This note is part of [[quantum/Practical Quantum Information System]]. > [!info] Course code > Use the companion repository for this lecture's runnable lab, helper functions, and regression checks: > - [notebooks/by_concept/entanglement_entropy.ipynb](https://github.com/montekkundan/quantum-code/blob/main/notebooks/by_concept/entanglement_entropy.ipynb) > - [qcourse/entanglement.py](https://github.com/montekkundan/quantum-code/blob/main/qcourse/entanglement.py) > - [tests/test_expanded_course_primitives.py](https://github.com/montekkundan/quantum-code/blob/main/tests/test_expanded_course_primitives.py) [TODO: add video - Entanglement Entropy] ## What This Concept Is For pure bipartite states, [[quantum/Glossary#Entanglement entropy|entanglement entropy]] is one of the cleanest quantitative measures in the subject. It tells you how much entanglement sits across a bipartition by taking the [[quantum/Glossary#Von Neumann entropy|Von Neumann entropy]] of either reduced state. The subtle part comes immediately after that clean definition. The same entropy language becomes much more delicate for mixed states. High subsystem entropy does not automatically mean high entanglement once classical mixing enters the story. ## Foundation Terms You Need First [[quantum/Glossary#Von Neumann entropy|Von Neumann entropy]] measures the uncertainty or mixedness of a density matrix. [[quantum/Glossary#Entanglement entropy|Entanglement entropy]] uses the reduced density matrix of a pure bipartite state. A [[quantum/Glossary#Reduced density matrix|reduced density matrix]] is obtained by tracing out the other subsystem. The crucial dependency order is: pure bipartite state first, reduced state second, entropy last. If the global state is mixed, you are already in a more complicated situation and should not assume the same interpretation survives unchanged. ## How The Idea Actually Works For a pure state $|\psi\rangle_{AB}$, let $ \rho_A = \mathrm{Tr}_B(|\psi\rangle \langle \psi |). $ Then the entanglement entropy is $ S(\rho_A) = -\mathrm{Tr}(\rho_A \log_2 \rho_A). $ Because the global state is pure, any mixedness in $\rho_A$ comes entirely from entanglement with subsystem $B$. That is what makes the measure so clean in the pure-state setting. The Schmidt decomposition makes this even simpler. If the Schmidt coefficients are $\lambda_i$, then those same $\lambda_i$ are the eigenvalues of the reduced state, so the entropy can be computed directly from them. This is why Schmidt decomposition and entanglement entropy are taught together. The caution comes when the global state is mixed. A subsystem can have high entropy for reasons that have nothing to do with bipartite entanglement. Classical uncertainty and environmental coupling can both increase entropy. So while entropy is a beautiful entanglement measure for pure states, it stops being a universal entanglement diagnostic once mixed states enter. ## Worked Example: Unequal Schmidt Weights For $ |\psi\rangle=\sqrt{\frac34}|00\rangle+\sqrt{\frac14}|11\rangle, $ the reduced-state eigenvalues are $3/4$ and $1/4$. The entanglement entropy is $ S=-\frac34\log_2\frac34-\frac14\log_2\frac14\approx0.811\text{ ebits}. $ This is more than zero because the state is entangled, but less than one because it is not maximally entangled. The Bell state has eigenvalues $1/2,1/2$ and entropy exactly $1$ ebit. Now compare a mixed separable state: $ \rho=\frac12|00\rangle\langle00|+\frac12|11\rangle\langle11|. $ Alice's reduced state is also $I/2$, so its local entropy is $1$. But the bipartite state is separable because it is a mixture of product projectors. This is the central warning: subsystem entropy alone measures pure-state entanglement only when the global bipartite state is pure. ## Source Reading Lens Use `qclec.pdf` Lecture 11 for the pure-state entropy story. Read it after [[concepts/Schmidt Decomposition]] and before Bell/nonlocal games so "entanglement amount" and "nonlocal-test performance" stay related but not conflated. ## Why It Matters - It gives a precise numerical measure of pure-state bipartite entanglement. - It links reduced density matrices, Schmidt coefficients, and