Entangled States

Entanglement is the failure of a composite quantum state to be separable across a specified subsystem split. For a bipartite pure state, separability is equivalent to factorization: once a split $A|B$ has been specified, the state is entangled precisely when it is not a product state. For a mixed state, nonfactorization alone is insufficient because a correlated convex mixture of product states can be separable.

That short definition has far-reaching consequences. An entangled joint state can be pure while each subsystem has a mixed local state; its correlations need not be reproducible by assigning a state vector to each part; and local unitary changes of basis preserve its entanglement. None of this makes entanglement a force or a faster-than-light communication channel.

Required background. State Vectors supplies pure-state representations and basis expansions; Density Operators supplies mixed and reduced states and trace-rule predictions. Familiarity with product bases, matrix rank, singular values, and partial contraction is assumed.

Helpful background. Correlations and Covariance supplies comparisons of joint correlations with products of marginal expectations.

Entanglement across a subsystem split

Let two distinguishable subsystems have Hilbert spaces $\mathcal H_A$ and $\mathcal H_B$. Their composite Hilbert space is

$$\mathcal H_{AB} = \mathcal H_A\otimes\mathcal H_B.$$

A nonzero pure state $\lvert\Psi\rangle_{AB}$ is a product state across $A|B$ if there are subsystem vectors $\lvert\psi\rangle_A$ and $\lvert\phi\rangle_B$ such that

$$\lvert\Psi\rangle_{AB} = \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B.$$

It is entangled across $A|B$ if no such factorization exists.

If $\lvert\Psi\rangle$ is normalized, its factors can also be chosen normalized. The factors are not unique as vectors because, for any nonzero complex number $z$,

$$\lvert\psi\rangle_A\otimes\lvert\phi\rangle_B = \bigl(z\lvert\psi\rangle_A\bigr) \otimes \bigl(z^{-1}\lvert\phi\rangle_B\bigr).$$

After normalization and the usual identification of global phase, however, a product pure state determines one pure ray for each subsystem.

Every bipartite pure state is therefore in exactly one of two classes relative to the chosen split:

$$\text{product} \qquad\text{or}\qquad \text{entangled}.$$

The word relative is essential. Entanglement is not a property of an abstract vector alone; it is a property of a vector together with a tensor-product decomposition. The split may represent two particles, two spatial regions, two modes, or two internal degrees of freedom. A different physically meaningful split can produce a different classification.

Coefficient-Matrix Test

Choose orthonormal bases $\{\lvert i\rangle_A\}$ and $\{\lvert j\rangle_B\}$. Any bipartite pure state can be expanded as

$$\lvert\Psi\rangle = \sum_{i,j} c_{ij}\, \lvert i\rangle_A\otimes\lvert j\rangle_B.$$

Arrange the amplitudes into a coefficient matrix $C$ with entries $C_{ij}=c_{ij}$. If the state is a product,

$$\lvert\psi\rangle_A = \sum_i a_i\lvert i\rangle_A, \qquad \lvert\phi\rangle_B = \sum_j b_j\lvert j\rangle_B,$$

then

$$c_{ij}=a_i b_j, \qquad C=\mathbf a\,\mathbf b^{\mathsf T}.$$

Such a nonzero matrix has rank one. Conversely, every rank-one matrix admits an outer-product factorization, so it defines a product state. Thus

$$\begin{aligned} \operatorname{rank}C&=1 &&\Longleftrightarrow& \lvert\Psi\rangle&\text{ is product},\\ \operatorname{rank}C&>1 &&\Longleftrightarrow& \lvert\Psi\rangle&\text{ is entangled}. \end{aligned}$$

For larger finite-dimensional systems, rank one is equivalent to the vanishing of every $2\times2$ minor of $C$.

Two-qubit determinant test

Write a normalized two-qubit state as

$$\lvert\Psi\rangle = a\lvert00\rangle +b\lvert01\rangle +c\lvert10\rangle +d\lvert11\rangle.$$

Its coefficient matrix is

$$C= \begin{pmatrix} a&b\\ c&d \end{pmatrix}.$$

The state is product exactly when

$$\det C=ad-bc=0,$$

and it is entangled exactly when $ad-bc\ne0$.

This test does not depend on the chosen local bases. Under active local unitaries $U_A\otimes U_B$ in a fixed basis, the coefficient matrix transforms as

$$C\longmapsto C' = U_A C U_B^{\mathsf T}.$$

Both unitary matrices are invertible, so $\operatorname{rank}C'=\operatorname{rank}C$. Local changes of basis cannot create or remove entanglement.

