Entangled Photon States

An entangled photon state is what you get when the tensor-product structure that field quantization hands you refuses to factor. The preceding articles build the machinery: quantizing the field gives each mode its own Fock ladder, and a two-photon state lives in the tensor product of two single-photon Hilbert spaces. Ordinarily a pair occupies a product of those spaces, $|\psi\rangle_1\otimes|\phi\rangle_2$ — each photon carries its own answer. Entanglement is the generic case where no such factorization exists in any local basis: the joint state is a coherent superposition of products, storing correlations that cannot be assigned to either photon separately. The paradigm realizations use polarization or path, but energy–time, frequency, and orbital-angular-momentum entanglement are equally physical. The operational signature is sharp. Trace out one photon and the survivor is left maximally mixed — completely random, carrying no polarization — even though the two-photon state is perfectly pure. The information has not vanished; it lives non-locally, in the correlations rather than in either marginal. This is what separates quantum entanglement from classical correlation, and John Bell turned the distinction into a number: any theory of local pre-existing values obeys the CHSH inequality $|S|\le2$ , while a shared entangled state pushes $S$ to the Tsirelson bound $2\sqrt2$ . Loophole-free experiments confirm the quantum value and close the door on local realism. Beyond foundations, entangled photons are the working resource of quantum key distribution, teleportation, dense coding, and photonic computing.

Key takeaways

An entangled two-photon state cannot be written as any product $|\psi\rangle_1\otimes|\phi\rangle_2$ ; the correlation is non-separable across polarization, path, or energy–time.
The canonical example is a polarization Bell state such as $|\Phi^+\rangle=(|HH\rangle+|VV\rangle)/\sqrt2$ — one of four maximally entangled basis states.
Each photon alone is maximally mixed, $\rho_1=\tfrac12\mathbb{1}$ : all the information sits in the correlations, none in the marginals.
Entanglement is certified, not merely suspected, by violating a Bell–CHSH inequality: the classical bound $|S|\le2$ is beaten up to the Tsirelson bound $2\sqrt2\approx2.828$ .
Spontaneous parametric down-conversion in a $\chi^{(2)}$ crystal is the workhorse source: a pump photon splits into a signal–idler pair, entangled where the which-path label is erased.
Key idea: entanglement is non-separability — the joint state is not a product of its parts, so all the correlation resides globally while each photon's marginal is maximally mixed, and Bell violation proves this cannot be faked by local pre-set values.

The core insight

Entanglement is non-separability: the joint state is not a product of its parts, so the whole correlation lives globally while each photon's reduced state is maximally mixed ( $\tfrac12\mathbb{1}$ ) — and Bell violation certifies the correlation is genuinely quantum, not a shared classical instruction set.

Where this lives today

The entangled photon has moved from foundational curiosity to laboratory instrument. The dominant modern source is still a nonlinear crystal — beta-barium borate (BBO), or a periodically poled waveguide in lithium niobate or KTP — pumped by a violet diode laser. A typical benchtop rig pumps BBO at 405 nm and collects degenerate pairs at 810 nm, compensating the H/V walk-off to reach visibilities above 0.98. These feed entanglement-based quantum key distribution, where a Bell test on a sample of pairs bounds any eavesdropper's information and the surviving correlated measurements become a shared secret key. At the frontier the resource is scaled up and sent farther: the Micius satellite distributed entangled pairs over more than a thousand kilometres for intercontinental key distribution and teleportation, loophole-free Bell tests in Delft, NIST, and Vienna closed the detection and locality loopholes simultaneously in 2015, and integrated photonic chips now generate and route many pairs on-chip — every one the object defined here, a pure two-photon state whose correlations refuse to factor and beat 2.

The physical picture

Picture two photons flying apart with a definite total polarization but each individually undecided. Measure either one along any axis and you instantly learn what the other will do along that same axis — not because a message flew between them, but because the joint state was never a pair of separate answers, only one shared correlation waiting to be read out. Nothing is sent: the correlation was prepared at the source and revealed only when the two records are later compared. The photons are born correlated inside a nonlinear crystal. A pump photon, unable to decay in free space, can split inside a medium with second-order susceptibility $\chi^{(2)}$ into two lower-frequency daughters — signal and idler — conserving the pump's energy and momentum. In the type-II geometry the daughters emerge orthogonally polarized on two cones; where the cones cross you cannot tell which photon went which way, the which-path label is erased, and the two indistinguishable amplitudes superpose into a polarization Bell state.

