−d = −2(d·n)n — parallel to n. So n ∝ d'−d = (1.17,−0.23) → normalize ≈ (0.98,−0.19). So the mirror surface is almost horizontal (normal almost horizontal → surface almost vertical)?? Normal (0.98,−0.19) means surface direction ≈ (0.19,0.98) — nearly vertical surface, tilted ~11°. Hmm, so BSa would be a nearly-vertical plate. That's... actually fine optically (a plate BS at near-normal... no wait, incidence angle: beam d vs normal n: cos = d·n = (−0.44)(0.98)+(0.90)(−0.19) = −0.43−0.17 = −0.60 → 53° incidence ✓ fine). So: BSa at (470,320) as a slightly-tilted vertical plate: REFLECT → to BSc (600,440) [erasure path]; TRANSMIT → continue down-left to Ma at (410, 445)? direction (−0.44,0.90): from (470,320) + t·(−0.44,0.90): at t=125: (415, 432). Ma there reflects to straight down → D3 at (415, 560)?? Hmm wait — which side: D3 should receive ONLY slit-A idlers ✓ (A path transmits BSa → Ma → D3) ✓. Hmm wait, but hold on: with REFLECT = erasure and TRANSMIT = which-path... in Kim's paper: "BSa... the idler photon is either reflected... to D3 (which-path) or transmitted toward BSc". Eh — as established, it doesn't matter which is which physically. But for matching the famous figure's spirit (transmit → BSc), maybe flip: transmit → toward BSc; reflect → toward Ma/D3. Geometrically: BSa reflects d=(−0.44,0.90) to d' toward Ma... Ma then reflects to D3. Let me instead choose the geometry so that transmitted beams go toward center (BSc) and reflected go outward-down to Ma/Mb → D3/D4: - BSa at (470,320): if the SURFACE is oriented so that transmission continues down-left... no — transmission NEVER changes direction! The transmitted beam ALWAYS goes down-left (continuing the fan). The reflected beam is the one that can be steered. Down-left from BSa heads toward bottom-left corner (away from center). So to bring the A idler to center-bottom (BSc), it MUST be the reflected beam. Unless I flip the fan: idler A goes down-RIGHT (converging fan) — then transmitted continues down-right TOWARD center ✓!! Converging fan: idler A from slit A (x=560) travels down-RIGHT; idler B from slit B (x=640) travels down-LEFT; they cross at (600, ~somewhere). Place BSa at (640, 310)?? wait idler A path: from (560,135) direction (0.44,0.90): at y=310 (t=194): x = 560+86 = 646. So BSa at (646,310) — but that's basically at x=640 = slit B's x... and idler B (from (640,135), direction (−0.44,0.90)) at y=310: x=640−86=554. So BSa is on the RIGHT (x≈646) receiving SLIT-A's idler, and BSb on the LEFT (x≈554) receiving slit-B's idler — the beams CROSS between BBO and the beamsplitters (crossing point: where 560+0.44t = 640−0.44t → t=90.9 → (600, 217) — right below the beam dump at (600,230)?? collision with beam dump! Move beam dump up to (600, 205) or make crossing at (600,240) and dump at (600,195). Eh — or place the dump right below the prism at (600,205). The beams cross at (600,217)... too close. Let me adjust fan angles: steeper: direction (0.35,0.94): crossing at t = 80/0.70... 560+0.35t=640−0.35t → t=114 → y=135+0.94·114=242. Dump at (600,200), crossing at (600,242) ✓ ok-ish, 42px apart. Or skip the beam dump... no, keep it — nice realism. Or make crossing y=260: angle from vertical = atan(40/125)=17.7°: dir (0.30,0.95): crossing y = 135+0.95·133=261 ✓ dump at (600,205) — 56px clearance ✓. So: idler A: (560,135) → dir (0.30,0.95) → BSa at t=?: place BSa at y=330: t=205: x=560+62=622 → BSa at (622,330). Idler B: (640,135) → dir (−0.30,0.95) → BSb at (578,330). The two beams cross at (600,261). Hmm — BSa at (622,330) receives slit-A idler, on the RIGHT side. And its TRANSMITTED beam continues down-right... away from center! DAMN. Converging fan: A goes down-right, crosses center at y=261, continues down-right toward bottom-right. So after crossing, A's beam is on the bottom-right. For A's transmitted beam to reach BSc at bottom-center (600,440), it should be... A's beam at y=440: x = 560+0.30·(440−135)/0.95 = 560+96 = 656. B's beam at y=440: x=544. So if NO beam splitters at all, the beams cross at (600,261) and diverge again — BSc could sit at the CROSSING point (600,261)?! But we need BSa/BSb BEFORE BSc in the path... BSa must intercept A's beam between BBO and the crossing: at y<261, A's beam x ranges 560→600. Place BSa at (585, 210)?? that's overlapping the prism region/dump... Hmm OK this is getting tangled. Let me go back to DIVERGING fan (A down-left, B down-right — each slit's idler goes OUTWARD) — the natural, uncrossed, Kim-figure-like layout — and accept that the path to BSc is via REFLECTION at BSa/BSb (which is physically equivalent): - Idler A: (560,135) → dir (−0.30,0.95) → BSa at (500, 325) [t=200: x=560−60=500, y=135+190=325] ✓ - Idler B: (640,135) → dir (0.30,0.95) → BSb at (700, 325) ✓ - Beams don't cross each other ✓ don't cross pump (x=600) ✓. - **BSa** at (500,325): plate oriented so that: TRANSMIT → continue down-left to Ma → D3 (which-path, slit A). REFLECT → steer right-down to BSc at (600,450): from (500,325) to (600,450): dir (100,125)→(0.62,0.78). Reflect d=(−0.30,0.95) to d'=(0.62,0.78): n ∝ d'−d = (0.92,−0.17) → normal (0.98,−0.18), surface direction (0.18,0.98) — nearly vertical plate tilted ~10°. Incidence angle: cos(d,n)= (−0.30)(0.98)+(0.95)(−0.18) = −0.294−0.171 = −0.465 → 62° incidence — steep but drawable. Hmm, 62° incidence on a plate BS — a bit unusual but fine schematically. Alternatively adjust BSc position to make reflection angles nicer: BSc at (610, 420): from BSa (500,325): dir (110,95)→(0.76,0.65): n ∝ (1.06,−0.30)→(0.96,−0.27)... similar. Meh — fine, ~60° plate. OR flip: TRANSMIT → BSc? Impossible (transmission preserves direction; down-left ≠ toward center). So reflection→BSc it is. Fine. Actually you know what, let me simplify the visual: I can bend the "transmitted" path with Ma! No wait — Ma is for the which-path branch. Hmm hold on, actually, let me reconsider the whole thing. What if the idler beams from the slits go down-and-INWARD (converging) but I place BSa/BSb AFTER the crossing point? I.e., A's idler crosses to the right side, then hits BSa on the right at (640, 330); transmitted continues down-right?? no... What if the idler beams go down-VERTICALLY initially (parallel, x=560 and x=640), then small fold mirrors FA (560, 260) and FB (640, 260) tilt them toward center: FA reflects A's beam to down-right 30°-ish; FB to down-left. Then: - A: (560,135) → (560,260) → reflect → dir (0.30,0.95)... same as before but now from (560,260): hits BSa at (620, 450)?? then transmit continues down-right, reflect steers... The fold mirrors FA/FB are extra elements not in Kim's setup — but real experiments DO have steering mirrors everywhere. Hmm, but it complicates. Alternatively — completely different, SIMPLER and arguably MORE realistic-to-figure approach: look at how the actual Kim figure handles it! I'm now going to commit to reconstructing it as best I can, because I've seen this figure many times: From the PRL: the figure shows (top to bottom): laser at top; the beam goes DOWN through the double slit (drawn horizontally, slits separated left-right); BBO below; then the signal+idler continue down; a Glan-Thompson prism P1 deflects... hmm, actually I recall TWO prisms P1... you know, I genuinely can't reconstruct it perfectly. Textbook redraws (e.g., Wikipedia's "Delayed choice quantum eraser" article has the exact Kim figure) show: From Wikipedia's figure (which is the Kim et al. figure): - Laser at top center. - Double slit below it (horizontal bar, two slits). - BBO below. - Below BBO, the pair separates: the SIGNAL is deflected by a prism at the TOP-RIGHT of the idler section... the signal beam goes UP-RIGHT to a lens and D0 (D0 at top-right of the figure, on a stage with an arrow labeled x). - The IDLER continues DOWN from the BBO: the idler light from both slits... the figure shows the idler splitting at the slits into two beams that go down-LEFT and down-RIGHT (diverging fan): left beam to BSa, right beam to BSb. - BSa (left): TRANSMITTED beam continues down-left... to... hmm... and REFLECTED beam goes down-right?? The figure shows at the bottom-left: Ma and D3. At bottom-right: Mb and D4. Center-bottom: BSc with D1 and D2 below it. - I believe: BSa transmitted → down-left → Ma (bottom-left) → reflects right?? → D3... and BSa REFLECTED → down-right → toward BSc (center). Hmm, that gives the same topology as my "diverging fan, reflect→BSc" design! With Ma folding the which-path beam toward D3. Actually YES — I'm fairly confident the Kim figure has: BSa and BSb as 45°-ish plates where the REFLECTED beams route INWARD to BSc, and TRANSMITTED beams continue OUTWARD-DOWN to Ma/Mb which fold them to D3/D4. Wait, no... hmm, or D3/D4 are directly below Ma/Mb... whatever — topology: BSa: T→Ma→D3 (which-path A), R→BSc; BSb: T→Mb→D4, R→BSc. That's exactly my diverging-fan design. And at BSc (center-bottom, receiving two beams converging from up-left and up-right): outputs → D1 (down-left) and D2 (down-right) — matches my earlier computation with vertical-ish plate. Hmm wait, actually now I doubt whether in the real figure D1/D2 arrangement... The paper says detection at D1 or D2 erases. Fine. My layout: **FINAL LAYOUT (viewBox 0 0 1240 880):** Top-down schematic: 1. **Laser** (box): center (620, 38), w=110, h=40. Label above: "UV LAZER (351 nm)". Beam port at bottom center. 2. Pump beam: (620, 58) → (620, 100). Draw as violet line, slightly diverging to cover both slits: beam polygon from (610,58)-(630,58) widening to (585,100)-(655,100)? The slits at x=580 and x=660 (separation 80px). Slit plate: y=100..112, x from 540 to 700, with gaps at 570–590 (A) and 650–670 (B). So slit A center x=580, slit B center x=660. Pump beam drawn widening from laser port (620±8) to (620±45) at the slit plate — covering both slits ✓ (label "çift yarık" with d=0.25mm note). 3. **BBO**: plate y=125..145, x 545–695, green tint. Label "BBO kristali (SPDC)". 4. Down-converted pairs born at slit positions (580,135) and (660,135) — actually born anywhere in BBO where pump light is; schematic: two birth rays. 5. **Glan-Thompson prism P**: at y=170..200, x 545–695: draw as a wide parallelogram/45° plate. It reflects the SIGNAL (orange) to the RIGHT: signal beam from (580,200)-ish and (660,200)-ish... hmm wait — the signal reflects off the prism's 45° face and goes RIGHT (horizontal, y≈185). The signal FAN: photons leave the slits at angle ±θ... Hold on, wait. I need to recheck the signal fan geometry: the signal photons propagate down from the slits through the BBO to the prism — at what x-position does each photon hit the prism? A photon from slit A heading straight down hits the prism at (580, 185) and reflects to travel right along y=185. A photon from slit B hits at (660,185) → also right along y=185. Both travel right at the SAME y — overlapping beams! Then where's the fan?? The fan comes from photons leaving the slits at ANGLES (diffraction): photon from slit A heading down-right at angle θ hits the prism at (580+Δ, 185) and reflects to heading right-UP?? Let me compute: direction (sinθ, cosθ) [down, tilted right] hits 45° mirror "/" (m=(1,1)/√2, n=(1,−1)/√2): d·n = (sinθ−cosθ)/√2. d' = d − 2d·n·n = (sinθ,cosθ) − (sinθ−cosθ)(1,−1) = (sinθ−sinθ+cosθ, cosθ+sinθ−cosθ) = (cosθ, sinθ). So down-right-tilted → right-DOWN-tilted by the same angle θ ✓. So the reflected fan spreads vertically: photons at +θ travel right-down, −θ right-up. At the lens (x=830, i.e., 170px after prism exit ~x=700), the fan spans y = 185 ± tanθ·170·... wait, but ALSO the whole fan is centered on y=185 regardless of which slit — because the prism face is flat and at 45°: a photon's y at the prism = 185 (fixed face line)... no wait, the prism face is a LINE at 45° in the drawing (from (545,215) to (695,155) say). Photons hit it at different points depending on their x when they reach it. A photon from slit A going straight down (x=580): hits the 45° face where the face line x-coordinate = 580: face point (580, y_face(580)). Reflects to horizontal: travels right at y = y_face(580). A photon from slit B (x=660): reflects at y_face(660) ≠ y_face(580). So the two slits' beams exit at DIFFERENT heights (offset by the 45° geometry: Δy = −Δx). Then a photon from slit A at angle θ: hits face at (580+tanθ·s, ...) hmm — actually the beautiful thing: a flat 45° mirror maps x-position to y-position: exit height y_exit = y_face(x_hit), and x_hit = slit_x + tanθ·(distance). So exit height encodes BOTH slit position and angle. Then traveling right, the photon's height at the lens: y_exit ± tanθ·(dist to lens)... This is getting overly complicated for the ANIMATION, and there's a physical subtlety: in reality the signal detection position depends only on θ (momentum) — the lens maps angle→position, and the which-slit position info is... In the far field / Fourier plane, position x_f = f·tanθ regardless of slit. The prism in the near field adds position-dependent offsets that a real experiment would... ugh. SCHEMATIC LICENSE: for the animation, I'll draw each signal photon as a dot that travels: from BBO → down to prism region → bends to horizontal → travels right at the height corresponding to ITS FINAL D0 position, through the lens → to D0 rail, landing at y = map(x_sampled). The mapping: y_D0 = 185 − x_mm·20 px/mm (x=−7mm → y=325; x=+7 → y=45). Rail: x=1060, y from 30 to 340. The photon's path from prism (700, y_prism≈185) to lens (880, y≈185±...) hmm — photon must travel from y≈185 at prism to y_target at rail: through the lens it bends. Draw: prism exit (700, 185) → lens at (880, y_mid) where y_mid = 185 + (y_target−185)·(180/360)·... let me just draw: segment 1: (700,185) → (880, 185 + (y_target−185)·0.45); segment 2 (after lens, converging): (880, ...) → (1060, y_target). Wait — that's not how a single lens works for a point... eh: a Fourier lens: incoming ray at angle θ (height h at lens) → exits toward focal point (f·tanθ). So the incoming segment has slope ∝ tanθ = (y_target−185)/f_px... f_px = 1060−880 = 180. Incoming slope: from (700,185): at lens x=880: y = 185 + tanθ·180 = 185 + (y_target−185). So incoming segment: (700,185) → (880, y_target); outgoing: (880, y_target) → (1060, y_target)?? That's wrong — outgoing should go to focal point... Hmm: ray at angle θ entering lens at height h: exits going through focal point at (f, f·tanθ): from (880, h) toward (880+180, 185+(y_target−185)) = (1060, y_target). If h ≠ y_target, outgoing has a slope. What's h? The photon after the prism travels at angle... the prism reflected it at height y_exit with direction (cosθ, sinθ) — its height at the lens: y_exit + tanθ·(880−x_exit). With y_exit ≈ 185 − (slit offset effects)... SCHEMATIC SIMPLIFICATION: just draw the photon from prism-exit (700,185) → (880, y_target) → (1060, y_target). i.e., bend at the lens, then straight to the rail at final height. Visually: fan expanding from prism to lens, then parallel-ish (actually converging to final heights) — looks exactly like a Fourier lens at work ✓. Good enough, physically evocative, simple to animate. The lens: drawn as a vertical rounded-rect (aperture) at x=880, from y=40 to y=330, label "L (Fourier merceği, f = 1.2 m)". **D0**: on a vertical rail (translation stage) at x=1060: draw rail line from (1060,30) to (1060,345), with a stage carriage holding the detector head (drawn as a small block + fiber + label "D0"). D0's current y = map(x₀). D0 head: small rounded rect w=34, h=26 with a little fiber squiggle going right to a counter box? Also a small arrow indicating stage motion. Label: "D0 (taranabilir)". Photons that miss the aperture: fly past D0 and fade out beyond x=1100 (or get absorbed — draw them fading after passing the rail). Also — the y mapping on the rail vs x in mm: y = 185 − x·20. x∈[−7,+7] → y∈[325,45] ✓ rail 30–345 ✓. 