Entropy of Pure and Mixed States

The von Neumann entropy draws a clean line between two kinds of quantum states. A state we know exactly — a single state vector — carries zero entropy. A state about which we are maximally ignorant carries the most entropy the system can hold. This lesson sharpens that contrast.

Pure states: zero entropy

A pure state is one that can be written as $\rho = |\psi\rangle\langle\psi|$ for some unit vector $|\psi\rangle$ . Its density matrix is a rank-one projector, so its eigenvalues are a single $1$ and the rest $0$ . The entropy is therefore

S(|\psi\rangle\langle\psi|) = -1\cdot\log_2 1 - 0 - \cdots = 0.

This holds for every pure state, including superpositions like $\tfrac{1}{\sqrt2}(|0\rangle + |1\rangle)$ . Superposition is not the same as ignorance: a qubit in a definite superposition is a state we know completely, hence $S = 0$ . A handy algebraic signature of purity is

\operatorname{Tr}(\rho^2) = 1 \iff \rho \text{ is pure}, \qquad \operatorname{Tr}(\rho^2) < 1 \iff \rho \text{ is mixed}.

The quantity $\operatorname{Tr}(\rho^2)$ is called the purity.

Mixed states: positive entropy

A mixed state is a probabilistic ensemble $\rho = \sum_i p_i |\psi_i\rangle\langle\psi_i|$ that cannot be reduced to a single vector. It has more than one nonzero eigenvalue and therefore $S > 0$ . Mixedness arises from genuine classical uncertainty about which state was prepared, or — more deeply — from entanglement with an inaccessible environment (the subject of the entanglement-entropy lesson).

The maximally mixed state

On a $d$ -dimensional system the maximally mixed state is

\rho_{\max} = \frac{\mathbb{1}}{d},

with all $d$ eigenvalues equal to $1/d$ . Its entropy saturates the upper bound:

S\!\left(\frac{\mathbb{1}}{d}\right) = -\sum_{k=1}^{d} \frac1d \log_2 \frac1d = \log_2 d.

For a single qubit ( $d=2$ ) this is $1$ bit; for two qubits ( $d=4$ ) it is $2$ bits; for $n$ qubits it is exactly $n$ bits. The maximally mixed state represents complete ignorance: every measurement in every basis yields uniformly random outcomes.

The spectrum of possibilities

Between these extremes lies everything else. For a single qubit, parameterize $\rho$ by its Bloch vector $\vec r$ with $|\vec r| \le 1$ : the eigenvalues are $\tfrac{1\pm|\vec r|}{2}$ , so

S(\rho) = h\!\left(\frac{1+|\vec r|}{2}\right),

which decreases monotonically from $1$ bit at the center ( $|\vec r| = 0$ , maximally mixed) to $0$ on the surface ( $|\vec r| = 1$ , pure). Entropy thus measures how far the state sits from the surface of the Bloch ball.

Try it

Return the von Neumann entropy, in bits, of the maximally mixed state of two qubits, $\rho = \mathbb{1}/4$ . You should find it equals $\log_2 4$ .

Run your code to see the quantum state.