Quantum Entanglement and Non-Separability

Entanglement is the feature Schrödinger called "not one but the characteristic trait of quantum mechanics, the one that enforces its entire departure from classical lines of thought." A composite state is separable if it factorizes, $|\Psi\rangle = |\alpha\rangle_A|\beta\rangle_B$ — the two parts have definite states of their own. It is entangled if it does not, and then the parts have no states of their own: you can know everything about the whole while knowing nothing local about either half. This article defines the distinction sharply and proves the theorem that makes it operational. The central result — owned here — is a three-way equivalence for pure bipartite states: $|\Psi\rangle$ is entangled $\iff$ its reduced state $\rho_A$ is mixed $\iff$ its Schmidt rank exceeds $1$ . Any one of the three is a certificate of the others, so the abstract "cannot be factored" becomes a computable test: reduce the state and check the purity. Product states leave $\rho_A$ pure; entanglement forces it mixed, and maximal entanglement makes it maximally mixed — half of a Bell pair is $\mathbb 1/2$ . We then exhibit the resource at its purest. The spin singlet $|\psi^-\rangle = (|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle)/\sqrt2$ is rotation-invariant — the same state in every basis — and shows perfect anti-correlation along every axis: measure both spins along any common direction and you get opposite results with certainty. This perfect correlation is the seed of the whole nonlocality story. A local-hidden-variable model of it is bounded by the classical CHSH value $2$ ; quantum mechanics reaches $2\sqrt2$ . We verify both numbers here so our statements are exact, but the inequality itself and its derivation are developed in Bell's Theorem and the CHSH Inequality by name; measures of entanglement in Entanglement Entropy and Measures of Entanglement by name; and the impossibility of copying entanglement (no-cloning) is stated with its source.

Key takeaways

Separable $\iff |\Psi\rangle = |\alpha\rangle_A|\beta\rangle_B$ ; entangled otherwise (the parts then have no local state).
Three-way equivalence (owned, proved): for pure $|\Psi\rangle$ , entangled $\iff\rho_A$ mixed $\iff$ Schmidt rank $> 1$ . Reduce and check purity — a computable entanglement test.
Not classical correlation: a separable mixed state $\sum_i p_i\rho_A^i\otimes\rho_B^i$ can be made by local operations and shared randomness; entanglement cannot (LOCC never creates it, stated with source).
The singlet $|\psi^-\rangle = (|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle)/\sqrt2$ : rotation-invariant ( $U\otimes U$ -invariant, verified), perfect anti-correlation $\langle\sigma_{\hat n}^A\sigma_{\hat n}^B\rangle = -1$ along every axis (verified) — the seed of Bell.
The classical ceiling: local-hidden-variable correlations satisfy the CHSH bound $2$ ; the singlet reaches $2\sqrt2$ (both verified; the inequality is owned by Bell's Theorem and the CHSH Inequality by name).
No-cloning (Wootters–Zurek/Dieks 1982): entangled states cannot be copied — which is why entanglement cannot be used to signal.
Key idea: entanglement is non-factorizability; for a pure state it is exactly the mixedness of the parts, and its cleanest witness — the singlet's perfect anti-correlation on every axis — beats every classical model.

The core insight

A pure bipartite state is entangled precisely when it cannot be written as a product — equivalently, when its reduced state is mixed, equivalently when its Schmidt rank exceeds one. The singlet realizes this maximally: rotation-invariant and perfectly anti-correlated along every axis, a correlation no local-hidden-variable model can match, capped classically at CHSH value $2$ and quantum-mechanically at $2\sqrt2$ .

Where this lives today

Entanglement is the fuel of quantum technology. Quantum computing speedups trace to entanglement across qubit registers — a product state of $n$ qubits is classically simulable, and the hard-to-simulate states are the entangled ones. Quantum cryptography (BB84, E91) turns the no-cloning theorem and entangled correlations into provably secure keys: an eavesdropper cannot copy the states, and tampering shows up as broken correlations. Quantum teleportation and networks consume shared entanglement (Bell pairs) as a resource to move quantum information, spent per use like a fuel. And in fundamental physics, entanglement entropy has become a probe of everything from topological order in condensed matter to the emergence of spacetime geometry in holography — the singlet's "the correlation is the object" lesson, scaled to the cosmos. The 2022 Nobel Prize recognized Aspect, Clauser, and Zeilinger for the experiments that turned Bell's inequality — the classical ceiling this article sets — from thought experiment into measured fact.

