Von Neumann Entropy

The von Neumann entropy is the single number quantum information theory cannot do without. It answers, for a quantum state, the same question Shannon's entropy answers for a probability distribution: how uncertain is it, and how much can it be compressed? The construction is a direct translation. Shannon writes $H=-\sum_x p_x\log p_x$ ; von Neumann replaces the distribution by a density operator $\rho$ — a positive, unit-trace operator on a Hilbert space $\mathcal{H}$ — and the sum by a trace, giving $S(\rho)=-\mathrm{Tr}(\rho\log\rho)$ . The magic is that this is not arbitrary: diagonalize $\rho$ and the trace collapses to exactly the Shannon entropy of its eigenvalues, which form a genuine probability distribution. Thus the spectrum of every density operator defines a classical probability distribution, and the von Neumann entropy equals the Shannon entropy of that eigenvalue distribution — measured in a basis the state itself picks out. Von Neumann wrote the formula in the late 1920s to give quantum statistical mechanics a proper notion of entropy, well before anyone spoke of "information". Its operational meaning arrived only in 1995, when Schumacher proved that a memoryless quantum source emitting independent signals with density operator $\rho$ can, in the asymptotic limit, be faithfully compressed to $S(\rho)$ qubits per signal (for base-2 logarithms), with no smaller asymptotic rate possible — the exact quantum analogue of Shannon's source-coding theorem — promoting $S$ from a thermodynamic bookkeeping device to the resource measure of quantum information. The deepest surprises are for composite systems. A pure entangled state $|\psi\rangle_{AB}$ has $S(\rho_{AB})=0$ while its parts satisfy $S(\rho_A)=S(\rho_B)>0$ : the whole is more certain than its pieces, something no classical joint distribution can do. Around this one fact orbit subadditivity, the Araki–Lieb inequality, strong subadditivity, and the whole edifice of quantum Shannon theory.

Key takeaways

The von Neumann entropy $S(\rho)=-\mathrm{Tr}(\rho\log\rho)$ is the unique quantum generalization of Shannon entropy; diagonalizing $\rho$ turns it into the Shannon entropy $-\sum_i\lambda_i\log\lambda_i$ of the eigenvalue spectrum.
It vanishes exactly for pure states, is maximized at $\log d$ by the maximally mixed state $I/d$ , and is unchanged by any unitary $\rho\mapsto U\rho U^\dagger$ because unitaries preserve the spectrum.
For a pure bipartite state the global entropy is $0$ while $S(\rho_A)=S(\rho_B)$ can be large — the entanglement entropy, and the sharpest break from classical intuition.
Operationally $S(\rho)$ is the qubit cost of Schumacher compression and the building block of the Holevo bound, quantum mutual information, and channel capacities.
Its structural inequalities — concavity, subadditivity, the Araki–Lieb bound, and strong subadditivity (Lieb–Ruskai) — are the load-bearing walls of quantum Shannon theory.
Key idea: diagonalize $\rho=\sum_i\lambda_i|i\rangle\langle i|$ and $S(\rho)=-\sum_i\lambda_i\log\lambda_i$ is the Shannon entropy of the eigenvalue distribution $\{\lambda_i\}$ ; every property of $S$ follows from that spectral picture together with operator convexity.

The core insight

Diagonalize $\rho=\sum_i\lambda_i|i\rangle\langle i|$ ; then $S(\rho)=-\sum_i\lambda_i\log\lambda_i$ is exactly the Shannon entropy of the eigenvalue distribution $\{\lambda_i\}$ . Every property of $S$ flows from this spectral picture combined with operator convexity.

Where this lives today

Von Neumann entropy is the currency in which modern quantum science is quoted. In quantum computing, von Neumann entropy quantifies the mixedness of a quantum state: an ideal register in a pure state has $S=0$ , while coupling to an environment can increase the entropy of the reduced state. In many-body physics the entanglement entropy of a region — the von Neumann entropy of its reduced state — is the central diagnostic of quantum phases: it typically obeys an area law in gapped ground states, while after a global quench it can grow in time toward volume-law scaling associated with thermalization, and its subleading term detects topological order and counts a critical point's central charge. In holographic quantum gravity, the Ryu–Takayanagi formula relates the entanglement entropy of a boundary region to the area of an associated minimal bulk surface, tying entropy to geometry and reframing the black-hole information problem. And experiments now measure it directly — cold-atom microscopes extract Rényi- $2$ entropies by interfering two copies of a state, and randomized "classical shadow" protocols estimate it on superconducting and trapped-ion processors — putting a once-abstract trace formula on the lab bench.

