Von Neumann entropy

Remarks

The base of the logarithm fixes the unit: base $2$ gives bits, base $e$ gives nats.

Example: for the maximally mixed state $\rho = I/d$, all eigenvalues equal $1/d$, so $S(\rho) = -d \cdot \tfrac1d \log \tfrac1d = \log d$. For a pure state $|\psi\rangle\langle\psi|$, the eigenvalues are $1, 0, \dots, 0$, so $S = 0$.

Non-example: the Shannon entropy of the diagonal entries is not $S(\rho)$ in general. For $\rho = \tfrac12 \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, a pure state, the diagonal entries $\tfrac12, \tfrac12$ give $\log 2$, while $S(\rho) = 0$.

Always $0 \le S(\rho) \le \log d$, with $S(\rho) = 0$ exactly for pure states and $S(\rho) = \log d$ exactly for $\rho = I/d$. The entropy is invariant under unitary conjugation, $S(U\rho U^*) = S(\rho)$, and it is concave in $\rho$.