Bell States

The four Bell states are

$$\begin{aligned} \lvert\Phi^\pm\rangle &= \frac{\lvert00\rangle\pm\lvert11\rangle}{\sqrt2},\\ \lvert\Psi^\pm\rangle &= \frac{\lvert01\rangle\pm\lvert10\rangle}{\sqrt2}. \end{aligned}$$

Each coefficient matrix has rank two, so every Bell state is entangled. For example,

$$C_{\Phi^+} = \frac{1}{\sqrt2} \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}, \qquad \det C_{\Phi^+} = \frac12.$$

A useful one-parameter family is

$$\begin{aligned} \lvert\Psi(\theta,\varphi)\rangle &= \cos\theta\,\lvert00\rangle + e^{i\varphi}\sin\theta\,\lvert11\rangle, \\ \theta&\in\left[0,\frac{\pi}{2}\right]. \end{aligned}$$

Its determinant is

$$\det C = e^{i\varphi}\cos\theta\sin\theta.$$

The state is product only at $\theta=0$ or $\theta=\pi/2$. It is entangled for every intermediate value. At $\theta=\pi/4$, its two terms have equal magnitude and it is locally equivalent to a Bell state; the Bell-state family above displays the corresponding maximally entangled two-qubit basis.

Schmidt Perspective

Every pure state in a finite-dimensional bipartite system admits a Schmidt decomposition,

$$\lvert\Psi\rangle = \sum_{r=1}^{R} s_r\, \lvert u_r\rangle_A \otimes \lvert v_r\rangle_B,$$

where

$$s_r>0, \qquad \sum_{r=1}^{R}s_r^2=1,$$

and both sets of Schmidt vectors are orthonormal. The Schmidt rank $R$ equals $\operatorname{rank}C$, so

$$\begin{aligned} R=1 &\quad\Longleftrightarrow\quad \text{product},\\ R>1 &\quad\Longleftrightarrow\quad \text{entangled}. \end{aligned}$$

The Schmidt coefficients $\{s_r\}$ do not change under local unitaries. They therefore contain basis-independent information about bipartite pure-state entanglement. In a coefficient-matrix representation, they are precisely the singular values, while local unitaries rotate the left and right singular vectors without changing those values.

Local States of an Entangled Pair

The density operator of a pure joint state is

$$\rho_{AB} = \lvert\Psi\rangle\langle\Psi\rvert.$$

Using the Schmidt form and tracing over $B$ gives

$$\rho_A = \operatorname{Tr}_B\rho_{AB} = \sum_{r=1}^{R} s_r^2\, \lvert u_r\rangle\langle u_r\rvert.$$

Similarly,

$$\rho_B = \sum_{r=1}^{R} s_r^2\, \lvert v_r\rangle\langle v_r\rvert.$$

The joint state is pure because $\rho_{AB}^2=\rho_{AB}$. Its reduced states have purity

$$\operatorname{Tr}\rho_A^2 = \operatorname{Tr}\rho_B^2 = \sum_{r=1}^{R}s_r^4.$$

Consequently,

$$\begin{aligned} \lvert\Psi\rangle\text{ product} &\quad\Longleftrightarrow\quad \operatorname{Tr}\rho_A^2=1,\\ \lvert\Psi\rangle\text{ entangled} &\quad\Longleftrightarrow\quad \operatorname{Tr}\rho_A^2<1. \end{aligned}$$

For $\lvert\Phi^+\rangle$,

$$\rho_A=\rho_B=\frac12 I.$$

No pure state vector of $A$ alone reproduces this local state. The joint state is known completely, yet the complete local description is mixed. It does not mean that the global preparation was an unknown member of an ensemble; the missing local purity is encoded in joint correlations.

The Reduced States gateway develops the subsystem interpretation, and Partial Trace gives the full calculation rule.

Correlations

For local measurement effects $E_a$ on $A$ and $F_b$ on $B$, the joint Born probability is

$$P(a,b) = \langle\Psi\rvert E_a\otimes F_b \lvert\Psi\rangle.$$

If $\lvert\Psi\rangle=\lvert\psi\rangle\otimes\lvert\phi\rangle$, then every pair of local measurements factorizes:

$$\begin{aligned} P(a,b) &= \langle\psi\rvert E_a\lvert\psi\rangle \langle\phi\rvert F_b\lvert\phi\rangle\\ &= P_A(a)P_B(b). \end{aligned}$$

Equivalently, all bounded local-observable expectation values factor:

$$\langle A\otimes B\rangle = \langle A\otimes I\rangle \langle I\otimes B\rangle.$$

An entangled pure state has local measurement choices that reveal nonfactorizing correlations. However, the absence of correlation in one chosen measurement setting does not prove that a state is product. Conversely, correlation in one setting does not by itself prove entanglement, because a separable mixed state can also be correlated.

