Von Neumann Entropy

Purpose

The von Neumann entropy of a density operator is

$$S(\rho) = - \operatorname{Tr} \left( \rho\log\rho \right).$$

If

$$\rho = \sum_i p_i \lvert i\rangle\langle i\rvert, \qquad p_i\ge0, \qquad \sum_i p_i=1,$$

then

$$S(\rho) = - \sum_i p_i\log p_i,$$

with

$$0\log0\equiv0.$$

Entropy depends only on the eigenvalues of $\rho$. It is the Shannon entropy of that spectrum, not the Shannon entropy of diagonal entries in an arbitrary measurement basis.

The canonical interpretation and derivations are at Entropy Overview. This card collects formulas, equality cases, subsystem rules, and calculation checks.

At a glance

Setting	Formula or result
Spectral entropy	$S(\rho)=-\sum_i p_i\log p_i$
Finite-dimensional bounds	$0\le S(\rho)\le\log d$
Rank-sensitive bound	$S(\rho)\le\log\operatorname{rank}\rho$
Pure state	$S(\rho)=0$
Maximally mixed state	$S(I_d/d)=\log d$
Qubit with Bloch radius $r$	$S(\rho)=h((1+r)/2)$
Unitary invariance	$S(U\rho U^\dagger)=S(\rho)$
Product additivity	$S(\rho_A\otimes\rho_B)=S(\rho_A)+S(\rho_B)$
Subadditivity	$S(\rho_{AB})\le S(\rho_A)+S(\rho_B)$
Araki–Lieb inequality	$\lvert S(\rho_A)-S(\rho_B)\rvert\le S(\rho_{AB})$
Pure bipartite state	$S(\rho_A)=S(\rho_B)$
Gibbs state, natural log	$S(\rho_\beta)=\beta\langle H\rangle_\beta+\ln Z$
Full-rank time derivative	$\dot S=-\operatorname{Tr}(\dot\rho\ln\rho)$

Here

$$h(p) = - p\log p - (1-p)\log(1-p)$$

is the binary entropy in the same logarithm base as $S$.

Logarithm convention and units

The logarithm base fixes the unit:

Logarithm	Entropy unit
$\log_2$	bits
$\ln$	nats
$k_{\mathrm B}\ln$	thermodynamic entropy units

Changing the base rescales the result:

$$\log_bx = \frac{\ln x}{\ln b}.$$

Unless otherwise stated, mathematical physics often uses the natural logarithm, while quantum information often uses base two. A numerical entropy without a stated base is incomplete.

The operator logarithm is defined spectrally on the support:

$$\log\rho = \sum_{p_i>0} (\log p_i) \lvert i\rangle\langle i\rvert.$$

Although $\log\rho$ diverges on the kernel, the product $\rho\log\rho$ has a continuous extension because

$$\lim_{x\to0^+} x\log x = 0.$$

Do not evaluate $\log\rho$ entry by entry.

Bounds and equality cases

For a density operator on a $d$-dimensional Hilbert space,

$$0 \le S(\rho) \le \log d.$$

The lower bound is saturated exactly by pure states:

$$S(\rho)=0 \quad\Longleftrightarrow\quad \rho=\lvert\psi\rangle\langle\psi\rvert.$$

The upper bound is saturated by the unique maximally mixed state:

$$\rho_*=\frac{I_d}{d}, \qquad S(\rho_*)=\log d.$$

If $R=\operatorname{rank}\rho$, the sharper bound is

$$S(\rho)\le\log R.$$

Equality holds when the state is uniform on its support:

$$\rho = \frac{\Pi_R}{R},$$

where $\Pi_R$ is the support projector.

Appending zero eigenvalues does not change $S(\rho)$. Therefore the rank-sensitive bound is often more informative than a bound based on an arbitrary embedding dimension.

Basis and ensemble independence

For every unitary $U$,

$$S(U\rho U^\dagger) = S(\rho),$$

because unitary conjugation preserves the spectrum.

A density operator can also have many ensemble decompositions:

$$\rho = \sum_j q_j \lvert\psi_j\rangle\langle\psi_j\rvert.$$

The preparation-label entropy

$$H(\{q_j\}) = - \sum_jq_j\log q_j$$

is not generally equal to $S(\rho)$. For a pure-state ensemble,

$$S(\rho)\le H(\{q_j\}),$$

with equality when the nonzero-probability states are mutually orthogonal. Redundant classical labels can have positive $H(\{q_j\})$ while all $\lvert\psi_j\rangle$ are the same and $S(\rho)=0$.