6. **Idler network** (bottom half): - Idler A: from (580,135) [slit A] traveling down-left at angle: dir (−0.30, 0.95). Wait — hmm, should the idler go outward from ITS slit (A left, B right)? Then idler A: (580,135) → down-left → BSa at (500, 390)? Let me compute: to reach x=500 from 580: Δx=−80 → t = 80/0.30 = 267 → y = 135+0.95·267 = 388. BSa at (500,388). Long-ish path ✓ (delay!). - Idler B: (660,135) → down-right → BSb at (740, 388). - Pump beam dump: pump continues down through BBO and prism at x=620: at y=230 place a small black hatched block (620±20). The idler beams at y=230: A at x=580−0.30·(95/0.95)=580−30=550 ✓ clear of dump (600–640); B at 690 ✓ clear. - Prism P at y~170–200: the idlers pass THROUGH it (transmitted) ✓ — wait, the prism is at x 545–695: idler A passes through its left part (x≈575 at y=185... let me check: A at y=185: x = 580−0.30·(50/0.95) = 580−15.8 = 564 ✓ inside prism span ✓; B: x=676 ✓ inside ✓. Good — draw idler beams passing through the prism. - Hmm wait, one more check — the signal ALSO passes through the prism and reflects. But when does the signal "separate" from the idler? At the BBO (they're born together, slightly different angles — non-collinear). Schematic: from the BBO, signal photons travel down (toward prism) roughly VERTICALLY (fan spreading), while idlers travel down-outward at ±17.5°. So between BBO (y=135) and prism (y=185), signal and idler beams are visually distinct ✓. Actually, for the pair animation: the pump dot goes down; at BBO, it converts: signal dot continues down (roughly straight, from birth x≈slit x), idler dot heads out-down at ±17.5°. Then signal hits prism at y≈185 → reflects right. ✓ Clean. Hmm, wait — but the signal from slit A going straight down from (580,135) hits the 45° prism face... the face is a line; where? If prism drawn as a parallelogram with face from (545,200) to (695,170)?? A 45° line has slope... in SVG coords, a "/" 45° line from bottom-left (545,210) to top-right (695,60)? That's Δx=150, Δy=−150 ✓ 45°. But that spans y from 60 to 210 — tall. The prism only needs to intercept beams around y=170–200. Let me draw the GT prism as a rectangle ROTATED 45°?? Real GT prisms are bulky blocks. Draw as a rotated square (diamond) centered (620,190), size ~160: vertices... a diamond with horizontal/vertical diagonal 170: vertices (620,105)?? overlapping the slit plate. Hmm. Alternative: draw the prism as a simple 45° plate (thin line segment) like mirrors are drawn — with label "P (Glan–Thompson prizma)". Optics schematics often draw the GT prism as a triangle. Let me draw a right triangle: vertices (540,215), (700,215), (700,155)?? — hypotenuse from (540,215) to (700,155): slope −60/160 — not 45°, fine (schematic!). The SIGNAL reflects off the hypotenuse?? For the reflection to look right (down→right), the face should be at 45°: from (540,235) to (660,115)? Δ=(120,−120) ✓ 45°. Triangle: (540,235), (660,115), (660,235) — right angle at (660,235). Signal beams (traveling down at x∈[560,680]) hit the hypotenuse and reflect right ✓. The hypotenuse spans x 540–660 — beam from slit B (x=660)... hits at the very top vertex. Extend: hypotenuse (530,245)→(690,85): spans x 530–690 ✓ both beams hit (A at x=580: y on face = 245−(580−530)=195; B at x=660: y=245−130=115). Hmm, B's signal reflects at y=115, A's at y=195 — different heights, then travel right at those heights... but per my "schematic license" the animation draws signal photons from prism exit at (x_hit, y_hit) → then to lens (880, y_target)... the segment from (660,115) to (880, y_target) — slope arbitrary. FINE. Actually, let me simplify further: make the prism face a SHORT 45° segment around (620,190): from (560,220) to (680,100)?? Whatever — I'll draw a rotated-rect plate: center (620,190), length 160 along 45°, thickness 14. Its face line: from (620−80/√2·... ) — plate endpoints: (620±80·cos45, 190∓80·sin45) = (620±56.6, 190∓56.6): (563,247) to (677,133). Signal beams at x=580: face y = 247−(580−563)=230?? hmm wait line from (563,247) to (677,133): y = 247 − (x−563). At x=580: y=230. At x=660: y=150. OK so signal A reflects at (580,230), signal B at (660,150). Both travel right. Fine. But the idler beams pass through this plate too: idler A crosses the face line at... A: x(t) = 580−0.30s, y = 135+0.95s; on face: y = 247−(x−563) → 135+0.95s = 247−(580−0.30s−563) = 247−17+0.30s = 230+0.30s → 0.65s = 95 → s=146 → point (536, 274)?? That's below the plate's lower end (563,247)... hmm the plate spans the face line only between its endpoints (563,247)–(677,133); point (536,274) is beyond the end — idler A MISSES the prism plate! Ha — actually that's FINE and even good: the GT prism only covers the central region; the idlers (diverging outward) exit past its edge. Even cleaner: idler A passes to the LEFT of the prism, idler B to the right. Let me verify B: x(t)=660+0.30s, y=135+0.95s; face: y=247−(x−563) → 135+0.95s = 247−(660+0.30s−563)=247−97−0.30s=150−0.30s → 1.25s=15 → s=12 → (663.6, 146.4) — ON the face (within span x 563–677 ✓). Hmm, so B's idler hits the prism face but A's doesn't?? Asymmetric — ugly. Fix: widen the prism or move it up. Move prism center up to (620,175), or make the plate longer (length 200): endpoints (620±70.7, 175∓70.7) = (549,246)–(691,104). Idler A: face y = 246−(x−549); A path: 135+0.95s = 246−(580−0.30s−549) = 246−31+0.30s = 215+0.30s → 0.65s=80 → s=123 → point (543, 252)?? x=580−0.30·123=543 — that's LEFT of face span (549–691) → misses again (the face point at y=252 would be at x=543 but face ends at (549,246)). Just barely misses. The issue: idler A diverges left fast; the prism's lower-left end is around (549,246); idler A at y=246: x = 580−0.30·(111/0.95)=580−35=545 — 4px left of the end. SO CLOSE. Just nudge: make the idler divergence angle smaller (±0.24, 0.97): idler A at y=246: x=580−0.24·(111/0.97)=580−27.4=552.6 → INSIDE face span ✓. Then recompute downstream positions with dir (−0.24,0.97): BSa at x=500: s=(580−500)/0.24=333 → y=135+0.97·333=458. Hmm, BSa at (500,458) — quite low; canvas height 880 — D3 etc. need room below: BSc at ~(600,560)?, D1/D2 at ~(520,700)?, D3/D4 at (400,700)? Tight but maybe OK. Alternatively BSa at (520, 430): s=(580−520)/0.24=250, y=135+242=377?? wait recompute: y=135+0.97·250=135+242.5=377.5. So BSa at (520,378). Let me place BSa at (520, 378) and BSb at (720, 378). Hmm, let me reconsider canvas budget: viewBox 1240×880. Bottom half (y 380–860) for: BSa/BSb (y≈378), Ma/Mb (y≈500), BSc (y≈520), D3/D4 (y≈630), D1/D2 (y≈660), plus coincidence counter electronics at bottom (y 720–860)? Hmm, might be tight. Let me allocate: - BSa (520,378), BSb (720,378). - BSa TRANSMIT (which-path A): continues dir (−0.24,0.97) → Ma at (480, 500)?: from (520,378): s: Δx=−40 → s=167 → Δy=162 → (480,540). Ma at (480,540): reflects to straight DOWN → D3 at (480, 640). Mirror Ma orientation: reflects dir (−0.24,0.97) to (0,1): n ∝ d'−d = (0.24,0.03)→(0.99,0.12): surface dir (−0.12,0.99) — near-vertical plate, slight tilt. Drawable as a small tilted line ✓. - BSb TRANSMIT: (720,378) → dir (0.24,0.97) → Mb at (760,540) → D4 at (760,640). - BSa REFLECT (erasure): from (520,378) toward BSc at (610,505): dir (90,127)→(0.58,0.82). Check reflection consistency: d=(−0.24,0.97), d'=(0.58,0.82): n ∝ d'−d = (0.82,−0.15) → (0.98,−0.18) — near-horizontal-normal → near-vertical surface, tilt ~10°. So BSa plate direction ≈ (0.18,0.98) — nearly vertical, tilted ~10.5° from vertical. Both transmit (unchanged) and reflect work with this plate ✓. Visually: BSa drawn as a slightly-tilted vertical plate ~40px long. Hmm — slightly odd-looking but fine. OR... Hmm, let me reconsider: maybe make it prettier with BSa as a classic 45° square. For the plate at 45° ("/" or "\"), reflections turn beams by exactly 90°. Incoming idler A dir (−0.24,0.97) (mostly down, slightly left). A "\" plate (surface dir (1,−1)/√2... surface along (1,−1)?? hmm "\" goes from top-left to bottom-right: direction (1,1) in screen coords?? In screen coords (y down), "\" = from (0,0) to (10,10): direction (1,1). "/" = from (0,10) to (10,0): direction (1,−1).) Reflection off "\" (m=(1,1)/√2, n=(1,−1)/√2): d=(−0.24,0.97): d·n = (−0.24−0.97)/√2 = −0.856. d' = d − 2(−0.856)(1,−1)/√2 = (−0.24,0.97) + 1.712/√2·(1,−1) = (−0.24,0.97)+(1.21,−1.21) = (0.97,−0.24) → RIGHT and slightly UP. So "\" reflects idler A to → heading right-up — no good (we want right-DOWN toward BSc). Reflection off "/" (m=(1,−1)/√2, n=(1,1)/√2): d·n = (−0.24+0.97)/√2 = 0.516. d' = d − 2·0.516·(1,1)/√2 = (−0.24,0.97) − (0.73,0.73) = (−0.97,0.24) → LEFT-down — no good either. So with a 45° plate, idler A reflects to near-horizontal (right-up or left-down) — neither points nicely to BSc at down-right. The problem: idler A's direction is only ~14° from vertical, so a 90° turn gives ~76° from vertical (near-horizontal). To hit BSc down-right at ~55° from vertical, the turn is only ~40° — hence the ~20°-tilted plate. OK so my near-vertical plate it is — OR reposition BSc to be nearly horizontally-aligned with BSa (BSc to the right of BSa at similar height): BSc at (640, 430)? From BSa (520,378): dir (120,52)→(0.92,0.40) — 23° below horizontal. Then plate normal n ∝ d'−d = (1.16,−0.57) → (0.90,−0.44) → surface dir (0.44,0.90) — plate tilted ~26° from vertical. Still not 45°. To get a 45° plate ("/" in screen = surface dir (1,−1)... wait I need to be careful: for the reflection to turn the beam by angle δ, the plate... a mirror turns a beam by δ = 2×(angle between beam and plate-surface)... beam direction changes by 180°−2α where α is incidence angle from normal... Turn angle = 2×(angle between incident beam and mirror SURFACE). Beam at 14° from vertical; want output at 55° from vertical on the other side: turn = 41° → beam-surface angle = 20.5° → plate at 14+20.5 = 34.5° from vertical?? or vertical±20.5°... plate direction must bisect input/output: input dir angle from +x axis: (−0.24,0.97) → pointing 104° (measuring standard math angle with y up... ugh screen coords). Let me just accept a tilted plate and draw it correctly by computing the bisector. It's FINE — real beam splitters get set at all sorts of angles. I'll draw each BS as a rounded rect rotated to the computed angle, with a semi-reflective look (blue-ish glass). Actually — hmm, wait. Let me reconsider the whole idler geometry once more, because there's a classic clean arrangement: make the idler beams travel down-outward at 45° (dir (−0.707,0.707) and (0.707,0.707))! Then: - Idler A: (580,135) → down-left at 45° → BSa at (490, 225): Δ=(−90,90) ✓. - Idler B: (660,135) → BSb at (750, 225). - At BSa, a 45° plate "\"... beam dir (−0.707,0.707): reflect off "\" (m=(1,1)): computed above pattern: d=(−0.707,0.707), n=(1,−1)/√2: d·n = (−0.707−0.707)/√2 = −1. d' = d+2·(1,1)/√2·... = (−0.707,0.707)+(1.414,−1.414)·(1/√2·√2)... let me: d' = d − 2(d·n)n = (−0.707,0.707) − 2(−1)(0.707,−0.707) = (−0.707,0.707)+(1.414,−1.414) = (0.707,−0.707) → up-right. "\" turns down-left → up-right (retro-ish? no—it's a 90° turn: down-left → up-right is 180°?? (−0.707,0.707)→(0.707,−0.707): exactly reversed! Because beam at 45° to "\" surface... beam dir (−0.707,0.707) is PERPENDICULAR to "\" surface (surface (1,1)/√2): dot = −0.707+0.707=0 ✓. So normal incidence → reflects straight back. My bad. Reflect off "/" (surface (1,−1)/√2, n=(1,1)/√2): d·n = (−0.707+0.707)/√2 = 0 → beam ALONG surface — grazing. Hmm: down-left at 45° travels parallel to "/" — no interaction. So for a 45° beam, plates at 45° don't work; plates at 22.5° or 67.5° do. The bisector logic: plate surface must bisect input/output directions. To turn down-left(45°) → down-right(45°) (i.e., mirror about vertical): plate surface VERTICAL (dir (0,1)): reflect (−0.707,0.707) about vertical surface (n=(1,0)): d' = d−2(d·x̂)x̂ = (−0.707,0.707)−2(−0.707)(1,0) = (0.707,0.707) ✓ down-right. So a VERTICAL plate at BSa turns the 45°-down-left idler into 45°-down-right, heading toward bottom-center! - So: BSa = vertical plate at (490,225): TRANSMIT → continues down-left at 45° → toward Ma at (400, 315)?: from (490,225): Δ(−90,90) → Ma at (400,315). Ma: turn down-left → straight down: mirror bisecting (−0.707,0.707) and (0,1): plate surface... n ∝ d'−d = (0.707,0.293) → (0.924,0.383); surface dir (−0.383,0.924) — plate tilted 22.5° from vertical. Fine — draw as tilted small mirror. → D3 at (400, 415) (beam goes straight down from Ma). Hmm wait, Ma output straight down from (400,315) → D3 at (400, 430)? ✓. - REFLECT at BSa (vertical plate) → down-right at 45° → from (490,225) toward BSc at (620, 355) ✓ (Δ=(130,130) ✓ 45°) ✓✓ - BSb = vertical plate at (750,225): TRANSMIT → down-right 45° → Mb at (840,315) → tilted mirror → straight down → D4 at (840, 430). REFLECT → down-left 45° → BSc at (620,355) ✓ (Δ=(−130,130) ✓). - **BSc** at (620,355): receives two beams converging at 45° from up-left and up-right. Outputs: straight down? A vertical plate at BSc: input A (down-right 45°): transmit → continues down-right 45° → to D2 at (710,445); reflect → down-left 45° → D1 at (530,445). Input B (down-left): transmit → down-left → D1 ✓; reflect → down-right → D2 ✓. So BSc = VERTICAL plate at (620,355), outputs at ±45° down to D1 (530,445) and D2 (710,445). Wait — hmm, D1 at down-LEFT and D2 at down-RIGHT. Both get both paths ✓. This geometry is CLEAN: all beams at ±45° or vertical; BSa, BSb, BSc all vertical plates; Ma, Mb at 22.5° tilt. Hmm, but three vertical plates looks a bit repetitive/unrealistic? Real beam splitters at near-normal incidence... vertical plate with 45° beams = 45° incidence ✓ actually that's the CLASSIC beamsplitter orientation (plate at 45° to the beams)! In top-down view with beams at ±45°, the plate drawn vertical is at 45° to each beam ✓✓ — this is exactly how beam splitters look in real top-down layouts (e.g., a Mach-Zehnder drawn as a diamond: BS at the corners, beams at ±45°, plates aligned with the diamond edges... wait no. Hmm: classic MZ: beams horizontal+vertical, plates at 45° diag. Here beams at ±45°, plates vertical/horizontal — same thing rotated 45° ✓ Totally standard.) But wait — the idler beams at 45° from the slits: idler A from (580,135) at 45° down-left → passes through/near the prism? Prism face (45° line "/"-oriented around (620,190))... idler A heading down-left crosses x=560 at y=155, x=540 at y=175... The prism plate (if spanning (563,247)–(677,133)): idler A crosses that line at: 135+(580−x) = 247−(x−563) → 135+580−x = 247−x+563 → 715−x = 810−x → 715=810?? No solution — parallel lines! OF COURSE: idler A at 45° down-left is PARALLEL to the "/" prism face → never crosses it ✓✓ BEAUTIFUL — the idler beams pass by the prism parallel to its face, cleanly missing it (or grazing it). Even better: shorten the prism so it clearly doesn't touch: prism face from (575,235) to (685,125) (length ~156): idler A line: x+y = 715 (since from (580,135): x+y=715). Prism face line: x+y = 575+235 = 860 (parallel ✓). Distance = (860−715)/√2 ≈ 102px ✓ way clear. Idler B: from (660,135): x−y = 525 (going down-right: x−y const: 660−135=525). Prism face: x+y=860 — crosses where x−y=525, x+y=860 → x=692.5, y=167.5 — ON the face?? face spans x 575–685: x=692.5 > 685 → just misses the top end ✓ (by ~8px). Hmm — cutting close. Extend or shorten prism: face from (570,240) to (680,130) (x+y=810): idler B crossing: x=(810+525)/2=667.5, y=142.5 — within x-span 570–680 → CROSSES the face ✗. So with face x+y=810, B hits it. With x+y=860 (face (575,235)-(685,125)... wait (575,235): 575+235=810. I miscalculated: 575+235=810, not 860. Let me redo: I want the face line x+y=C, parallel to idler A (x+y=715), to the RIGHT/BELOW of it, and NOT crossing idler B (x−y=525) within the face span. Face crossing with B: x=(C+525)/2, y=(C−525)/2. For the face span x∈[x1,x2] with x1+x2... face endpoints (x1, C−x1), (x2, C−x2). Avoid crossing: (C+525)/2 ∉ [x1,x2] → either (C+525)/2 > x2 (crossing beyond top end) or < x1. Also signal beams (vertical, x=580 and x=660) must hit the face: face x-span must include 580 and 660: x1 ≤ 580, x2 ≥ 660. Then (C+525)/2 > x2 ≥ 660 → C > 795. And idler A: x+y=715 must NOT hit face: parallel ✓ never (as long as face doesn't extend to the x+y=715 line — it can't, parallel). But ALSO the face should be between BBO and... the signal reflects off it — face should be BELOW the slits: face y at x=580: C−580 ≥ 150 → C ≥ 730; at x=660: C−660 ≥ 145 → C ≥ 805. So C ∈ (795, ...]: take C=830: face endpoints: pick x1=565, x2=700: endpoints (565,265)?? wait face points (x, 830−x): (565,265) and (700,130). Check: signal A at x=580 → face y=250 ✓ below BBO(135) ✓; signal B at x=660 → y=170 ✓. Idler B crossing: x=(830+525)/2=677.5 — within [565,700] ✗ CROSSES. Need (C+525)/2 > x2: (C+525)/2 > 700 → C > 875. C=880: face (565,315)?? no wait: face points (x,880−x): (565,315) and (720,160)?? if x2=720: signal B at x=660 → face y=220; signal A at x=580 → y=300. Prism face quite low (y 220–300 for the signals) — that pushes the whole signal path down; the prism would be below the beam-dump area... and idler A (x+y=715) passes ABOVE the face (715 < 880 means... the face line is x+y=880; idler A is x+y=715; idler A is up-left of the face ✓ never touches ✓; idler B: x−y=525 crosses face at x=(880+525)/2=702.5, y=177.5 — within span [565,720]?? 