Entanglement Entropy

For a pure state of a composite system, the von Neumann entropy of the whole is zero — yet the parts can be deeply uncertain. That residual uncertainty in a subsystem is entanglement entropy: the canonical measure of how entangled a pure bipartite state is.

The reduced state and the partial trace

Given a pure state $|\psi\rangle_{AB}$ , the state of subsystem $A$ alone is obtained by the partial trace over $B$ :

\rho_A = \operatorname{Tr}_B\big(|\psi\rangle\langle\psi|_{AB}\big).

This is the unique operator reproducing every measurement statistic local to $A$ . Even though $|\psi\rangle_{AB}$ is pure, $\rho_A$ is generally mixed — and it is mixed precisely because of entanglement with $B$ . There is no extra ignorance about $A$ : the mixedness is forced by the correlations.

Definition

The entanglement entropy of a pure bipartite state is

E(|\psi\rangle_{AB}) = S(\rho_A) = -\operatorname{Tr}(\rho_A \log_2 \rho_A).

A foundational fact (proved via the Schmidt decomposition in the next lesson) is that the two reduced states share identical nonzero eigenvalues, so

S(\rho_A) = S(\rho_B),

and the measure does not care which side we trace out. This symmetry is special to pure global states.

Reading off the extremes

Product (unentangled) state. $|\psi\rangle = |\phi\rangle_A \otimes |\chi\rangle_B$ gives a pure $\rho_A = |\phi\rangle\langle\phi|$ , so $E = 0$ . Zero entanglement entropy iff the state factorizes.
Maximally entangled state. A Bell state has $\rho_A = \mathbb{1}/2$ , so $E = 1$ ebit (one unit of entanglement). In dimension $d$ , the maximal value is $\log_2 d$ .
In between. A partially entangled state interpolates smoothly between these.

A worked example

Consider

|\psi\rangle = \frac{\sqrt3}{2}\,|00\rangle + \frac12\,|11\rangle.

The coefficients are already in "Schmidt form" — each basis label on $A$ pairs with a distinct label on $B$ . Tracing out $B$ ,

\rho_A = \left|\tfrac{\sqrt3}{2}\right|^2 |0\rangle\langle0| + \left|\tfrac12\right|^2 |1\rangle\langle1| = \begin{pmatrix} \tfrac34 & 0 \\ 0 & \tfrac14 \end{pmatrix}.

Its eigenvalues are $\tfrac34$ and $\tfrac14$ , so

E = h(\tfrac14) = -\tfrac34\log_2\tfrac34 - \tfrac14\log_2\tfrac14 \approx 0.8113\text{ ebits}.

The state is entangled but not maximally so: $0 < 0.8113 < 1$ .

Why it is the right measure

For pure states, entanglement entropy is the unique entanglement measure (up to normalization) that is invariant under local unitaries and non-increasing under local operations and classical communication (LOCC). Operationally, $E$ ebits is the asymptotic rate at which copies of $|\psi\rangle_{AB}$ can be reversibly converted to and from Bell pairs — the gold standard of entanglement as a resource.

Try it

Compute the entanglement entropy, in bits, of $|\psi\rangle = \tfrac{\sqrt3}{2}|00\rangle + \tfrac12|11\rangle$ , by forming $\rho_A$ and taking its von Neumann entropy. Return the value.

Run your code to see the quantum state.