Entropy therefore characterizes the density operator, not a particular story about how it was prepared.

Measurement entropy

A rank-one projective measurement in basis $\{\lvert m\rangle\}$ has probabilities

$$q_m = \langle m\rvert\rho\lvert m\rangle.$$

Their Shannon entropy satisfies

$$H(\{q_m\}) \ge S(\rho).$$

Equality holds when the measurement basis diagonalizes $\rho$, up to freedom within degenerate eigenspaces.

Equivalently, the unread projective measurement produces

$$\Delta(\rho) = \sum_m \lvert m\rangle\langle m\rvert \rho \lvert m\rangle\langle m\rvert,$$

whose eigenvalues are the $q_m$. Since $\Delta$ is unital,

$$S(\Delta(\rho)) = H(\{q_m\}) \ge S(\rho).$$

A pure state can therefore give random measurement outcomes even though its von Neumann entropy is zero.

Qubit shortcut

Write a qubit state as

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right), \qquad r=\lVert\mathbf r\rVert\le1.$$

Its eigenvalues are

$$p_\pm = \frac{1\pm r}{2}.$$

Therefore

$$S(\rho) = h\left( \frac{1+r}{2} \right).$$

Only the Bloch radius matters. The direction changes eigenvectors but not the entropy.

For

$$\rho = \begin{pmatrix} a&c\\ c^*&1-a \end{pmatrix},$$

the radius is

$$r = \sqrt{ (2a-1)^2 + 4\lvert c\rvert^2 }.$$

The physicality condition is $0\le r\le1$. At $r=1$ the state is pure; at $r=0$ it is maximally mixed.

Products and correlations

For a product state,

$$S(\rho_A\otimes\rho_B) = S(\rho_A)+S(\rho_B).$$

For a general joint state,

$$S(\rho_{AB}) \le S(\rho_A)+S(\rho_B).$$

The difference defines the quantum mutual information:

$$I(A:B) = S(\rho_A) + S(\rho_B) - S(\rho_{AB}) \ge0.$$

It measures total correlations, classical and quantum, and vanishes exactly for a product state in finite dimension.

The Araki–Lieb inequality is

$$\lvert S(\rho_A)-S(\rho_B) \rvert \le S(\rho_{AB}).$$

Together, subadditivity and Araki–Lieb constrain any computed triplet $S_A,S_B,S_{AB}$. Violating either inequality signals an invalid state, inconsistent logarithm units, or numerical error.

See Mutual Information for correlation identities and data-processing properties.

Entanglement entropy

Let the joint state be pure with Schmidt decomposition

$$\lvert\Psi\rangle_{AB} = \sum_i \sqrt{\lambda_i} \lvert a_i\rangle \lvert b_i\rangle.$$

The reduced states have the same nonzero eigenvalues:

$$\rho_A = \sum_i \lambda_i \lvert a_i\rangle\langle a_i\rvert, \qquad \rho_B = \sum_i \lambda_i \lvert b_i\rangle\langle b_i\rvert.$$

Therefore

$$S(\rho_A) = S(\rho_B) = - \sum_i \lambda_i\log\lambda_i.$$

For a pure bipartite joint state, this common value is the entanglement entropy across the $A$–$B$ split. It vanishes exactly when the state is a product.

For a mixed $\rho_{AB}$, $S(\rho_A)$ is local mixedness, not an entanglement measure by itself. Classical correlations, local noise, and entanglement with an external environment can all contribute.

The canonical subsystem treatment is at Subsystem Entropy and Entanglement Entropy.

Bell-state check

For

$$\lvert\Phi^+\rangle = \frac{ \lvert00\rangle+\lvert11\rangle }{ \sqrt2 },$$

the joint density operator is pure:

$$S(\rho_{AB})=0.$$

Either reduced state is

$$\rho_A=\rho_B=\frac{I_2}{2},$$

so, in bits,

$$S(\rho_A) = S(\rho_B) = 1.$$

A pure whole can have mixed parts. The local entropy records entanglement across this pure-state bipartition, not ignorance about a classical preparation of the whole.

Concavity and channel behavior

Entropy is concave:

$$S\left( \sum_jq_j\rho_j \right) \ge \sum_j q_jS(\rho_j).$$

Forgetting a classical label cannot lower the average entropy. Concavity does not mean every quantum channel raises entropy.

Unitary channels preserve entropy:

$$S(U\rho U^\dagger)=S(\rho).$$

For a dimension-preserving unital CPTP map,

$$\Phi(I)=I,$$

one has

$$S(\Phi(\rho)) \ge S(\rho).$$

A general nonunital channel can decrease entropy. Cooling, reset, and amplitude damping can transfer entropy from the system to an environment. Conditioning on a measurement outcome can also reduce entropy because the observer has gained information.