For a concrete comparison, $|\Phi^+\rangle$ and the separable mixture $\rho_{\mathrm{cc}}=(|00\rangle\langle00|+|11\rangle\langle11|)/2$ both have perfectly correlated $Z$ outcomes. Their $X\otimes X$ expectations are respectively $1$ and $0$. The calculation in complementary bases explains how these coherences distinguish the two states and how to turn correlations into an entanglement test.

Local measurements and signalling

If a trace-preserving local operation acts on $B$ and its outcome is ignored, $A$‘s reduced state is unchanged. Thus $A$ cannot read a message from its local statistics by learning whether $B$ measured something. The proof for projective measurements is exercise 7; the general channel argument is in local measurement statistics.

Conditioning on a particular result $b$ with $p_b>0$ can change the assigned state $\rho_{A|b}$. These selected states average to the original one:

$$\sum_{b:p_b>0}p_b\rho_{A|b}=\rho_A.$$

The outcome is random, and $A$ cannot sort data by $b$ until the result is communicated. Conditional-state change therefore does not imply controllable faster-than-light signalling.

Creating and preserving entanglement

An interaction describes dynamics; entanglement describes a state. A joint gate can generate entanglement: controlled-$Z$ changes $|++\rangle$ into a state with coefficient determinant $-1/2$. Exercise 6 checks its Schmidt coefficients. The coupling calculation explains which Hamiltonian generates this gate.

The same gate leaves $|00\rangle$ unchanged, so an entangling operation need not entangle every input. Conversely, once created, an entangled state can persist while the subsystems cease interacting. Local unitaries preserve its Schmidt coefficients; local noise or measurements can reduce its entanglement.

Why Entanglement Matters

Entanglement changes how a composite system can store and distribute quantum information:

• A pure joint state may have no pure state vector for either subsystem.
• Correlations can carry information that is absent from each marginal state.
• Local operations and classical communication cannot create entanglement from an initially separable state.
• Shared entanglement enables protocols such as quantum teleportation and superdense coding, together with the required classical or quantum communication.
• Suitable entangled states and measurement choices can violate Bell inequalities, sharpening the conflict between quantum predictions and local hidden-variable models.
• In many-body physics, entanglement structure helps characterize correlations, phases, and the difficulty of classical simulation.

These statements should not be collapsed into the slogan that entanglement is automatically useful. Different tasks require different states, measurements, communication resources, and noise tolerances. Entanglement is necessary for Bell nonlocality, but not every entangled mixed state violates a given Bell inequality. The Bell locality sequence states the additional assumptions and measurements needed to turn entanglement into a Bell-inequality test.

Mixed-state entanglement

In finite dimensions, a mixed state is separable across $A|B$ when it has at least one convex-product decomposition

$$\begin{aligned} \rho_{AB} &= \sum_k p_k\, \rho_A^{(k)}\otimes\rho_B^{(k)},\\ p_k&\ge0, & \sum_k p_k&=1. \end{aligned}$$

It is entangled when no such decomposition exists. The correlated state $\rho_{\mathrm{cc}}$ compared with the Bell state above is nonproduct but separable, so “not a product” is not a mixed-state criterion. Because ensemble decompositions are not unique, establishing mixed-state entanglement generally requires an invariant criterion or an entanglement witness rather than inspection of one convenient preparation.

For two qubits and qubit–qutrit systems, positivity of the partial transpose is a complete separability criterion. In higher dimensions it is only a one-sided test: some entangled states have positive partial transpose. See negativity and the PPT criterion for the calculation. In infinite dimensions, separability uses the trace-norm closure of finite convex product mixtures; separable mixed states states this extension.

A classical pair of coins prepared as “both heads” or “both tails” with equal probability is correlated but not entangled. The density-operator analogue is precisely a separable convex mixture such as $\rho_{\mathrm{cc}}$.

Entanglement depends on the subsystem decomposition

Entanglement is relative to a tensor-product structure. The same abstract Hilbert space can sometimes be factored in more than one physically meaningful way.