Figure

Type-II spontaneous parametric down-conversion: a 405 nm pump photon in a χ⁽²⁾ crystal splits into a signal and idler pair near 810 nm on two orthogonally polarized cones, and where the H and V cones intersect the which-path label is erased so the pair emerges polarization-entangled

The reduced state of one photon in a Bell state is $\tfrac12\mathbb{1}$ — perfectly random, carrying no polarization. That is not a defect of preparation; it is maximal entanglement. The "missing" polarization is stored in the correlation with the partner, so looking at one photon tells you nothing while comparing the two records tells you everything — entanglement is information that cannot be localized to either subsystem. Bell's genius was to make this testable. Under any local-realistic account — each photon carrying pre-set instructions, neither influenced by the far analyzer — a certain combination of four correlation measurements is bounded by 2. Quantum mechanics predicts up to $2\sqrt2\approx2.828$ , and entangled photons deliver it: the correlations are too strong to have come from instructions written at the source. The second figure is what an experimentalist measures — the singlet correlation versus relative analyzer angle, with the four Bell settings marked.

Figure

The singlet correlation E = −cos 2Δ versus the relative analyzer angle Δ: sampled at Δ = 22.5° and 67.5° the four CHSH terms combine to S = 2√2 ≈ 2.828, well above the local-realistic bound of 2, which is the Tsirelson maximum

The core equations

The natural basis for two-qubit polarization entanglement is the four maximally entangled Bell states. They span the same space $\mathbb{C}^2\otimes\mathbb{C}^2$ as the products $|HH\rangle,|HV\rangle,|VH\rangle,|VV\rangle$ , but each is a coherent superposition no product can reproduce.

|\Phi^\pm\rangle=\tfrac{1}{\sqrt2}\big(|HH\rangle\pm|VV\rangle\big),\qquad |\Psi^\pm\rangle=\tfrac{1}{\sqrt2}\big(|HV\rangle\pm|VH\rangle\big) \qquad (1)

where: $|H\rangle,|V\rangle$ — horizontal and vertical single-photon polarization; $|HH\rangle$ — shorthand for $|H\rangle_1\otimes|H\rangle_2$ ; $\pm$ — relative phase distinguishing symmetric from antisymmetric. Each Bell state predicts a perfect (anti-)correlation in every basis — impossible for any product state. The formal definition of entanglement is the negation of separability: a pure state is entangled precisely when it cannot be split into a state for photon 1 times a state for photon 2 in any local basis.

|\psi\rangle\neq|\phi\rangle_1\otimes|\chi\rangle_2\quad\text{for every }|\phi\rangle_1,|\chi\rangle_2 \qquad (2)

where: $\otimes$ — the tensor product joining the two Hilbert spaces; $|\phi\rangle_1,|\chi\rangle_2$ — arbitrary normalized single-photon states. If such a factorization existed the photons would carry independent answers; its impossibility forces the correlation to be shared, not owned. The cleanest operational test is to discard one photon. Partial-tracing photon 2 out of a Bell state leaves photon 1 proportional to the identity — maximally mixed (Derivation A).

\rho_1=\mathrm{Tr}_2\,|\Phi^+\rangle\langle\Phi^+|=\tfrac12\mathbb{1} \qquad (3)

where: $\mathrm{Tr}_2$ — partial trace over photon 2; $\rho_1$ — reduced density operator of photon 1; $\mathbb{1}$ — the $2\times2$ identity on the polarization qubit. The reading is stark: a pure global state whose parts are maximally mixed can only be entangled, since a product pure state leaves both marginals pure. To certify entanglement, Bell's inequality is recast as the CHSH combination of four correlation measurements, two settings per side. Any local-hidden-variable theory bounds it by 2 — a theorem about correlations, independent of quantum mechanics.

S=E(a,b)-E(a,b')+E(a',b)+E(a',b'),\qquad |S|\le2\ \ \text{(local realism)} \qquad (4)

where: $E(a,b)$ — correlation (mean product of $\pm1$ outcomes) at angle $a$ on photon 1 and $b$ on photon 2; $a,a'$ and $b,b'$ — the two settings per side. $S$ packages how the correlations shift as each observer flips between two questions; local instructions cannot push it past 2. For the polarization singlet $|\Psi^-\rangle$ quantum mechanics gives a correlation depending only on the relative analyzer angle — a consequence of the singlet's rotational invariance (Derivation B).