Under global unitary system–environment evolution, $S(\rho_{SE})$ remains constant, while reduced entropies can increase or decrease as correlations and local mixedness change.

Time derivative

If $\rho(t)$ is differentiable, full rank, and trace preserving, then

$$\frac{dS}{dt} = - \operatorname{Tr} \left( \dot\rho\log\rho \right).$$

The term $-\operatorname{Tr}\dot\rho$ vanishes because $\operatorname{Tr}\rho=1$. The same formula applies when the support remains fixed and the logarithm is interpreted on that support.

For unitary dynamics,

$$\dot\rho = - \frac{i}{\hbar}[H,\rho],$$

cyclicity gives

$$\dot S=0.$$

For an open-system generator, the sign of $\dot S$ is not universal. A unital semigroup cannot lower entropy, while a nonunital relaxation process can. Near eigenvalues crossing zero, differentiating $\log\rho$ requires support care even when $S(\rho)$ itself is continuous in finite dimension.

Gibbs-state shortcut

For

$$\rho_\beta = \frac{e^{-\beta H}}{Z(\beta)}, \qquad Z(\beta) = \operatorname{Tr} e^{-\beta H},$$

one has

$$\ln\rho_\beta = -\beta H-\ln Z.$$

Therefore, in nats,

$$S(\rho_\beta) = \beta \langle H\rangle_\beta + \ln Z.$$

Multiplication by $k_{\mathrm B}$ gives thermodynamic entropy. This interpretation depends on equilibrium, the Hamiltonian, and the temperature; an arbitrary mixed state is not automatically thermal.

For a two-level Hamiltonian with energies $0$ and $\epsilon$,

$$Z=1+e^{-\beta\epsilon},$$

and the excited-state probability is

$$p_e = \frac{e^{-\beta\epsilon}}{1+e^{-\beta\epsilon}}.$$

The entropy is simply

$$S(\rho_\beta)=h(p_e).$$

It approaches zero at zero temperature for a nondegenerate ground state and $\log2$ at infinite temperature.

Relation to purity and Rényi entropy

Purity is

$$\gamma(\rho) = \operatorname{Tr}(\rho^2).$$

The order-two Rényi entropy is

$$S_2(\rho) = - \log \operatorname{Tr}(\rho^2).$$

With the same logarithm base,

$$S(\rho)\ge S_2(\rho).$$

For a qubit, purity fixes the Bloch radius and therefore fixes $S(\rho)$. In dimensions $d\ge3$, states can have equal purity and different von Neumann entropy. The two quantities are related spectral summaries, not interchangeable names.

Use the Purity Formula Card for rank bounds, swap measurements, and thermal shortcuts.

Relative-entropy identity

The quantum relative entropy is

$$D(\rho\Vert\sigma) = \operatorname{Tr} \left[ \rho \left( \log\rho-\log\sigma \right) \right],$$

provided

$$\operatorname{supp}\rho \subseteq \operatorname{supp}\sigma.$$

Otherwise it is $+\infty$. For the maximally mixed state,

$$D\left( \rho\middle\Vert\frac{I_d}{d} \right) = \log d-S(\rho).$$

This identity gives the upper entropy bound from positivity of relative entropy and quantifies the entropy deficit from complete mixing. See Relative Entropy for monotonicity and support conventions.

Numerical workflow

1. Verify $\rho=\rho^\dagger$, $\operatorname{Tr}\rho=1$, and $\rho\ge0$ within a stated tolerance.
2. Use a Hermitian eigensolver and keep eigenvalue multiplicities.
3. Distinguish tiny negative values caused by roundoff from a materially nonpositive reconstruction; do not silently repair physical errors.
4. If clipping roundoff-scale negatives, renormalize and report the procedure.
5. Evaluate $-p\log p$ with a stable branch that returns zero at $p=0$.
6. State the logarithm base.
7. Check $0\le S\le\log R\le\log d$.
8. For bipartite states, compare $S_A,S_B,S_{AB}$ against subadditivity and Araki–Lieb.
9. Test unitary invariance by diagonalization or a random basis rotation.

Near-pure states are sensitive to small eigenvalues because the derivative of $-p\log p$ grows as $p\to0^+$. Entropy estimates from finite-data state tomography can therefore be biased by physicality constraints and spectral thresholds.

