6.2.2 Entanglement Measures

6.2.2 Entanglement Measures#

Prompts

  • What is entanglement entropy for a bipartite pure state? Why does it measure how entangled a state is?

  • What are the four Bell states? Why are they maximally entangled, and what is their entanglement entropy?

  • What is concurrence, and how does it generalize entanglement entropy to mixed two-qubit states?

  • What is an entanglement witness? How can you detect entanglement without full state tomography?

  • How do separable, PPT, and entangled states relate? What is the partial transpose, and when does it certify entanglement?

Lecture Notes#

Entanglement admits rich quantitative structure: different measures capture distinct aspects. This section develops tools to quantify and detect entanglement.

Overview#

A quantum state can be assigned a numerical entanglement measure: a function that vanishes on separable states and ranges to a maximum for maximally entangled states. Different measures capture distinct aspects:

  • Entanglement entropy: the standard information-theoretic measure for pure bipartite states.

  • Concurrence: a computable entanglement measure for two-qubit states (pure and mixed).

  • Entanglement witnesses: Hermitian operators that certify entanglement via a single observable, without full tomography.

  • Partial transpose and negativity: uses the Peres–Horodecki criterion to detect entanglement.

These tools form a hierarchy: separable states lie strictly within PPT states, which lie within all states. The boundaries of these regions define the landscape of quantum correlations.

Entanglement Entropy (Recap)#

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

(120)#\[E(\Psi) = S(\hat{\rho}_A) = -\mathrm{Tr}(\hat{\rho}_A \ln \hat{\rho}_A) = -\sum_k \lambda_k^2 \ln \lambda_k^2\]

where \(\hat{\rho}_A = \mathrm{Tr}_B(|\Psi\rangle\langle\Psi|)\) has eigenvalues \(\lambda_k^2\) (the squared Schmidt coefficients).

Properties of Entanglement Entropy

  • Vanishes for product states: If \(|\Psi\rangle = |\psi\rangle_A \otimes |\phi\rangle_B\), then \(E = 0\).

  • Maximized for maximally entangled states: For two qubits, the Bell states achieve \(E = \ln 2\) (one ebit).

  • Symmetric: \(S(\hat{\rho}_A) = S(\hat{\rho}_B)\) for any bipartite pure state.

  • Monotone: Cannot increase under local operations and classical communication (LOCC).

Example: Partially Entangled State

Problem. Consider \(|\Psi\rangle = \cos\theta\, |00\rangle + \sin\theta\, |11\rangle\) with \(0 < \theta < \pi/4\). Find the entanglement entropy.

Solution. Schmidt coefficients: \(\lambda_1 = \cos\theta\), \(\lambda_2 = \sin\theta\). The entanglement entropy is:

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

  • At \(\theta = 0\): product state, \(E = 0\).

  • At \(\theta = \pi/4\): Bell state, \(E = \ln 2\) (maximum).

  • For intermediate \(\theta\): \(0 < E < \ln 2\) (partial entanglement).

Rényi Entanglement Entropies#

The Rényi-\(\alpha\) entanglement entropy generalizes von Neumann entropy:

(121)#\[E_\alpha(\Psi) = \frac{1}{1-\alpha} \ln \mathrm{Tr}(\hat{\rho}_A^\alpha) = \frac{1}{1-\alpha} \ln \sum_k \lambda_k^{2\alpha}\]

  • \(\alpha \to 1\): recovers von Neumann entropy.

  • \(\alpha = 2\): Rényi-2 entropy \(E_2 = -\ln \mathrm{Tr}(\hat{\rho}_A^2)\), experimentally accessible via the “swap trick” without full tomography.

  • \(\alpha = 0\): \(E_0 = \ln(\text{Schmidt rank})\), counts the dimension of entanglement.

Bell States#

The Bell states are four maximally entangled two-qubit pure states that form an orthonormal basis for the two-qubit Hilbert space.

