Von Neumann Entropy
Quantum analogue of Shannon entropy — measures the uncertainty in a quantum state.
Formula
Variables
Example
A maximally mixed single qubit: S = 1 bit
How it works
Von Neumann entropy quantifies how spread out a quantum state is over its possible outcomes, using the density matrix rho rather than a single probability list. The formula works by taking the matrix logarithm of the state and weighting it by rho itself, so eigenvalues that are close to 1 contribute little uncertainty while small, evenly distributed eigenvalues increase it. For a pure state one eigenvalue is 1 and the rest are 0, giving zero entropy; for mixed states the value rises as the state becomes less certain.
When to use it
Use this when you need to measure mixedness, uncertainty, or information content of a quantum state in quantum information, thermodynamics, or entanglement theory. It is especially useful for comparing reduced states of subsystems, where a nonzero entropy often signals entanglement with the rest of the system.
Worked example, step by step
Take a qubit in the mixed state rho = diag(0.75, 0.25). Step 1: Find the eigenvalues of rho, which are already 0.75 and 0.25. Step 2: Apply S( ho) = -\sum_i \lambda_i \log_2 \lambda_i. Step 3: Compute each term: -0.75 \log_2(0.75) \approx -0.75(-0.4150) = 0.3113 -0.25 \log_2(0.25) = -0.25(-2) = 0.5000 Step 4: Add them: S( ho) = 0.3113 + 0.5000 = 0.8113 bits Final answer: S( ho) \approx 0.811 bits.
Common mistakes to avoid
- Using the state vector |\psi\rangle directly instead of the density matrix rho
- Forgetting that the logarithm is base 2, so the entropy is measured in bits
- Trying to take the logarithm of the matrix entries instead of diagonalizing rho and using its eigenvalues
Did You Know?
A pure quantum state has S = 0 — zero uncertainty — while a maximally mixed n-qubit state has S = n.
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