Infinite-dimensional caveat

For a trace-class density operator on a separable Hilbert space,

$$S(\rho) = - \sum_i p_i\log p_i$$

may be finite or $+\infty$. There is no normalizable maximally mixed state on an unrestricted infinite-dimensional space, and entropy is not uniformly continuous without constraints such as an energy bound.

In quantum field theory, spatial entanglement entropies are generally ultraviolet regulator dependent, and local algebras need not correspond to a simple tensor factor with a density matrix. Finite-dimensional formulas remain guides, not automatic regulator-independent observables.

Common mistakes

• Applying $-x\log x$ to matrix entries instead of eigenvalues.
• Omitting the logarithm base.
• Treating measurement-outcome entropy as state entropy.
• Assigning entropy to one ensemble decomposition rather than to $\rho$.
• Calling $S(\rho_A)$ an entanglement measure when $\rho_{AB}$ is mixed.
• Assuming every noisy or Lindblad evolution increases entropy.
• Forgetting that a nonunital reset or cooling channel can reduce entropy.
• Treating entropy as the expectation value of one fixed one-copy observable.
• Confusing von Neumann entropy with purity or linear entropy.
• Dropping small physical eigenvalues without an error analysis.
• Using $\log d$ as an upper bound when $d$ is an arbitrary truncation rather than the modeled Hilbert-space dimension.
• Assuming finite-dimensional continuity and maximum-entropy statements hold unchanged in infinite dimension or quantum field theory.

Exercises

1. A qubit has Bloch radius $r$. Derive its entropy and evaluate the pure and maximally mixed limits.

Solution

The eigenvalues are

$$p_\pm = \frac{1\pm r}{2}.$$

Therefore

$$S(\rho) = - \frac{1+r}{2} \log\frac{1+r}{2} - \frac{1-r}{2} \log\frac{1-r}{2}.$$

At $r=1$, the spectrum is $(1,0)$ and $S=0$. At $r=0$, the spectrum is $(1/2,1/2)$ and $S=\log2$, or one bit for base-two logarithms.

2. Compute the global and local entropies of $\lvert\Phi^+\rangle=(\lvert00\rangle+\lvert11\rangle)/\sqrt2$.

Solution

The global density operator is rank one, so

$$S(\rho_{AB})=0.$$

Tracing out either qubit gives

$$\rho_A=\rho_B=\frac{I_2}{2}.$$

Hence

$$S(\rho_A)=S(\rho_B)=\log2.$$

This common reduced entropy is the entanglement entropy because the joint state is pure.

3. For a thermal two-level system with energies $0$ and $\epsilon$, derive the entropy and its zero- and infinite-temperature limits.

Solution

The probabilities are

$$p_g = \frac{1}{1+e^{-\beta\epsilon}}, \qquad p_e = \frac{e^{-\beta\epsilon}}{1+e^{-\beta\epsilon}}.$$

Therefore

$$S(\rho_\beta) = - p_g\log p_g - p_e\log p_e = h(p_e).$$

As $\beta\to\infty$, $(p_g,p_e)\to(1,0)$ and $S\to0$. As $\beta\to0$, $(p_g,p_e)\to(1/2,1/2)$ and $S\to\log2$.

4. Pure dephasing turns $\lvert+\rangle\langle+\rvert$ into

$$\rho(c) = \frac12 \begin{pmatrix} 1&c\\ c^*&1 \end{pmatrix}, \qquad \lvert c\rvert\le1.$$

Find the entropy as a function of $\lvert c\rvert$.

Solution

The Bloch radius is $r=\lvert c\rvert$, so the eigenvalues are

$$p_\pm = \frac{1\pm\lvert c\rvert}{2}.$$

Thus

$$S(\rho(c)) = h\left( \frac{1+\lvert c\rvert}{2} \right).$$

At $\lvert c\rvert=1$, the state is pure and $S=0$. At $c=0$, dephasing is complete and $S=\log2$. The phase of $c$ rotates the Bloch vector but does not change its entropy.

Canonical links

• Entropy Overview
• Subsystem Entropy
• Entanglement Entropy
• Mutual Information
• Relative Entropy
• Purity Formula Card

References

• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• A. Wehrl, “General Properties of Entropy,” Reviews of Modern Physics 50, 221–260 (1978).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010, Ch. 11.
• M. M. Wilde, Quantum Information Theory, 2nd ed., Cambridge University Press, 2017, Chs. 10–11.
• M. Ohya and D. Petz, Quantum Entropy and Its Use, 2nd ed., Springer, 2004.
• I. Bengtsson and K. Życzkowski, Geometry of Quantum States, 2nd ed., Cambridge University Press, 2017, Ch. 12.