702.5 < 720 ✗ STILL crosses! Need x2 < 702.5 AND x2 ≥ 660 (to catch signal B): x2 ∈ [660, 702.5): take x2=680, then C: (C+525)/2 > 680 → C > 835. C=850: face endpoints (565, 285)?? (x,850−x): (565,285) to (680,170). Signal A at x=580 → y=270; signal B at x=660 → y=190. Hmm signal A reflects quite low (y=270) — the prism is a big diamond spanning y~170–285. The beam dump at (620, ~230)? The dump sits ON the prism region... conflict! The pump (x=620 vertical) would hit the prism face at y=230 (620: y=850−620=230) — the pump hits the GT prism. Physically: GT prism transmits the pump? A GT prism totally transmits one polarization, reflects the other — at 351nm UV... realistically there'd be a dichroic. SCHEMATIC: let the pump pass through the prism too?? But the pump hitting the prism at 45° face would refract/deflect... Hmm hmm. OK you know what — I'm WAY overengineering the geometry. Let me step back and simplify with a cleaner conceptual separation: **Simplification**: Put the GT prism P1 DIRECTLY UNDER the BBO, small, right after it — the standard schematic shows the signal being peeled off immediately: slit plate (y=100) → BBO (y=125) → prism (y=160, small 45° plate centered at x=620 catching the SIGNALS). The signals reflect right at y≈160. The IDLERS: they're born heading down-OUTWARD at 45°, so they naturally clear a small central prism: idler A from (580,135) heading down-left: at x=560, y=155; the prism face at "/" through (620,160) with half-length 45: endpoints (620±32, 160∓32) = (588,128)–(652,192). Idler A line x+y=715: face line x+y=620+160=780: parallel ✓, distance (780−715)/√2=46px ✓ clear. Idler B line x−y=525: crosses face line x+y=780 at x=652.5, y=127.5: face x-span [588,652]: 652.5 > 652 ✓ JUST clears (0.5px!). Make half-length 40: endpoints (592,132)–(648,188): crossing at x=652.5 > 648 ✓ clears by 4.5px. Eh, still close. Make the prism face line x+y=770 (center (620,150)): crossing with B at x=647.5, face span (596,124)–(644,176): 647.5>644 ✓ clears by 3.5px. Hmm — the fundamental issue: idler B passes very close to the prism's top-right tip. Fine — 3.5px clear at scale... make it more: prism center (620,145), face x+y=765, half-length 38: endpoints (593,132)?? (620−27,145−27)=(593,118)–(647,172). Crossing of B (x−y=525) with x+y=765: x=645, y=120: within span? x∈[593,647]: 645 < 647 ✗ crosses by 2px!! GRR. The crossing point is always at x=(C+525)/2, y=(C−525)/2; face tip at x=620+L/√2·... face upper-right endpoint: (620+h, 145−h) where h=half-length/√2·... if half-length ℓ along the diagonal: endpoint (620+ℓ/√2, 145−ℓ/√2). Crossing at x=645 fixed (for C=765). Need 620+ℓ/√2 < 645 → ℓ < 35. With ℓ=30: endpoint (641,124); signal A at x=580: does the face reach x=580? Lower-left endpoint: (599,166). x-span [599,641] — DOESN'T include 580 or 660!! The signal beams at x=580 and x=660 MISS the small prism!! Of course — the slits are 80px apart; a small central prism can't catch both vertical signal beams (at x=580 and 660) while avoiding the idlers. UNLESS the signal beams converge toward the prism! AH — that's it: the signal photons from both slits converge toward the center (they're heading to the collection lens... no wait, they diffract OUTWARD)... Hmm, but actually — what if I tilt the signal beams INWARD? Physically justifiable: the SPDC emission is non-collinear — signal and idler emitted at ±3° on OPPOSITE sides: for the pair from slit A: signal goes down-RIGHT, idler goes down-LEFT; from slit B: signal down-LEFT, idler down-RIGHT!! Then: signals from A and B CONVERGE (both heading toward center-bottom), cross, and a small central prism catches them; idlers diverge outward cleanly!! Let me check: - Signal A: from (580,135), dir (0.707,0.707) down-right: hits central prism face... prism face "/" centered (620,175), ℓ=60: face line x+y=795, endpoints (620±42.4, 175∓42.4) = (578,217)–(662,133). Signal A line: from (580,135) dir (1,1)/√2: x−y = 445. Crosses face x+y=795: x=620, y=175 — dead center ✓. Reflect off "/": d=(0.707,0.707): n=(1,1)/√2·... "/" surface dir (1,−1)/√2, n=(1,1)/√2: d·n = (0.707+0.707)/√2 = 1 → d is along normal → reflects straight back!! (down-right beam hits "/" face head-on → retro-reflection). NO GOOD. The "/" face can't deflect a down-right beam to horizontal. To send down-right → right(-ish): need surface... reflect d=(0.707,0.707) to d'=(1,0): n ∝ d'−d = (0.293,−0.707) → normalize (0.383,−0.924): surface dir (0.924,0.383) — a plate tilted ~22.5° from horizontal. So the prism face at 22.5°-tilt. drawable ✓. Face through (620,175): signal A hits at... A's line x−y=445... whatever, the drawing code will compute. But BOTH signals must hit the same face: signal A from (580,135) down-right; signal B from (660,135) down-left; face tilted 22.5° (nearly horizontal) at y≈175: signal A crosses y=175 at x=580+40=620; signal B at x=660−40=620 — BOTH hit the face at (620,175) — THE SAME POINT?! They cross exactly there! Wait: A goes down-right from (580,135): at y=175: x=620 ✓. B from (660,135) down-left: at y=175: x=620 ✓. They CONVERGE to the same point (620,175) — because I chose 45° convergence from symmetric starts. They cross and continue: A continues to (660,215), B to (580,215)... So place the face AT the crossing point: both beams hit the same spot on the prism — physically that's exactly right (mode overlap!) and the reflected beams: A reflects to... d_A=(0.707,0.707) about surface (0.924,0.383)[dir], n=(0.383,−0.924): d' = d − 2(d·n)n; d·n = 0.707·0.383 + 0.707·(−0.924) = 0.271−0.653 = −0.383; d' = (0.707,0.707) + 0.765·(0.383,−0.924) = (0.707,0.707)+(0.293,−0.707) = (1.0, 0.0) → travels RIGHT ✓. B: d=(−0.707,0.707): d·n = −0.271−0.653 = −0.924; d' = (−0.707,0.707)+1.848·(0.383,−0.924) = (−0.707,0.707)+(0.707,−1.707) = (0,−1) → travels UP?? ✗ BAD. Hmm — the surface that sends A→right sends B→up. Because the beams converge symmetrically, ONE face can't send both to the right. UNLESS the face is at the crossing point with BOTH reflected to right?? A→right and B→right: reflect d_A=(0.707,0.707)→(1,0) needs the 22.5°-tilt face; reflect d_B=(−0.707,0.707)→(1,0): n ∝ (1.707,−0.707) → (0.924,−0.383): surface dir (0.383,0.924) — tilted the OTHER way (+22.5°). Two different faces. So a single flat face can't do it — need TWO faces (a prism with a roof?) or... Alternatively, DON'T converge the signals; keep signals VERTICAL (down) from both slits and make the prism WIDE (as I had originally: face spanning x 565–685), and route the IDLERS to miss it: idlers diverge outward at 45°... idler B from (660,135) down-right CROSSES any face spanning x≥660 at y ≤ 135+(x−660)... face "/" line x+y=C: crossing at x=(C+525)/2. For face with x-span [565,685] catching signals at x=580,660: need C such that face y at x=580 = C−580 ∈ [150,200] → C ∈ [730,780]; at x=660: y=C−660 ∈ [150,200] → C ∈ [810,860]. CONTRADICTION — a 45° face can't be at y≈160-190 at BOTH x=580 and x=660 (45° slope: Δy=−Δx=−80: if y(580)=190, y(660)=110 — above the BBO!). So a 45° face catching both vertical signals sits at y(580)=y(660)+80. E.g., face x+y=C=830: y(580)=250, y(660)=170. Idler B crossing: x=(830+525)/2=677.5 ∈ [565,685] ✗ crosses. To avoid: face x-span upper end < 677.5 → face from x=560 to 675: catches signal A (580 ✓) and signal B (660 ✓); idler B crossing at 677.5 > 675 ✓ CLEARS by 2.5px. TOO TIGHT. Reduce slit separation in drawing: slits at x=590 and x=650 (60px apart): idler B line: x−y = 650−135 = 515. Face x+y=C: crossing x=(C+515)/2. Face spans [x1,x2] ⊇ [590,650]. y(590)=C−590, y(650)=C−650, want both in [155,230]: C∈[745,820]∩[805,880] → C∈[805,820]. C=812: y(590)=222, y(650)=162. Crossing with idler B: x=(812+515)/2=663.5; need x2 < 663.5: face x2=660 ✓ clears by 3.5px. STILL TIGHT. The idlers at 45° hug the prism corner unavoidably when slits are close. Reduce idler angle to 30° from vertical: idler B: dir (0.5,0.866): line: x = 650+0.577(y−135). At the prism face x+y=812: 650+0.577y−77.9+y = 812 → 1.577y = 239.9 → y=152.1, x=659.9: face span ends at x2=660 → crossing AT the corner (659.9≈660) ✗✗ ARGH. 20° from vertical (dir (0.342,0.94)): x=650+0.364(y−135); face: 650+0.364y−49.1+y=812 → 1.364y=211.1 → y=154.8, x=655.6: face span [560,660]: 655.6 < 660 ✗ crosses. Hmm!! The problem: idler B (from x=650 heading right-down) inevitably passes through the region right where the prism's right portion sits (the prism must span x=650 to catch signal B, at y≈162; idler B at y=162 is at x=650+0.364·27=659.8 — just past 650... within the face span if the face extends to 660). To clear, face must end before x≈655 at y≈155... but face needs to reach x=650 AT y=162 (to catch signal B). The face endpoint at x=650 is (650,162); idler B passes (659.8,162) — 9.8px to the right ✓; and at y=155: face is at x=657 (since x=812−155=657), idler B at x=650+0.364·20=657.3 ✗ CROSS AT (657,155). The face's upper portion (x>650, y<162) sticks out into idler B's path! SOLUTION: TRUNCATE the face: the face doesn't need to extend beyond x=650! Face x-span: [560, 650] exactly (catches signal A at x=580→y=232?? wait C=812: y(580)=232, y(650)=162). Face from (560,252) to (650,162). Idler B at y≥162 is at x≥650... wait idler B: x=650+0.364(y−135): at y=162: x=659.8; the face at x≤650 → idler B always RIGHT of face's right edge? At y=135 (BBO): idler B at x=650; face at x=812−135=677 — the face's extension would be at 677 but truncated at 650 ✓. So idler B passes to the right of the truncated prism ✓✓. And idler A (from x=590, dir (−0.342,0.94)): x=590−0.364(y−135): at y=232 (where face is at x=580): idler A at x=590−35.3=554.7 < 580 ✓ left of face ✓; face's left edge (560,252): idler A at y=252: x=590−42.5=547.5 < 560 ✓ clears ✓. And the pump (x=620 vertical): hits face at y=192 (620: y=812−620=192) — pump passes THROUGH the prism?? Physically the GT prism would deflect/absorb the pump... schematic: draw the pump beam STOPPING at the prism? But then the beam dump is unnecessary — even cleaner! But hmm, the GT prism story: it separates signal (reflected) from idler (transmitted) — the pump also hits it. Real setups: after the BBO, the remaining pump is dumped — often the GT prism itself or a dichroic handles it. SIMPLEST SCHEMATIC: pump beam ends at the prism (absorbed/blocked — label the prism "P (Glan–Thompson prizması — sinyali ayırır, pompayı bloklar)"... or just let the pump visually terminate at the prism with a small "×" or fade). Hmm, actually — a cleaner idea: put a small pump BLOCK right under the BBO... but the signals/idlers emanate from the BBO downward through that area... The signals (vertical at x=590, x=650) and the pump (x=620) — a narrow block at (620, 150–158) between the two signal beams, BEFORE the prism?? But the signal beams are only 60px apart (590, 650); a block at x=620±6 fits between them ✓!! And the idlers at y=150: idler A at x=590−0.364·15=584.5 (left of 614 ✓), idler B at 655.5 (right of 626 ✓). So: small beam dump at (620,150), w=12,h=14 — pump stops there. Then the prism below catches signals only, and pump is gone before the prism ✓✓ CLEAN! Wait, but is the pump physically blocked there? In reality the pump would continue... but there IS often a beam block or the pump walks off... eh — schematic license, it's fine and looks intentional. Hmm, actually, hmm: the dump between the two down-converted beams is a bit odd (real experiments: the pump is removed by dichroic filters at each detector, or the geometry separates it). Honestly for the schematic it's fine and tidy: "pomp ışını bloğu". OR even simpler: let the pump terminate AT the BBO (absorbed/transmitted harmlessly — actually! Type-I/II BBO is transparent to the pump; the pump continues and hits the prism...). You know what — dump at (620,150) it is. Tidy. Hmm wait, actually — I realize I should double-check the signal verticality choice: signals travel VERTICALLY down from slits at x=590 and x=650 to the prism, then reflect RIGHT. The reflected signal beams: signal A reflects at (590,222), signal B at (650,162). Reflected: both travel right: A along y=222, B along y=162 — two parallel horizontal beams 60px apart, both heading to the lens region! But PHYSICALLY the two signals should be COHERENT/OVERLAPPING (they interfere!). Drawing them as two separate parallel beams is... actually FINE and common in schematics — hmm, but no wait: the whole point is the signal from A and B OVERLAP and interfere at D0. Two separate beams to D0?? D0 is ONE detector on a rail. If beam A is at y=222 and beam B at y=162, they'd hit the rail at different spots — WRONG. The diffraction fan means each signal photon has an angle; the reflected beam should be a FAN spreading vertically. The two discrete beams (A at y=222, B at y=162) represent the "straight-through" rays of each slit; the FAN spans between and beyond: after reflection, the fan is bounded... Let me think again: signal photons leave slit A at all angles θ_A ∈ [−θmax, θmax] (relative to vertical), same for B. The PRISM reflects those heading down (all of them, it's wide enough... no wait — the prism is finite: face x∈[560,650]. A photon from slit A at x=590 with angle θ: hits the face plane... its trajectory: x=590+tanθ·(y−135); hits face when x+y=812: 590+tanθ·(y−135)+y=812 → y(1+tanθ)=222+135·tanθ → y=(222+135 tanθ)/(1+tanθ). For tanθ=0.3: y=(222+40.5)/1.3=202 → x=590+0.3·67=610 ✓ within face ✓. For tanθ=−0.3 (leftward): y=(222−40.5)/0.7=259 → x=590−0.3·124=553 — face left edge is 560: 553<560 ✗ MISSES the prism (passes left of it). So leftward-angled photons from A miss the prism! The prism can't catch the whole fan when it's offset... SCHEMATIC, PEOPLE. It's a schematic! No real schematic ray-traces the full fan through every element. The fan is drawn as a soft-shaded wedge; the animated PHOTON DOTS follow clean idealized paths (from BBO center-ish → prism center → lens → D0 rail position). NOBODY will ray-trace my photons' incidence points. DECISION: - Draw a soft FAN wedge from the slit pair through the prism bending rightward toward the lens (two wedge edges, symmetric about the horizontal centerline y≈192), purely decorative. - Animate signal photons: born at (slit_x, 135) → travel down to prism face along... honestly, animate: (slit_x,135) → prism hit point computed on the face line (for the born-slit's straight ray: A→(590,222), B→(650,162)) → reflect → travel right at y_hit → lens at x=880 → after lens, head to (1060, y_target) → D0 rail. The photon's y_target depends on its sampled diffraction angle θ: y_target = 192 − (λf mapping) i.e., y_target = 192 − x_mm·20. And between prism and lens, the photon travels at y_hit (its slit's exit height), then the lens redirects it to y_target. WAIT — that's WRONG physics: the lens maps ANGLE to position, not "bends from slit height to target". The slit height (near-field position) should NOT affect the far-field position. If I draw A-exit photons at y=222 and B-exit at y=162 then lens-bending them to arbitrary y_target — visually implies the lens bends rays by different amounts depending on target — which is... actually EXACTLY what a lens does with a fan: each incoming ANGLE maps to an output height. The problem: I'm drawing A-photons always entering the lens at y=222 regardless of θ, then exiting to y_target. The incoming angle would differ per θ... whatever — a viewer sees: photons leave the prism at two heights (the two slits' beams — labeled "A sinyali", "B sinyali"?), pass through the lens, and get focused to their target height on the D0 rail. Hmm, but two horizontal input beams at fixed heights with outputs at varying heights = the lens looks like it's doing magic redirection. Alternatively: merge visually — draw the signal photon AFTER the prism at height y_hit + small θ-based offset that GROWS toward the lens: incoming height at lens = y_hit + tanθ_draw·(880−x_hit) where tanθ_draw = (y_target−192)/180·k... SIMPLEST CLEAN OPTION: animate the signal photon as: born at (slit_x,135) → to prism point (x_face(slit_x), y_face) → reflect → travel to lens point (880, y_mid) where y_mid = y_hit·0.35 + y_target·0.65?? ... hmm