Figure

The Bell gap: the singlet's correlation beats every local model

The physical picture

Classically, if you know the state of a composite system, you know the state of each part — a pair of coins, however correlated, is always "this coin is heads, that one tails." Entanglement breaks this. The Bell state $(|00\rangle + |11\rangle)/\sqrt2$ is a completely definite, pure state of two qubits, yet neither qubit has a state: reduce to one and you get $\mathbb 1/2$ , maximal local ignorance. The information that makes the pair definite is not in either qubit — it is in the correlation between them, and correlation is not a property any single part carries. This is the whole mystery in one sentence: the best possible knowledge of a whole need not include the best possible knowledge of its parts (Schrödinger's phrasing). The three-way equivalence turns this into a tool. Since entanglement of a pure whole equals mixedness of a part, and mixedness is measured by purity, you can detect and quantify entanglement by a local computation: trace out one side, compute $\mathrm{Tr}\rho_A^2$ , and read off the answer — $1$ for a product state, below $1$ for entanglement, $1/d$ for maximal entanglement. The singlet shows the extreme: perfectly anti-correlated on every axis simultaneously, a feat that looks paradoxical until you accept that the correlation, not the parts, is the physical object. Whether that correlation can be explained by local pre-agreed values is the question Bell answered — no — and the singlet is the state that settles it.

Figure

Separable versus entangled: the parts have states, or they don't

Separability and the three-way equivalence

A bipartite pure state $|\Psi\rangle\in\mathcal H_A\otimes\mathcal H_B$ is separable (a product) if $|\Psi\rangle = |\alpha\rangle_A\otimes|\beta\rangle_B$ for some local states, and entangled otherwise. The definition is clean but hard to check directly — you cannot try all factorizations. The equivalence makes it decidable.

Derivation AEntangled ⟺ reduced state mixed ⟺ Schmidt rank > 1

We prove that for a pure bipartite state the three conditions — non-separable, $\rho_A$ mixed, Schmidt rank $> 1$ — are equivalent, giving a computable test for entanglement. The engine is the Schmidt decomposition (proved in The Schmidt Decomposition by name), used here as the canonical form; the equivalence itself is owned here.

Step 1. Put the state in Schmidt form. Any pure $|\Psi\rangle$ can be written $|\Psi\rangle = \sum_{k=1}^{r}\sqrt{\lambda_k}\,|k\rangle_A|k\rangle_B$ with $\lambda_k > 0$ , $\sum_k\lambda_k = 1$ , orthonormal Schmidt bases, and $r$ the Schmidt rank (the number of terms) — the decomposition and its uniqueness belong to The Schmidt Decomposition by name. Everything about bipartite entanglement of a pure state is encoded in the list $\{\lambda_k\}$ .

Step 2. Reduce. Tracing out $B$ (The Reduced Density Matrix and Partial Trace, recalled), $\rho_A = \sum_k\lambda_k|k\rangle\langle k|_A$ — the Schmidt weights are its eigenvalues. So $\mathrm{Tr}\,\rho_A^2 = \sum_k\lambda_k^2$ , and $\rho_A$ is pure ( $=1$ ) iff a single $\lambda_k = 1$ , mixed ( $< 1$ ) iff two or more $\lambda_k$ are nonzero.

Step 3. Close the loop. Separable means $r = 1$ : exactly one term, $|\Psi\rangle = |1\rangle_A|1\rangle_B$ , a product — and then $\rho_A = |1\rangle\langle1|$ is pure. Conversely $r = 1$ forces the product form. So $r = 1\iff$ separable $\iff\rho_A$ pure, and negating, $r > 1\iff$ entangled $\iff\rho_A$ mixed. Why the three collapse into one: the Schmidt weights simultaneously count the terms (rank), fill the reduced spectrum (mixedness), and decide factorizability (separability) — one list, three readings. The test is now mechanical: reduce, compute purity, done.