The physical picture

A density operator is a state of incomplete knowledge, but the incompleteness can have two very different sources. It can arise from classical mixing — you prepared $|0\rangle$ or $|1\rangle$ by a coin flip and forgot which — or from entanglement — the system is one half of a pure global state, definite as a whole but indefinite on its own. The von Neumann entropy does not care which story is true; it reads only the spectrum of $\rho$ . The most spectacular consequence of that indifference is the entanglement picture below: a Bell pair whose two halves each look maximally random, yet whose joint state is completely certain.

Figure

Entanglement in a Bell pair: the joint state |Φ⁺⟩ = (|00⟩ + |11⟩)/√2 is pure with S(ρ_AB) = 0, yet each half is maximally mixed with S(ρ_A) = S(ρ_B) = 1 bit, so the whole is more certain than either part — a genuinely quantum possibility, since for classical random variables H(X,Y) ≥ H(X) and H(X,Y) ≥ H(Y).

Diagonalize $\rho$ in its eigenbasis and it becomes a probability distribution over orthogonal eigenstates — a weighted coin — and $S$ is nothing more than the Shannon entropy of that coin. This is why a pure state has $S=0$ (the coin always lands the same) and $I/2$ has $S=1$ bit (a fair coin). Everything in between is the single-parameter entropy curve of the second figure, which also fixes the number for the worked qubit $\rho=\mathrm{diag}(\tfrac34,\tfrac14)$ used below.

Figure

The qubit entropy curve

S=-\lambda\log_2\lambda-(1-\lambda)\log_2(1-\lambda)

peaks at

1

bit for the maximally mixed state

\lambda=1/2

, falls to

0

for pure states

\lambda=0

1

, and gives

S\approx0.811

bit for the mixed state

\rho=\operatorname{diag}(3/4,1/4)

\lambda=3/4

The core equations

Von Neumann entropy carries Shannon's formula word-for-word into operator language: a probability distribution becomes a density operator, the sum becomes a trace. Because a trace is basis-independent, the definition is a property of the state alone, not of any chosen measurement.

S(\rho)=-\mathrm{Tr}(\rho\log\rho)=-\sum_i\lambda_i\log\lambda_i \qquad(1)

where: $\rho$ — density operator, $\rho\ge0$ and $\mathrm{Tr}\,\rho=1$ ; $\lambda_i$ — eigenvalues of $\rho$ , forming a probability distribution; $\log$ — base 2 for bits or natural for nats, fixed once by convention (we use base 2). The right-hand equality is the whole story: once diagonalized, $S$ is the Shannon entropy of the spectrum, with the convention $0\log 0\equiv 0$ . The entropy is squeezed between two extremes fixed only by the dimension of the space. Perfect knowledge sits at the bottom; maximal ignorance at the top, one $\log d$ per qudit.

0\le S(\rho)\le\log d \qquad(2)

where: $d$ — dimension of $\mathcal{H}$ . The lower bound is saturated by pure states (one eigenvalue $1$ , rest $0$ ); the upper bound only by $\rho=I/d$ (all eigenvalues $1/d$ ). For a qubit $d=2$ , so $0\le S\le 1$ bit. Because $S$ reads only the spectrum, any transformation that shuffles the eigenvectors but leaves the eigenvalues alone cannot change it. Unitary evolution is exactly such a transformation.

S(U\rho U^\dagger)=S(\rho) \qquad(3)

where: $U$ — any unitary. Conjugation rotates the eigenbasis but preserves every eigenvalue, so the entropy of a closed system is constant under unitary dynamics; the entropy of a subsystem can change through interactions and correlations with other degrees of freedom or through measurement. Mixing states can only raise uncertainty: if you forget which of several states you were handed, you know less than you did on average beforehand. This is concavity, the quantum face of "mixing increases entropy".

S\!\left(\sum_i p_i\rho_i\right)\ge\sum_i p_i\,S(\rho_i) \qquad(4)

where: $\{p_i,\rho_i\}$ — an ensemble of states $\rho_i$ with probabilities $p_i$ . Equality holds iff every $\rho_i$ with $p_i>0$ is the same operator. The gap is a sum of relative entropies and reappears as the Holevo $\chi$ . For a joint system the entropy of the whole cannot exceed the sum of its parts — but, unlike the classical case, it can be strictly less, even zero, while the parts are highly entropic.