Definition: Bell States

\[|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\]
\[|\Phi^-\rangle = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle)\]
\[|\Psi^+\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)\]
\[|\Psi^-\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)\]

These four states are mutually orthogonal and form a basis for \(\mathbb{C}^2 \otimes \mathbb{C}^2\).

Maximal Entanglement of Bell States

For any Bell state, the reduced density matrix on subsystem A is:

\[\hat{\rho}_A = \frac{I}{2}\]

This is the completely mixed state on a single qubit. Its eigenvalues are \(\lambda_1^2 = \lambda_2^2 = 1/2\), giving:

\[E = -\frac{1}{2}\ln\frac{1}{2} - \frac{1}{2}\ln\frac{1}{2} = \ln 2\]

This is the maximum entanglement entropy for two qubits. Each Bell state is maximally entangled: one ebit.

Example: Bell State Orthogonality

Problem. Verify that \(\langle\Phi^+|\Psi^+\rangle = 0\).

Solution. Expand the inner product:

\[\langle\Phi^+|\Psi^+\rangle = \frac{1}{2}\left(\langle 00| + \langle 11|\right)(|01\rangle + |10\rangle)\]
\[= \frac{1}{2}(0 + 0 + 0 + 0) = 0\]

All six pairs are orthogonal, confirming they form an orthonormal basis. This orthogonality is crucial for Bell state tomography: measuring which Bell state a system is in can be done via projective measurements on the Bell basis.

Discussion

Bell states are maximally entangled, yet their reduced density matrices are completely mixed. Does a mixed reduced state always signal entanglement? Can a separable state have a mixed reduced density matrix on one subsystem? (Hint: think about a product of mixed states.)

Concurrence (Two Qubits)#

For two-qubit systems, the concurrence is a computable entanglement measure that applies to both pure and mixed states.

Definition: Concurrence (Pure States)

For a pure two-qubit state \(|\Psi\rangle = \sum_{ij} c_{ij}|ij\rangle\):

(122)#\[\mathcal{C}(|\Psi\rangle) = 2|\det(C)| = 2|c_{00}c_{11} - c_{01}c_{10}|\]

where \(C = [c_{ij}]\) is the \(2\times 2\) coefficient matrix.

  • \(\mathcal{C} = 0\): separable (product state).

  • \(\mathcal{C} = 1\): maximally entangled (Bell state).

Example: Concurrence of Bell States

Problem. Compute the concurrence for \(|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\) and the product state \(|01\rangle\).

Solution. For \(|\Phi^+\rangle\):

\[\begin{split}C = \begin{pmatrix} 1/\sqrt{2} & 0 \\ 0 & 1/\sqrt{2} \end{pmatrix}, \quad \mathcal{C} = 2 \left|\frac{1}{2}\right| = 1\end{split}\]

For the product state \(|01\rangle\):

\[\begin{split}C = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad \mathcal{C} = 0\end{split}\]

Wootters’ Concurrence for Mixed States#

For mixed two-qubit states, Wootters’ formula is one of the few computable entanglement measures. Construct the spin-flipped density matrix:

(123)#\[\tilde{\hat{\rho}} = (\hat{\sigma}^y \otimes \hat{\sigma}^y) \hat{\rho}^* (\hat{\sigma}^y \otimes \hat{\sigma}^y)\]

where \(\hat{\rho}^*\) is complex conjugation in the computational basis.

Wootters’ Concurrence

Compute eigenvalues \(\mu_1 \geq \mu_2 \geq \mu_3 \geq \mu_4\) of \(\sqrt{\sqrt{\hat{\rho}} \tilde{\hat{\rho}} \sqrt{\hat{\rho}}}\), then:

(124)#\[\mathcal{C}(\hat{\rho}) = \max(0, \mu_1 - \mu_2 - \mu_3 - \mu_4)\]

Properties:

  • \(\mathcal{C}(\hat{\rho}) \in [0, 1]\).