Step 4. Quantify with a family. For $|\Psi(\theta)\rangle = \cos\theta|00\rangle + \sin\theta|11\rangle$ , the Schmidt weights are $\cos^2\theta, \sin^2\theta$ , so $\rho_A = \mathrm{diag}(\cos^2\theta, \sin^2\theta)$ and $\mathrm{Tr}\,\rho_A^2 = 1 - \tfrac12\sin^2 2\theta$ : purity $1$ (separable) at $\theta = 0, \pi/2$ , dipping to $\tfrac12$ (maximal entanglement) at $\theta = \pi/4$ — a continuous dial from product to Bell. Verified numerically: a product state has Schmidt rank $1$ and pure $\rho_A$ ; the entangled $\sqrt{0.7}|00\rangle + \sqrt{0.3}|11\rangle$ has rank $2$ and mixed $\rho_A$ ; and $(|00\rangle + |01\rangle)/\sqrt2$ factors as $|0\rangle(|0\rangle+|1\rangle)/\sqrt2$ (rank $1$ , separable) while $(|00\rangle + |11\rangle)/\sqrt2$ has rank $2$ and $\rho_A = \mathbb 1/2$ .

Entanglement is not classical correlation

A subtle point separates entanglement from mere correlation. A separable mixed state $\rho_{AB} = \sum_i p_i\,\rho_A^i\otimes\rho_B^i$ is correlated — measuring $A$ tells you about $B$ — but every such state can be prepared by two distant parties using only local operations and a shared random number $i$ (each prepares $\rho_A^i$ or $\rho_B^i$ on hearing the shared value): this is classical correlation, and no quantum resource is spent. An entangled state is exactly one that cannot be prepared this way. The precise statement, entanglement cannot be created by local operations and classical communication (LOCC), is the operational definition of the resource (stated with source: Werner 1989 introduced the separable/entangled split for mixed states). So "entangled" means "correlated beyond what shared randomness can explain" — and it is that excess that Bell's inequality detects and that discord (a still-finer notion of non-classicality, owned by Quantum Discord and Non-Classical Correlations by name) refines further.

The singlet: the entanglement resource at its purest

Go deeper · Derivation B · 2 minThe singlet is rotation-invariant and perfectly anti-correlated on every axis

Derivation BThe singlet is rotation-invariant and perfectly anti-correlated on every axis

We prove two properties of the spin singlet $|\psi^-\rangle = (|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle)/\sqrt2$ that make it the canonical entangled resource and the seed of Bell's theorem: it is invariant (up to phase) under $U\otimes U$ for any SU(2) rotation $U$ , and it gives perfectly anti-correlated spin measurements along every common axis. The singlet's construction as the total-spin-zero state is owned by Addition of Angular Momenta by name; the correlations are owned here.

Step 1. Rotation invariance. The singlet is the unique total-spin- $0$ state of two spin- $\tfrac12$ 's, and a spin- $0$ object is a rotational scalar: applying the same rotation $U$ to both spins, $U\otimes U$ , returns $|\psi^-\rangle$ up to a global phase. Why: $|\psi^-\rangle$ carries zero total angular momentum, so the rotation generators $\mathbf S_A + \mathbf S_B$ annihilate it, and exponentiating gives $(U\otimes U)|\psi^-\rangle = |\psi^-\rangle$ (phase absorbed). Consequently the singlet looks identical in every basis: written in $x$ -, $y$ -, or $z$ -eigenstates it is always $(|+\rangle|-\rangle - |-\rangle|+\rangle)/\sqrt2$ for that axis.

Step 2. Perfect anti-correlation along a fixed axis. Measure $\sigma_z$ on both spins. The singlet has support only on $|\!\uparrow\downarrow\rangle$ and $|\!\downarrow\uparrow\rangle$ , so the outcomes are always opposite: $\langle\sigma_z^A\sigma_z^B\rangle = -1$ . Whenever $A$ reads $+$ , $B$ reads $-$ , with certainty.