E(a,b)=-\cos\!\big[2(a-b)\big] \qquad (5)

where: $a-b$ — relative angle between the two linear polarizers. At parallel analyzers the singlet gives $E=-1$ (perfect anticorrelation); at $45^\circ$ apart, $E=0$ ; the cosine's steepness at $22.5^\circ$ beats the classical bound. Feeding (5) into (4) at the optimal angles saturates the largest CHSH value quantum theory allows — the Tsirelson bound.

|S|_{\max}=2\sqrt2\approx2.828 \qquad (6)

where: $|S|_{\max}$ — greatest CHSH value permitted by quantum theory, a factor $\sqrt2$ above 2 — wide enough that imperfect experiments still clear it by many standard deviations. The states are made by parametric down-conversion: expanding the two-mode squeezing interaction to first order in the weak coupling gives, beside the vacuum, a small amplitude for one photon each in signal and idler.

|\psi\rangle_{\text{SPDC}}\approx|0,0\rangle+\xi\,\hat a_s^\dagger\hat a_i^\dagger\,|0,0\rangle+\mathcal{O}(\xi^2) \qquad (7)

where: $\hat a_s^\dagger,\hat a_i^\dagger$ — creation operators for signal and idler; $\xi$ — effective coupling, proportional to $\chi^{(2)}$ , pump amplitude, and crystal length; $|0,0\rangle$ — vacuum in both modes. Pairs are emitted rarely (small $\xi$ ), so heralding one photon by its partner is standard. For anything short of maximal entanglement one needs a measure. The concurrence runs from 0 (separable) to 1 (Bell state).

C(\rho)=\max\{0,\ \lambda_1-\lambda_2-\lambda_3-\lambda_4\} \qquad (8)

where: $\lambda_i$ — square roots of the eigenvalues of $\rho\tilde\rho$ in decreasing order, with $\tilde\rho=(\sigma_y\otimes\sigma_y)\rho^*(\sigma_y\otimes\sigma_y)$ ; $C$ — concurrence. It measures the coherence between the correlated branches: a classical mixture with the same correlations has $C=0$ . For a pure state there is a more fundamental measure — the von Neumann entropy of the reduced state, in ebits, computed in Derivation C from the Schmidt decomposition.

E(|\psi\rangle)=-\mathrm{Tr}\big(\rho_1\log_2\rho_1\big) \qquad (9)

where: $\rho_1$ — reduced density matrix of either subsystem; $E$ — entropy of entanglement, in ebits. A product state has a pure marginal and $E=0$ ; a Bell state has $\rho_1=\tfrac12\mathbb{1}$ and $E=1$ ebit, maximal for a photon pair.

Derivation ATracing out one photon leaves the other maximally mixed

The most operational fingerprint of entanglement is that discarding one photon of a pure Bell pair leaves the other in complete ignorance — a maximally mixed density operator carrying no polarization. The goal here is to prove that from first principles: form the projector onto $|\Phi^+\rangle$ , take the partial trace over photon 2, and turn "the information lives in the correlations" into a concrete $2\times2$ matrix equal to half the identity — the calculation every no-signaling claim rests on.

Step 1. Write the projector as $|\Phi^+\rangle\langle\Phi^+|=\tfrac12\sum_{i,j\in\{H,V\}}|ii\rangle\langle jj|$ . Why: the double sum exposes both the diagonal terms $i=j$ that survive tracing and the coherence terms $i\neq j$ that will not.

Step 2. Partial-trace over photon 2 by sandwiching with its basis states: $\rho_1=\sum_k\langle k|_2\,|\Phi^+\rangle\langle\Phi^+|\,|k\rangle_2=\tfrac12\sum_i|i\rangle\langle i|$ . Why: orthonormality $\langle j|i\rangle_2=\delta_{ij}$ kills every off-diagonal term — the coherence between $|HH\rangle$ and $|VV\rangle$ is exactly what is discarded when the partner is ignored.

Step 3. Sum the survivors: $\rho_1=\tfrac12(|H\rangle\langle H|+|V\rangle\langle V|)=\tfrac12\mathbb{1}$ . Why: proportional to the identity, so every measurement on photon 1 gives 50/50 outcomes in any basis — the photon alone is a coin, all structure moved into the discarded correlations.