S(\rho_{AB})\le S(\rho_A)+S(\rho_B) \qquad(5)

where: $\rho_{AB}$ — a joint state on $\mathcal{H}_A\otimes\mathcal{H}_B$ ; $\rho_A=\mathrm{Tr}_B\,\rho_{AB}$ , $\rho_B=\mathrm{Tr}_A\,\rho_{AB}$ — its reduced states. Equality holds iff $\rho_{AB}=\rho_A\otimes\rho_B$ . The difference between the two sides is precisely the quantum mutual information, $I(A:B)=S(\rho_A)+S(\rho_B)-S(\rho_{AB})\geq 0$ . The classical inequality $H(X)\le H(X,Y)$ — "the whole is at least as uncertain as a part" — simply fails in quantum mechanics. What survives is a two-sided triangle bound on the marginal entropies.

|S(\rho_A)-S(\rho_B)|\le S(\rho_{AB}) \qquad(6)

This is the Araki–Lieb inequality. For a globally pure state $S(\rho_{AB})=0$ , so the bound forces $S(\rho_A)=S(\rho_B)$ — the equal marginal entropies of Derivation 2, the closest quantum echo of a classical fact that no longer holds outright. Strong subadditivity states that the entropy of the whole plus that of the middle subsystem never exceeds the sum of the two overlapping bipartite entropies.

S(\rho_{ABC})+S(\rho_B)\le S(\rho_{AB})+S(\rho_{BC}) \qquad(7)

where: $\rho_{ABC}$ — a tripartite state, all marginals by partial trace. This is strong subadditivity (Lieb–Ruskai, 1973), equivalent to the monotonicity of relative entropy under channels — the keystone from which the data-processing inequality, the Holevo bound, and channel-capacity theorems descend. For independent copies the entropy is extensive — the fact that makes the resource counting behind Schumacher compression work.

S(\rho^{\otimes n})=n\,S(\rho) \qquad(8)

where: $\rho^{\otimes n}$ — $n$ identical, uncorrelated copies. So $S(\rho)$ is a rate — qubits per signal — exactly as $H$ is bits per symbol. Finally, the von Neumann entropy is the central member of a one-parameter family of Rényi entropies, recovered as the order tends to one.

S(\rho)=\lim_{\alpha\to1}\frac{1}{1-\alpha}\log\mathrm{Tr}(\rho^\alpha) \qquad(9)

where: $\alpha$ — Rényi order; $S_\alpha(\rho)=\tfrac{1}{1-\alpha}\log\mathrm{Tr}\,\rho^\alpha$ . Special orders include $S_0=\log\operatorname{rank}\rho$ , $S_2=-\log\operatorname{Tr}\rho^2$ , and $S_\infty=-\log\lambda_{\max}$ ; each captures different spectral information, while the limit $S_1=S$ gives the von Neumann entropy.

Derivation AWhy the entropy equals the Shannon entropy of the eigenvalues

The definition $S(\rho)=-\mathrm{Tr}(\rho\log\rho)$ contains a function of an operator, $\log\rho$ , and at first sight looks as though it might depend on the basis used to compute the trace or on how the logarithm of a matrix is even defined. The goal here is to show that neither worry is real: for any density operator the whole expression reduces to the Shannon entropy $-\sum_i \lambda_i \log \lambda_i$ of its eigenvalue distribution, with the convention $0\log 0=0$ .

Step 1. Because $\rho$ is Hermitian and positive semidefinite with unit trace, the spectral theorem gives an orthonormal eigenbasis and $\rho=\sum_i\lambda_i|i\rangle\langle i|$ with $\lambda_i\ge0$ and $\sum_i\lambda_i=1$ . Why: the eigenvalues are automatically a probability distribution, which is what makes a Shannon entropy of them meaningful in the first place.

Step 2. A function of a Hermitian operator acts eigenvalue-by-eigenvalue, so $\log\rho=\sum_i(\log\lambda_i)\,|i\rangle\langle i|$ , and hence $\rho\log\rho=\sum_i\lambda_i\log\lambda_i\,|i\rangle\langle i|$ , using the convention $0\log0\equiv0$ (justified by $\lambda\log\lambda\to0$ as $\lambda\to0^+$ ). Why: this is the definition of a matrix function via the spectral decomposition — it is not an approximation but an identity.

