Von Neumann Entropy

The classical Shannon entropy measures the uncertainty in a probability distribution. A quantum state is described not by a distribution but by a density operator $\rho$ — a positive semidefinite operator with $\operatorname{Tr}\rho = 1$. John von Neumann's entropy is the direct generalization, obtained by replacing the probabilities with the operator $\rho$ itself.

Definition

The von Neumann entropy of a density operator $\rho$ is

Here $\log_2 \rho$ is the matrix logarithm, defined through the spectral decomposition. Writing $\rho$ in its eigenbasis,

the operator $\rho\log_2\rho$ is diagonal in the same basis with entries $\lambda_k\log_2\lambda_k$, and the trace collects them:

So the von Neumann entropy is precisely the Shannon entropy of the eigenvalues of $\rho$, again with $0\log_2 0 = 0$. The eigenvalues form a genuine probability distribution because $\rho$ is a positive operator of unit trace.

Why the eigenvalues, and nothing else

The entropy is basis-independent: it depends only on the spectrum of $\rho$, not on how $\rho$ is written. This is forced by the trace, which is invariant under unitary conjugation. If $\rho' = U\rho U^\dagger$ for any unitary $U$, then $\rho$ and $\rho'$ have the same eigenvalues and hence $S(\rho') = S(\rho)$. Physically: applying a reversible (unitary) transformation never changes the information content of a state.

Basic properties

• Non-negativity. $S(\rho) \geq 0$, with $S(\rho) = 0$ iff $\rho$ is pure (a single eigenvalue $1$, the rest $0$).
• Maximum. On a $d$-dimensional system, $S(\rho) \leq \log_2 d$, with equality iff $\rho = \mathbb{1}/d$ is maximally mixed.
• Unitary invariance. $S(U\rho U^\dagger) = S(\rho)$, as shown above.
• Additivity on products. $S(\rho \otimes \sigma) = S(\rho) + S(\sigma)$.

A worked qubit example

Take the density matrix

It is Hermitian, positive, and has unit trace. Its eigenvalues solve $\lambda^2 - \lambda + \det\rho = 0$ with $\det\rho = \tfrac14 - \tfrac1{16} = \tfrac{3}{16}$, giving

Therefore

Note that $S(\rho) < 1$: although the off-diagonal coherences make $\rho$ look "spread out", its eigenvalues are unequal, so it is less mixed than the maximally mixed qubit $\mathbb{1}/2$ (which has $S = 1$).

Try it

Compute the von Neumann entropy in bits of $\rho = \begin{pmatrix} 0.5 & 0.25 \\ 0.25 & 0.5\end{pmatrix}$ by diagonalizing it and applying $S = -\sum_k \lambda_k\log_2\lambda_k$. Return the value.

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