hmm. OR EVEN SIMPLER: animate signal photon: born (slit_x,135) → prism (face point) → SINGLE SEGMENT from prism face point directly to (1060, y_target), passing THROUGH the lens (lens just drawn on the path; the segment bends slightly AT the lens): segment 1: prism point → (880, y_lens_in) with y_lens_in = interpolate(prism_y, y_target, (880−x_prism)/(1060−x_prism))... i.e., just draw the straight line prism→D0-target and place the lens ON it — the "kink" at the lens: to make it look lensy, make segment 1 have slope = (y_target−y_prism)/(1060−x_prism)·0.7 and segment 2 steeper... ENOUGH. FINAL: draw TWO segments: prism→lens-edge and lens-edge→D0, where lens-edge point y = y_prism + (y_target − y_prism)·(880−x_prism)/(1060−x_prism) — i.e., EXACTLY on the straight line (no kink), and add a subtle glow at the lens crossing. Clean, simple, looks like the photon goes straight (as rays do through the center of a lens!). Actually a ray through the lens CENTER goes straight! So if the geometry roughly passes near the lens center region — perfect, that's the chief ray. FINE. Actually hmm, the fan: photons to y_target=45 (top) from prism y=162: the line from (650,162) to (1060,45) — crosses x=880 at y = 162−(45−162)·(230/410)= 162−117·0.561=96. OK inside lens aperture (40–330) ✓. From A (590,222) to y=325: at x=880: y=222+(325−222)·(290/470)=222+103·0.617=285 ✓ within lens. ✓ All good. NOW — what about the two discrete signal beams from A and B (y=222 and y=162) — do I draw them as static beams? YES: draw both static faint beams (labeled subtly or not), plus the DECORATIVE FAN wedge... hmm, actually drawing both discrete beams AND a fan is redundant/conflicting. Just draw the two static beams from slits→prism (vertical segments at x=590, x=650 from y=135 to face), then from prism: two horizontal beams at y=222 and y=162 to the lens, and from the lens a soft FAN spreading to the full rail extent. This reads as: "two slits' signals, collimated through, lens spreads them over the scan range". Physically eh (the fan should exist before the lens too), but visually CLEAR and standard-looking. GOOD. Wait, one more consideration — actually, physically with the Fourier lens: the two collimated input beams (from A and B, parallel) enter the lens; parallel beams at different heights... focus to different points?? A Fourier lens maps input ANGLE to focal position; input HEIGHT maps to output angle. Two parallel beams at different heights → both focus at the SAME focal point (angle=0 → x_f=0) but exit at different... hmm, so physically after the lens they'd converge to the center?? THAT'S WRONG for our story — the photons must land at various x! The landing x is set by their DIFFRACTION ANGLE from the slits, which in my static-drawing got collapsed (I drew only the θ=0 rays as static beams, but the animated photons carry the true angles). The static beams are just decoration at θ=0; the animated photons show the real fan (each at its own angle→target). The lens fan-drawing after the lens = the spread of targets ✓. I think visually it all hangs together: static θ=0 beams + animated photons fanning to their targets + a soft fan wedge after the lens. Hmm, OK wait, I also want to double check the IDLER static beams and make sure the full idler network is consistent with slit positions x=590 (A) and x=650 (B), BBO y=135, idler angle 20° from vertical (dir (±0.342, ±0.94)): - Idler A: from (590,135), dir (−0.342,0.94). Path: param (590−0.342s, 135+0.94s). - Idler B: from (650,135), dir (0.342,0.94). - Prism face: x+y=812, truncated x∈[560,650]: verified idlers clear ✓. - Where do idlers hit BSa/BSb? Choose BSa at y=340: s=(340−135)/0.94=218 → x=590−74.6=515.4 → BSa at (515,340). BSb at (725,340) (symmetric: 650+74.6=724.6 ✓). - BSa plate: vertical (to turn the 20°-off-vertical beam): we want REFLECT to route toward BSc. BSc position: let me place at (620, 480). From BSa (515,340) to BSc (620,480): dir (105,140) → norm (0.6,0.8). Reflect d=(−0.342,0.94) → d'=(0.6,0.8): n ∝ d'−d = (0.942,−0.14) → (0.989,−0.147) — plate surface dir (0.147,0.989): plate tilted ~8.5° from vertical. Incidence: cos = |d·n| = |−0.338−0.138|=0.476 → 61.5° — quite oblique for a plate BS but OK schematically (drawn as a tilted vertical plate). Hmm — alternatively make BSc farther right-down to reduce the turn... The turn from d (200° ... ugh). Turn angle between d and d': cos = d·d' = −0.205+0.752=0.547 → 56.8° turn. Plate bisects: 28.4° from each. Plate vs vertical: d is 20° from vertical; plate is 28.4° from d on the other side → plate at 8.4° from vertical on the... yes ~8.5° ✓ as computed. FINE — draw it. Hmm, wait — should I double-check the TRANSMITTED beam at BSa (which-path to Ma→D3)? Transmit: continue dir (−0.342,0.94) from (515,340): → Ma at y=480: s=149 → x=515−51=464 → Ma at (464,480). Ma reflect → straight down (0,1): n ∝ (0,1)−(−0.342,0.94) = (0.342,0.06) → (0.985,0.173): surface dir (−0.173,0.985): tilted ~10° from vertical. → D3 at (464, 590) (beam down from Ma (464,480) to (464,590)) ✓. Hmm — Ma at (464,480) and BSc at (620,480): same height — and D3 at (464,590), D1/D2 below BSc at y≈590 too?? Layout getting crowded but symmetric. Hmm, wait — actually, let me reconsider: it might be cleaner to give the which-path branch MORE separation from the erasure branch: Ma/D3 further out-left: Ma at (420, 530)? transmit beam from BSa (515,340) dir (−0.342,0.94): at x=420: s=278 → y=340+261=601 — too far. Ma at (464,480) is what the geometry gives with straight-down output. Alternatively Ma reflects the transmitted beam to down-left at steeper angle... keep simple: Ma→straight down→D3. Actually hmm, wait. Let me reconsider the idler beam angle: 20° from vertical gives beams that separate slowly. What if idlers go at 45° (dir (±0.707,0.707))? Then BSa at... idler A from (590,135) at 45° down-left: BSa at (480,245)? (Δ(−110,110) ✓). BSc could be at (620, 400)? From BSa (480,245): dir to BSc (140,155) — not 45° (49°). Reflect d=(−0.707,0.707)→(0.67,0.74): n ∝ (1.38,0.03)→(1,0.02): surface ~VERTICAL (tilt 1°)! ✓✓ Because the beams are symmetric about vertical, a vertical plate reflects left-going to right-going. So: BSa = VERTICAL plate at (480,245): transmit → continue 45° down-left → Ma at (390,335) → reflect to straight down?? n ∝ (0,1)−(−0.707,0.707)=(0.707,0.293)→(0.924,0.383): surface dir (−0.383,0.924): 22.5° tilt ✓ → D3 at (390, 445). REFLECT at BSa → 45° down-right → BSc at (620, 385): from (480,245): Δ(140,140) ✓ EXACTLY 45° ✓✓ BEAUTIFUL. Symmetric for B: BSb vertical plate at (760,245): transmit → Mb (850,335) → D4 (850,445); reflect → 45° down-left → BSc (620,385) ✓ (Δ(−140,140) ✓). - **BSc**: at (620,385), receives A-beam (dir (0.707,0.707)) from up-left and B-beam (dir (−0.707,0.707)) from up-right. Vertical plate: A transmit → continues down-right → D2 at (710,475) ✓ (Δ(90,90)); A reflect → down-left → D1 at (530,475) ✓. B transmit → down-left → D1 ✓; B reflect → down-right → D2 ✓. PERFECT — all 45° beams, vertical plates at BSa/BSb/BSc, 22.5°-tilted mirrors at Ma/Mb. CLEAN GEOMETRY. ✓✓✓ But now re-check prism clearance with 45° idlers: idler A: x+y=725 (590+135); idler B: x−y=515 (650−135). Prism face x+y=C truncated [x1,x2]: idler A parallel ✓ never crosses ✓. Idler B crosses at x=(C+515)/2: need outside [x1,x2] (above x2): C such that... face must catch signal A (x=590) and signal B (x=650): x1≤590, x2≥650; face y at 590: C−590 ≥ ~150 → C≥740; y at 650: C−650 ≥ 150 → C ≥ 800; y at 590 ≤ ~260 → C ≤ 850. Idler B crossing x=(C+515)/2 ≥ (800+515)/2 = 657.5 > 650 = x2 ✓✓ CLEARS if C≥800! Take C=805: face from (560,245) to (655,150)?? endpoints (x,805−x): choose x1=560 → (560,245); x2=655 → (655,150). Signal A hits face at (590,215); signal B at (650,155) ✓. Idler B crossing at x=660 > 655 ✓ clears by 5px... want more margin: x2=652 → face (560,245)–(652,153); crossing at 660 ✓ 8px. Hmm, and idler A: parallel line x+y=725 vs face x+y=805: separation 80/√2=56.6px ✓ but the face's LOWER-LEFT END (560,245): idler A passes (x+y=725): at y=245: x=480 — far left of 560 ✓ no conflict. ✓. And the pump beam dump at (620,150): pump from laser x=620 hits face at y=805−620=185 — the face is AT (620,185); dump must be ABOVE: at (620,148)?? — but the BBO is at y≈125–140... dump between BBO (140) and face-at-620 (185): dump at (620,158), h=12: y∈[152,164] ✓ fits. Idler A at y=158: x=590−23=567 — left of dump (620±7=613–627) ✓; idler B: x=650+23=673 ✓ right ✓. Signals (vertical at 590,650) pass the dump's y-range at x=590/650 ✓ clear of 613–627 ✓✓. Hmm, the dump at (620,158) sitting right below the BBO — visually: pump beam from laser (620,58)→slits(620,100)... wait the pump goes through the SLIT PLATE at x=620: the slit plate spans x 545–695 with gaps at x=580–600 (A) and x=640–660 (B)?? Hold on — slit positions: A at x=590, B at x=650. Slit gaps: [580,600] and [640,660]. The pump beam at x=620±? — the pump must pass through the PLATE (blocked!) at x=620?? NO WAIT — the pump must go THROUGH THE SLITS! The pump illuminates both slit gaps: pump beam centered x=620, wide enough to cover both gaps: draw pump beam as a wedge from laser port (620±10) at y=58 expanding to (620±45) at y=100 — covering gaps 580–600 and 640–660 ✓ (wedge edges at y=100: x∈[575,665] ✓ covers both). The center of the plate (x∈[600,640] at y=100–112) blocks the pump's center — FINE, the wedge shows light passing through both gaps and the plate blocking the middle ✓ realistic! After the slits, the transmitted pump: two beamlets through gaps A and B... which then hit the BBO (y=125–140) creating pairs at both slits ✓✓ and the remaining pump continues down through the dump?? The pump beamlets at x=590 and x=650 continue down past the BBO... they'd overlap the SIGNAL beams (also at x=590,650)! Physically true (pump co-propagates with down-converted light until separated). Schematic: draw faint pump continuation to the dump — but the dump is at x=620, not 590/650... The pump beamlets at 590/650 would hit the PRISM FACE (at (590,215),(650,155))... and the GT prism reflects the pump too?! In reality: the GT prism separates by polarization — type-II SPDC: signal & idler have orthogonal pols; pump has... the pump (351nm) — GT prism transmits UV... honestly, REAL LAYOUT: after BBO there's typically a dichroic or the GT prism + filters handle the pump; schematic figures often just don't draw the pump downstream. DECISION: draw the pump beam ending at the BBO with a subtle "absorbe/filtered" fade (label the BBO region: pump is converted/absorbed; add tiny "pomp filtresi" note). Cleanest: pump wedge laser→slits, faint continuation slits→BBO, then STOPS (absorbed by BBO mount/filters — add a tiny gray block right under the BBO spanning x 570–670?? that would block the signal beams too!). NO BLOCK: just fade the pump out below the BBO with a small label "pomp filtrelenir". DONE. Actually — simpler visual trick: draw the pump beam with dashed continuation through BBO ending in a subtle "⊗"-ish fade + label. FINE. Moving on!! Wait, one more geometry check — the SIGNAL beams and the IDLER beams near the BBO: signal A: vertical from (590,135) down to prism face (590,215). Idler A: 45° from (590,135) to BSa (480,245) — wait: from (590,135) going down-left at 45°: at y=215: x=590−80=510. Signal A at x=590 vertical ✓ no overlap ✓. Idler B from (650,135) down-right: at y=215: x=730 ✓ clear of signal B (x=650) ✓. And BSa at (480,245): the idler A beam from (590,135) to (480,245): Δ(−110,110) ✓ 45° ✓. Prism face lower end (560,245) — vs BSa (480,245): 80px apart ✓. BSb (760,245) vs face upper end (652,153): fine ✓. And what about the SIGNAL beams reflected rightward at y=215 (A) and y=155 (B) — do they cross the IDLER beams? Idler B from (650,135) down-right at 45°: crosses y=155 at x=670, y=215 at x=730. Signal B horizontal beam at y=155 from x=652 rightward: crosses idler B?? Idler B at y=155 is at x=670: signal B beam (y=155, x from 652 to lens 880) passes through (670,155) — CROSSES idler B ✓ (beams crossing in free space — fine, happens in every optical layout; draw idler beams distinct color). Signal A beam y=215 from x=590→880 crosses idler B at (730,215) ✓ crossing OK. And idler A crosses signal beams? Idler A: x+y=725: signal A at y=215: crossing x=510 — signal A beam starts at x=590 → no ✓. Signal B y=155: crossing at x=570 < 652 ✓ no. ✓ Only two clean crossings. Also check: BSa reflected beam (erasure) from (480,245) to BSc (620,385) at 45°: crosses anything? The transmitted which-path beam from BSb... BSb transmit: from (760,245) down-right 45° to Mb (850,335) — right side, no. BSb REFLECT: (760,245)→(620,385) down-left 45°: parallel to... does it cross BSa's reflect beam? They CONVERGE at BSc (620,385) ✓ meet exactly there ✓. Does BSb's reflect beam cross BSa's TRANSMIT beam (from (480,245) down-left)? No (left side) ✓. Cross the Ma→D3 vertical beam (x=390)? No ✓. Do BSa/BSb reflected beams cross the SIGNAL beams? Signal beams are at y=155/215, x∈[590,880]; BSa-reflect beam from (480,245) to (620,385): y∈[245,385] — below signal beams ✓ NO ✓. - **BSc outputs**: D1 at (530,475), D2 at (710,475): beams from BSc (620,385) at 45° down-left/right ✓. Cross Ma→D3 beam (x=390, y 335–445)? D1 beam from (620,385) to (530,475): x∈[530,620] — no ✓. - **Detectors D1–D4**: draw as semicircle detectors (classic optic symbol: D-label with a semicircle facing the incoming beam): - D3 at (390, 445+): incoming from above (Ma→D3 vertical): semicircle facing UP at (390,460). Label "D3 (yol A)". - D4 at (850, 460): facing up. "D4 (yol B)". - D1 at (530−?, 475+): incoming from up-right (45°): face up-right: place D1 at (515, 495) semicircle facing up-right, i.e., rotated 45°. Label "D1 (silgi)". - D2 at (725, 495) facing up-left. "D2 (silgi)". Hmm — wait, I had D1 at (530,475) along the beam from BSc; let me set: BSc→D1: from (620,385) 45° down-left, length ~127: D1 at (530,475) ✓; BSc→D2: (710,475) ✓. - **Coincidence counter / electronics**: bottom of canvas (y 560–760): draw a box labeled "Çakışma sayacı (coincidence counter)" with lines (cables) from all D0–D4 into it, and outputs to the histogram display. REALISM ✓ (Kim's setup: all detectors feed a coincidence circuit). Cables: thin gray curves from each detector to the box. The box: x 300–940, y 620–700? Draw simple: rounded rect with "ÇAKIŞMA SAYACI" + small LED-style channel indicators D0D1..D0D4 that FLASH when a coincidence registers!! Nice interactive feedback ✓✓. Hmm wait, D3 at (390,460) and the coincidence box at y=620+: D1/D2 at y≈475–495, D3/D4 at ~460. Box below at y 580–680, x 250–990. Cables from detectors down into the box ✓. Let me now also reconsider: is there ROOM for the histogram/graph panel? I'll make the layout: LEFT/TOP = optical table (SVG ~1240×760?), RIGHT or BOTTOM = data panel. Better: overall page: header (title), main flex: optical-table SVG (left, ~62%) + control/data column (right ~38%) with: (a) controls, (b) histogram canvas with channel buttons, (c) counts + explanation text. On narrow screens stack vertically. The optical table SVG viewBox maybe 1240×760 — with the coincidence box at its bottom. Let me re-budget vertical space in the SVG: - y 0–60: laser + labels. - y 60–140: pump, slits, BBO. - y 140–260: prism, dump, idler beams start, signal beams out. - y 240–500: idler network (BSa/BSb y=245, Ma/Mb y=335, BSc y=385, D1–D4 y~455–500). - y 560–680: coincidence counter box + cables. - The D0 rail/lens: x 880–1120, y 40–345 (right side, vertically spanning) — the signal beams region y 150–260 extends right into x 880+. Lens at x=880 (vertical bar y 40–345), rail at x=1060. D0 head on rail. Hmm — but wait: the idler network (BSb at (760,245), Mb (850,335), D4 (850,460)) — and the SIGNAL beams pass overhead at y=155/215 through x 590–880 — BSb at (760,245) is BELOW signal beam A (y=215)?? BSb at y=245 — 30px below signal beam A's line ✓ no overlap ✓. Mb at (850,335), D4 (850,460): the LENS is at x=880, y 40–345!! Mb at (850,335) is 30px left of the lens bottom ✓ CLOSE but OK. D4 at (850, 460) vs lens (x=880, y≤345): clear ✓. Hmm, but the D0 rail at x=1060, y 30–345 — and D4 at (850,460) below-left ✓ fine. The coincidence box (x 250–990, y 560–680) — below everything ✓. Actually — wait, D2 at (710,475), D1 (530,475), D3 (390,460), D4 (850,460): box y 560 ✓ clear. Hmm, let me reconsider widths: elements: laser (620), slits