Step 3. The operator $\rho\log\rho$ is already diagonal in the eigenbasis, so its trace is the sum of diagonal entries: $\mathrm{Tr}(\rho\log\rho)=\sum_i\lambda_i\log\lambda_i$ , giving $S(\rho)=-\sum_i\lambda_i\log\lambda_i$ . Why: the trace is basis-independent, so we may evaluate it in the most convenient basis — the eigenbasis — where the answer is manifestly the Shannon entropy $H(\{\lambda_i\})$ . *For example, $\rho=\mathrm{diag}(\tfrac34,\tfrac14)$ gives $-\tfrac34\log_2\tfrac34-\tfrac14\log_2\tfrac14=2-\tfrac34\log_2 3=0.8113$ bit, and $I/2$ gives $\log_2 2=1$ bit — the two marked points on the qubit entropy curve.*

Go deeper · Derivation B · 1 minSubadditivity from the non-negativity of relative entropy

Derivation BSubadditivity from the non-negativity of relative entropy

Subadditivity — that a joint state is no more entropic than the sum of its parts — is the workhorse behind quantum mutual information, and it follows in a few lines from one master inequality: Klein's inequality states that $D(\rho\Vert\sigma)\geq 0$ , with equality iff $\rho=\sigma$ ; when $\operatorname{supp}\rho\subseteq\operatorname{supp}\sigma$ , $D(\rho\Vert\sigma)=\operatorname{Tr}(\rho\log\rho)-\operatorname{Tr}(\rho\log\sigma)$ , and otherwise $D(\rho\Vert\sigma)=+\infty$ . The goal is to apply this with a well-chosen reference state so the relative entropy collapses into exactly the subadditivity deficit, revealing that deficit to be the quantum mutual information and therefore non-negative.

Step 1. Take the joint state $\rho_{AB}$ and compare it to the product of its own marginals, $\sigma=\rho_A\otimes\rho_B$ . Why: the product of marginals is the "no-correlation" reference, so measuring $\rho_{AB}$ against it should quantify exactly how correlated the two parts are.

Step 2. Expand the relative entropy: $D(\rho_{AB}\|\rho_A\otimes\rho_B)=-S(\rho_{AB})-\mathrm{Tr}\big(\rho_{AB}\log(\rho_A\otimes\rho_B)\big)$ , using $\mathrm{Tr}(\rho_{AB}\log\rho_{AB})=-S(\rho_{AB})$ . Why: this isolates the joint entropy in the first term; the second term still needs to be reduced to marginal entropies.

Step 3. Since $\rho_A\otimes I$ and $I\otimes\rho_B$ commute, the logarithm of the product splits: $\log(\rho_A\otimes\rho_B)=\log\rho_A\otimes I+I\otimes\log\rho_B$ . Tracing each piece against $\rho_{AB}$ uses only the marginals, e.g. $\mathrm{Tr}\big(\rho_{AB}(\log\rho_A\otimes I)\big)=\mathrm{Tr}(\rho_A\log\rho_A)=-S(\rho_A)$ . Why: partial trace is precisely the operation that turns a joint expectation of a one-sided operator into a marginal expectation, so the cross terms become marginal entropies.

Step 4. Assemble the pieces: $D(\rho_{AB}\|\rho_A\otimes\rho_B)=S(\rho_A)+S(\rho_B)-S(\rho_{AB})$ . Why: the relative-entropy distance from a state to the product of its marginals is literally the entropy deficit — this identity is what names the deficit.

Step 5. Apply Klein's inequality $D\ge0$ : $S(\rho_A)+S(\rho_B)-S(\rho_{AB})\ge0$ , which is subadditivity, with equality iff $\rho_{AB}=\rho_A\otimes\rho_B$ . Why: the deficit is a relative entropy and relative entropies are non-negative, so the inequality is forced; the equality condition identifies product states as precisely the states that saturate subadditivity. *The deficit $S(\rho_A)+S(\rho_B)-S(\rho_{AB})$ equals the quantum mutual information $I(A{:}B)$ ; for any product state $I=0$ and (5) holds with equality.*

Go deeper · Derivation C · 1 minEqual marginal entropies of a pure bipartite state

Derivation CEqual marginal entropies of a pure bipartite state

The most quantum feature of $S$ is that a globally pure — hence zero-entropy — state can have highly entropic parts, and moreover the two parts always carry identical entropy no matter how lopsided their dimensions. The goal is to prove $S(\rho_A)=S(\rho_B)$ for any pure $|\psi\rangle_{AB}$ using the Schmidt decomposition, and to see that this common value — the entropy of entanglement — measures how strongly the two subsystems are correlated.