Step 3. Anti-correlation along every axis. By Step 1 the singlet is the singlet in the $\hat n$ -basis too, so the same argument gives $\langle\sigma_{\hat n}^A\sigma_{\hat n}^B\rangle = -1$ for every direction $\hat n$ — perfect anti-correlation is not a $z$ -axis accident but a rotational invariant. Why this is startling: the two spins are perfectly anti-aligned along $x$ AND along $y$ AND along $z$ simultaneously, though no single spin can have definite components along non-commuting axes. The correlation lives in the pair, not the parts. More generally $\langle\sigma_{\hat a}^A\sigma_{\hat b}^B\rangle = -\hat a\cdot\hat b = -\cos\theta_{ab}$ .

Step 4. Set the classical ceiling. This perfect same-axis anti-correlation, plus the smooth $-\cos\theta_{ab}$ falloff, is exactly what a local-hidden-variable model tries and fails to reproduce at intermediate angles. Feeding the four-setting CHSH combination the LHV rule (each spin carries pre-agreed $\pm1$ values) caps the correlation sum at $2$ ; the quantum $-\cos\theta_{ab}$ at the optimal angles reaches $2\sqrt2$ . The gap between $2$ and $2\sqrt2$ is the experimental signature of entanglement's non-classicality — the inequality, its derivation, and the Tsirelson bound are owned by Bell's Theorem and the CHSH Inequality by name; here we only certify the two numbers. Verified numerically: $|\psi^-\rangle$ is invariant under $U\otimes U$ (up to phase) for 50 random SU(2) rotations; $\langle\sigma_{\hat n}^A\sigma_{\hat n}^B\rangle = -1$ over a $7\times7$ grid of directions; the CHSH classical bound is $2$ (enumerating all $\pm1$ assignments) and the singlet reaches $2\sqrt2 = 2.8284$ at the optimal angles $a = 0°, a' = 90°, b = 45°, b' = -45°$ .

No-cloning: why entanglement cannot signal

Entanglement's perfect correlations tempt a signalling scheme — but it fails, and the reason is the no-cloning theorem (Wootters–Zurek and Dieks, 1982): no unitary can copy an unknown quantum state, $|\psi\rangle|0\rangle\to|\psi\rangle|\psi\rangle$ for all $|\psi\rangle$ , because cloning would be nonlinear while quantum evolution is linear. If Alice could clone her half of an entangled pair, she could read out which basis Bob measured in by making many copies and doing statistics — instant signalling. No-cloning forbids the copies, and the no-signalling theorem (owned by The Reduced Density Matrix and Partial Trace by name) forbids the effect directly: whatever Bob does, Alice's $\rho_A$ is unchanged. The two protections are consistent, and together they place entanglement exactly where relativity allows: strong enough to violate every classical correlation bound, powerless to carry a message.

Limiting cases

state	Schmidt rank	$\rho_A$	entangled?
product $\|\alpha\rangle\|\beta\rangle$	$1$	pure	no
$\cos\theta\|00\rangle + \sin\theta\|11\rangle$ , $0 < \theta < \pi/2$	$2$	mixed	yes (partial)
Bell / singlet	$2$	$\mathbb 1/2$	yes (maximal)
separable mixed $\sum_i p_i\rho_A^i\otimes\rho_B^i$	—	$\sum_i p_i\rho_A^i$	no (classical correlation)
singlet correlation $\langle\sigma_{\hat a}\sigma_{\hat b}\rangle$	—	—	$-\hat a\cdot\hat b$
CHSH sum	—	—	classical $\le 2$ , quantum $\le 2\sqrt2$

Common trapReduced-state mixedness detects entanglement only for PURE global states

The three-way equivalence — entangled $\iff\rho_A$ mixed — is a theorem about pure $|\Psi\rangle$ , and applying it to a mixed $\rho_{AB}$ is a genuine error. A classically correlated separable state like $\tfrac12|00\rangle\langle00| + \tfrac12|11\rangle\langle11|$ has $\rho_A = \mathbb 1/2$ — maximally mixed — yet is not entangled (it is preparable by shared randomness). So a mixed reduced state does not certify entanglement once the global state is itself mixed: local mixedness there could be either entanglement or plain classical ignorance. Detecting entanglement in mixed states needs stronger tools — the Peres–Horodecki partial-transpose (PPT) criterion, entanglement witnesses, negativity — developed in Entanglement Entropy and Measures of Entanglement by name. The clean rule "reduce and check purity" is pure-state only; reach for it on a mixed $\rho_{AB}$ and you will call classical correlation "entanglement."

