6.2.1 Product and Entangled States#

Prompts

What is a product state and what is an entangled state in a bipartite quantum system? How can one determine whether a given two-qubit state is entangled?
Why does entanglement require both many-body structure and quantum superposition? Give an example where each ingredient alone is insufficient.
What is the Schmidt decomposition of a bipartite pure state? How is it related to the singular value decomposition?
How does the Schmidt rank classify the entanglement of a bipartite pure state? What additional information do the Schmidt coefficients carry beyond the rank?
Why does an entangled pure state necessarily have a mixed reduced density matrix? What does this imply about local measurements on one subsystem?

Lecture Notes#

Overview#

When two quantum systems combine, their joint Hilbert space is the tensor product \(\mathcal{H}_A \otimes \mathcal{H}_B\) (see §2.1.1 Tensor Product for the construction). Most states in this composite space cannot be decomposed into independent states of the parts — they are entangled. Entanglement is a uniquely quantum phenomenon that arises from quantum superposition applied to many-body systems: it requires both many-body structure and superposition, but neither alone is sufficient. This section defines product and entangled states precisely, introduces the Schmidt decomposition as the canonical tool for revealing entanglement structure, and motivates quantitative entanglement measures.

Product States and Entangled States#

Definition: Product State

A state \(\vert\Psi\rangle_{AB} \in \mathcal{H}_A \otimes \mathcal{H}_B\) is a product state (separable pure state) if it can be written as:

(222)#\[ \vert\Psi\rangle_{AB} = \vert\psi\rangle_A \otimes \vert\phi\rangle_B \]

for some \(\vert\psi\rangle_A \in \mathcal{H}_A\) and \(\vert\phi\rangle_B \in \mathcal{H}_B\).

In a product state, each subsystem has a definite quantum state of its own. Measuring subsystem \(A\) reveals nothing about subsystem \(B\): the two are statistically independent.

Definition: Entangled State

A state \(\vert\Psi\rangle_{AB}\) that cannot be written as a single product \(\vert\psi\rangle_A \otimes \vert\phi\rangle_B\) is called entangled.

Entanglement arises from quantum superposition of many-body (tensor product) basis states. Consider two qubits with basis \(\{\vert 0\rangle, \vert 1\rangle\}\) on each side. The general state is:

\[ \vert\Psi\rangle = c_{00}\vert 00\rangle + c_{01}\vert 01\rangle + c_{10}\vert 10\rangle + c_{11}\vert 11\rangle \]

Arrange the coefficients into a matrix:

\[\begin{split} C = \begin{pmatrix} c_{00} & c_{01} \\ c_{10} & c_{11} \end{pmatrix} \end{split}\]

The state is a product state if and only if \(\mathrm{rank}(C) = 1\) (equivalently \(\det(C) = c_{00}c_{11} - c_{01}c_{10} = 0\)). When \(\mathrm{rank}(C) \geq 2\), no single product can reproduce the superposition — the state is entangled.

Attention: Not Every Superposition is Entangled

Consider \(\vert\Psi\rangle = \frac{1}{2}(\vert 00\rangle + \vert 01\rangle + \vert 10\rangle + \vert 11\rangle)\). This is a superposition of four product basis states, but:

\[\begin{split} C = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \quad \det(C) = 0 \end{split}\]

so \(\vert\Psi\rangle = \vert +\rangle_A \otimes \vert +\rangle_B\) is a product state! Superposition alone does not create entanglement — it must be a superposition that cannot be factored.

Example: The Bell State

Problem. Show that \(\vert\Phi^+\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\) is entangled.

Solution. The coefficient matrix is:

\[\begin{split} C = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{split}\]

Since \(\det(C) = \frac{1}{2} \neq 0\) (equivalently \(\mathrm{rank}(C) = 2\)), the state cannot be written as a product. It is entangled.

Discussion: what does it take to entangle?

Entanglement requires both many-body structure and quantum superposition. A single-particle superposition \(\alpha\vert 0\rangle + \beta\vert 1\rangle\) is not entangled (it lives in a single Hilbert space). A product of two particles \(\vert 0\rangle_A \otimes \vert 0\rangle_B\) involves many-body structure but no entangling superposition. Only when superposition creates correlations that cannot be factored do we get entanglement. Can you think of a classical analogy that captures the “correlation” aspect but fails to capture the full quantum strangeness?