A(590)/B(650), prism (560–652), dump (620,158), BSa (480,245), BSb (760,245), Ma (390,335), Mb (850,335), D3 (390,460), D4 (850,460), BSc (620,385), D1 (530,475), D2 (710,475), lens (880), rail (1060), D0 head (1060), plus labels sticking out. X-range roughly 330–1120. viewBox 0 0 1240 760 ✓. Signal center height y≈185 (between 155 and 215). Hmm wait, one thing: the D0 rail spans y 30–345 at x=1060, and the coincidence box at y 560–680 spans x 250–990... the cable from D0 (1060, y_D0) must route to the box: draw a cable from D0 head right/down around... from (1080, y_D0) → right to x=1150 → down to y=620 → left into box at (990,620)?? The box right edge at x=990; cable path (1080,y)→(1160,y)→(1160,630)→(990,630)?? hmm 1160 > viewBox 1240 OK ✓. Fine — polyline cable. Hmm, actually, let me reconsider the overall canvas ratio: viewBox 1240×760 is wide (1.63). The optical table area is fine. Let me also double check BSc height: BSc at (620,385), but Ma/Mb at y=335 and the which-path beams from BSa: (480,245)→(390,335) 45° ✓; BSc receive beams (480,245)→(620,385): hmm wait — that's Δ(140,140) ✓ 45° ✓ and (760,245)→(620,385) Δ(−140,140) ✓ ✓. And D1/D2 beams: (620,385)→(530,475) Δ(−90,90) ✓; →(710,475) ✓. WAIT — I should double-check something physical about BSc: A-photon arriving at BSc going down-right (0.707,0.707); vertical plate; transmit → continues (0.707,0.707) → down-right → D2 (710,475)?? From (620,385): continuing down-right at 45°: (620+t·0.707, 385+t·0.707): at t=127: (710,475) ✓ D2 is down-RIGHT. Reflect → (−0.707,0.707)?? reflection of (0.707,0.707) about vertical plate (n=(1,0)): d' = d−2(d·n)n = (0.707,0.707)−(1.414,0)=(−0.707,0.707) ✓ down-left → D1 (530,475) ✓. And B-photon arriving down-left (−0.707,0.707): transmit → D1 ✓; reflect → (0.707,0.707) → D2 ✓. So: D1 receives: A-reflected + B-transmitted; D2: A-transmitted + B-reflected ✓✓ both mixed ✓ ERASURE ✓. And at BSa (vertical plate at (480,245)): idler A arrives going down-left (−0.707,0.707): transmit → continues down-left → Ma (390,335) ✓ [Δ(−90,90) ✓]; reflect → (0.707,0.707) down-right → BSc ✓. So which-path (D3) = TRANSMITTED at BSa; erasure = REFLECTED. As discussed — physically equivalent to Kim's (his had erasure on transmit, doesn't matter — actually hmm, wait: does it matter for the "erasure" condition that both paths to D1 have the same... the erasure requires indistinguishability: path A→BSa(reflect)→BSc(reflect)→D1 vs path B→BSb(reflect)→BSc(transmit)→D1 — different reflection sequences — but in the quantum description the beamsplitter transformations make the path states at D1 identical up to phase ✓ standard. FINE.) NOW — the ANIMATION photon paths (idler): - Pair born at slit A (590,135) [or B (650,135)]. - Idler dot: (590,135) → BSa (480,245) [45°]. At BSa: - If which-path (D3): → Ma (390,335) → D3 (390,460). - If erasure: → BSc (620,385). At BSc: → D1 (530,475) or D2 (710,475). - For erasure outcomes, ALSO animate the "partner ghost" from the other slit: ghost from B (650,135) → BSb (760,245) → BSc (620,385) — arriving at BSc simultaneously with the real one — then ONE dot continues to D1/D2. This shows the superposition/erasure beautifully: TWO ghost idlers (from A and B) merge at BSc ✓✓. And for which-path outcomes: single idler dot (the true path), maybe with a subtle "yol bilgisi" tag. LOVE IT. Wait, hmm — but for which-path (D3): the idler went through slit A — draw single dot from A. But the pair was born at slit A — so also the SIGNAL photon came from slit A... but we don't know that until... NO WAIT — for D3 outcome, the pair REALLY was at slit A (path entanglement: |A,A⟩+|B,B⟩; detecting idler at D3 projects to |A⟩). The signal photon "was" at slit A too. But the D0 detection happened EARLIER in time — the animation shows the signal landing at x with the correct conditional distribution P(x|D3) ∝ |ψ_A(x)|² (single-slit envelope) — no fringes ✓. We animate the signal the same way regardless (dot from BBO area → prism → lens → D0 at y_target) — sampled from P(x|k) ✓. The signal dot's birth point: at the slit where the pair was born (A or B) — hmm, but for erasure outcomes the signal is in superposition of both slits... ANIMATE signal as TWO ghost dots from both slits converging at the prism?? Then one dot continues to the lens/D0. For which-path outcomes: single dot from the true slit. This nicely mirrors the idler treatment and shows superposition consistently. But again — the signal animation must NOT reveal the future outcome... if for D3/D4 outcomes the signal is drawn single-slit and for D1/D2 double-ghost, then during the signal's flight (before idler detection) the viewer ALREADY knows — bad pedagogy? Hmm — actually NO WAIT: even for D3/D4 outcomes, at the time of the signal's flight, the outcome isn't determined... but in a Monte Carlo we've pre-sampled it. The REAL question is pedagogical: showing the signal's path through a specific slit for D3-outcomes is "retro-causal" storytelling — exactly the misconception to avoid!! CORRECT approach: ALWAYS animate the signal as two ghosts (superposition through both slits — this is the true state until detection at D0!) converging then single detection at y_target. And the idler: two ghosts from both slits... wait — the idler state is ALSO always superposed (|A⟩+|B⟩) until ITS detection. So ALWAYS: pair birth: ghost pair at BOTH slits (2 ghosts × ... hmm, the STATE after SPDC: (|A,A⟩+|B,B⟩)/√2: one signal + one idler, both at slit A — or both at B. So the ANIMATION of a single pair: 2 signal ghosts (A,B) + 2 idler ghosts (A,B) = 4 dots, born simultaneously at the two slits. Then: signal ghosts → prism → merge → single signal dot → lens → D0 (click at y_target). Idler ghosts → BSa/BSb → IF erasure outcome: both ghosts continue to BSc, merge, single dot → D1/D2 (click). IF which-path outcome: one ghost FADES (the wrong-slit amplitude — "collapses away") and the true-slit ghost continues to Ma/Mb → D3/D4 (click). Hmm — when does the wrong idler ghost fade? At BSa/BSb: for D3 outcome, the B-ghost at BSb: in reality the B-amplitude also reflects/transmits... the "collapse" happens upon DETECTION at D3. So the B-ghost should travel to BSc and split to D1/D2... and then FADE when D3 clicks (no detection there). Hmm — that's the honest picture: all amplitudes propagate everywhere; detection at ONE detector kills the others. But visually chaotic (ghosts flying to D1–D4 that then fade). SIMPLIFICATION: fade the non-detected amplitudes AT the moment of the true detection (they vanish with a subtle "poof"), but let them fly the full path until then. Hmm — but then during the flight, ghosts are heading to D1/D2 AND a ghost heading to D3... when D3 clicks, the D1/D2-bound ghosts poof ✓. This is actually physically honest AND dramatic ("the idler's fate decided at the beam splitters — 50/50 quantum chance"). Let me do exactly this: - Idler ghosts: A-ghost: slit A → BSa: SPLITS (50/50) into: ghost-T (transmit → Ma → D3) and ghost-R (reflect → BSc). B-ghost: slit B → BSb → splits: ghost-T (→ Mb → D4), ghost-R (→ BSc → splits again → D1, D2). TOO MANY GHOSTS (5 idler ghost-paths per pair: A→D3, A→BSc→{D1,D2}, B→D4, B→BSc→{D1,D2}) — visual chaos! PRAGMATIC COMPROMISE (final): animate per-pair with the PRE-SAMPLED outcome but keep the signal ALWAYS superposed (2 ghosts → merge at prism → 1 dot → D0): - D3 outcome: idler = 2 ghosts from both slits → at BSa/BSb: A-ghost continues (transmit → Ma → D3 click); B-ghost fades at BSb (subtle dissolve + tiny "?" ... just fade). - D4: mirror-symmetric. - D1/D2 outcome: 2 ghosts → both continue (transmit... wait NO: for erasure the idlers REFLECT at BSa/BSb (my geometry): both ghosts reflect at BSa/BSb → converge at BSc → merge into 1 dot → to D1 or D2 (click). So: idler always starts as 2 ghosts (superposition ✓); at BSa/BSb either both continue (erasure) or one fades (which-path = "measurement" happens at the beam splitter pair — pedagogically defensible: the which-path info becomes AVAILABLE there). Hmm — pedagogically: the "fade at BSb" for D3 outcomes represents the amplitude for slit-B dying because D3 can only be reached from A... eh — close enough; add a caption explaining. OK. Hmm — actually WAIT, simpler idea for which-path: for D3/D4 outcomes, do the idler ghosts even need to start as 2? The idler state IS superposed at birth. But to reduce clutter, I could show for D3: 2 ghosts; at BSa, A-ghost transmits onward; B-ghost hits BSb and REFLECTS toward BSc but fades halfway (dissolving) — showing "that amplitude existed but the universe chose otherwise"... nah — fade AT BSb immediately (with a soft shimmer). Keep it simple: at the moment ghosts hit BSa/BSb (both arrive simultaneously — same path length ✓ symmetric geometry ✓), decide: erasure → both reflect toward BSc; which-path → the true-slit ghost transmits toward Ma/Mb, the other ghost dissolves AT its BS with a small "yok oldu" fade effect. GOOD. TIMELINE per pair (with all path lengths): - Pump: laser (620,58) → slit plate (620,100): length ~42px... the pump dot travels to... hmm — the pump photon goes through the slit — WHICH slit (superposition through both!)? Draw pump as SINGLE dot to the plate center (620,100), then it "splits" into the pair ghosts at BOTH slits?? Physically the pump diffracts through both slits and converts at the BBO... ANIMATE: pump dot (620,58)→(620,100) → continues to BBO (620,130)?? but pairs are born AT the slit positions (590/650,135)... simplest visual: pump dot travels to (620, 130) (BBO center) and "bursts" into 4 ghosts (2 signal at (590,135),(650,135) + 2 idler at same)? Hmm — the burst spawning ghosts at slit positions from a center impact: draw a subtle ring/sparkle at (620,130), then the 4 ghosts appear at (590,135) & (650,135). Close enough — OR the pump dot itself splits into two ghost pump-dots that pass through the two slits then hit the BBO and each converts... TOO MUCH. FINAL: pump dot: (620,58) → (620,100): at the slit plane, it becomes TWO pump ghosts (590,100),(650,100) (through the gaps — showing diffraction/superposition!), each → its BBO point (590,135)/(650,135), each converts: SPARKLE + signal ghost + idler ghost born there. So after conversion: signal ghosts at (590,135),(650,135); idler ghosts at same points. Signal ghosts: travel DOWN to the prism face (590,215)/(650,155)... wait — signal ghosts from the two slits travel vertically down, hit the prism face at DIFFERENT points (590,215) and (650,155), reflect, travel right at y=215 (A) and y=155 (B) — then MERGE where? They're parallel — never merge!! Hmm. For the merge, let them converge at the LENS: A-ghost: (590,215)→(880, y?) and B: (650,155)→(880,?)... they can converge at the lens center (880,185)?? then single dot → (1060, y_target). The kink: A-ghost travels right along y=215 then bends up to lens center?? — just draw A-ghost: (590,215) → (880,185)-ish with a slight slope; B-ghost: (650,155) → (880,185). So they converge at the lens center, merge → single dot exits toward (1060,y_target) — THROUGH the lens, bending to the target height ✓. And for the static beams drawing: static A/B beams drawn horizontally at y=215/155 — hmm, inconsistent with the sloping ghosts... just draw the static beams ALSO sloping toward the lens center: A-beam from (590,215) to (880,192); B: (650,155) to (880,178)?? Meh — you know what, draw the static signal beams as a soft WEDGE from the prism (between (590,215)-(650,155) edge... the prism emits a fan) toward the lens — a shaded quadrilateral from prism face region to the full lens height, PLUS the two thin "principal rays" if needed. DECISION: static: soft wedge prism→lens (subtle, orange, 12% opacity) + thin lines for the two principal rays ((590,215)→(880,~200)→ hmm...). KISS: two thin static lines: (590,215)→(880,192) and (650,155)→(880,178) — slightly converging at the lens — fine. And after the lens: static fan wedge from lens (880, y 40–345 aperture... the wedge from lens center (880,185) spreading to the rail (1060, 45–325)) ✓. These are decorative; photons animate on top. Hmm, hold on — actually wait. I keep going back and forth; let me LOCK the signal animation: signal ghosts: ghostA: (590,135)→(590,215) [down]; ghostB: (650,135)→(650,155) [down]; then ghostA: (590,215)→(875,190); ghostB: (650,155)→(875,180) — both arriving near lens center (875,185) — MERGE (they overlap in the last stretch) → single signal dot: (880,185) → (1060, y_target) [straight line through lens — chief-ray style] → D0 head at (1060, y_target): if |y_target − y_D0head| ≤ aperture_px: CLICK (flash + count) else: dot continues past (1100+, fades out) = "missed D0" (absorbed by the dark box). ✓ LOCKED. Wait — aperture: D0 aperture in px: x bins: ±7mm over rail y 45–325 (280px → 20px/mm); bin width 0.5mm = 10px; aperture = 10px. The dot lands at y_target (its sampled x); D0 head spans y_D0 ± 5px. Detection iff |y_target − y_D0| ≤ 5px. For the ANIMATED detected events (we only animate detections at higher speeds), y_target will be within the head — draw the click at the head ✓. For animated MISSES (only at slow speed): dot flies past ✓. Hmm — but WAIT: if I only animate detections, then y_target is ALWAYS ≈ y_D0 — the "landing at various heights" only shows in miss animations. FINE — at default speed, animate ALL pairs (including misses) up to a cap; at high speeds animate only detections (or a sample). Actually hmm, do misses even matter pedagogically? "D0 only sees a tiny slice of the pattern at a time — that's why it must SCAN" — yes, worth showing occasionally. Speed tiers: "yavaş (her foton)", "normal", "hızlı", "turbo": - yavaş: animate every pair incl. misses; rate ~2 pairs/s. - normal: animate every pair but skip some misses (animate 1 of 4 misses); rate ~6/s. - hızlı: animate ~ every detection + rare miss; statistical processing for the rest; rate ~60 pairs/s. - turbo: animate a detection every ~0.3s; rate ~2000 pairs/s. Implementation: each frame, advance statistical sim by (rate × dt) pairs (fractional accumulation); maintain an animation queue of "display events" (a detected pair every so often, or a miss every so often), spawned with proper random sampling (channel k by marginal P(k), x by P(x|k)) — for misses, sample x from... hmm, misses are x sampled OUTSIDE the aperture: sample x from P(x|k) conditioned on |x−x₀|>aperture... approximately: sample x from P(x|k) full; if inside aperture → it's a detection (counts + maybe animate); else miss (maybe animate). So the statistical loop literally simulates each pair: sample k (marginal: depends on BS configuration), sample x ~ P(x|k) (rejection or tabulated CDF), check aperture, accumulate. With rate 2000/s × 60fps = 33 pairs/frame — trivial JS ✓. Even 100k/frame fine. Rejection sampling with tabulated inverse CDF per channel (4096-entry tables) — fast ✓. MARGINALS for k: given BS states: - All BS in: P(D1)=P(D2)=P(D3)=P(D4)=1/4. - BSa out (idler A can't reach D3; A's idler... hmm wait — CAREFUL: if BSa is REMOVED, what happens to idler A? It continues straight through where BSa was → toward Ma → D3!! NO WAIT — BSa removed means no reflection: the idler A goes straight (transmit direction) → Ma → D3?? Then ALL A-idlers go to D3?! Hmm — that's geometrically what "BS removed" means in my layout: the beam continues along the transmit path. But the TRANSMIT path in my layout leads to Ma→D3 (which-path)! So removing BSa → all A idlers → D3 (always which-path for A). And removing BSc → the erasure-bound beams pass through BSc's position un-deflected → they'd continue 45° down-left/down-right to... D1/D2 positions?! A-erasure beam continues down-right 45° past BSc toward... (710,475) = D2's position!! Hmm — so BSc removed → A-erasure beam goes straight to D2, B-erasure beam straight to D1?? WAIT NO: A's reflect-beam from BSa (480,245) heads down-right 45°: passes BSc (620,385), continues to (710,475) — that's D2 ✓. B's beam from BSb (760,245) heads down-left: continues past BSc to (530,475) = D1 ✓. So BSc removed: D1 receives ONLY B-path light, D2 ONLY A-path light → which-path info available at D1/D2!! (D1 = "slit B", D2 = "slit A" — path info NOT erased!) — physically CORRECT! Without the final beamsplitter, the "erasure" detectors become which-path detectors! ✓✓ This is a REAL feature of the experiment (in Kim's paper: removing BSc... they discuss delayed choice via BSc insertion). PERFECT — the toggles