Step 1. Every pure bipartite state admits a Schmidt decomposition $|\psi\rangle_{AB}=\sum_k c_k\,|a_k\rangle_A|b_k\rangle_B$ with orthonormal $\{|a_k\rangle\}$ , $\{|b_k\rangle\}$ and Schmidt coefficients $c_k\ge0$ , $\sum_k|c_k|^2=1$ . Why: this single-sum form (a consequence of the singular-value decomposition of the coefficient matrix) is what makes both reduced states diagonal at once.

Step 2. Trace out $B$ : $\rho_A=\mathrm{Tr}_B|\psi\rangle\langle\psi|=\sum_k|c_k|^2\,|a_k\rangle\langle a_k|$ , a diagonal operator with eigenvalues $|c_k|^2$ . Why: the orthonormality of the Schmidt vectors kills all cross terms, so the reduced state is already in spectral form.

Step 3. Trace out $A$ instead: $\rho_B=\sum_k|c_k|^2\,|b_k\rangle\langle b_k|$ , with the same eigenvalues $|c_k|^2$ , only carried by different eigenvectors. Why: the Schmidt coefficients are shared symmetrically between the two sides, so the two marginals differ only by a basis change.

Step 4. Since $S$ depends only on the spectrum, $S(\rho_A)=-\sum_k|c_k|^2\log|c_k|^2=S(\rho_B)$ , and this vanishes iff a single $c_k=1$ , i.e. the state factorizes. Why: equal spectra force equal entropy, so the entanglement entropy is a genuinely symmetric, basis-independent measure of correlation. *For instance, $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$ has Schmidt coefficients $|c_k|^2=\tfrac12$ , giving $S(\rho_A)=S(\rho_B)=-2\cdot\tfrac12\log_2\tfrac12=1$ bit while $S(\rho_{AB})=0$ — exactly the values shown in the first figure.*

Limiting cases

The four corners below pin down the range of $S$ and characterize what it does and does not measure.

Case	Result	Meaning
Pure state	$\rho=\vert\psi\rangle\langle\psi\vert \Rightarrow S=0$	one eigenvalue $1$ , all others $0$ with $0\log 0=0$ ; the state is pure and the entropy attains its minimum value
Maximally mixed	$\rho=I/d \Rightarrow S=\log d$	all $d$ eigenvalues equal $1/d$ ; saturates the upper bound, maximal ignorance
Classical / diagonal	$\rho=\sum_x p_x\vert x\rangle\langle x\vert \Rightarrow S=H(\{p_x\})$	on commuting states $S$ collapses to the Shannon entropy — the classical embedding
Bipartite pure state	$\vert\psi\rangle_{AB}\Rightarrow S(\rho_A)=S(\rho_B)$	equal Schmidt spectra give equal marginal entropy — the entropy of entanglement

Common trapThe entropy is a property of the state, not of any ensemble that prepares it

It is tempting to say "I prepared $|0\rangle$ or $|+\rangle$ with equal probability, so the entropy is $1$ bit." It is not. Many different ensembles $\{p_i,|\psi_i\rangle\}$ produce the very same $\rho$ , and they are physically indistinguishable. The von Neumann entropy reads only $\rho$ , i.e. only its eigenvalues, and is generally strictly smaller than the Shannon entropy $H(\{p_i\})$ of your mixing probabilities, with equality only when the $|\psi_i\rangle$ are mutually orthogonal. For the equal mixture of $|0\rangle$ and $|+\rangle$ the true entropy is $0.60$ bit, not $1$ bit (worked below), because the two states overlap. The Shannon entropy $H(\{p_i\})$ quantifies uncertainty in the preparation labels, whereas $S(\rho)$ depends only on the resulting density operator; nonorthogonality can therefore make $S(\rho)<H(\{p_i\})$ .

Worked examples

1 · Entropy of a mixed qubit (confidence builder)

Compute $S(\rho)$ in bits for $\rho=\tfrac34|0\rangle\langle0|+\tfrac14|1\rangle\langle1|$ , and compare it to the maximally mixed qubit.

Step 1. The state is already diagonal, so its eigenvalues are read straight off: $\lambda_1=\tfrac34$ , $\lambda_2=\tfrac14$ . By equation (1), $S(\rho)=-\tfrac34\log_2\tfrac34-\tfrac14\log_2\tfrac14$ .