Why We Need Entanglement Measures#

The rank of the coefficient matrix tells us whether a state is entangled, but it does not capture how entangled it is — or even match our naive intuition about superposition. Consider the state \(\frac{1}{2}(\vert 00\rangle + \vert 01\rangle + \vert 10\rangle + \vert 11\rangle)\): it is a superposition of all many-body basis states, yet it is a product state \(\vert +\rangle_A \otimes \vert +\rangle_B\) with no entanglement at all. More superposition does not mean more entanglement. Meanwhile, the Bell state \(\frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\) involves only two terms but is maximally entangled.

What matters is not how many terms appear, but whether the superposition can be factored. We need a quantitative measure \(E(\vert\Psi\rangle)\) that captures this — one that vanishes for product states, is maximal for maximally entangled states, and cannot increase under local operations and classical communication (LOCC monotone). The Schmidt decomposition provides the natural framework for constructing such measures.

Schmidt Decomposition#

Theorem: Schmidt Decomposition

Any bipartite pure state \(\vert\Psi\rangle_{AB} \in \mathcal{H}_A \otimes \mathcal{H}_B\) can be written as:

(223)#\[ \vert\Psi\rangle_{AB} = \sum_{k=1}^{r} \lambda_k \vert u_k\rangle_A \otimes \vert v_k\rangle_B \]

where:

\(r \leq \min(d_A, d_B)\) is the Schmidt rank
\(\lambda_k > 0\) are the Schmidt coefficients, normalized: \(\sum_k \lambda_k^2 = 1\)
\(\{\vert u_k\rangle_A\}\) and \(\{\vert v_k\rangle_B\}\) are orthonormal sets in their respective spaces

The Schmidt decomposition is a diagonal representation: unlike the general expansion \(\vert\Psi\rangle = \sum_{ij} c_{ij} \vert i\rangle_A \vert j\rangle_B\) which uses \(d_A \cdot d_B\) coefficients, the Schmidt form uses only \(r\) terms, each pairing one basis state on \(A\) with one on \(B\).

Derivation: Schmidt Decomposition via SVD

Write \(\vert\Psi\rangle = \sum_{i,j} c_{ij} \vert i\rangle_A \vert j\rangle_B\). Form the coefficient matrix \(C\) with elements \(c_{ij}\) (dimensions \(d_A \times d_B\)).

By singular value decomposition (SVD):

\[ C = U \Lambda V^\dagger \]

where \(U\) is \(d_A \times d_A\) unitary, \(V\) is \(d_B \times d_B\) unitary, and \(\Lambda\) is \(d_A \times d_B\) with non-negative diagonal entries \(\lambda_1 \geq \lambda_2 \geq \cdots \geq 0\).

Substituting:

\[ \vert\Psi\rangle = \sum_{i,j,k} U_{ik} \lambda_k (V^\dagger)_{kj} \vert i\rangle_A \vert j\rangle_B = \sum_k \lambda_k \left(\sum_i U_{ik} \vert i\rangle_A\right) \left(\sum_j V_{jk} \vert j\rangle_B\right) \]

Defining \(\vert u_k\rangle_A = \sum_i U_{ik}\vert i\rangle_A\) and \(\vert v_k\rangle_B = \sum_j V_{jk}\vert j\rangle_B\) gives the Schmidt form. Unitarity of \(U\) and \(V\) guarantees orthonormality. The number of nonzero singular values is \(r = \mathrm{rank}(C)\).

Schmidt Rank as Entanglement Classifier#

Schmidt Rank and Entanglement

The Schmidt rank \(r\) classifies entanglement:

\(r = 1\): Product state — \(\vert\Psi\rangle = \vert\psi\rangle_A \otimes \vert\phi\rangle_B\), no entanglement.
\(r \geq 2\): Entangled — the state cannot be written as a single product.
\(r = \min(d_A, d_B)\) with all \(\lambda_k\) equal (\(\lambda_k = 1/\sqrt{r}\)): Maximally entangled.