Worked examples

Example 1 — is it entangled? the factoring test (confidence builder). Two candidate two-qubit states; decide which is entangled by attempting to factor, then by the purity test.

Step 2. Try to factor $(|00\rangle + |11\rangle)/\sqrt2$ . No grouping works: the first qubit being $|0\rangle$ forces the second to $|0\rangle$ , and $|1\rangle$ forces $|1\rangle$ — the two qubits' identities are locked together, which no product can express. It is entangled (Schmidt rank $2$ ). Confirm: $\rho_A = \tfrac12(|0\rangle\langle0| + |1\rangle\langle1|) = \mathbb 1/2$ , maximally mixed, $P = \boxed{\tfrac12}$ .

Step 3. Read the boundary. Same superposition structure ("a plus of two terms"), opposite verdicts — the difference is whether the second qubit's value is determined by the first. When it is not (both terms share $A = |0\rangle$ ), the state factors and is separable; when it is (each term pairs a distinct $A$ with a distinct $B$ ), it is entangled. Purity is the machine test: $P = 1$ separable, $P < 1$ entangled (all machine-verified).

Example 2 — the singlet's impossible correlations (exam level). Show that the singlet is perfectly anti-correlated along $x$ , $y$ , and $z$ at once, and state the classical bound it violates.

Step 1. Along $z$ . In the $z$ -basis $|\psi^-\rangle = (|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle)/\sqrt2$ , so both outcomes are opposite: $\langle\sigma_z^A\sigma_z^B\rangle = \boxed{-1}$ . Alice's $+$ guarantees Bob's $-$ .

Step 2. Along $x$ and $y$ . By rotation invariance the singlet has the identical form in the $x$ - and $y$ -bases, so $\langle\sigma_x^A\sigma_x^B\rangle = \langle\sigma_y^A\sigma_y^B\rangle = \boxed{-1}$ as well (all machine-verified). The two spins are perfectly anti-aligned along three mutually incompatible axes simultaneously — impossible for any pair of classical arrows, which can be anti-aligned along at most one axis, yet routine for the entangled pair because the anti-alignment is a property of the correlation, not of either spin's (nonexistent) local direction.

Step 3. The classical ceiling. A local-hidden-variable model assigns each spin pre-existing $\pm1$ values for every axis. The best such model matches the perfect anti-correlation at equal angles but cannot track the quantum $-\cos\theta_{ab}$ at intermediate angles: the four-setting CHSH combination it produces is bounded by $|S|\le 2$ (enumerating all $\pm1$ assignments gives max $2$ ). The singlet gives $|S| = 2\sqrt2 \approx 2.83$ at the optimal angles $a = 0°, a' = 90°, b = 45°, b' = -45°$ (machine-verified) — a $41\%$ violation. That gap is where entanglement stops being explainable by any local story; the inequality and its proof are owned by Bell's Theorem and the CHSH Inequality by name.

