6.2.1 Composite Systems#

Prompts

What is the tensor product of two Hilbert spaces? Why is it natural for composite quantum systems?
How does the partial trace extract information about a subsystem? What does it mean physically to “trace out” degrees of freedom?
What is the key difference between a product state and an entangled state? How does Schmidt rank reveal this?
For a bipartite pure state, how does entanglement entropy quantify entanglement? When is it zero, and when is it maximum?
Why can a pure bipartite state have a mixed reduced density matrix? What does this tell us about the structure of entanglement?

Lecture Notes#

Overview#

When two quantum systems combine, their joint state lives in a tensor product Hilbert space—exponentially larger than either subsystem alone. Yet quantum entanglement reveals that the system cannot always be decomposed into independent pieces: a pure composite state can have a mixed reduced state. This section develops the mathematical framework to understand and quantify entanglement.

Tensor Product Hilbert Space#

Definition: Tensor Product Hilbert Space

For two quantum systems \(A\) and \(B\) with Hilbert spaces \(\mathcal{H}_A\) (dimension \(d_A\)) and \(\mathcal{H}_B\) (dimension \(d_B\)), the composite system \(AB\) lives in the tensor product space:

\[\mathcal{H}_{AB} = \mathcal{H}_A \otimes \mathcal{H}_B\]

with dimension \(\dim(\mathcal{H}_{AB}) = d_A \cdot d_B\).

Basis and Product States#

If \(\{\vert i\rangle_A : i=1,\ldots,d_A\}\) and \(\{\vert j\rangle_B : j=1,\ldots,d_B\}\) are bases for \(\mathcal{H}_A\) and \(\mathcal{H}_B\), then \(\{\vert i\rangle_A \otimes \vert j\rangle_B\}_{i,j}\) is a basis for \(\mathcal{H}_{AB}\). We abbreviate \(\vert i\rangle_A \otimes \vert j\rangle_B\) as \(\vert ij\rangle\) or \(\vert i\rangle\vert j\rangle\).

Any state can be expanded:

\[\vert\Psi\rangle_{AB} = \sum_{i,j} c_{ij} \vert i\rangle_A \vert j\rangle_B\]

A state is separable (product state) if:

\[\vert\Psi\rangle_{AB} = \vert\psi\rangle_A \otimes \vert\phi\rangle_B\]

For two qubits, this means the coefficient matrix \(C = [c_{ij}]\) has rank 1, equivalently \(\det(C) = 0\).

Tensor Products of Operators#

For operators \(\hat{A} \in \text{End}(\mathcal{H}_A)\) and \(\hat{B} \in \text{End}(\mathcal{H}_B)\):

\((\hat{A} \otimes \hat{B})(\vert\psi\rangle_A \otimes \vert\phi\rangle_B) = (\hat{A}\vert\psi\rangle_A) \otimes (\hat{B}\vert\phi\rangle_B)\)
\(\text{Tr}(\hat{A} \otimes \hat{B}) = \text{Tr}(\hat{A}) \cdot \text{Tr}(\hat{B})\)

The expectation value of an observable \(\hat{O}_A \otimes \hat{I}_B\) in state \(\vert\Psi\rangle_{AB}\) depends only on the reduced state of subsystem A.

Partial Trace and Reduced Density Matrix#

Definition: Partial Trace

The partial trace over subsystem \(B\) is the linear map \(\text{Tr}_B : \text{End}(\mathcal{H}_{AB}) \to \text{End}(\mathcal{H}_A)\) defined on basis states by:

\[\text{Tr}_B(\vert i\rangle\langle i'\vert \otimes \vert j\rangle\langle j'\vert) = \delta_{jj'} \vert i\rangle\langle i'\vert\]

For a general state \(\hat{\rho}_{AB}\), the reduced density matrix is:

\[\hat{\rho}_A = \text{Tr}_B(\hat{\rho}_{AB})\]