For a single spin-$1/2$ particle with position and spin,

$$\mathcal H = L^2(\mathbb R^3)\otimes\mathbb C^2.$$

A spinor wavefunction

$$\Psi(\mathbf x) = \begin{pmatrix} \psi_\uparrow(\mathbf x)\\ \psi_\downarrow(\mathbf x) \end{pmatrix}$$

is product across position and spin only if it can be written as

$$\Psi(\mathbf x) = \psi(\mathbf x)\chi$$

for one spatial wavefunction $\psi$ and one fixed spinor $\chi$. If the spin direction depends on position, the state is entangled across the position-spin split, even though there is only one particle.

With three or more subsystems, the partition remains part of the claim. A state can be product across one bipartition and entangled across another; full separability is stronger than separability across any single split, and pairwise reduced states do not exhaust genuinely multipartite structure.

For identical particles, the caution is stronger. The formal labels in $\Psi(x_1,x_2)$ are coordinate arguments, not observable particle names. Entanglement questions must specify a physical split, such as modes, spatial regions, spin sectors, species, or an algebra of observables. In continuous-variable systems, ideal Einstein–Podolsky–Rosen states are non-normalizable limits; physical implementations use finite squeezing and require convergence control beyond the finite-dimensional treatment here.

Common pitfalls

The pure-state criterion needs a pure joint state. Mixed marginals characterize entanglement of a pure bipartite state. They do not distinguish classical mixing from entanglement in a mixed joint state.

Off-diagonal matrix elements are basis dependent. Even a product state can have many nonzero off-diagonal entries. Entanglement tests concern a fixed subsystem split and are invariant under local basis changes.

Bell violation requires a measurement scenario. Entanglement is necessary for a Bell violation, but it is not equivalent to violating an arbitrary Bell inequality with fixed settings.

References

• John S. Bell, “On the Einstein Podolsky Rosen paradox,” Physics Physique Fizika 1, 195–200 (1964). DOI.
• Ryszard Horodecki, Paweł Horodecki, Michał Horodecki, and Karol Horodecki, “Quantum entanglement,” Reviews of Modern Physics 81, 865–942 (2009). DOI.
• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, section 2.5.
• John Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chapter 6. Author-hosted text.

Exercises

1. Determine whether

$$\lvert\Psi\rangle = \frac12 \bigl( \lvert00\rangle +i\lvert01\rangle -\lvert10\rangle -i\lvert11\rangle \bigr)$$

is product or entangled. If it is product, factor it.

Solution

The coefficient matrix is

$$C = \frac12 \begin{pmatrix} 1&i\\ -1&-i \end{pmatrix},$$

and

$$\det C = \frac14 \bigl( -i+i \bigr) =0.$$

The state is product. A factorization is

$$\lvert\Psi\rangle = \frac{ \lvert0\rangle-\lvert1\rangle }{\sqrt2} \otimes \frac{ \lvert0\rangle+i\lvert1\rangle }{\sqrt2}.$$

2. Consider

$$\lvert\Psi_\lambda\rangle = \frac{ \lvert00\rangle+\lambda\lvert11\rangle }{ \sqrt{1+\lvert\lambda\rvert^2} },$$

where $\lambda$ is finite. For which values of $\lambda$ is the state entangled?

Solution

The coefficient matrix is diagonal:

$$C_\lambda = \frac{1}{ \sqrt{1+\lvert\lambda\rvert^2} } \begin{pmatrix} 1&0\\ 0&\lambda \end{pmatrix}.$$

Its determinant is

$$\det C_\lambda = \frac{\lambda}{ 1+\lvert\lambda\rvert^2 }.$$

Therefore the state is product at $\lambda=0$ and entangled for every $\lambda\ne0$.

3. Prove that local unitaries preserve the coefficient-matrix rank of a bipartite pure state.

Solution

If $C$ is the original coefficient matrix, applying $U_A\otimes U_B$ gives

$$C'=U_A C U_B^{\mathsf T}.$$

Unitary matrices are invertible, as is $U_B^{\mathsf T}$. Multiplication by an invertible matrix on either side does not change rank:

$$\operatorname{rank}C' = \operatorname{rank}C.$$

Thus local unitaries preserve the product-versus-entangled classification.

4. Find the reduced states and their purities for

$$\lvert\Psi(\theta,\varphi)\rangle = \cos\theta\,\lvert00\rangle + e^{i\varphi}\sin\theta\,\lvert11\rangle.$$

Take $0\le\theta\le\pi/2$ and real $\varphi$. Use the result to recover the entanglement condition.