Step 4. Compare purity: $\mathrm{Tr}(\rho^2)=1$ (pure global) versus $\mathrm{Tr}(\rho_1^2)=\tfrac12$ (mixed part). Why: a pure whole with mixed parts is impossible for a product state, so this inequality is a rigorous entanglement witness — and entropy 1 ebit says the entanglement is maximal.

Go deeper · Derivation B · 1 minThe singlet state violates the CHSH bound

Derivation BThe singlet state violates the CHSH bound

Local realism caps the CHSH combination at $|S|\le2$ ; the whole content of Bell's theorem is that quantum mechanics does not respect that cap. The goal here is to compute $S$ for the polarization singlet $|\Psi^-\rangle$ — using its rotational invariance to reduce every correlation to a function of the relative angle, then choosing the four settings that maximize the combination — and show the result reaches $2\sqrt2$ , a clean factor $\sqrt2$ above the classical limit and the number that ruled out local hidden variables.

Step 1. Express each correlation as $E(a,b)=\langle\Psi^-|\hat\sigma_a\otimes\hat\sigma_b|\Psi^-\rangle=-\cos2(a-b)$ , with $\hat\sigma_a$ returning $\pm1$ along analyzer angle $a$ . Why: the singlet is invariant under identical rotations of both photons, so only the difference $a-b$ survives — the compact cosine correlation.

Step 2. Choose the optimal settings $a=0^\circ,\ a'=45^\circ,\ b=22.5^\circ,\ b'=67.5^\circ$ . Why: three of the four relative angles are $22.5^\circ$ (each $-\cos45^\circ$ ) and the fourth is $67.5^\circ$ ( $+\cos45^\circ$ ), aligning every term to reinforce in the sum.

Step 3. Evaluate: $E(a,b)=-\tfrac{1}{\sqrt2}$ , $E(a,b')=+\tfrac{1}{\sqrt2}$ , $E(a',b)=-\tfrac{1}{\sqrt2}$ , $E(a',b')=-\tfrac{1}{\sqrt2}$ . Why: $E(a,b')$ carries the plus sign but is subtracted in CHSH, so it too adds constructively — the classic sign slip to avoid.

Step 4. Combine: $S=-\tfrac{1}{\sqrt2}-\tfrac{1}{\sqrt2}-\tfrac{1}{\sqrt2}-\tfrac{1}{\sqrt2}=-2\sqrt2$ , so $|S|=2\sqrt2\approx2.828$ . Why: four terms of magnitude $1/\sqrt2$ sum to the Tsirelson maximum — comfortably above the local-realistic 2.

Go deeper · Derivation C · 1 minSchmidt decomposition and the entropy of entanglement

Derivation CSchmidt decomposition and the entropy of entanglement

For a general bipartite pure state one wants a canonical form that both diagonalizes each reduced density matrix and reads off "how entangled" it is. The Schmidt decomposition does exactly this: every pure two-party state is a single sum over matched orthonormal pairs, with non-negative weights whose spread measures the entanglement. The goal here is to build that form from a singular-value decomposition, show the reduced matrix's eigenvalues are the Schmidt weights, and define the entropy of entanglement.

Step 1. Expand in local bases as $|\psi\rangle=\sum_{ij}c_{ij}|i\rangle_1|j\rangle_2$ , collecting amplitudes into a coefficient matrix $c$ . Why: entanglement is encoded in $c$ — a product state has rank-1 $c$ , an entangled state higher rank.

Step 2. Apply the singular-value decomposition $c_{ij}=\sum_k U_{ik}\sqrt{p_k}\,V_{kj}$ , with non-negative singular values $\sqrt{p_k}$ . Why: the SVD always exists, so the decomposition is guaranteed and basis-independent — the $\sqrt{p_k}$ are intrinsic.

Step 3. Absorb $U,V$ into Schmidt bases $|u_k\rangle_1=\sum_i U_{ik}|i\rangle$ , $|v_k\rangle_2=\sum_j V_{kj}|j\rangle$ , giving $|\psi\rangle=\sum_k\sqrt{p_k}\,|u_k\rangle_1|v_k\rangle_2$ . Why: this pairs each Schmidt vector of photon 1 with exactly one of photon 2 — the number of nonzero $p_k$ separates separable (rank 1) from entangled.