The characteristic trait

Origins · 1 minquantum entanglement and how this was discovered

quantum entanglement

1935 · Oxford, Princeton

The word and the concept were fixed in a single remarkable year. In 1935, reacting to the Einstein–Podolsky–Rosen challenge (owned by The EPR Paradox and Local Realism by name), Erwin Schrödinger — then in Oxford — wrote a pair of papers that named the phenomenon: Verschränkung in German, "entanglement" in his own English, which he called "not one but the characteristic trait of quantum mechanics, the one that enforces its entire departure from classical lines of thought." His deepest observation was the one this chapter is built on: for an entangled pair, "the best possible knowledge of a whole does not include the best possible knowledge of its parts" — maximal knowledge of the joint state, none of the reduced states, the paradox that becomes the theorem "entangled $\iff$ reduced state mixed." The same papers introduced the cat (a macroscopic superposition entangled with a decaying atom) and "steering" (a distant measurement fixing a local conditional state). For decades entanglement was a philosophers' puzzle; John Bell turned it into an experiment in 1964, and the 2022 Nobel to Aspect, Clauser, and Zeilinger closed the arc — the "characteristic trait" became a measured, engineered, sellable resource. Schrödinger the man appears across this track's dynamics chapters; here the emphasis is his 1935 naming and the whole-versus-parts insight that organizes entanglement theory.

History

1935

— Schrödinger coins "entanglement" (Verschränkung); "the characteristic trait" (Proc. Camb. Phil. Soc. 31, 555)

1935

— the cat and "steering" appear in the companion Naturwissenschaften paper

1964

— Bell turns the whole-versus-parts puzzle into a testable inequality

1989

— Werner defines the separable/entangled split for mixed states

2022

— Nobel Prize to Aspect, Clauser and Zeilinger for the Bell-test experiments

Going further

References. Nielsen & Chuang, §2.3–2.5, §12.5 (entanglement, LOCC) · Sakurai & Napolitano, §3.10–3.11 · Preskill, Lecture Notes, Ch. 4 · Horodecki, Horodecki, Horodecki & Horodecki, Rev. Mod. Phys. 81, 865 (2009) (entanglement review) · Schrödinger, Proc. Camb. Phil. Soc. 31, 555 (1935) · Werner, Phys. Rev. A 40, 4277 (1989) · Wootters & Zurek, Nature 299, 802 (1982); Dieks, Phys. Lett. A 92, 271 (1982) (no-cloning).

Provenance. All results are standard. The reduced-density-matrix machinery and no-signalling are owned by The Reduced Density Matrix and Partial Trace, recalled by name; the purity criterion by Pure versus Mixed States and the Purity Criterion; the singlet's construction as the total-spin- $0$ state by Addition of Angular Momenta. Owned and derived here: (1) the three-way equivalence entangled $\iff\rho_A$ mixed $\iff$ Schmidt rank $> 1$ , machine-verified on product and entangled states and on the $\theta$ -family $\rho_A = \mathrm{diag}(\cos^2\theta,\sin^2\theta)$ , purity $1 - \tfrac12\sin^2 2\theta$ ; (2) the singlet's rotation invariance ( $U\otimes U$ -invariant up to phase over 50 random SU(2) elements) and perfect anti-correlation $\langle\sigma_{\hat n}^A\sigma_{\hat n}^B\rangle = -1$ over a $7\times7$ direction grid; and the classical-versus-quantum CHSH numbers $2$ versus $2\sqrt2$ (verified: LHV enumeration max $2$ ; singlet $2.8284$ at $a=0°,a'=90°,b=45°,b'=-45°$ ) — the inequality and its derivation owned by Bell's Theorem and the CHSH Inequality by name. Worked numbers recomputed: $(|00\rangle+|01\rangle)/\sqrt2$ factors (separable, $\rho_A$ pure); $(|00\rangle+|11\rangle)/\sqrt2$ has $\rho_A = \mathbb 1/2$ (entangled); singlet anti-correlated $-1$ along $x,y,z$ . The Schmidt decomposition is recalled by name from The Schmidt Decomposition; the entanglement measures from Entanglement Entropy and Measures of Entanglement; the EPR argument from The EPR Paradox and Local Realism; GHZ multiparty nonlocality from The GHZ State and All-or-Nothing Nonlocality; discord from Quantum Discord and Non-Classical Correlations — all by name.

Builds on: The Reduced Density Matrix and Partial Trace · Pure versus Mixed States and the Purity Criterion · Addition of Angular Momenta · Spin-1/2 and the Pauli Matrices

Leads to: The Schmidt Decomposition · Entanglement Entropy and Measures of Entanglement · The Bell States and Maximally Entangled Pairs · Bell's Theorem and the CHSH Inequality

Is entanglement the same as correlation?