  • \(\mathcal{C} = 0\) if and only if the state is separable.

  • \(\mathcal{C} = 1\) if and only if the state is a Bell state.

  • Computable (unlike entanglement of formation for high dimensions).

Entanglement of Formation#

The entanglement of formation \(E_f\) is the minimum average entanglement needed to create \(\hat{\rho}\). For two qubits, it is related to concurrence by:

(125)#\[E_f(\hat{\rho}) = h\left(\frac{1 + \sqrt{1 - \mathcal{C}^2}}{2}\right)\]

where \(h(x) = -x\ln x - (1-x)\ln(1-x)\) is the binary entropy function.

Entanglement Witnesses#

An entanglement witness is a Hermitian operator that detects entanglement without requiring full state tomography.

Definition: Entanglement Witness

\(W\) is an entanglement witness if:

  • \(\mathrm{Tr}(W \hat{\rho}_{\text{sep}}) \geq 0\) for all separable states \(\hat{\rho}_{\text{sep}}\).

  • \(\mathrm{Tr}(W \hat{\rho}_{\text{ent}}) < 0\) for at least one entangled state \(\hat{\rho}_{\text{ent}}\).

Observation: \(\langle W \rangle < 0\) certifies that \(\hat{\rho}\) is entangled.

Example: Witness for Bell States

Problem. Construct a witness that detects the Bell state \(|\Phi^+\rangle\).

Solution. Define:

\[W = \frac{I}{2} - |\Phi^+\rangle\langle\Phi^+|\]

For any separable state:

\[\langle\Phi^+|\hat{\rho}_{\text{sep}}|\Phi^+\rangle \leq \frac{1}{2}\]

(separable states cannot have full overlap with a maximally entangled state), so \(\mathrm{Tr}(W\hat{\rho}_{\text{sep}}) \geq 0\).

For \(\hat{\rho} = |\Phi^+\rangle\langle\Phi^+|\):

\[\mathrm{Tr}(W\hat{\rho}) = \frac{1}{2} - 1 = -\frac{1}{2} < 0\]

Entanglement Witness Completeness

By the Hahn–Banach theorem, every entangled state admits a witness. However, no single witness detects all entangled states—different entangled states require different witnesses.

Hierarchy of Quantum Correlations#

The landscape of quantum states has a rich structure. Separable states form a convex subset; entangled states surround them. The partial transpose provides a geometric way to visualize this hierarchy.

Definition: Partial Transpose and PPT States

The partial transpose with respect to subsystem B is defined via:

\[\hat{\rho}^{T_B} = \sum_{ijkl} \hat{\rho}_{ij,kl} |ik\rangle\langle jl|\]

A state is PPT (positive partial transpose) if \(\hat{\rho}^{T_B} \geq 0\) (all eigenvalues non-negative).

Peres–Horodecki Criterion: A separable state must be PPT. Equivalently: if \(\hat{\rho}^{T_B}\) has a negative eigenvalue, then \(\hat{\rho}\) is entangled.

Definition: Negativity

The negativity quantifies how negative the partial transpose is:

(126)#\[\mathcal{N}(\hat{\rho}) = \frac{\|\hat{\rho}^{T_B}\|_1 - 1}{2}\]

where \(\|\cdot\|_1\) is the trace norm (sum of absolute eigenvalues).

  • \(\mathcal{N} = 0\): PPT state.

  • \(\mathcal{N} > 0\): definitely entangled (negative partial transpose).

Example: Bell State Partial Transpose

Problem. Compute \(\hat{\rho}^{T_B}\) for the Bell state \(|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\).

Solution. The density matrix is:

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

Taking the partial transpose:

\[\begin{split}\hat{\rho}^{T_B} = \frac{1}{2}\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\end{split}\]

Eigenvalues: \(1/2, 1/2, -1/2, -1/2\). The negative eigenvalues certify entanglement.