Step 4. Trace out photon 2: $\rho_1=\sum_k p_k|u_k\rangle\langle u_k|$ , so the weights $p_k$ are the reduced-state eigenvalues, and $E=-\sum_k p_k\log_2 p_k$ . Why: for $|\Phi^+\rangle$ the weights are $p_1=p_2=\tfrac12$ , giving $E=1$ ebit — provably maximal for a photon pair.

Limiting cases

The four rows span the range from a product state to a maximally entangled Bell state, and contrast a genuine Bell state against its classically correlated impostor.

Case	Result	Meaning
Product (separable) pure state	$\vert\psi\rangle=\vert H\rangle_1\otimes\vert V\rangle_2$	No entanglement; concurrence $C=0$ ; both reduced states pure; obeys $\vert S\vert\le2$ .
Maximally entangled Bell state	$\vert\Phi^+\rangle=\tfrac{1}{\sqrt2}(\vert HH\rangle+\vert VV\rangle)$	Concurrence $C=1$ ; each reduced state $\tfrac12\mathbb{1}$ ; saturates the Tsirelson bound $\vert S\vert=2\sqrt2$ ; $E=1$ ebit.
Partially entangled state	$\vert\psi\rangle=\cos\theta\,\vert HH\rangle+\sin\theta\,\vert VV\rangle$	Concurrence $C=\vert\sin2\theta\vert$ ; entanglement tunable from 0 (at $\theta=0$ ) to 1 (at $\theta=\pi/4$ ).
Classically correlated (separable mixed)	$\rho=\tfrac12(\vert HH\rangle\langle HH\vert+\vert VV\rangle\langle VV\vert)$	Perfect HV-basis correlation but $C=0$ ; cannot violate any Bell inequality; the incoherent shadow of $\vert\Phi^+\rangle$ .

Common trapEntanglement lets you signal faster than light

It does not — and the reason is quantitative. When Alice measures her photon, Bob's is described from his side by $\rho_2=\mathrm{Tr}_1|\Phi^+\rangle\langle\Phi^+|=\tfrac12\mathbb{1}$ , completely independent of which axis Alice chose and what result she got. Bob's local statistics are 50/50 in every basis no matter what Alice does, so nothing at her end changes any measurable distribution at his — no channel carries a message. The correlations are real, but become visible only after Alice and Bob compare their outcome lists over a classical channel, itself limited to light speed — the no-signaling theorem. A companion trap: perfect correlation is not entanglement. The separable mixture $\tfrac12(|HH\rangle\langle HH|+|VV\rangle\langle VV|)$ is perfectly correlated in the HV basis yet has $C=0$ and never beats $|S|=2$ ; only the coherent superposition does.

Worked examples

1 · The Bell number a real experiment reports (confidence builder)

Compute the CHSH parameter $S$ for the singlet at $a=0^\circ,\ a'=45^\circ,\ b=22.5^\circ,\ b'=67.5^\circ$ , then say what a real source with visibility $V=0.97$ reports and whether it still refutes local realism.

Step 1. Use $E(a,b)=-\cos2(a-b)$ . The four relative angles are $-22.5^\circ$ , $-67.5^\circ$ , $22.5^\circ$ , $-22.5^\circ$ , so the doubled angles are $\mp45^\circ$ and $-135^\circ$ : $E(a,b)=-\cos45^\circ=-0.7071$ , $E(a,b')=-\cos135^\circ=+0.7071$ , $E(a',b)=-0.7071$ , $E(a',b')=-0.7071$ .

Step 2. Assemble the CHSH sum, watching signs: $S=(-0.7071)-(+0.7071)+(-0.7071)+(-0.7071)=-2.8284$ , so $|S|=2\sqrt2=\mathbf{2.828}$ — exactly the Tsirelson bound, a full $\sqrt2$ above the local-realistic ceiling of 2.

Step 3. Real sources are imperfect: noise and residual which-path distinguishability drop the fringe visibility below unity, and for a Werner-type state CHSH scales linearly, $S=2\sqrt2\,V$ . Violation survives only while $V>1/\sqrt2=0.7071$ — a forgiving threshold.

Step 4. For $V=0.97$ , $S=2\sqrt2\times0.97=\mathbf{2.744}$ , still clearly above 2 — a violation by dozens of sigma, which is why Aspect's 1982 measurement of $S\approx2.70$ was already decisive. Ideal $|S|=2\sqrt2\approx2.828$ ; at $V=0.97$ , $S\approx2.744>2$ — local realism is excluded.