Physical meaning: \(\hat{\rho}_A\) contains all information about subsystem \(A\) alone. It extracts what remains when you “trace out” (ignore) the degrees of freedom in \(B\). Any expectation value of observables acting only on \(A\) can be computed from \(\hat{\rho}_A\):

\[\langle \hat{O}_A \rangle = \text{Tr}_A(\hat{\rho}_A \hat{O}_A) = \text{Tr}_{AB}(\hat{\rho}_{AB} (\hat{O}_A \otimes \hat{I}_B))\]

Computing the Partial Trace#

To compute \(\hat{\rho}_A = \text{Tr}_B(\hat{\rho}_{AB})\), insert a complete basis for \(B\):

\[\hat{\rho}_A = \sum_j (I_A \otimes \langle j\vert_B) \hat{\rho}_{AB} (I_A \otimes \vert j\rangle_B)\]

Example: Two-Qubit Bell State

Problem. Find the reduced density matrix \(\hat{\rho}_A\) for the Bell state \(\vert\Phi^+\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\).

Solution. The pure state density matrix is:

\[\begin{split}\hat{\rho}_{AB} = \vert\Phi^+\rangle\langle\Phi^+\vert = \frac{1}{2}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix}\end{split}\]

Apply the partial trace over \(B\) using basis \(\{\vert 0\rangle_B, \vert 1\rangle_B\}\):

\[\hat{\rho}_A = \sum_{j=0}^{1} (\langle j\vert_B \otimes I_A) \hat{\rho}_{AB} (\vert j\rangle_B \otimes I_A)\]

For \(j=0\): \((\langle 0\vert_B \otimes I) \hat{\rho}_{AB} (\vert 0\rangle_B \otimes I) = \frac{1}{2} \vert 0\rangle\langle 0\vert\)

For \(j=1\): \((\langle 1\vert_B \otimes I) \hat{\rho}_{AB} (\vert 1\rangle_B \otimes I) = \frac{1}{2} \vert 1\rangle\langle 1\vert\)

Thus:

\[\hat{\rho}_A = \frac{1}{2}(\vert 0\rangle\langle 0\vert + \vert 1\rangle\langle 1\vert) = \frac{I}{2}\]

The reduced state is maximally mixed, even though the total state \(\vert\Phi^+\rangle\) is pure!

Key Property: Eigenvalues from Schmidt Coefficients#

For a bipartite pure state \(\vert\Psi\rangle_{AB}\), the eigenvalues of the reduced density matrices are the squares of the Schmidt coefficients. This will be shown below via the Schmidt decomposition.

Schmidt Decomposition#

Theorem: Schmidt Decomposition

Any bipartite pure state \(\vert\Psi\rangle_{AB}\) can be written in the form:

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

where:

\(r = \text{rank}(C)\) is the Schmidt rank (number of nonzero terms)
\(\lambda_k > 0\) are the Schmidt coefficients (normalized: \(\sum_k \lambda_k^2 = 1\))
\(\{\vert u_k\rangle_A\}\) and \(\{\vert v_k\rangle_B\}\) are orthonormal bases

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:

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

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

Rewrite the state:

\[\vert\Psi\rangle = \sum_{i,j,k} U_{ik} \lambda_k (V^\dagger)_{kj} \vert i\rangle_A \vert j\rangle_B\]

Define orthonormal states:

\[\vert u_k\rangle_A = \sum_i U_{ik} \vert i\rangle_A, \quad \vert v_k\rangle_B = \sum_j V_{jk} \vert j\rangle_B\]

Then:

\[\vert\Psi\rangle = \sum_k \lambda_k \vert u_k\rangle_A \vert v_k\rangle_B\]

where the sum runs over nonzero \(\lambda_k\) (Schmidt rank \(r\)).