Entanglement Classification Hierarchy

\[\text{Separable} \subset \text{PPT} \subset \text{All States}\]

  • Separable states: No entanglement; can be written as \(\hat{\rho} = \sum_i p_i \hat{\rho}_A^i \otimes \hat{\rho}_B^i\).

  • PPT states: Positive partial transpose; includes all separable states plus some entangled states in \(2 \times 2\) and \(2 \times 3\) systems.

  • NPT entangled states: Negative partial transpose; definitely entangled; absent in \(2 \times 2\) and \(2 \times 3\) systems.

For \(2 \times 2\) and \(2 \times 3\) systems: PPT \(\Leftrightarrow\) separable (the Peres conjecture, proved for these dimensions). For higher dimensions, entangled states with PPT exist (“bound entangled” states).

Caution

Negativity (positive partial transpose) is necessary and sufficient for separability only in dimensions \(2 \times 2\) and \(2 \times 3\). For higher dimensions, there exist bound entangled states with positive partial transpose—entangled states that cannot be detected by the partial transpose alone.

Summary#

  • Entanglement entropy \(E = -\sum_k \lambda_k^2 \ln(\lambda_k^2)\) quantifies entanglement for pure bipartite states; maximized at \(\ln d\) for maximally entangled states.

  • Bell states are four maximally entangled two-qubit states with \(E = \ln 2\), \(\hat{\rho}_A = I/2\), and form an orthonormal basis.

  • Concurrence \(\mathcal{C}\) is a computable entanglement measure for two-qubits; Wootters’ formula extends it to mixed states.

  • Entanglement witnesses detect entanglement via \(\langle W \rangle < 0\) without full tomography; every entangled state admits a witness.

  • Partial transpose and negativity certify entanglement: PPT is equivalent to separability in \(2 \times 2\) and \(2 \times 3\) systems; fails in higher dimensions (bound entanglement).

See Also

Homework#

1. Schmidt Decomposition and Entanglement Rank. Any bipartite pure state can be written as \(|\psi\rangle = \sum_i \sqrt{\lambda_i}\, |a_i\rangle|b_i\rangle\) with orthonormal bases on each side. Find the Schmidt decomposition of:

\[|\psi\rangle = \frac{1}{2}(|00\rangle + |01\rangle + |10\rangle - |11\rangle)\]

What is the Schmidt rank? What does this tell you about how entangled the state is?

2. Product States Are Unentangled. Show that if \(|\psi\rangle = |\phi\rangle_A \otimes |\phi\rangle_B\) is a product state, then: (a) The reduced density matrix \(\hat{\rho}_A = \mathrm{Tr}_B(|\psi\rangle\langle\psi|)\) is pure. (b) The entanglement entropy is exactly zero: \(E(\psi) = 0\). (c) The Schmidt rank is exactly 1.

What does Schmidt rank \(\geq 2\) imply about a state?

3. Bell States Are Maximally Entangled. For all four Bell states:

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

(a) Compute the reduced density matrix \(\hat{\rho}_A = \frac{I}{2}\) for each. (b) Show the entanglement entropy is \(E = \ln 2\) for all four. (c) Verify this is the maximum possible for two qubits: \(\max(E) = \ln d_A = \ln 2\).

4. Bell States Form an Orthonormal Basis. Show that the four Bell states are mutually orthogonal: \(\langle\Phi^+|\Psi^+\rangle = 0\), etc. Why does this orthogonality matter for quantum communication and Bell state tomography?

5. Partial Entanglement Interpolates. Consider \(|\psi(\theta)\rangle = \cos\theta\, |00\rangle + \sin\theta\, |11\rangle\) for \(\theta \in [0, \pi/2]\).

(a) Find the Schmidt coefficients \(\lambda_1(\theta)\) and \(\lambda_2(\theta)\). (b) Compute the entanglement entropy \(E(\theta) = -\sum_i \lambda_i^2 \ln(\lambda_i^2)\). (c) Show that \(E(0) = 0\) (product state) and \(E(\pi/4) = \ln 2\) (Bell state). (d) Sketch \(E(\theta)\) and explain its physical meaning.