Beyond the rank, the distribution of Schmidt coefficients quantifies the degree of entanglement. Two states with the same Schmidt rank can have very different entanglement: \(\lambda = (1/\sqrt{2}, 1/\sqrt{2})\) (Bell state) versus \(\lambda = (\sqrt{0.99}, \sqrt{0.01})\) (weakly entangled). This motivates the entanglement measures developed in §6.2.2 Entanglement Entropy.

Example: Schmidt Decomposition of a Two-Qubit State

Problem. Find the Schmidt decomposition of \(\vert\psi\rangle = \frac{1}{\sqrt{3}}(\vert 00\rangle + \vert 01\rangle + \vert 11\rangle)\).

Solution. The coefficient matrix is:

\[\begin{split} C = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \end{split}\]

Since \(\det(C) = \frac{1}{3} \neq 0\), the state is entangled with Schmidt rank \(r = 2\). By SVD, the singular values are:

\[ \lambda_1 = \frac{1}{\sqrt{3}}\frac{1+\sqrt{5}}{2}, \quad \lambda_2 = \frac{1}{\sqrt{3}}\frac{\sqrt{5}-1}{2} \]

(after normalization). The Schmidt decomposition is \(\vert\psi\rangle = \lambda_1 \vert u_1\rangle_A\vert v_1\rangle_B + \lambda_2 \vert u_2\rangle_A\vert v_2\rangle_B\) where \(\vert u_k\rangle\), \(\vert v_k\rangle\) are the left and right singular vectors.

Purity as a Simple Entanglement Indicator#

The Schmidt coefficients connect directly to the purity of a subsystem. For a bipartite pure state with Schmidt decomposition \(\vert\Psi\rangle = \sum_k \lambda_k \vert u_k\rangle_A \vert v_k\rangle_B\), the reduced density matrix of subsystem \(A\) is (as will be shown in §6.2.2 via the partial trace):

(224)#\[ \hat{\rho}_A = \sum_k \lambda_k^2 \vert u_k\rangle_A\langle u_k\vert \]

The purity of this reduced state is:

(225)#\[ \mathrm{Tr}(\hat{\rho}_A^2) = \sum_k \lambda_k^4 \]

This equals 1 if and only if \(r = 1\) (product state), and is minimized when all \(\lambda_k\) are equal (maximally entangled). The purity thus serves as a simple entanglement indicator: a pure composite state whose subsystem is mixed must be entangled.

A Pure State with a Mixed Subsystem Signals Entanglement

For a bipartite pure state \(\vert\Psi\rangle_{AB}\):

\[ \vert\Psi\rangle \text{ is a product state} \iff \hat{\rho}_A \text{ is pure} \iff \mathrm{Tr}(\hat{\rho}_A^2) = 1 \]

If \(\hat{\rho}_A\) is mixed (\(\mathrm{Tr}(\hat{\rho}_A^2) < 1\)), then \(\vert\Psi\rangle\) is necessarily entangled.

Multipartite Systems#

For \(N\) systems, the Hilbert space is \(\mathcal{H}_{1\cdots N} = \mathcal{H}_1 \otimes \cdots \otimes \mathcal{H}_N\) with dimension \(\prod_i d_i\). Multipartite entanglement is richer than bipartite: a state can be entangled across one bipartition but not another, or it can be genuinely multipartite entangled (not expressible as a product across any bipartition). The GHZ state \(\frac{1}{\sqrt{2}}(\vert 000\rangle + \vert 111\rangle)\) and the W state \(\frac{1}{\sqrt{3}}(\vert 001\rangle + \vert 010\rangle + \vert 100\rangle)\) exemplify two inequivalent classes of tripartite entanglement — they cannot be converted into each other by local operations.

Summary#

Product vs entangled: A bipartite pure state is a product state iff its coefficient matrix has rank 1. Entanglement arises from superpositions that cannot be factored into independent subsystem states.
Entanglement requires both ingredients: Many-body (tensor product) structure and quantum superposition — neither alone suffices.
Schmidt decomposition: Any bipartite pure state has the diagonal form \(\vert\Psi\rangle = \sum_k \lambda_k \vert u_k\rangle_A \vert v_k\rangle_B\) via SVD of the coefficient matrix.
Schmidt rank: \(r = 1\) means product; \(r \geq 2\) means entangled; equal coefficients at \(r = \min(d_A, d_B)\) means maximally entangled.
Purity criterion: A pure bipartite state is entangled iff its reduced state is mixed (\(\mathrm{Tr}(\hat{\rho}_A^2) < 1\)).