2 · How much entanglement is in a partially entangled state?

A down-conversion source is tuned to produce $|\psi\rangle=\cos\theta\,|HH\rangle+\sin\theta\,|VV\rangle$ with $\theta=30^\circ$ . Find its entropy of entanglement and its concurrence, and compare them to a Bell state.

Step 1. Trace out one photon. The state is already in Schmidt form, so the reduced matrix is diagonal with eigenvalues $p=\cos^2\theta$ and $1-p=\sin^2\theta$ . At $\theta=30^\circ$ : $p=0.75$ , $1-p=0.25$ .

Step 2. Entropy (equation 9): $E=-0.75\log_2 0.75-0.25\log_2 0.25$ . With $\log_2 0.75=-0.4150$ and $\log_2 0.25=-2$ , $E=0.3113+0.5=\mathbf{0.811}$ ebits.

Step 3. Concurrence for this pure state: $C=|\sin2\theta|=|\sin60^\circ|=\mathbf{0.866}$ . Both sit between the extremes — a product state ( $\theta=0$ ) gives $E=C=0$ ; a Bell state ( $\theta=45^\circ$ ) gives $E=1$ , $C=1$ .

Step 4. This source carries about four-fifths of an ebit — usable but not maximal, and its Bell violation cannot reach $2\sqrt2$ ( $C=0.866<1$ ), which is why key-distribution sources are engineered as close to $\theta=45^\circ$ as the optics allow. $E=0.811$ ebits, $C=0.866$ — partial entanglement, short of the Bell maximum of 1.

Physics of the portrait

Entangled photons are where a philosophical objection became experimental science — Einstein's discomfort, Bell's inequality, and the crystals that measured the violation. No one embodies the experimental end of that arc more cleanly than the physicist who first made a Bell test airtight with light.

Origins · 1 minAlain Aspect and how this was discovered

Alain Aspect

1947– · the experimentalist who turned Bell's theorem into a measurement with entangled photons

Alain Aspect made the Bell inequality a laboratory quantity rather than a thought experiment. Working at Orsay around 1980, he built a source of correlated photon pairs from an atomic calcium cascade and measured their polarization correlations against the CHSH inequality with unprecedented precision; his 1982 results reported $S\approx2.70$ , violating the local-realistic bound of 2 by many standard deviations and matching the quantum prediction $2\sqrt2$ . The decisive refinement came the same year: Aspect, Jean Dalibard, and Gérard Roger switched each analyzer's orientation while the photons were in flight, using fast acousto-optic deflectors, so no signal at or below light speed could carry one side's setting to the other. That closed the "locality loophole" and made local hidden variables untenable even to sceptics, seeding the quantum-information era of down-conversion sources, teleportation, and cryptography. In 2022 he shared the Nobel Prize in Physics with John Clauser and Anton Zeilinger for experiments with entangled photons establishing the violation of Bell inequalities.

History

1935

— Einstein, Podolsky and Rosen argue quantum mechanics is incomplete; Schrödinger names entanglement the same year

1964

— John Bell derives his inequality, converting local realism into a testable bound

1969

— Clauser, Horne, Shimony and Holt recast Bell's bound as the experiment-ready CHSH inequality

1972

— Freedman and Clauser report the first experimental violation of a Bell inequality

1982

— Aspect, Dalibard and Roger switch analyzers in flight, closing the locality loophole

2022

— Aspect, Clauser and Zeilinger share the Nobel Prize for entangled-photon experiments

Going further

Texts. Loudon, The Quantum Theory of Light, 3rd ed., Ch. 6 · Saleh and Teich, Fundamentals of Photonics, 2nd ed., Ch. 13 and 21 · Nielsen and Chuang, Quantum Computation and Quantum Information, Ch. 2 · Foot, Atomic Physics, Bell tests · Hecht, Optics, 5th ed.

Provenance. Bell-state definitions, the CHSH combination, and the Schmidt decomposition follow Nielsen and Chuang Ch. 2 and Loudon Ch. 6; the SPDC output and phase matching follow Saleh and Teich Ch. 21; the experimental context follows Foot and the original papers of Aspect, Grangier and Roger, Phys. Rev. Lett. 49, 91 (1982), and Kwiat et al., Phys. Rev. Lett. 75, 4337 (1995) for the bright type-II source. A verifier should recompute: (1) the partial trace $\rho_1=\tfrac12\mathbb{1}$ ; (2) the CHSH sum at $0^\circ,45^\circ,22.5^\circ,67.5^\circ$ giving $|S|=2\sqrt2\approx2.828$ ; (3) $E=1$ ebit for $|\Phi^+\rangle$ and $E=0.811$ ebits, $C=0.866$ at $\theta=30^\circ$ .

