6.2.2 Entanglement Entropy#

Prompts

What is the partial trace, and how does it produce the reduced density matrix \(\hat{\rho}_A\)? Why are its eigenvalues the squared Schmidt coefficients?
How does the von Neumann entropy of \(\hat{\rho}_A\) become an entanglement measure for pure bipartite states? When does it vanish, and when is it maximal?
Why is \(S(A) = S(B)\) a natural consistency requirement, and what does this symmetry tell us about pure-state entanglement?
How do Rényi entanglement entropies \(S_\alpha\) relate to the von Neumann entropy, and why is \(S_2\) experimentally easier to access?

Lecture Notes#

Overview#

In §6.2.1 we defined entanglement via the Schmidt rank: a bipartite pure state is entangled when its coefficient matrix has rank \(\geq 2\). But rank alone cannot distinguish a barely-entangled state from a maximally-entangled one. To quantify how much entanglement a state carries, we need the reduced density matrix (obtained by the partial trace) and the von Neumann entropy (introduced in §6.1.2). Combining these gives the entanglement entropy — the central measure for pure bipartite states. The section also introduces Rényi entanglement entropies (generalizations that probe the full Schmidt spectrum) and the four Bell states (the maximally entangled two-qubit basis used throughout Chapter 6).

Partial Trace and Reduced Density Matrix#

Definition: Partial Trace

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

\[ \mathrm{Tr}_B(\hat{A} \otimes \hat{B}) = \hat{A} \cdot \mathrm{Tr}(\hat{B}) \]

Tracing over \(A\) is defined analogously, \(\mathrm{Tr}_A(\hat{A} \otimes \hat{B}) = \mathrm{Tr}(\hat{A})\,\hat{B}\), and extends by linearity to all operators on \(\mathcal{H}_{AB}\).

Example: Partial trace in the two-qubit \(4\times 4\) picture

Work in the computational product basis \(|ij\rangle\) with \(i,j\in\{0,1\}\) for qubits \(A\) and \(B\), ordered lexicographically \(|00\rangle,|01\rangle,|10\rangle,|11\rangle\). Index rows and columns by \(m,n\in\{0,1,2,3\}\) with \(m=2i+j\) labeling \(|ij\rangle\) (and the same for \(n\)). Then

\[\begin{split} \hat{\rho}_{AB} = \begin{pmatrix} \rho_{00} & \rho_{01} & \rho_{02} & \rho_{03} \\ \rho_{10} & \rho_{11} & \rho_{12} & \rho_{13} \\ \rho_{20} & \rho_{21} & \rho_{22} & \rho_{23} \\ \rho_{30} & \rho_{31} & \rho_{32} & \rho_{33} \end{pmatrix}. \end{split}\]

Trace out \(B\) (\(\mathrm{Tr}_B\), leaving a \(2\times 2\) matrix on \(A\)). With compound indices \(m = 2i+j\) and \(n = 2i'+j\),

\[ (\rho_A)_{ii'} = \sum_{j=0}^{1} \rho_{\,(2i+j),\,(2i'+j)}. \]

Equivalently, view \(\hat{\rho}_{AB}\) as a \(2\times 2\) block matrix in the \(A\) indices; then \((\rho_A)_{ii'}\) is the trace of the \((i,i')\) block (a \(2\times 2\) matrix in \(B\)):

\[\begin{split} \hat{\rho}_A = \begin{pmatrix} \rho_{00}+\rho_{11} & \rho_{02}+\rho_{13} \\ \rho_{20}+\rho_{31} & \rho_{22}+\rho_{33} \end{pmatrix}. \end{split}\]

Trace out \(A\) (\(\mathrm{Tr}_A\), leaving a \(2\times 2\) matrix on \(B\)). With \(m = 2i+j\) and \(n = 2i+j'\),

\[ (\rho_B)_{jj'} = \sum_{i=0}^{1} \rho_{\,(2i+j),\,(2i+j')}. \]

In the same \(4\times 4\) labeling,

\[\begin{split} \hat{\rho}_B = \begin{pmatrix} \rho_{00}+\rho_{22} & \rho_{01}+\rho_{23} \\ \rho_{10}+\rho_{32} & \rho_{11}+\rho_{33} \end{pmatrix}. \end{split}\]

Definition: Reduced Density Matrix

For a state \(\hat{\rho}_{AB}\), the reduced density matrix on subsystem \(A\) is

(226)#\[ \hat{\rho}_A = \mathrm{Tr}_B(\hat{\rho}_{AB}) \]

Physical meaning: \(\hat{\rho}_A\) encodes all predictions for measurements on subsystem \(A\) alone:

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

To compute \(\hat{\rho}_A\) in practice, insert a complete basis \(\{\vert j\rangle_B\}\) for subsystem \(B\):

\[ \hat{\rho}_A = \sum_j \langle j\vert_B\, \hat{\rho}_{AB}\, \vert j\rangle_B \]

Example: Partial Trace of a 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 full density matrix is \(\hat{\rho}_{AB} = \vert\Phi^+\rangle\langle\Phi^+\vert = \frac{1}{2}(\vert 00\rangle\langle 00\vert + \vert 00\rangle\langle 11\vert + \vert 11\rangle\langle 00\vert + \vert 11\rangle\langle 11\vert)\).