Homework#

1. Product state expansion. Show that the state \(\vert\psi\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle_A + \vert 1\rangle_A) \otimes \vert 0\rangle_B\) is a product state. Write it in the standard two-qubit basis \(\{\vert 00\rangle, \vert 01\rangle, \vert 10\rangle, \vert 11\rangle\}\) and verify directly that its coefficient matrix has rank 1.

2. Entanglement test. For each of the following two-qubit states, determine whether it is a product state or entangled by computing the determinant of the coefficient matrix \(C\).

(a) \(\vert\psi_1\rangle = \frac{1}{2}(\vert 00\rangle + \vert 01\rangle + \vert 10\rangle + \vert 11\rangle)\)

(b) \(\vert\psi_2\rangle = \frac{1}{\sqrt{2}}(\vert 01\rangle + \vert 10\rangle)\)

3. Schmidt rank criterion. Show that a bipartite pure state \(\vert\psi\rangle_{AB}\) with Schmidt decomposition \(\vert\psi\rangle = \sum_k \lambda_k \vert u_k\rangle_A \vert v_k\rangle_B\) is a product state if and only if its Schmidt rank is 1 (only one nonzero \(\lambda_k\)).

4. Schmidt decomposition. Find the Schmidt decomposition of \(\vert\psi\rangle = \frac{1}{\sqrt{3}}(\vert 00\rangle + \vert 01\rangle + \vert 11\rangle)\) by computing the SVD of its coefficient matrix. What is the Schmidt rank?

5. Maximally entangled states. A bipartite state on \(\mathbb{C}^d \otimes \mathbb{C}^d\) is called maximally entangled if all Schmidt coefficients are equal: \(\lambda_k = 1/\sqrt{d}\) for \(k = 1, \ldots, d\).

(a) Show that the Bell state \(\vert\Phi^+\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\) is maximally entangled for \(d = 2\).

(b) Construct a maximally entangled state for two qutrits (\(d = 3\)).

6. Purity from Schmidt coefficients. For a bipartite pure state \(\vert\Psi\rangle_{AB}\) with Schmidt coefficients \(\{\lambda_k\}\), the reduced density matrix is \(\hat{\rho}_A = \sum_k \lambda_k^2 \vert u_k\rangle\langle u_k\vert\).

(a) Show that \(\mathrm{Tr}(\hat{\rho}_A^2) = \sum_k \lambda_k^4\).

(b) Using \(\sum_k \lambda_k^2 = 1\), prove that \(\mathrm{Tr}(\hat{\rho}_A^2) \leq 1\) with equality iff the state is a product.

7. Separable decomposition. A density matrix \(\hat{\rho}_{AB}\) is called separable if it can be written as \(\hat{\rho}_{AB} = \sum_i p_i \hat{\rho}_A^{(i)} \otimes \hat{\rho}_B^{(i)}\) with \(p_i \geq 0\), \(\sum_i p_i = 1\). Show that the maximally mixed state \(\hat{\rho} = \hat{I}/4\) on two qubits is separable by explicitly constructing such a decomposition.

8. GHZ and W states. Consider the 3-qubit GHZ state \(\vert\mathrm{GHZ}\rangle = \frac{1}{\sqrt{2}}(\vert 000\rangle + \vert 111\rangle)\) and the W state \(\vert W\rangle = \frac{1}{\sqrt{3}}(\vert 001\rangle + \vert 010\rangle + \vert 100\rangle)\).

(a) For each state, compute the Schmidt decomposition across the bipartition \(A \vert BC\) (where \(A\) is the first qubit and \(BC\) is the remaining two).

(b) Compare the Schmidt ranks. Are both states entangled across this bipartition?

(c) Trace out qubit \(A\) from each state. Is the resulting two-qubit state entangled or separable? (Hint: check the rank of the reduced state.)