Prerequisites. Quantization of the Electromagnetic Field · Fock States and the Photon Number Basis · Coherence Functions and the Degree of Coherence. Leads to. Quantum Key Distribution · Quantum Teleportation · Hong–Ou–Mandel Interference.

Ask the tutor: ask me to walk through the CHSH sum term by term, or to show exactly how erasing which-path information turns two SPDC cones into a Bell state.

What is the difference between correlation and entanglement?

Ordinary correlation just means two records tend to agree; you can make it classically by flipping one coin and copying the result to a distant partner — the separable mixture $\tfrac12(|HH\rangle\langle HH|+|VV\rangle\langle VV|)$ , perfectly correlated in the HV basis but with concurrence zero. Entanglement is stronger: the correlation holds coherently in every basis at once, which no shared classical instruction reproduces. The test is a Bell inequality — a classical state never pushes $|S|$ above 2, an entangled one reaches $2\sqrt2$ . Entanglement is the correlation that survives a change of basis.

Does measuring one photon really change the other instantly?

The distant photon's description updates the instant you measure yours, but nothing physical is transmitted and no experiment on the far photon alone can detect it. Before your measurement it is $\tfrac12\mathbb{1}$ ; after, a definite state correlated with your result — yet its local statistics stay 50/50 in every basis. The "change" is in your knowledge and in the correlations, not in any measurable quantity at the far end. Only when the two outcome lists are compared over a classical channel does the correlation appear, and that comparison is bounded by light speed — exactly why entanglement respects no-signaling.

How are entangled photons actually made?

Overwhelmingly by spontaneous parametric down-conversion. A pump photon in a crystal with second-order nonlinearity $\chi^{(2)}$ occasionally splits into signal and idler photons conserving energy ( $\omega_p=\omega_s+\omega_i$ ) and momentum ( $\mathbf{k}_p=\mathbf{k}_s+\mathbf{k}_i$ , phase matching). In the type-II geometry the two emerge orthogonally polarized on two cones, and where the cones intersect you cannot tell which is which, so the pair comes out in a polarization Bell state. Alternatives include atomic cascades, quantum dots, and four-wave mixing in fibres, but SPDC remains the workhorse for its brightness and fidelity.

Why is each photon of a Bell pair maximally mixed?

Because all the structure lives in the correlation, not the individual photons. Tracing out photon 2 from $|\Phi^+\rangle$ discards the coherence between the $|HH\rangle$ and $|VV\rangle$ branches, leaving $\tfrac12(|H\rangle\langle H|+|V\rangle\langle V|)=\tfrac12\mathbb{1}$ — a perfect 50/50 coin in every basis, no polarization, no pattern. This is not ignorance a better apparatus could remove; the photon genuinely has no polarization of its own. It exists only as a relationship — the meaning of "the information lives non-locally in the correlations".

What does the Tsirelson bound 2√2 actually mean?

It is the quantum ceiling on Bell correlations, the analogue of the classical "2". Local hidden variables cap CHSH at $|S|\le2$ ; quantum mechanics relaxes that to $|S|\le2\sqrt2\approx2.828$ , saturated by a maximally entangled state. Crucially, quantum theory does not allow arbitrarily strong correlations — you cannot reach the algebraic maximum $|S|=4$ , even though no-signaling alone would permit it. That intermediate value is a deep fingerprint of the theory, and "super-quantum" theories reaching $|S|>2\sqrt2$ are studied precisely to understand why nature stops here.

Why does indistinguishability matter for making entanglement?

Entanglement is coherence between alternatives, and coherence dies the moment the alternatives become distinguishable. In type-II SPDC the Bell state forms only where the two pathways — signal-H with idler-V, and signal-V with idler-H — are indistinguishable, so their amplitudes superpose rather than add as probabilities. Anything that tells the pathways apart, such as the temporal walk-off between the differently polarized photons, tags the paths and degrades the superposition toward the classical mixture. This is why real sources insert compensating crystals to erase walk-off, and why "which-path erasure" is the operational heart of every entanglement source.