Tracing over \(B\):

\[ \hat{\rho}_A = \langle 0\vert_B \hat{\rho}_{AB} \vert 0\rangle_B + \langle 1\vert_B \hat{\rho}_{AB} \vert 1\rangle_B = \frac{1}{2}\vert 0\rangle\langle 0\vert + \frac{1}{2}\vert 1\rangle\langle 1\vert = \frac{\hat{I}}{2} \]

The reduced state is maximally mixed, even though the total state \(\vert\Phi^+\rangle\) is pure — a hallmark of maximal entanglement.

Connection to Schmidt Decomposition#

For a bipartite pure state in Schmidt form \(\vert\Psi\rangle = \sum_k \lambda_k \vert u_k\rangle_A \vert v_k\rangle_B\) (see §6.2.1), the reduced density matrices are diagonal in the Schmidt bases:

\[ \hat{\rho}_A = \sum_k \lambda_k^2 \vert u_k\rangle_A\langle u_k\vert, \quad \hat{\rho}_B = \sum_k \lambda_k^2 \vert v_k\rangle_B\langle v_k\vert \]

The eigenvalues \(\lambda_k^2\) are the squared Schmidt coefficients — equal on both sides.

Entanglement Entropy#

The von Neumann entropy \(S(\hat{\rho}) = -\mathrm{Tr}(\hat{\rho}\ln\hat{\rho})\) was introduced in §6.1.2 as a measure of how mixed a quantum state is. Applied to the reduced density matrix, it becomes the entanglement entropy.

Definition: Entanglement Entropy

For a bipartite pure state \(\vert\Psi\rangle_{AB}\) with Schmidt coefficients \(\{\lambda_k\}\), the entanglement entropy is:

(227)#\[ S(A) = S(\hat{\rho}_A) = -\sum_k \lambda_k^2 \ln \lambda_k^2 \]

It quantifies the entanglement across the \(A\vert B\) bipartition.

Key properties:

\(S(A) = 0\) iff the state is a product (Schmidt rank 1; \(\hat{\rho}_A\) is pure).
\(S(A) = \ln\min(d_A, d_B)\) for a maximally entangled state (all Schmidt coefficients equal).
\(S(A) = S(B)\) for any bipartite pure state (both reduced states share the same eigenvalue spectrum).
\(S\) is an LOCC monotone: it cannot increase under local operations and classical communication.

Example: Partially Entangled State

Problem. Find the entanglement entropy of \(\vert\Psi(\theta)\rangle = \cos\theta\,\vert 00\rangle + \sin\theta\,\vert 11\rangle\).

Solution. Schmidt coefficients: \(\lambda_1 = \cos\theta\), \(\lambda_2 = \sin\theta\).

\[ S(\theta) = -\cos^2\theta\,\ln(\cos^2\theta) - \sin^2\theta\,\ln(\sin^2\theta) \]

At \(\theta = 0\): \(S = 0\) (product). At \(\theta = \pi/4\): \(S = \ln 2\) (Bell state, maximally entangled).

Poll: Entanglement entropy for pure states

For a pure bipartite state \(\vert\psi\rangle_{AB}\), the entanglement entropy is \(S = -\mathrm{Tr}(\hat{\rho}_A \ln \hat{\rho}_A)\) where \(\hat{\rho}_A = \mathrm{Tr}_B(\vert\psi\rangle\langle\psi\vert)\). When is \(S = 0\)?

(A) Only when \(\vert\psi\rangle = \vert a\rangle_A \otimes \vert b\rangle_B\) (product state).

(B) When both subsystems are in pure states.

(D) Never — all entangled states have \(S > 0\).

Rényi Entanglement Entropies#

Entanglement entropy captures the average information lost when tracing out a subsystem. The Rényi entropies generalize this by probing the full spectrum of Schmidt coefficients at different “resolutions.”

(228)#\[ S_\alpha(A) = \frac{1}{1-\alpha} \ln \sum_k \lambda_k^{2\alpha} \]

Special cases:

\(\alpha \to 1\): recovers the von Neumann entanglement entropy.
\(\alpha = 2\): \(S_2 = -\ln\mathrm{Tr}(\hat{\rho}_A^2)\), experimentally accessible via the “swap trick” (measuring the overlap of two copies of \(\hat{\rho}_A\)).
\(\alpha = 0\): \(S_0 = \ln r\) where \(r\) is the Schmidt rank.
\(\alpha \to \infty\): \(S_\infty = -\ln\lambda_{\max}^2\), dominated by the largest Schmidt coefficient.