now have genuine physical meaning: - BSa in: A-idler 50% → D3 (which-path), 50% → BSc. - BSa out: A-idler → always Ma → D3 (which-path always). - BSb similarly for B. - BSc in: erasure-bound idlers split 50/50 to D1/D2, paths MIXED → erasure ✓. - BSc out: erasure-bound idlers go straight: A→D2, B→D1 → NO mixing → which-path at D1/D2 ✓. - All in: D1,D2 = erased (fringes/anti-fringes); D3,D4 = which-path (blobs). - BSa+BSb out, BSc in: ALL idlers → BSc → D1/D2 (all erased; D3/D4 dark). - All out: A→D3... wait BSa out AND BSb out: A→Ma→D3, B→Mb→D4: all which-path. And BSc irrelevant. And the x-distributions: - D3 coincidence: P(x|D3) ∝ |ψ_A(x)|² (slit A envelope). - D4: |ψ_B(x)|² — with my symmetric geometry |ψ_A|²=|ψ_B|² — hmm, should I make them slightly different? Physically: slit A and B envelopes are identical (same width) but CENTERED at the same place (both slits near axis, far-field centers coincide). In the real Kim experiment D3/D4 patterns are two slightly different diffraction patterns (offsets due to geometry). For the sim: identical single-slit envelopes — fine, and note D3 = "slit A only", D4 = "slit B only". Hmm — wait, actually in my formulas: ψ_A(x) = sinc(β)e^{iφ/2}, ψ_B = sinc(β)e^{−iφ/2} — |ψ_A|²=|ψ_B|²=sinc²(β). OK identical. To make D3 vs D4 visually distinct I could offset the slits slightly... no — keep physically clean: both show the same single-slit envelope. The teaching: "no fringes" ✓. Hmm, but then D3 and D4 histograms look IDENTICAL — meh, acceptable (and true for symmetric layout). Actually — hmm, let me add a tiny realistic touch: give slit A and B slightly different... NO. Keep it honest: identical. The labels explain. - D1: P(x|D1) ∝ |ψ_A + e^{iα}ψ_B|²/... with α=0: 4sinc²cos²(φ/2) → fringes; D2: |ψ_A−ψ_B|² = 4sinc²sin²(φ/2) → anti-fringes ✓ (when BSc in and mixed). - BSc OUT: D1 = |ψ_B|² (B only!), D2 = |ψ_A|² — wait recheck: BSc out → D1 receives B-path only → P(x|D1) ∝ |ψ_B(x)|² = single-slit blob ✓; D2 → A only ✓. INTERESTING: so with BSc removed, D1/D2 become which-path (blob) too! And the TOTAL D0 still no-fringes always ✓. Hmm wait, one more: with BSa removed (A always → D3) and BSb in: erasure channel only gets B... B's idler reflects at BSb → BSc → D1/D2 — but ONLY B-path light at BSc → no mixing → D1/D2 = |ψ_B|² blobs too. So channels: D3: |ψ_A|², D1/D2: |ψ_B|², D4: dark (BSb reflects 50%... wait BSb IN: B idler 50% → D4, 50% → BSc). Hmm right: BSa out, BSb in: A→D3 (100%), B→ 50% D4, 50% BSc→{D1,D2}. D1/D2 receive B-only → blobs. OK — the simulation logic handles all this generically: for each pair, determine the idler's ROUTE by the BS states (random where applicable), and the x-distribution = the correct coherent sum for that detector's MIXTURE of paths. GENERAL RULE: for detector Dk, let S_k = set of slits whose amplitude can reach Dk (given BS states): if |S_k|=2 (mixed): P(x|Dk) ∝ |ψ_A(x) + e^{iθ_k}ψ_B(x)|² where θ_k is the relative phase for that detector (D1: 0; D2: π; when BSc out: D1/D2 single-path anyway); if |S_k|=1: P(x|Dk) ∝ |ψ_{s}(x)|². Wait — careful with BSc IN and both paths to D1: relative phase: A→BSa(ref)→BSc(ref)→D1: reflections: 2 (BSa, BSc) + ... B→BSb(ref)→BSc(trans)→D1: 1 reflection. Relative phase from reflections: each reflection adds i (π/2) for a symmetric BS... The standard result: D1 and D2 get fringe/anti-fringe — I'll set θ_D1 = 0, θ_D2 = π. ✓ (matches sin²/cos² complementarity). Hmm hold on, also the case BSa in, BSb out, BSc in: A: 50% D3, 50% → BSc; B: 100% → Mb → D4. BSc gets A-only → D1/D2: |ψ_A|² blobs. D3: |ψ_A|², D4: |ψ_B|². ✓ handled by the generic rule. NICE — the generic rule handles everything. Also P(k) marginals: each pair: - A-route: BSa in: 50% D3-path else BSc-path; BSa out: 100% D3-path. - B-route: BSb in: 50% D4-path else BSc-path; BSb out: 100% D4-path. - BSc-path (from slit s): BSc in: 50% D1, 50% D2 (MIXED if both slits present, single if one); BSc out: 100% → (s==A ? D2 : D1). Then the pair's slit is chosen 50/50 (A or B) — wait, NO — hmm: each PAIR is born at one slit?? NO!! The pair is born in SUPERPOSITION of both slits (that's the entanglement!). The idler ROUTE isn't "slit A's idler goes..." — the idler's wavefunction covers BOTH routes simultaneously. The DETECTION at Dk collapses. Monte Carlo: the joint probability P(k, x) = ... Let me define amplitudes properly: State: (|A⟩_s|A⟩_i + |B⟩_s|B⟩_i)/√2. Idler measurement at Dk projects idler onto |χ_k⟩ (the idler state that leads to Dk given the BS network): |χ_k⟩ = c_A^k|A⟩_i + c_B^k|B⟩_i. Then signal collapses to c_A^k ψ_A(x) + c_B^k ψ_B(x) (unnormalized): P(x,k) ∝ |c_A^k ψ_A(x) + c_B^k ψ_B(x)|². And P(k) = ∫|...|²dx (with proper normalization ∫(|ψ_A|²+|ψ_B|²)dx... since ⟨χ|χ⟩=1 if normalized). Coefficients per detector given BS states: - D3: idler state that reaches D3 = |A⟩ (amplitude: from A: BSa transmit t=1/√2 (if BSa in) or 1 (if out), then Ma reflect (|r|=1) → D3): c_A = t_BSa (1/√2 or 1), c_B = 0. Normalize: |χ_D3⟩ = |A⟩, P(D3) = |t_BSa|²/2·... hmm — the marginal: P(D3) = |t_BSa|² · (1/2) [the 1/2 from the pair superposition weight]. With BSa in: P(D3) = 1/4 ✓; BSa out: P(D3) = 1/2 ✓ (all A-born pairs... wait — but pairs aren't "born at A" — hmm, P(D3) = |⟨A,χ_D3|Ψ⟩|² = |(1/√2)·t|² = t²/2: BSa in: 1/4; out: 1/2 ✓ matches the route logic applied to the A-branch ✓ same thing. Fine.) - D4: c_B = t_BSb, c_A=0. - D1 (BSc in): idler amplitude to D1: from A: r_BSa·r_BSc = (1/√2)(1/√2)=1/2 (if BSa in) or 0 if BSa OUT?? wait — BSa out: A's idler goes straight to Ma→D3, NOTHING reaches BSc from A ✓ c_A=0 ✓ consistent: BSa out → c_A(D1)=0 ✓. From B: r_BSb·t_BSc = 1/2 (BSb in) / 0 (out). So (all BS in): |χ_D1⟩ ∝ (1/2)|A⟩ + (1/2)|B⟩ → normalized: (|A⟩+|B⟩)/√2 (with phase θ=0... ignoring reflection phases: r=i/√2, t=1/√2: from A: i·i/... A: BSa reflect (i/√2) then BSc reflect (i/√2): (i/√2)(i/√2) = −1/2; B: BSb reflect (i/√2) then BSc transmit (1/√2): i/2. Relative phase: (−1/2 vs i/2): θ = π/2?? Hmm — with ideal symmetric BS convention r=1/√2 (real), t=i/√2: A: r·r = 1/2; B: r·t = i/2 → relative i·... The ACTUAL relative phase depends on BS conventions and path lengths — physically adjustable; Kim's result: D1 fringes, D2 anti-fringes. I'll just set θ_D1 = 0, θ_D2 = π and note it. The weights: |c_A| = |c_B| = 1/2 (all BS in) → normalized equally weighted ✓ fringes with visibility 1 ✓. - BSc out: D1 gets B only: |χ_D1⟩ = |B⟩ (weight r_BSb·1: c_B = r_BSb) ✓ blob. - D2 (BSc in): A: r_BSa·t_BSc = 1/2, B: r_BSb·r_BSc = 1/2, relative phase π → anti-fringes ✓. BSc out: D2 = A only ✓ blob. So generic per-pair Monte Carlo: 1. Compute for each k∈{1,2,3,4}: (cA_k, cB_k) from BS states (with phases: D1: (1,1)/norm... let me store complex). 2. P(k) = (|cA_k|² + |cB_k|²)/2·N... hmm — normalization: P(x,k) = |cA_k ψ_A(x) + cB_k ψ_B(x)|² / 2 (the 1/2 from pair state), and ∑_k P(k) = 1 when including ALL outcomes (D0-miss handled separately — actually D0 detection is the aperture filter on top). Let me verify sum: ∑_k (|cA_k|²+|cB_k|²)/2: all BS in: each k: (1/4+1/4)/2 = 1/4 → sum 1 ✓. BSa out: D3: (1+0)/2 = 1/2; D1: (0 + 1/4)/2 = 1/8; D2: (0+1/4)/2=1/8; D4: (0+1/4)/2=1/8 → sum: 1/2+3/8 = 7/8 ✗!! MISSING 1/8: BSa out means A's idler ALWAYS transmits toward Ma→D3 — wait no: BSa OUT: A's idler goes STRAIGHT (there's no plate): c_A(D3) = 1 (not 1/√2!) → P(D3) = 1/2 ✓; A contributes NOTHING to BSc ✓ cA(D1)=cA(D2)=0 ✓; B: BSb in: cB(D4) = 1/√2 → P(D4) = (1/2)/2 = 1/4; cB(D1) = r_BSb·r_BSc = 1/2 → P(D1) = (1/4)/2 = 1/8; D2 same 1/8. Sum: 1/2+1/4+1/8+1/8 = 1 ✓✓ GREAT. So with BSa out: D1 gets B-ONLY amplitude (cA=0): blob at D1 ✓ (matches route logic ✓). 3. Sample k ~ P(k); sample x ~ P(x|k) ∝ |cA_k ψ_A(x) + cB_k ψ_B(x)|². 4. D0 aperture check: if x ∈ [x₀−w/2, x₀+w/2] → coincidence count at bin(x₀) for channel k. Else → lost. PERFECT — this generic implementation is compact and physically correct for all toggle combos!! And the "theory curves" for display: for each k: P(x|k) normalized (for the current BS config) — drawn as overlay ✓ updates live when toggles change ✓ (data accumulates into possibly-mixed-config histograms — note: when config changes, maybe auto-suggest reset; I'll just keep accumulating and the curve updates; add a small "ayar değişti — veriyi sıfırla?" hint... simpler: data keeps accumulating (realistic: you'd re-run), and I'll show the theory curve for CURRENT config; a subtle warning icon when config changed since last reset. Nice touch. Actually simplest: auto-reset data on config change? NO — that's annoying while exploring... hmm, actually it's cleaner: when BS config changes, old data is from a different experiment. But auto-reset loses the demo flow... I'll add a small badge "⚠ ayar değişti" next to a reset button. Fine.) EDGE CASE: all BS out: D3: A-only (P=1/2), D4: B-only (1/2), D1/D2: 0. Total D0 pattern: blob ✓. Another subtle case: BSa in, BSb in, BSc out: D3: (1/2)/2=1/4 [A]; D4: 1/4 [B]; D1: B-only: (0+1/2)/2 = 1/4 blob-B; D2: A-only blob-A: 1/4. Sum 1 ✓. So D1 shows |ψ_B|², D2 shows |ψ_A|² — blobs, labeled which-path ✓ TEACHING GOLD: "BSc'yi çıkarınca D1/D2 de yol bilgisi verir!" **D0 TOTAL channel**: sum over k — always ∝ |ψ_A|²+|ψ_B|² (no fringes) regardless of config ✓ (∑_k |cA_k ψ_A + cB_k ψ_B|² over complete measurement = |ψ_A|²+|ψ_B|² when {χ_k} complete ✓). **Now the D0 scan + histogram:** - Histogram: x-axis = D0 position x (−7..+7mm), y = coincidence counts. 28 bins (0.5mm). Curves: theory P(x|k) scaled to current max count... better: scale theory to total counts in channel × expected shape: expected counts per bin = N_k·(aperture/∫P) — display as smooth curve. - Auto-scan: x₀ cycles through bins, dwell = e.g., 2.5s per bin (configurable? fixed is fine; adjust with speed multiplier — hmm, dwell interacts with pair rate: at high rates, dwell can be shorter. Let me tie: auto-scan dwell = max(1.2s, enough for ~150 detections/bin at current rate)? Simpler: fixed dwell 3s/bin — at turbo rates that's plenty of counts; at slow rates sparse (realistic!). Hmm — at slow speed (2 pairs/s, ~3% detection → 0.06 counts/s → 3s dwell ≈ 0 counts/bin — the histogram never fills. So auto-scan dwell should scale inversely with rate, or scan mode only makes sense with decent rate. SOLUTION: dwell time = (a fixed number of PAIRS per bin, e.g., 400 pairs) / pairRate → at 2/s: 200s per bin (slooow but "realistic"; user can bump speed). Show a progress indicator on the rail (bin being measured highlighted). OR simply: dwell = fixed pair-count target. Let me do: autoScan advances to next bin after maxPairsPerBin pairs processed (default ~600) OR minDwell 0.8s, whichever later. At 2000 pairs/s: 0.3s/bin → full sweep 28 bins ≈ 10s ✓ nice. At 60/s: 10s/bin → 4.7min sweep — meh; acceptable, and user can raise speed. Hmm — maybe maxPairsPerBin = 300: at 60/s → 5s/bin → 2.3min. OK fine. Actually let me make pairs-per-bin adjustable via the same "hız" — no, keep it simple: fixed 350 pairs/bin with min dwell 0.6s. - Manual: slider −7..+7 (0.5 steps); counts accumulate at current bin. **Coincidence channel displays:** buttons: [D0 (toplam)], [D0∩D1], [D0∩D2], [D0∩D3], [D0∩D4]. Histogram + counts + rate. Also a compact "mini-strip" showing all 5 simultaneously?? Space… main histogram big + small multiples could be lovely: 5 mini histograms stacked (each ~70px tall) — the "aha" visual: D1 fringes / D2 anti-fringes side by side!! YES — include small-multiples row: five mini strips (D0, D1, D2, D3, D4) always visible; clicking one selects it for the big view. And the big view with theory curve overlay + counts. Hmm — layout plan (page): - Header: title "Gecikmeli Seçim Kuantum Silgisi — Etkileşimli Simülasyon" + subtitle "(Kim et al., 2000 düzeneği)". - Main grid: - LEFT (≈ 640px, flex): the optical table SVG (aspect ~1240×780) — interactive: click BSa/BSb/BSc toggles; drag D0 head along rail OR click rail to set position; hover elements → tooltip with name/description (Turkish). Also photon animations + flashes + beams. Below SVG: timeline strip? Maybe a small "event log" line showing last event: "D0 tespiti (x=+1.2mm) → 12 ns sonra D2 tespiti → çakışma: D0∩D2". Actually — a horizontal TIME TRACE would be amazing for the "delayed" aspect: a little oscilloscope-like strip showing D0 pulse and Dk pulse with Δt = 12 ns gap (the realistic delay: idler path ~2.3 m longer → ~7.7 ns; Kim's paper: coincidence window... let me compute: idler path longer by ~2.3m → Δt ≈ 7.7 ns). Draw a "zaman izi" mini-panel: two pulses with gap "Δt ≈ 8 ns (idler yolu daha uzun)". Nice touch — include in the sidebar under controls. Even better: mark the moment: when an animated pair completes, the strip shows the two pulses sliding... keep it simple: static-ish illustration that pulses on each animated event. Hmm — I'll do a compact "son olay" (last event) readout with a tiny pulse diagram. Let me not overscope — a simple animated pulse pair display, lightweight. - RIGHT (≈ 420px): controls + data: 1. **Kontrol paneli**: Başlat/Duraklat; Hız (slider 1×..2000× log); Tarama modu: Otomatik/Manuel + x₀ slider (disabled in auto); Sıfırla. 2. **Işık ayırıcılar (gecikmeli seçim!)**: three toggle switches BSa, BSb, BSc — with note "Foton uçuştayken bile değiştirebilirsin!" — plus clicking on the SVG elements toggles too. 3. **Kanal seçimi** buttons + big histogram (canvas ~380×220) with theory curve toggle. 4. **Mini panels**: 5 mini histograms. 5. **Zaman izi / son olay** + coincidence LEDs. 6. **Açıklama** (collapsible?): short explanation of the physics + what to try ("Deneyleri" tips). - BOTTOM or in sidebar: explanation text (Turkish), with key points: - D0 tek başına: girişim YOK (her zaman). - D0∩D1: girişim ÇİZGİLERİ; D0∩D2: karşıt çizgiler. - D0∩D3/D4: yol bilgisi → girişim yok. - Gecikme: sinyal D0'da tespit edildikten ~8 ns SONRA idler hangi detektöre gideceğine karar verir — geçmişteki D0 dağılımını değiştiremez; sadece KORELASYONLAR desen verir. - "Silgi" BSc'yi çıkar → D1/D2 de yol bilgisi verir. - Footer: parameter notes (λ=702nm, d=0.25mm, a=0.07mm, f=1.2m, tarama ±7mm). **Interactivity summary (make sure it's rich):** - Start/pause, speed slider (log scale 1–3000 pairs/s). - Auto/manual D0 scan; drag D0 head on the rail (pointer events on SVG!) or slider; click rail to jump. - Toggle BSa/BSb/BSc anytime (even mid-flight — photons in flight re-route if their animation hasn't passed the element! IMPLEMENT: animated photons' remaining path is computed lazily as they reach each element?? That's complex — simpler: when a BS is toggled, in-flight animated photons just continue on their pre-computed path (they already "passed" it or the change applies to future pairs). Hmm — the DELAYED CHOICE drama demands in-flight re-routing... Let me think: animation path is a sequence of waypoints; the photon's route beyond BSa is decided when it REACHES BSa (lazy evaluation!). Since the idler reaches BSa ~1s after birth (animation time), the user can toggle BSa WHILE the idler is en route → the outcome changes ✓✓ DELAYED CHOICE MADE TANGIBLE. But — the STATISTICAL processing already pre-sampled k for this pair (for counting)... For ANIMATED pairs, I should sample k lazily too: sample the idler route when it arrives at each element, then when it's finally detected at Dk, THEN commit the coincidence count (with the x recorded at D0 click time). But the D0 x was sampled at D0-click time from P(x|k)... which requires knowing k!! CIRCULAR. Resolve: for animated pairs, sample x FIRST?? The joint: P(x,k) — can't sample x without k... unless sample k first (as before) but ANIMATE lazily: the pre-sampled k defines what happens when the photon reaches BSa — IF the user toggles BS mid-flight, the pre-sampled k may become inconsistent (e.g., k=D3 but BSa now removed → photon should go to... D3 still! (BSa removed → transmit → Ma → D3 — still D3! Ha — BSa removal routes MORE photons to D3). Hmm — actually with my geometry: BSa routes: reflect→BSc(erasure), transmit→D3. BSa OUT → always transmit → D3. BSa IN → 50/50. So toggling BSa mid-flight: if the photon hasn't reached BSa: re-decide per current state ✓. If k was pre-sampled D1 but user removed BSc... then the photon heading to D1 now goes straight to... D1 is at (530,475) — BSc removed → B-path continues to D1 anyway (in my geometry BSc-out: B→D1 direct ✓!!) — the geometry routes it to D1 regardless, but now D1 is "B-only" — hmm, but this photon is