Reduced Mixed States and Schmidt Coefficients#

For a bipartite pure state in Schmidt form:

\[\vert\Psi\rangle = \sum_k \lambda_k \vert u_k\rangle_A \vert v_k\rangle_B\]

the reduced density matrices are:

\[\hat{\rho}_A = \text{Tr}_B(\vert\Psi\rangle\langle\Psi\vert) = \sum_k \lambda_k^2 \vert u_k\rangle_A\langle u_k\vert\]

\[\hat{\rho}_B = \text{Tr}_A(\vert\Psi\rangle\langle\Psi\vert) = \sum_k \lambda_k^2 \vert v_k\rangle_B\langle v_k\vert\]

Key insight: The eigenvalues of \(\hat{\rho}_A\) are \(\lambda_k^2\) (Schmidt coefficients squared), and the eigenvectors are \(\vert u_k\rangle_A\).

Schmidt Rank and Entanglement#

Schmidt Rank Quantifies Entanglement

The Schmidt rank is the minimal number of product states needed to represent a bipartite pure state:

\(r = 1\): Separable (product state); \(\vert\Psi\rangle = \vert\psi\rangle_A \otimes \vert\phi\rangle_B\)
\(r \geq 2\): Entangled (cannot be written as a single product)
\(r = \min(d_A, d_B)\): Maximally entangled; all Schmidt coefficients equal: \(\lambda_k = 1/\sqrt{r}\)

The purity of the reduced state:

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

equals 1 iff \(r=1\) (product), and is minimized when all \(\lambda_k\) are equal (maximally entangled).

Example: Bell State Schmidt Rank

Problem. Find the Schmidt rank of the Bell state \(\vert\Phi^+\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\).

Solution. The coefficient matrix is:

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

By SVD: \(\lambda_1 = \lambda_2 = 1/\sqrt{2}\), so Schmidt rank \(r=2\) (maximally entangled for two qubits).

The Schmidt decomposition is:

\[\vert\Phi^+\rangle = \frac{1}{\sqrt{2}}\vert 0\rangle_A\vert 0\rangle_B + \frac{1}{\sqrt{2}}\vert 1\rangle_A\vert 1\rangle_B\]

The reduced state \(\hat{\rho}_A = \frac{1}{2}(\vert 0\rangle\langle 0\vert + \vert 1\rangle\langle 1\vert) = \frac{I}{2}\) is maximally mixed.

Entanglement Entropy#

Definition: Entanglement Entropy

For a bipartite pure state \(\vert\Psi\rangle_{AB}\) with Schmidt decomposition \(\vert\Psi\rangle = \sum_k \lambda_k \vert u_k\rangle_A \vert v_k\rangle_B\), the entanglement entropy is the von Neumann entropy of the reduced state:

\[S(A) = -\text{Tr}(\hat{\rho}_A \log_2 \hat{\rho}_A) = -\sum_k \lambda_k^2 \log_2(\lambda_k^2)\]

Physical meaning: \(S(A)\) measures how much the pure bipartite state is “entangled across the \(A|B\) cut.” It quantifies the amount of classical information needed to describe the reduced state \(\hat{\rho}_A\).

Properties of Entanglement Entropy#

For a pure bipartite state:

\[0 \leq S(A) \leq \min(\log_2 d_A, \log_2 d_B)\]

\(S(A) = 0\): iff the state is separable (product state); \(\hat{\rho}_A\) is pure.
\(S(A) = \log_2 r\): For equal Schmidt coefficients \(\lambda_k = 1/\sqrt{r}\); maximally entangled.
\(S(A) = S(B)\): Always equal for bipartite pure states (the reduced states have the same spectrum).

Discussion

Entanglement entropy measures how many “Bell pair units” of entanglement exist in a state. For two qubits, maximum entanglement (one of the four Bell states) gives \(S = 1\) bit. How does this relate to the Schmidt rank, and what does it mean for quantum information?