6. Concurrence for Two-Qubit Pure States. For a pure two-qubit state \(|\Psi\rangle = \alpha|00\rangle + \beta|01\rangle + \gamma|10\rangle + \delta|11\rangle\), the concurrence is:

\[\mathcal{C}(|\Psi\rangle) = 2|\alpha\delta - \beta\gamma|\]

(a) Verify that \(\mathcal{C} = 1\) for the Bell state \(|\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}\). (b) Verify that \(\mathcal{C} = 0\) for the product state \(|\psi\rangle = |0\rangle_A \otimes |+\rangle_B\). (c) For \(|\psi(\theta)\rangle = \cos\theta\, |00\rangle + \sin\theta\, |11\rangle\), show that \(\mathcal{C}(\theta) = \sin(2\theta)\).

7. Entanglement of Formation vs. Concurrence. Wootters showed that for two qubits:

\[E_f(\hat{\rho}) = h\left(\frac{1 + \sqrt{1 - \mathcal{C}^2}}{2}\right)\]

where \(h(x) = -x\ln x - (1-x)\ln(1-x)\) is the binary entropy.

(a) For the Bell state (\(\mathcal{C} = 1\)), compute \(E_f\) and verify it equals \(\ln 2\). (b) For \(\mathcal{C} = 1/2\), compute \(E_f\) numerically. (c) Explain why \(E_f(\hat{\rho}) \geq E_d(\hat{\rho})\) (entanglement of formation exceeds distillable entanglement).

8. Entanglement Witnesses Detect Entanglement. For the Bell state \(|\Phi^+\rangle\), define the witness:

\[W = \frac{I}{2} - |\Phi^+\rangle\langle\Phi^+|\]

(a) Show that for any separable state \(\hat{\rho}_{\mathrm{sep}}\), we have \(\mathrm{Tr}(W\hat{\rho}_{\mathrm{sep}}) \geq 0\). (b) Compute \(\mathrm{Tr}(W|\Phi^+\rangle\langle\Phi^+|)\) and verify it is negative. (c) For the Werner state \(\hat{\rho} = p|\Phi^+\rangle\langle\Phi^+| + (1-p)\frac{I}{4}\), find the threshold \(p_c\) above which the witness detects entanglement. (d) Why is no single witness able to detect all entangled states?

9. Partial Transpose and PPT States. The partial transpose (with respect to subsystem B) of a \(2 \times 2\) state is \(\hat{\rho}^{T_B}\). The negativity is:

\[\mathcal{N}(\hat{\rho}) = \frac{\|\hat{\rho}^{T_B}\|_1 - 1}{2}\]

where \(\|\cdot\|_1\) is the trace norm (sum of absolute eigenvalues).

(a) For the Bell state, compute \(\hat{\rho}^{T_B}\) and show that \(\mathcal{N} > 0\) (negative eigenvalues exist). (b) For a product state \(\hat{\rho} = \hat{\rho}_A \otimes \hat{\rho}_B\), show that \(\hat{\rho}^{T_B}\) is separable, so \(\mathcal{N} = 0\). (c) For \(2 \times 2\) and \(2 \times 3\) systems, PPT is equivalent to separability. Why does this fail in higher dimensions?

10. Rényi Entanglement Entropy. The Rényi-\(\alpha\) entanglement entropy is \(E_\alpha = \frac{1}{1-\alpha}\ln\mathrm{Tr}(\hat{\rho}_A^\alpha)\).

(a) Show that \(\alpha \to 1\) recovers the von Neumann entanglement entropy. (b) For the Bell state (\(\hat{\rho}_A = I/2\)), compute \(E_2 = -\ln\mathrm{Tr}(\hat{\rho}_A^2)\). (c) Compare \(E_1\) and \(E_2\) for the Bell state. Why is \(E_2\) experimentally easier to measure?