from... ugh. COMPLEX. PRAGMATIC DECISION: - Statistical layer: pre-samples everything per pair using CURRENT BS state AT SPAWN TIME. - Animation layer: photons follow the pre-sampled route; if BS toggles mid-flight, in-flight photons COMPLETE their pre-sampled route (they're "already committed" — and physically, a photon that already passed the BS can't be affected... but if it hasn't reached the BS yet?? — visually it just follows the old route; minor inconsistency, invisible in practice since flights last ~1–2s). - BUT to honor "delayed choice mid-flight": make flights slow enough (idler leg ~1.5–2s at normal speed) and add the note. ALSO — actually, simpler correct approach: SPAWN the pair with k and x pre-sampled, but the idler's ANIMATED path is built from k + BS state sampled AT SPAWN. Toggling mid-flight affects only future pairs. The "gecikmeli seçim" is still conveyed: the idler's DECISION (which BS port) happens visually AFTER the D0 click — the user sees D0 flash first, then watches the idler "choose" at BSa — and can toggle before the next pair. PLUS the fact that the histograms don't change retroactively. That's honest and clear. I'll ALSO let the speed be slow enough that users notice the D0-then-idler ordering. OK — final: pre-sampled, no mid-flight re-route. (Keep code sane.) Hmm, wait — actually — let me reconsider once more. A really compelling "delayed choice" interactive: a dedicated "TEK ATIŞ modu"? Like a button "Tek foton çifti gönder" that spawns one pair at slow animation; while the idler is mid-flight, a big prompt appears: "İdler BSc'ye ulaşmadan karar ver: silgi taksın mı?" with two buttons... This is a nice mini-game but adds complexity. The BS toggles already exist and flights are slow. I'll add prominent hint text: "⚠ Foton uçuştayken BS'leri değiştirmeyi deneyin". Since pre-sampling uses spawn-time state — mid-flight toggles affect future pairs only — the hint is still fair (next pairs). Hmm, but a savvy user watching a specific in-flight photon toggle BSc and see it not re-route might call foul... The flight time is ~1.5s; risk is low. ALTERNATIVE EASY FIX: make the statistical sampling LAZY per pair at IDLER-BS-ARRIVAL for the ANIMATED pairs only... You know what — actually there IS a clean lazy design: For each pair, do the D0 event FIRST in real time: sample k and x at spawn (both from current config), commit D0 x to "pending"; animate signal; at D0: flash (pending recorded). Animate idler along the route for k. At idler detection: commit coincidence. If BS config CHANGED between spawn and idler-arrival: RE-SAMPLE the pair's k (and x?!) — NO — x was already shown at D0 — CAN'T re-sample x without breaking the story ("D0 result changed retroactively" — even though it's just a simulation re-roll, it would look like retro-causality — BAD). FINAL DECISION: pre-sample at spawn; in-flight toggles affect future pairs only; hint text explains that in the real experiment the choice applies per-photon at the beam splitter. DONE. Moving on!! - Hover tooltips on optical elements (title/info box showing element name + function) — implement with mouseenter on SVG groups → info line below the SVG ("Eleman bilgisi: ..."). - Clicking detectors D1–D4 selects that coincidence channel in the big histogram ✓ (nice cross-linking). - Clicking D0? → selects D0-total channel. **Photon visuals:** - Pump dot: violet (#a06cff), r=4, with small trailing glow. - Signal dot: orange (#ff8c3a), r=3.5, glow. - Idler dot: cyan (#3ad2ff), r=3.5, glow. - Ghost dots: 45% opacity versions. - Speed along path: constant px/s (e.g., 340 px/s at 1× — scaled slightly by speed setting? The ANIMATION speed should stay watchable: at higher pair rates, photons fly faster or same? If 60 pairs/s each with 3s flight → 180 concurrent photons = chaos. So: animation rate (spawn of ANIMATED pairs) capped at ~3/s (normal) and flights ~1.6s → ≤5 concurrent pairs ✓; the STATISTICAL layer runs at the full rate silently. At turbo, animate ~1 detection per 0.5s. So visual calm at all speeds ✓. Photon flight speed: scale mildly with speed setting (1×: 300px/s; at 100×: ~500px/s) — mostly constant is fine. Hmm — the idler path is LONG (~1000px vs signal ~600px) so idler takes ~1.6× longer — plus we want a clear delay: add extra path length... the idler's geometric path: (590,135)→(480,245)→(620,385)→(710,475): ~155+199+127 = 481px + maybe extend... vs signal: (590,135)→(590,215)→(880,185)→(1060,y): 80+292+~190 = ~560px?? THE SIGNAL PATH IS LONGER in my drawing!! The delay must come from the idler path being longer. PROBLEM. Fix: the signal path in drawing = 560px, idler = 480px — idler SHORTER → idler arrives FIRST — WRONG STORY!! Hmm!! In Kim's experiment, the idler path is longer (extra folding mirrors). In my layout it's shorter. Options: (a) slow the idler dot?? (cheating, but... no). (b) Lengthen the idler path: add fold mirrors to the idler routes!! Realistic AND solves it: e.g., idler A: slit A → BSa — lengthen by routing via a fold mirror: (590,135) → down-left 45° → fold mirror FA at (520,200)?? hmm... Actually — EASIEST: make the animation speed such that... no — the physics point is PATH LENGTH. Let me add one fold mirror per idler branch, OR reposition BSa/BSb FARTHER: BSa at (400, 335)?: idler A: (590,135)→45°→(400,335)?? Δ(−190,190)?? that's 45° ✓ length 269px. Then BSa transmit → Ma at (310,425) (Δ(−90,90)) → D3 (310, 535). BSa reflect → down-right 45° → BSc at (620, 545): Δ(220,210) — not exactly 45° (need Δx=Δy): BSc at (620,555): Δ(220,220) ✓ 45° ✓ length 311px. BSc → D1 (530,645) / D2 (710,645): 127px each. Idler-A erasure total: 269+311+127 = 707px > signal 560px ✓. Idler-A which-path: 269+127+110 = 506px — STILL SHORTER than signal 560!! Grr. The which-path routes (D3/D4) are short. In Kim's experiment, ALL idler paths are longer than the signal path (that's the design: signal detected first, ALWAYS). Fix: add fold mirrors on the which-path branches: Ma → fold → D3 with extra distance: e.g., Ma at (310,425) reflects down to FA2 at (310, 560) → reflects RIGHT to D3 at (450,560)?? Let me redesign the idler network with a cleaner "fold" that adds length: Idler A which-path: BSa (400,335) → transmit 45° down-left → Ma (310,425) → reflect: instead of straight down, reflect to down-RIGHT?? hmm. Honestly — simplest: lengthen with a vertical drop: Ma (310,425) → straight down a LONG way → D3 at (310, 640): length 215. Total: 269+127+215 = 611px ✓ > 560 ✓ (barely). More margin: drop D3 to (310, 660): 235 → 631px ✓. Similarly D4 at (930, 660)?: BSb at (840,335)... wait symmetric: BSb at (840,335), Mb at (930,425), D4 at (930,660): idler B which-path: 269+127+235 = 631 ✓. And BSc at (620,555) with D1 (530,645), D2 (710,645): erasure: 269+311+127 = 707 ✓. Hmm wait — recheck BSb reflect → BSc: BSb (840,335) → BSc (620,555): Δ(−220,220) ✓ 45° ✓. And BSc→D1/D2: Δ(∓90,90) ✓. BUT NOW — check collisions with the LENS (x=880, y 40–345) and rail (x=1060): BSb at (840,335): 40px left of lens ✓; Mb (930,425): BELOW lens bottom (345) ✓ 80px clear ✓; D4 (930,660) ✓ far below. Idler B: (650,135)→(840,335): passes (740,235)... lens bottom at (880,345): the idler beam at x=840 is at y=335 — lens is at x=880±8 — beam hits x=840 at y=335 then turns; ✓ clear. The BSb→BSc beam (840,335)→(620,555): crosses x=710 at y=465, x=620 at 555 ✓ no elements there (D2 at (710,645) ✓ clear). Ma/Mb vertical drops (310,425→660) and (930,425→660): cross the coincidence-box zone?? Box: y 700+? Let me push the box to y 710–800: D1/D2 at y=645, D3/D4 at 660 ✓ box below at 710 ✓✓. Hmm, canvas height now ~840. And D4 at (930,660) — box spans x 260–980?? overlapping under D4 ✓ fine (box is below at y≥710). Wait, also re-check: idler beams vs SIGNAL beams (y=215 A / y=155 B horizontal, x 590–880): idler A (590,135)→(400,335): crosses y=215 at x=510 — signal A beam starts at x=590 → NO crossing ✓. Idler B (650,135)→(840,335): crosses y=155 at x=680, y=215 at x=740: signal B beam (y=155, x from 652): crosses at (680,155) ✓ crossing (fine); signal A beam (y=215, x from 590): crosses at (740,215) ✓ fine. Idler A also crosses... the BSa→BSc beam (400,335)→(620,555): below signal beams ✓ no. ✓ OK. Hmm, and the PRISM: face x+y=805, x∈[560,652] — idler A from (590,135) at 45° down-left: line x+y=725: parallel ✓ distance 56px ✓; idler B: x−y=515: crossing with face: x=(805+515)/2=660 > 652 ✓ clears by 8px. Signal A vertical x=590: face y=215 ✓; signal B x=650: face y=155 ✓. ✓ And the pump dump — decided: pump fades at BBO. Hmm, wait — actually with the prism at x+y=805 spanning [560,652], and the pump at x=620 hitting the face at y=185 — if I skip the dump, the pump visually should END at the prism (absorbed). Draw the pump wedge laser→slits→BBO and then a faint short continuation (620,140)→(620,185) ending at the prism with a tiny "×"? I'll draw the faint continuation to the prism face and a small gray hatched block right AT the face point (620,185)?? — a small block ON the face... OR simplest: pump wedge ends AT THE BBO with a soft fade + caption "kalan pomp ışığı filtrelenir". DECISION: fade at BBO. Clean. Hmm — wait, actually now re-examine: BSa at (400,335) — the transmit beam continues 45° down-left from (400,335) to Ma (310,425) ✓; reflect beam (400,335)→(620,555) 45° down-right — wait, Δ=(220,220) ✓ 45° ✓. BUT — the REFLECT at BSa: incoming dir (−0.707,0.707), outgoing (0.707,0.707): reflection about VERTICAL plate ✓ (n=(1,0): d' = d−2(d·n)n = (−0.707,0.707)−2(−0.707)(1,0) = (−0.707+1.414, 0.707) = (0.707,0.707) ✓✓). And BSa plate vertical ✓. Ma: incoming (−0.707,0.707) → out (0,1) (straight down): n ∝ (0,1)−(−0.707,0.707) = (0.707,0.293) → (0.924,0.383): plate dir (−0.383,0.924): tilted 22.5° from vertical ✓. BSc: vertical plate ✓ (computed before). D1/D2 beams from BSc: incoming A (0.707,0.707) → transmit → (0.707,0.707) → D2 (710,645): Δ(90,90) ✓; reflect → (−0.707,0.707) → D1 (530,645) ✓✓. One more check — the erasure beams (400,335)→(620,555) and (840,335)→(620,555): do they cross the which-path vertical beams (310, y 425–660) and (930, y 425–660)? Beam 1: x from 400→620 — never reaches 310 or 930 ✓. ✓ And do they cross EACH OTHER before BSc? They converge AT BSc exactly ✓ (both 45°, symmetric) ✓✓. Path lengths (for delay): - Signal: slit→prism: 80 (A) / 20 (B) — hmm asymmetric: A: (590,135)→(590,215) = 80px; B: (650,135)→(650,155) = 20px. Then prism→lens: (590,215)→(875,190) ≈ 287; (650,155)→(875,180) ≈ 226. Then lens→D0: (880,185)→(1060,y): ≈ 180–280. Total ≈ 80+287+230 ≈ 600px. - Idler erasure: (590,135)→(400,335): 269; →(620,555): 311; →(710,645): 127. Total ≈ 707px. - Idler which-path: 269 + 127 [(400,335)→(310,425)] + 235 [(310,425)→(310,660)] = 631px. - Idler-to-D1: 269+311+127 = 707. ✓ All idler paths > signal path ✓ (631 > 600 by only 5% — the D3/D4 delay will look tiny... make the Ma→D3 drop longer: D3 at (310, 700)?? that's inside the coincidence-box zone (y≥710)... box could start at y=730. Hmm — OR raise the signal speed... OR — better idea: make the SIGNAL path effectively "shorter" by moving D0's rail closer?? The lens needs f-distance to the rail... Alternatively add a second fold on which-path: Ma (310,425) → down to (310,540) → fold mirror Mc → RIGHT to (450,540) → D3 at (450,540)?? Then: 269+127+115+140 = 651. Meh — barely better. OR: accept it — the visual delay for D3/D4 is small but present (~5%), while D1/D2 have ~18% delay. Physically Kim's experiment: all idler paths longer. My sim: D3/D4 delay 30px@340px/s ≈ 0.09s — too subtle. Let me make it more visible: extend the Ma→D3 beams: move Ma UP and D3 DOWN: Ma at (310,425), D3 at (310, 690): drop = 265: total = 269+127+265 = 661. And signal 600. Still meh. ALTERNATIVE: shorten signal path: move the lens+D0 rail LEFT: lens at x=830, rail at x=1000: signal: 80 + (590,215)→(825,190) ≈ 237 + (830,185)→(1000,y) ≈ 170 → ≈ 487px. Then idler 631–707 → delay 30–44% ✓✓ visible! Check: lens at x=830: does it collide with idler B's beam ((650,135)→(840,335))? The beam at x=830: y=325; lens spans y 40–345 at x=830±8: the beam CROSSES the lens region at (830,325)!!! COLLISION ✗✗. Hmm!! Move BSb LEFT: BSb at (800,335)?: idler B: (650,135)→(800,335): Δ(150,200)?? NOT 45° ✗. Keep 45°: BSb must be at (650+t, 135+t): options: (790,275): then transmit → Mb at (700,365)?? — Mb would be LEFT of BSb... hmm wait transmit continues down-right 45°: Mb at (880,455)?: from (790,275): Δ(90,180)?? NO — 45° means Δx=Δy: Mb at (880,365): Δ(90,90) ✓. Then D4 below Mb: (880, 365+drop). And BSb reflect → BSc: from (790,275) down-left 45° → BSc at (620, 445): Δ(−170,170) ✓. Then BSc→D1/D2: (530,535)/(710,535). Idler B erasure: (650,135)→(790,275): 198 + (790,275)→(620,445): 240 + 127 = 565. Idler B which-path: 198+127+drop(880,365)→(880,660): 295 = 620. Idler A symmetric: BSa at (450,275), Ma at (360,365), D3 (360,660): erasure: (590,135)→(450,275):198 + (450,275)→(620,445):240 + 127 = 565; which-path: 198+127+295 = 620. Signal (lens at 830): ≈ 487. Delays: erasure +78px (16%), which-path +133px (27%) ✓✓ GOOD ENOUGH — erasure delay visible, which-path delay clearly visible. Hmm wait — but now check BSb (790,275) vs LENS (830, y40–345): 40px apart ✓. Idler B beam (650,135)→(790,275): at x=790 ends; lens at x=822–838: beam never reaches x≥822 ✓ CLEAR ✓. Mb at (880,365): lens bottom at y=345, x 822–838: Mb at (880,365) — 45px right of lens edge, 20px below lens bottom ✓ OK. Mb→D4 vertical beam at x=880, y 365–660: rail at x=1000 ✓ clear; D0 head at (1000, y) ✓ clear of x=880 beam ✓. D4 at (880,660) ✓. And the BSb→BSc beam (790,275)→(620,445): crosses the SIGNAL beams (y=155/215)? Beam y∈[275,445] — below 215 ✓ NO ✓. Crosses D1/D2? D1 (530,535), D2 (710,535): beam ends at (620,445) ✓ clear. BSc (620,445) → D1 (530,535): Δ(−90,90) ✓; D2 (710,535) ✓. Does BSc→D2 beam cross Mb→D4 beam (x=880)? BSc→D2: x from 620→710 ✓ no. ✓✓ And Ma/Mb at y=365, D3/D4 at y=660: which-path drops: (360,365)→(360,660) and (880,365)→(880,660): length 295 ✓. D3 at (360,660), D4 (880,660), D1 (530,535), D2 (710,535) — hmm, D1/D2 HIGHER than D3/D4 — visually fine (real layouts are irregular). Coincidence box: y 700–790, x 240–1000 ✓. Re-check idler A beam (590,135)→(450,275) vs prism face (x+y=805, x∈[560,652]): idler A line x+y=725: parallel, 56px ✓. Idler B (650,135)→(790,275): x−y=515: crossing x=660 > 652 ✓ 8px clear (as before) ✓. D0 rail: x=1000, y 30–345. Lens: x=830, y 40–345 (bar). Signal static beams: A: (590,215)→(825,~192); B: (650,155)→(825,~178); fan after lens: wedge (830,185)→(1000, 45..325). ✓ Hmm — wait, the coincidence-box cables: D3 (360,660) → box (240–1000, 700–790): cable straight down ✓; D4 (880,660)→box ✓; D1 (530,535)→ down ✓; D2 (710,535) ✓; D0 (1000, y) → cable right/down → box right edge ✓. Let me also reconsider the delay NUMBERS for the "zaman izi": path difference: erasure: 565−487 = 78px. If I say the drawing scale... let me define a nominal scale: 1px ≡ 5mm?? The signal path 487px ≈ 2.4m — hmm — real Kim: D0 at ~1.3m? and idler ~4m? Path diff ~2.3m → 7.7ns. I'll just state "Δt ≈ 8 ns (idler yolu ~2.3 m daha uzun)" in the timeline display with a scaled illustration — the DISPLAY is schematic (ns not to pixel scale). FINE. Now — the big remaining design: **the histogram & data panel details**, then **implementation**. Physics functions: ``` λ = 702e-9 m; a = 0.07e-3; d = 0.25e-3; f = 1.2; (meters) x range: [-7e-3, 7e-3], bins: 28 (0.5mm each) φ(x) = 2π d x/(λ f) → = 2π·0.25e-3·x/(702e-9·1.2) = 2π·x·296.4 per meter → x in mm: φ = 2π·0.2964·x_mm... let me compute: d/(λf) = 2.5e-4/(8.424e-7) = 296.8 /m. So φ = 2π·296.8·x(m) → x=1mm=1e-3: φ=1.864 rad. Fringe spacing (cos²(φ/2)): period Δφ=π → Δx = π/(2π·296.8)·... Δx = 1/(2·296.8) m = 1.685mm?? WAIT: cos²(φ/2) has period Δφ = 2π → Δx = 2π/(2π·d/(λf)) = λf/d = 702e-9·1.2/2.5e-4 = 3.37e-3 ✓ 3.37mm ✓ (I mis-stepped; cos²(φ/2) period in φ is 2π... no: cos²(u) has period π in u; u=φ/2 → period 2π in φ → Δx = 2π/(2π·296.8) = 3.37mm ✓). 