Example: Entanglement Entropy Calculations#

For the Bell state \(\vert\Phi^+\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\):

\[S(A) = -2 \cdot \frac{1}{2} \log_2\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{2} \cdot 1 = 1 \text{ bit}\]

For a product state \(\vert\psi\rangle_A \otimes \vert\phi\rangle_B\):

\[\lambda_1 = 1, \quad \lambda_k = 0 \text{ for } k > 1 \implies S(A) = 0\]

For the 3-qubit GHZ state \(\vert\text{GHZ}\rangle = \frac{1}{\sqrt{2}}(\vert 000\rangle + \vert 111\rangle)\), the entanglement entropy for a single qubit depends on which qubit we trace out. By symmetry, \(S = 1\) bit when we trace out one qubit (leaving a Bell pair).

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: states can have entanglement that cannot be localized to a single bipartition, giving rise to genuinely multipartite phenomena. This is a frontier of current research in quantum information and many-body physics.

Summary#

Tensor product: \(\mathcal{H}_{AB} = \mathcal{H}_A \otimes \mathcal{H}_B\); composite Hilbert space with dimension \(d_A \cdot d_B\).
Partial trace: \(\hat{\rho}_A = \text{Tr}_B(\hat{\rho}_{AB})\); extracts information about subsystem \(A\). Key property: eigenvalues of \(\hat{\rho}_A\) are squared Schmidt coefficients. Physically, it “traces out” degrees of freedom you don’t observe.
Schmidt decomposition: Any bipartite pure state is \(\vert\Psi\rangle = \sum_k \lambda_k \vert u_k\rangle_A \vert v_k\rangle_B\) with orthonormal bases and normalized coefficients \(\sum_k \lambda_k^2 = 1\).
Schmidt rank: \(r = \text{number of nonzero Schmidt coefficients}\). Distinguishes entanglement: \(r=1\) ⟹ product; \(r \geq 2\) ⟹ entangled.
Entanglement entropy: \(S(A) = -\sum_k \lambda_k^2 \log_2(\lambda_k^2)\). Zero for product states; maximum (\(\log_2 r\)) when all Schmidt coefficients are equal.

Homework#

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

2. The Schmidt decomposition of a bipartite pure state \(|\psi\rangle_{AB}\) is \(|\psi\rangle = \sum_k \sqrt{\lambda_k} |u_k\rangle_A |v_k\rangle_B\) where \(\lambda_k \geq 0\) and \(\sum_k \lambda_k = 1\). Show that \(|\psi\rangle\) is a product state if and only if it has Schmidt rank 1 (only one nonzero \(\lambda_k\)).

3. Find the Schmidt decomposition of \(|\psi\rangle = \frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)\). What is the Schmidt rank? Is this state entangled?

4. Find the Schmidt decomposition of the state \(|\psi\rangle = \frac{1}{\sqrt{3}}(|00\rangle + |01\rangle + |11\rangle)\). (Hint: write the coefficient matrix and find its singular value decomposition.)

5. For the Bell state \(|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\), compute the reduced density matrix \(ho_A = ext{Tr}_B(|\Phi^+\rangle\langle\Phi^+|)\) explicitly. What is its purity? Is it pure or mixed?

6. Compute the entanglement entropy \(S(A)\) for the Bell state \(|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\). Then compute it for the product state \(|\psi\rangle = |0\rangle_A \otimes |0\rangle_B\).

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

8. For a pure state \(|\psi\rangle_{AB}\) with Schmidt decomposition \(|\psi\rangle = \sum_k \sqrt{\lambda_k}|u_k\rangle_A|v_k\rangle_B\), show that \( ext{Tr}(ho_A^2) = \sum_k \lambda_k^4 \leq 1\), with equality only when the state is a product state.

9. Show that for any bipartite pure state, \(S(A) = S(B)\), where \(S(X)\) denotes the entanglement entropy of subsystem \(X\). What does this symmetry tell us about entanglement?

10. Consider the 3-qubit GHZ state \(| ext{GHZ}\rangle = \frac{1}{\sqrt{2}}(|000\rangle + |111\rangle)\). Compute the entanglement entropy when you trace out one qubit. Is the remaining two-qubit state a Bell pair?