Solution

Tracing out either qubit removes the cross terms:

$$\rho_A=\rho_B = \cos^2\theta\,\lvert0\rangle\langle0\rvert + \sin^2\theta\,\lvert1\rangle\langle1\rvert.$$

Their purity is

$$\begin{aligned} \operatorname{Tr}\rho_A^2 &= \cos^4\theta+\sin^4\theta\\ &= 1-\frac12\sin^2(2\theta). \end{aligned}$$

The purity equals one only when $\theta=0$ or $\theta=\pi/2$. For $0<\theta<\pi/2$, the reduced states are mixed and the joint pure state is entangled. The phase $\varphi$ does not affect the Schmidt coefficients.

5. Show that $\lvert\Phi^+\rangle$ gives equal outcomes when both qubits are measured in the $x$ basis. Then find the four $x$-basis probabilities for $\rho_{\mathrm{cc}}$.

Solution

Using

$$\lvert0\rangle = \frac{\lvert+\rangle+\lvert-\rangle}{\sqrt2}, \qquad \lvert1\rangle = \frac{\lvert+\rangle-\lvert-\rangle}{\sqrt2},$$

one obtains

$$\lvert\Phi^+\rangle = \frac{ \lvert++\rangle+\lvert--\rangle }{\sqrt2}.$$

Thus

$$\begin{aligned} P(+,+)=P(-,-)&=\frac12,\\ P(+,-)=P(-,+)&=0. \end{aligned}$$

For either product component $\lvert00\rangle$ or $\lvert11\rangle$ of $\rho_{\mathrm{cc}}$, all four joint $x$-basis outcomes have probability $1/4$. Their mixture therefore also gives

$$\begin{aligned} P(+,+)=P(+,-)&=\frac14,\\ P(-,+)=P(-,-)&=\frac14. \end{aligned}$$

6. Verify that controlled-$Z$ entangles $\lvert+\rangle\lvert+\rangle$, and find the Schmidt coefficients of the output.

Solution

The output coefficient matrix is

$$C = \frac12 \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}.$$

Since $\det C=-1/2$, the state is entangled. Moreover,

$$CC^\dagger = \frac12 I.$$

The eigenvalues of $CC^\dagger$ are both $1/2$, so the singular values of $C$, and hence the Schmidt coefficients, are

$$s_1=s_2=\frac{1}{\sqrt2}.$$

The output is maximally entangled.

7. Let $\{P_b\}$ be a complete projective measurement on $B$. Prove directly that discarding its outcome leaves $\rho_A$ unchanged.

Solution

After the nonselective measurement,

$$\rho'_{AB} = \sum_b \bigl(I_A\otimes P_b\bigr) \rho_{AB} \bigl(I_A\otimes P_b\bigr).$$

The partial trace is cyclic with respect to operators acting only on the traced subsystem. Since $P_b^2=P_b$,

$$\begin{aligned} \rho'_A &= \sum_b \operatorname{Tr}_B \left[ \rho_{AB} \bigl(I_A\otimes P_b^2\bigr) \right]\\ &= \operatorname{Tr}_B \left[ \rho_{AB} \left( I_A\otimes\sum_b P_b \right) \right]\\ &= \operatorname{Tr}_B\rho_{AB} = \rho_A. \end{aligned}$$

Therefore $A$ cannot determine from local statistics whether the nonselective measurement was performed.

8. For the three-qubit GHZ state

$$\lvert\mathrm{GHZ}\rangle = \frac{ \lvert000\rangle+\lvert111\rangle }{\sqrt2},$$

classify the state across the split $A|BC$, and find $\rho_{AB}$ after tracing out $C$.

Solution

Across $A|BC$, the displayed expression is already a Schmidt decomposition:

$$\lvert\mathrm{GHZ}\rangle = \frac{1}{\sqrt2} \lvert0\rangle_A\lvert00\rangle_{BC} + \frac{1}{\sqrt2} \lvert1\rangle_A\lvert11\rangle_{BC}.$$

Its Schmidt rank is two, so it is entangled across $A|BC$. Tracing out $C$ removes the off-diagonal terms because $\langle0\vert1\rangle_C=0$:

$$\rho_{AB} = \frac12\lvert00\rangle\langle00\rvert + \frac12\lvert11\rangle\langle11\rvert.$$

This two-qubit reduced state is correlated but separable. The example shows why multipartite entanglement cannot be inferred solely from pairwise reduced states.