information-theoretic language. - It teaches one of the most important mixed-state cautionary lessons in the course. ## Related Questions > [!question] Why do both sides of a pure bipartite state have the same entanglement entropy? >> [!answer] The Schmidt decomposition gives both reduced states the same nonzero eigenvalues. Since Von Neumann entropy depends only on those eigenvalues, $S(\rho_A)=S(\rho_B)$ for a pure bipartite state. > [!question] What entropy should a product state have across the chosen bipartition? >> [!answer] Zero. The reduced state of either side is pure, so there is no entanglement across that bipartition. > [!question] Why can high entropy be misleading for mixed states? >> [!answer] Mixed-state entropy can come from classical uncertainty, environmental noise, or entanglement with an outside system. For a mixed bipartite state, subsystem entropy alone does not isolate entanglement between the two named subsystems. ## Additional Study Notes **Check yourself: entropy of unequal two-term entanglement** For $\sqrt{2/3}|10\rangle-\sqrt{1/3}|01\rangle$, the reduced-state eigenvalues are $2/3$ and $1/3$, so the entanglement entropy is $-\frac23\log_2\frac23-\frac13\log_2\frac13\approx0.918$ ebits. **Key fact: both sides agree for pure bipartite states** For any pure bipartite state, Alice's and Bob's reduced density matrices have the same nonzero eigenvalues. Their Von Neumann entropies are therefore equal. **Check yourself: local basis changes do not change entanglement entropy** Applying a unitary to Bob's subsystem can change Bob's matrix representation, but it does not change the Schmidt coefficients or Alice's reduced-state eigenvalues. **Question: Why does the Bell state have one ebit of entanglement?** Tracing out either qubit gives $I/2$, whose eigenvalues are $1/2$ and $1/2$. The entropy is $-\frac12\log_2\frac12-\frac12\log_2\frac12=1$. ## Study Checks Use these after the explanation, not before it. ### Quick Checks - MCQ: Entanglement entropy is a clean entanglement measure for which case? A. Any mixed state B. Any noisy hardware sample C. Pure bipartite states D. Classical probability distributions only **Answer:** C. Pure bipartite states. - What is one ebit? **Answer:** The entanglement entropy of a maximally entangled two-qubit Bell pair, equal to 1 bit. ### Oral Exam Anchors > [!question] Oral exam anchor > Explain when entropy is and is not a clean entanglement measure. A good answer should say: For a pure bipartite state |psi>_AB, the entanglement entropy is the Von Neumann entropy of either reduced state: S(rho_A) = -Tr(rho_A log_2 rho_A), where rho_A = Tr_B(|psi><psi|). In this pure-state setting, any mixedness of rho_A comes from entanglement with B, so the entropy cleanly measures bipartite entanglement. For mixed bipartite states, the same shortcut fails. A subsystem or joint state can have entropy because of classical randomness, noise, or entanglement with an external environment. For example, the separable mixture (1/2)|00><00| + (1/2)|11><11| has local entropy, but it is not entangled between the two named qubits. Entropy is therefore a clean entanglement measure only in the pure bipartite setting without additional caveats. ## Practical Lab Use a small recurring set of states here so the comparison stays interpretable. - Compute reduced density matrices for a product state, a Bell state, and a mixed separable state. - Compute the Von Neumann entropy of each reduced state. - Record one case where entropy tracks entanglement cleanly and one case where mixed-state entropy becomes misleading. ## Homework Make the pure-state and mixed-state distinction explicit. - Define entanglement entropy for a pure bipartite state. - Explain why high subsystem entropy does not automatically imply entanglement for mixed states. - Relate Schmidt coefficients to entanglement entropy for a pure state. ## References - Scott Aaronson, [Introduction to Quantum Information Science](https://www.scottaaronson.com/qclec.pdf), Lecture 11. - [[Resources]] - QuTiP documentation, [main docs](https://qutip.org/). - IBM Quantum Learning, [learning portal](https://quantum.cloud.ibm.com/learning/en).