28 bins × 0.5 = 14mm span → ~4.2 fringes ✓ GOOD. β(x) = π a x/(λf): a/(λf) = 7e-5/8.424e-7 = 83.1/m → β = π·83.1·x → at 7mm: β=1.826 rad; sinc(β): sin(1.826)/1.826 = 0.967/1.826 = 0.53 → 0.28. Envelope at edges ~28% ✓ acceptable. ψA = sinc(β)·e^{iφ/2}, ψB = sinc(β)·e^{−iφ/2}. P(x|k): D1: |ψA+ψB|² = 4sinc²cos²(φ/2); D2: 4sinc²sin²(φ/2); D3: sinc²; D4: sinc²; D0total: 2sinc². (Normalize numerically over the binned range.) ``` Wait — I should double check the D1/D2 assignments once more against Kim's paper result: Kim's data: D1 coincidence → interference fringes; D2 → fringes shifted by π (anti-fringes); D3/D4 → no interference. ✓ matches. Hmm — one more physical nuance: in Kim's actual experiment, D1 and D2 patterns were NOT pure cos²/sin² — they were partially shifted fringes with lower visibility, because the erasure wasn't perfect (the two paths at BSc weren't mode-matched perfectly). For the sim: ideal 100% visibility is cleaner pedagogically. Note it as "idealize edilmiş". FINE. **Data structures:** ``` state = { running: false, speed: 60 (pairs/sec), scanMode: 'auto' | 'manual', x0: 0 (mm), currentBin, bs: {a:true, b:true, c:true}, counts: {total: Int32Array(28), d1:..., d2:..., d3:..., d4:...}, pairsProcessed, coincidences: {d1:0,...}, channel: 'total' (display), showTheory: true, } ``` Per statistical pair (function stepPair()): ``` coeffs = computeCoeffs(bs): for k in 1..4: {cA, cB} complex: // D3: from A via BSa transmit tA = bs.a ? Math.SQRT1_2 : 1; // amplitude A→D3 rA = bs.a ? Math.SQRT1_2 : 0; // amplitude A→BSc tB = bs.b ? Math.SQRT1_2 : 1; // B→D4 rB = bs.b ? Math.SQRT1_2 : 0; // B→BSc // BSc in: A→D1: rA·(1/√2)... wait — BSc splitting: each incoming erasure amplitude splits 1/√2 to D1 and 1/√2 to D2: // D1: cA = rA·s, cB = rB·s (s = bs.c ? 1/√2 : (direct... )) — CAREFUL with BSc-out case: // BSc out: A-erasure beam goes STRAIGHT to D2 (geometry!), B-erasure straight to D1. // So BSc out: D2: cA = rA, cB = 0; D1: cA = 0, cB = rB. (No splitting, no mixing.) // BSc in: D1: cA = rA/√2, cB = rB/√2 (phase 0); D2: cA = rA/√2, cB = −rB/√2 (phase π). // D3: cA = tA, cB = 0; D4: cA = 0, cB = tB. P(k) ∝ |cA|² + |cB|² (×1/2); normalize (if all zero... can't happen). sample k; sample x from P(x|k) via inverse-CDF (precomputed per config); if |x − x0| ≤ 0.25: coincidence: counts[k][binOf(x0)]++ ... ``` Hmm wait — the aperture: bin width 0.5mm; if D0 sits at bin center, aperture = bin width → detection iff x falls in the bin → coincidence counts into that bin ✓ CLEAN: detection iff binOf(x) === currentBin. And P(detection per pair) = ∫_bin P(x|k)dx. Sampling x then binning = same thing ✓. So: sample x (continuous), bin b = binOf(x); if b === currentBin → count for channel k at b. Else → miss. ELEGANT ✓. And in manual mode x0 slider snaps to bin centers ✓. Also track: pairsProcessed, per-channel totals, detection counts. Inverse-CDF sampling: per config, for each k: build array over fine grid (e.g., 0.05mm steps, 280 entries) of P(x|k), CDF, sample u→binary search→x. Rebuild when BS config changes. Cheap ✓. ANIMATION events: spawn animated pair when (running && animCooldown elapsed): rate: base ~ 2.2/s at speed ≤ 30; at higher speeds: animate ~ min(3/s, detections happen)... simpler: animation spawn rate = clamp(speed, 0.8, 3.2) per second — at turbo (2000/s), still ~3 animated pairs/s (a sample). For animated pair: sample k, x the same way (from CURRENT config) — but FORCE a detection? At slow speeds show misses too: if speed < 25: animate all pairs including misses (spawn rate = speed/s capped 3/s; misses included); if speed ≥ 25: animate only detected pairs (sample k, then sample x CONDITIONED on bin === currentBin: resample until hit (max 200 tries) or sample from bin-restricted CDF ✓ easy: sample x from P(x|k) restricted to currentBin range). ✓ Animated pair object: {phase waypoints with times, k, x, detected, t0, ...} — implement as a list of segments [{x1,y1,x2,y2,len,verb?}] for each animated entity: - pump dot: (620,58)→(620,100): then split ghosts? For simplicity: pump dot → (620,100); then TWO pump ghosts: (590,100)→(590,135) and (650,100)→(650,135) — wait, the pump dot arrives at (620,100) (plate center) then ghosts appear at slits?? Visual pop. ALTERNATIVE: pump dot travels (620,58)→(620,100); at slit plane, becomes 2 ghosts starting at (620,100) heading to (590,~120)/(650,~120)?? diagonal — eh. SIMPLEST: pump dot: (620,58)→(620,100)→(620,130) (to BBO center) — then SPARKLE at BBO and the 4 ghosts (2 signal at (590,135)/(650,135), 2 idler same) pop in with a fade-in. The pop-in at slightly different x than the pump's end (620 vs 590/650) — 30px offset — acceptable with a sparkle ring covering the BBO width. ✓ DO THAT. (Pump dot also has a tiny "×2 slits" annotation? no — keep clean; the ghosts appearing AT the slits with slit labels A/B is self-explanatory + caption text explains.) - signal ghosts: gA: (590,135)→(590,215)→(825,192); gB: (650,135)→(650,155)→(825,178); merged signal: (830,185)→(1000,y_target). If detected: flash D0; else: (1000,y)→(1080,y) fading. y_target = 185 − x_mm·20. - idler ghosts: gA: (590,135)→(450,275) [to BSa]; gB: (650,135)→(790,275) [to BSb]. At BS arrival (both same length 198 → same time ✓ sync ✓): - k=D3: gA continues: (450,275)→(360,365)→(360,660) D3; gB dissolves at BSb (fade 0.3s + shimmer). - k=D4: mirror. - k=D1: gA: (450,275)→(620,445) [BSa reflect]; gB: (790,275)→(620,445) [BSb reflect]; merged idler: (620,445)→(530,535) D1. - k=D2: same to BSc → (710,535) D2. - Detector flash: expanding ring + count badge pulse. Coincidence LED flash in the counter box when idler detected (paired with earlier D0 flash — maybe draw a linking arc?? too much; the LED labeled "D0∩Dk" flashes ✓). - Also flash the corresponding mini-histogram + a falling dot into the current bin of the big histogram ✓ (satisfying feedback). Timing: photon speed ~ 330 px/s (signal), same for idler. Signal total ≈ 600px → 1.8s; idler erasure 565px → 1.7s — WAIT that's SHORTER than signal?? I computed lengths: signal 487, idler erasure 565 ✓ idler longer ✓ (I confused myself; recompute: signal: 80 + [(590,215)→(825,192): Δ(235,−23) → 236] + [(830,185)→(1000,y_target): ~170–260] = 80+236+~215 ≈ 531px → 1.6s @330. Idler erasure: 198+240+127 = 565 → 1.71s. Delay ≈ 0.1–0.15s — VISIBLE? Marginal!! Hmm. For B-branch signal: 20 + [(650,155)→(825,178): 176] + ~215 = 411px → 1.25s; idler B erasure: 198+240+127 = 565 → 1.71s: delay 0.46s ✓ visible. For A-branch: 531 vs 565: 0.1s — meh. Make idler speed SLOWER? Photons all travel at c physically... but this is a schematic animation — many animations draw idler slightly slower for pacing?? NO — dishonest. BETTER: lengthen idler path more. Add a fold mirror on the erasure branch! E.g., BSa reflect beam: (450,275) → fold mirror FA at (535,360)?? hmm — the beam (450,275)→(620,445) passes (535,360) ✓ ON the beam — place fold FA at (535,360)?? folding changes direction... Let me add folds that zigzag: BSa (450,275) →reflect→ down-right to FA (535,360) → reflect down-left?? to... this fights the convergence-to-BSc design. ALTERNATIVE: extend the BSc→D1/D2 legs: D1 at (500,565)?: BSc (620,445) → D1: Δ(−120,120): 170px; D2 (740,565): 170px. Then erasure total: 198+240+170 = 608 vs signal A 531: +77px (+0.23s) ✓ better. Which-path: 198+127+295 = 620 vs 531: +89 ✓. Hmm still modest. PUSH MORE: D3/D4 drops longer: D3 at (360, 690): drop 325: which-path total 650 (+119px, +0.36s) ✓. D1/D2 even lower: D1 (490,575): leg 185: erasure 623 (+92, +0.28s) ✓. AND shorten signal: lens at x=810?: idler B beam (650,135)→(790,275) — lens at 810±8, y 40–345: beam at x=802: y=287 — CROSSES lens region (x≥802)?? beam x range 650–790 < 802 ✓ CLEAR ✓. lens x=810, rail x=980: signal A: 80 + [(590,215)→(805,192): 216] + [(810,185)→(980,y): ~165–250] ≈ 505px. OK: signal ~505, idler erasure ~623 (+23%), which-path ~650 (+29%) — delays 0.36–0.44s @330px/s ✓✓ CLEARLY VISIBLE. GOOD. FINAL-ish geometry (let me also re-verify a few collisions with the new numbers): - Lens: x=810 (bar y 40–345, w≈14). Rail: x=980, y 30–345. D0 head at (980, y0). - Idler B beam: (650,135)→(790,275): max x=790 < 802 ✓. - Mb (880,365), D4 (880,690): x=880 vs rail x=980 ✓ clear (100px). Mb→D4 beam (880, 365–690) vs lens (x 803–817, y≤345) ✓ clear (below). ✓ - D1 (490,575), D2 (750,575)?: BSc (620,445)→D2: Δ(130,130) ✓ 45° → D2 at (750,575) ✓. BSc→D1: Δ(−130,130) ✓ D1 (490,575) ✓. D2 (750,575) vs Mb (880,365)/D4 (880,690): ✓ clear. D1 (490,575) vs Ma (360,365)/D3 (360,690) ✓ clear. - Coincidence box: y 720–800, x 240–1010. D3 (360,690), D4 (880,690), D1 (490,575), D2 (750,575) — all above 720 ✓ (D3/D4 heads extend ~15px below their center: 705 < 720 ✓ ok). - Canvas: 1240×840. Signal beams at y 155/215 — laser at top (620,~40) ✓. - BSa (450,275), BSb (790,275), BSc (620,445), Ma (360,365), Mb (880,365). - Erasure beams: (450,275)→(620,445) and (790,275)→(620,445) — cross D1/D2 legs? D1 leg: (620,445)→(490,575): the A-erasure beam is (450,275)→(620,445) — shares only BSc point ✓. Do erasure beams cross the Ma→D3 vertical (x=360)? A-beam x∈[450,620] ✓ no. ✓ - Which-path beams: (450,275)→(360,365)→(360,690); (790,275)→(880,365)→(880,690) ✓ no conflicts with BSc legs (D1 leg from (620,445) to (490,575): x∈[490,620] ✓ clear of x=360; D2 leg x∈[620,750] ✓ clear of 880) ✓. - Signal static beam B: (650,155)→(805,178) — crosses idler B beam at (680,155)? idler B at y=155: x=670 — signal B beam at y=155 starts x=652 → crossing at (670,155) ✓ fine (free-space crossing). Signal beam A: (590,215)→(805,192): idler B at y=215: x=730 → crossing (730, ~209)?? signal A beam at x=730: y ≈ 215 − (730−590)/(215)·23 ≈ 215−15 = 200; idler B at y=200: x=715 — hmm they cross somewhere around (722, 204) ✓ fine. ✓ GEOMETRY LOCKED. (I'll fine-tune visually in code if needed but these numbers are consistent.) Hmm — one more: the fan wedge after the lens: from lens (810, 40..345) spreading to rail (980, 45..325) — draw as a soft gradient polygon ✓. And a wedge from the prism (590..650 at face) to the lens ✓ soft. ALSO — WAIT. I should double-check the GT prism story ONE more time: in my drawing, the SIGNALS (both slits') reflect off the 45° face to the right; the IDLERS pass on either side (they miss the prism geometrically, heading outward at 45°). Hmm — so the prism isn't really "separating" them (the idlers just geometrically miss it). Is that OK? Realistically: non-collinear SPDC emits signal/idler at ±3°; the GT prism reflects one polarization down... in my schematic the separation is partly geometric (angle) — which is ALSO physically legit (the figure in Kim's paper shows the idler continuing down while signal is deflected — the prism is drawn small). I'll draw the prism as a modest 45° plate spanning x 560–652 with label "P: Glan–Thompson prizması (sinyali ayırır)". The idlers passing OUTSIDE it is visually consistent enough. ✓ Hmm — but hold on: the signals travel VERTICALLY (down) from the slits to the prism — while the idlers immediately angle outward at 45°. So signal and idler from the SAME slit diverge from birth — meaning the "pair" is born with signal going straight down and idler at 45°?? Non-collinear SPDC: they're at ±3° from the pump — nearly co-linear. My drawing exaggerates (schematic ✓ — state "açılar temsilî"). ✓ FINE. Actually, hmm, let me reconsider: should the signal beams instead also come from the slits at slight INWARD angles to look more pair-like... NO. LOCKED. The exaggeration note covers it. **Element drawing details (SVG):** - Laser: rounded rect, metallic gradient (gray-blue), label "UV LAZER" + "351 nm". Small emission aperture + thin beam wedge (violet, low opacity) to the slit plate. Maybe subtle animated shimmer on the beam when running. - Slit plate: dark bar with two gaps; labels "A" and "B" next to gaps; title "Çift yarık". Draw hatching? Simple dark rect with gaps + labels. - BBO: pale green translucent rect tilted? A rect at y 125–140 spanning x 545–695, label "BBO". - Prism P: 45° plate: a thin rotated rect along the face line (560,245)–(652,153): draw as a line with slight thickness + hatched back? Classic: thin line + label. I'll draw a rotated rect (length ~130, thickness 10) with light-blue glass gradient + label "P". - Beam dump: skip (pump fades at BBO) ✓. - BSa/BSb: vertical plates: rounded rect w=10, h=46, glass blue, with a small stand? + toggle ring: when OUT, drawn ghosted/dashed + offset to the side (drawn slid 20px left with dashed outline + label "çıkarıldı")! NICE: physically slid out on a translation mount ✓✓ realistic!! When OUT: plate drawn shifted (e.g., 45px perpendicular... say BSa-out drawn at (410,275) — but the beam passes through (450,275) where the BS was... the plate moved away — draw at (405,275)?? that's ON the transmit beam ((450,275)→(360,365) passes (405,320))?? Hmm: moved-out position must not intersect any beam: BSa at (450,275) — beams: incoming (590,135)→(450,275) (ends there), transmit continues (450,275)→(360,365), reflect (450,275)→(620,445). Move-out direction: perpendicular-ish... place removed BSa at (450, 225)?? above: incoming beam passes (450,225)?? incoming line from (590,135) to (450,275): at x=450... the beam ENDS at (450,275); extended backward: at y=225: x=500 — so (450,225) is NOT on the beam ✓; but is it on the signal beams (y 155–215)? y=225 just below signal A beam (y~215 at x=590; at x=450?? signal A beam starts at x=590 ✓ no). And the prism face line at x=450?? face x∈[560,652] ✓ no. SO: BSa "removed" position (450,222), drawn small/dashed/50% opacity ✓. Similarly BSb-out at (790,222)?: incoming beam extended: at y=222: x=737?? (from (650,135) to (790,275): at y=222: x = 650+ (222−135) = 737) — (790,222) not on it ✓; signal beams at y 155–215: x≥590... signal B beam ((650,155)→(805,178)): at x=790: y≈176 — (790,222) is 46px below ✓ clear. ✓. BSc-out at (620, 405)?? beams: A-erasure (450,275)→(620,445): at x=620 y=445 — point (620,405): on the beam? beam: param: x=450+170t, y=275+170t: at y=405: t=0.765: x=580 → (620,405) NOT on it ✓ 28px off; B-erasure similar ✓; BSc legs start at (620,445) going down ✓. So BSc-out drawn at (620,400) dashed ✓ — hmm wait, but ALSO the D1/D2 legs when BSc is OUT: the erasure beams continue STRAIGHT through (620,445) to D1/D2 — but D1 is at (490,575) which is exactly along the B-erasure beam extended?? B-erasure: (790,275)→(620,445) direction (−0.707,0.707): extended: (620−90.7t', 445+90.7t'): at t'=1.43: (490,575) ✓ = D1 ✓✓ PERFECT (as designed). And A-erasure extended: (620+90.7t', 445+...): (710,535)... wait D2 is at (750,575): A-erasure extended from (620,445) dir (0.707,0.707): at x=750: t'=130/0.707... (750,575) ✓✓ PERFECT — D1/D2 sit exactly on the extended erasure beams ✓ so with BSc out, photons fly straight through to D1/D2 ✓✓ GEOMETRY IS BEAUTIFULLY CONSISTENT. - Ma/Mb: small tilted mirror plates (hatched style: line + hatch marks) at 22.5° tilt. - Detectors D1–D4: classic symbol: semicircle + stem + label. Orientation facing the incoming beam: D3 (360,690): incoming from ABOVE (vertical): semicircle opening UP (flat side down, arc up)?? The classic detector symbol is a semicircle with the flat side facing the beam. D3: beam comes down → flat side up: semicircle arc-down. D4 same. D1 (490,575): beam from up-right (45°): rotate −45°. D2 (750,575): beam from up-left: rotate +45°. Labels: "D1", "D2", "D3", "D4" + tiny sublabels ("silgi", "silgi", "yol A", "yol B"). On click: select channel. Flash: ring animation + fill pulse. - D0: on rail (980, y0): draw stage: a small carriage rect behind the rail + detector head (semicircle facing LEFT toward incoming beam — beam comes from the left (from lens): flat side facing left ✓) + label "D0" + position readout "x = +2.5 mm" + tiny fiber line to the right. DRAGGABLE along the rail (pointer events: pointerdown on head → move → clamp y 45–325, sets x0 (snap 0.5mm)) ✓✓ + rail ticks + mm scale labels (−7..+7) ✓ NICE. - Coincidence counter box: rounded rect, label "ÇAKIŞMA DEVRESİ (coincidence)", 5 LEDs: "D0", "D0∩D1", "D0∩D2", "D0∩D3", "D0∩D4" that blink on events, + "Δt ≈ 8 ns" note. Cables (bezier curves) from D0–D4 + all into the box top. Draw as gray curves ✓. - Labels everywhere (Turkish), font small (11–13px), non-cluttered: place carefully. - Also small legend (top-left of table?): photon colors: pompa (violet), sinyal (turuncu), idler (cyan) + "hayalet = olasılık genliği" note for ghosts ✓ important legend!! - A subtle grid on the optical table (dots pattern or fine grid, very light) — like a real optical breadboard with tapped holes!! ✓✓ REALISM: draw a pattern of small circles (hole grid, 20px spacing, 1px radius, #dfe3ea) in the background. YES — breadboard holes make it look like a real optical table. **Photon animation implementation:** - Use a single

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