No — entanglement is correlation of a specific, non-classical kind. Two coins glued to always land the same way are perfectly correlated but not entangled: their correlation is explainable by shared classical information (which way they were glued). A separable state, even a highly correlated one, is exactly the "shared classical randomness" kind — preparable by local operations plus a common random number (LOCC). Entanglement is the correlation that cannot be so prepared, the excess over what shared randomness allows. Bell's inequality is the quantitative line: classical correlation obeys it, entanglement can violate it. A still finer notion, quantum discord, finds non-classicality even in some separable states — owned by Quantum Discord and Non-Classical Correlations by name.

If the singlet is perfectly anti-correlated on every axis, why isn't that a signal?

Because the perfect correlation is only visible after the two parties compare results, and the comparison needs classical communication (limited by light speed). Locally, Alice sees random $\pm$ outcomes with no pattern (her $\rho_A = \mathbb 1/2$ ), whatever axis Bob chose and whatever he got. Only when she later learns Bob's axis and outcomes does the $-1$ anti-correlation appear in the joint data. Bob choosing a different axis changes the correlation structure but not Alice's local statistics — the no-signalling theorem. The correlation is real and non-classical; it just cannot be used to transmit a choice.

Can a mixed state be entangled, and does the purity test still work?

Yes, mixed states can be entangled — but the "reduce and check purity" test does not work for them. That test is a pure-state theorem: a classically correlated separable mixed state can have a maximally mixed reduced state yet no entanglement (e.g. $\tfrac12|00\rangle\langle00| + \tfrac12|11\rangle\langle11|$ ). Detecting mixed-state entanglement requires stronger criteria — most usefully the Peres–Horodecki partial-transpose test (a state is entangled if its partial transpose has a negative eigenvalue, sufficient for two qubits) and its quantifier, the negativity. These are owned by Entanglement Entropy and Measures of Entanglement by name; here we flag only that the clean purity test is pure-state exclusive.

How is "cannot be created by LOCC" a useful definition if it sounds circular?

It is not circular — it is operational, and it makes entanglement a resource. LOCC (local operations and classical communication) is the set of things two distant parties can do for free: manipulate their own systems and phone each other. The theorem is that no amount of LOCC turns a product state into an entangled one; entanglement can only be consumed or distributed, never conjured locally. This is exactly what makes it valuable — like a fuel that must be shipped, not manufactured on site. It also gives entanglement measures their meaning: a good measure never increases under LOCC, so it quantifies "how much of the non-manufacturable resource you hold." The measure theory is owned by Entanglement Entropy and Measures of Entanglement by name.

Does entanglement require the two parties to be far apart, or is it about the Hilbert-space structure?

It is about the tensor-product structure, not distance. Entanglement is defined by a partition of the system into subsystems $A$ and $B$ ; whether they are across a lab or across a galaxy is irrelevant to whether $|\Psi\rangle$ factors. Two electrons in a helium atom are entangled in their spin-position structure though they overlap in space. Distance matters only for the dramatic consequences (spacelike-separated correlations that stress locality), and even those are about the partition being into causally separated regions. The core notion — non-factorizability across a chosen cut — is purely algebraic, which is why entanglement entropy of a spatial region is a meaningful quantity in field theory and gravity, where "the other party" is just the complementary region.

Why the minus sign in the singlet — does |↑↓⟩ + |↓↑⟩ behave differently?

Yes, importantly. The minus combination $(|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle)/\sqrt2$ is the total-spin- $0$ singlet: rotation-invariant, anti-correlated on every axis, the unique rotational scalar. The plus combination $(|\!\uparrow\downarrow\rangle + |\!\downarrow\uparrow\rangle)/\sqrt2$ is one of the three total-spin- $1$ triplet states — also maximally entangled and anti-correlated along $z$ , but not rotation-invariant, so its correlations are axis-dependent (along $x$ it is correlated, not anti-correlated). The four Bell states are exactly the singlet plus the three triplets, a maximally entangled orthonormal basis owned by The Bell States and Maximally Entangled Pairs by name. The sign encodes the exchange symmetry that Addition of Angular Momenta builds.