Bell States#

Definition: Bell States

The four Bell states form an orthonormal basis for \(\mathbb{C}^2 \otimes \mathbb{C}^2\) consisting entirely of maximally entangled states:

\[ \vert\Phi^\pm\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle \pm \vert 11\rangle), \quad \vert\Psi^\pm\rangle = \frac{1}{\sqrt{2}}(\vert 01\rangle \pm \vert 10\rangle) \]

Each has Schmidt rank 2 with \(\lambda_1 = \lambda_2 = 1/\sqrt{2}\), giving \(S = \ln 2\) and \(\hat{\rho}_A = \hat{I}/2\).

Discussion: mixed reduced state does not always mean entanglement

Bell states are maximally entangled, yet their reduced density matrices are completely mixed. Does a mixed reduced state always signal entanglement? Can a separable mixed state \(\hat{\rho}_{AB}\) have a mixed reduced density matrix on one subsystem? (Hint: consider a product of mixed states, \(\hat{\rho}_A \otimes \hat{\rho}_B\).)

Summary#

Partial trace: \(\hat{\rho}_A = \mathrm{Tr}_B(\hat{\rho}_{AB})\) extracts subsystem information; eigenvalues are the squared Schmidt coefficients for pure bipartite states.
Entanglement entropy: \(S(A) = -\sum_k \lambda_k^2 \ln\lambda_k^2\) — the von Neumann entropy of the reduced state; zero iff product, maximum (\(\ln r\)) iff maximally entangled.
Rényi entropies: \(S_\alpha\) generalize the von Neumann entropy and probe the full Schmidt spectrum; \(S_2\) is experimentally accessible via the swap trick.
Bell states: Four maximally entangled two-qubit states forming an orthonormal basis; \(S = \ln 2\) each.

Homework#

1. Partial trace computation. For the Bell state \(\vert\Phi^+\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\), compute the reduced density matrix \(\hat{\rho}_A = \mathrm{Tr}_B(\vert\Phi^+\rangle\langle\Phi^+\vert)\) explicitly by inserting a complete basis for \(B\). What is the purity \(\mathrm{Tr}(\hat{\rho}_A^2)\)?

2. Entropy symmetry. Show that for any bipartite pure state, \(S(A) = S(B)\), where \(S(X) = -\mathrm{Tr}(\hat{\rho}_X \ln \hat{\rho}_X)\) is the von Neumann entropy of subsystem \(X\). Explain why this symmetry is a necessary condition for \(S\) to be a good entanglement measure.

3. Entanglement entropy. Compute the entanglement entropy \(S(A)\) for:

(a) The Bell state \(\vert\Phi^+\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\).

(b) The product state \(\vert\psi\rangle = \vert 0\rangle_A \otimes \vert 0\rangle_B\).

(c) The partially entangled state \(\vert\psi(\theta)\rangle = \cos\theta\,\vert 00\rangle + \sin\theta\,\vert 11\rangle\) as a function of \(\theta\).

4. Rényi entropy. The Rényi-\(\alpha\) entanglement entropy is \(S_\alpha = \frac{1}{1-\alpha}\ln\mathrm{Tr}(\hat{\rho}_A^\alpha)\).

(a) Show that \(\alpha \to 1\) recovers the von Neumann entanglement entropy.

(b) For the Bell state (\(\hat{\rho}_A = \hat{I}/2\)), compute \(S_2 = -\ln\mathrm{Tr}(\hat{\rho}_A^2)\).

5. Bell basis orthonormality. Show that the four Bell states \(\vert\Phi^\pm\rangle\), \(\vert\Psi^\pm\rangle\) are mutually orthogonal and form a complete basis for the two-qubit Hilbert space. Why does this orthogonality matter for quantum communication protocols?

6. GHZ entanglement entropy. Consider the 3-qubit GHZ state \(\vert\mathrm{GHZ}\rangle = \frac{1}{\sqrt{2}}(\vert 000\rangle + \vert 111\rangle)\).

(a) Compute the reduced density matrix \(\hat{\rho}_{AB}\) by tracing out qubit \(C\).

(b) Compute the entanglement entropy \(S(C)\) for the bipartition \(AB\vert C\).

(c) Show that \(\hat{\rho}_{AB}\) can be written as a classical mixture of product states, and conclude that \(\hat{\rho}_{AB}\) is separable even though the global GHZ state is genuinely tripartite-entangled.