Step 2. Evaluate: $\log_2\tfrac34=\log_23-2=-0.415$ and $\log_2\tfrac14=-2$ , so $S=-\tfrac34(-0.415)-\tfrac14(-2)=0.311+0.500$ . In closed form $S=2-\tfrac34\log_2 3=\mathbf{0.811\ bit}$ .

Step 3. Compare to $I/2$ , eigenvalues both $\tfrac12$ : $S(I/2)=-2\cdot\tfrac12\log_2\tfrac12=\mathbf{1\ bit}$ . The biased state is more predictable, so it carries less entropy — the amber point below the green peak on the entropy curve. A qubit's entropy runs from 0 (pure) to 1 bit (I/2); this state sits at 0.811 bit.

2 · Entanglement entropy of a partially entangled pair

A source emits the pure state $|\psi\rangle_{AB}=\sqrt{0.8}\,|00\rangle+\sqrt{0.2}\,|11\rangle$ . Find $S(\rho_{AB})$ , the entanglement entropy $S(\rho_A)$ , and check the Araki–Lieb inequality.

Step 1. The global state is pure, so $S(\rho_{AB})=\mathbf{0}$ by definition — a single Schmidt sum means one nonzero eigenvalue.

Step 2. The expression is already in Schmidt form with coefficients $c_0=\sqrt{0.8}$ , $c_1=\sqrt{0.2}$ , so tracing out $B$ gives $\rho_A=0.8\,|0\rangle\langle0|+0.2\,|1\rangle\langle1|$ (Derivation 2). Its eigenvalues are $0.8$ and $0.2$ .

Step 3. Entanglement entropy: $S(\rho_A)=-0.8\log_2 0.8-0.2\log_2 0.2=0.2575+0.4644=\mathbf{0.722\ bit}$ . By symmetry $S(\rho_B)=0.722$ bit as well — the two halves share one spectrum.

Step 4. Araki–Lieb check: $|S(\rho_A)-S(\rho_B)|=0\le S(\rho_{AB})=0$ , satisfied with equality. This has no classical analogue: for classical variables $H(X,Y)\geq H(X)$ , whereas here $S(\rho_{AB})=0<S(\rho_A)=0.722$ bit, corresponding to a negative quantum conditional entropy. The pair is perfectly known ( $S_{AB} = 0$ ) yet each qubit carries 0.722 bit of entanglement entropy.

3 · The Holevo χ of two non-orthogonal states

Alice sends $|0\rangle$ or $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$ , each with probability $\tfrac12$ . How many bits about her choice can Bob extract? Compute the Holevo bound $\chi=S(\rho)-\sum_x p_xS(\rho_x)$ .

Step 1. Each signal is pure, so $S(\rho_x)=0$ and the bound reduces to $\chi=S(\rho)$ , where $\rho=\tfrac12|0\rangle\langle0|+\tfrac12|+\rangle\langle+|$ is the average state.

Step 2. In the computational basis $\rho=\tfrac12\begin{pmatrix}1&0\\0&0\end{pmatrix}+\tfrac12\begin{pmatrix}\tfrac12&\tfrac12\\\tfrac12&\tfrac12\end{pmatrix}=\begin{pmatrix}\tfrac34&\tfrac14\\\tfrac14&\tfrac14\end{pmatrix}$ . This has trace $1$ and determinant $\tfrac{3}{16}-\tfrac1{16}=\tfrac18$ .

Step 3. Eigenvalues solve $\lambda^2-\lambda+\tfrac18=0$ , giving $\lambda_\pm=\tfrac12\big(1\pm\sqrt{1-\tfrac12}\big)=\tfrac12\big(1\pm\tfrac1{\sqrt2}\big)$ , i.e. $\lambda_+=0.8536$ , $\lambda_-=0.1464$ .

Step 4. Entropy: $S(\rho)=-0.8536\log_2 0.8536-0.1464\log_2 0.1464=\mathbf{0.601\ bit}$ , so $\chi=0.601$ bit. Bob can never recover a full bit about Alice's choice: because $|0\rangle$ and $|+\rangle$ overlap, the accessible information is capped at 0.60 bit, not 1 — the mixing entropy is $1$ bit, but the accessible classical information is bounded above by the Holevo quantity $\chi=0.601$ bit.

Physics of the portrait

The formula, the density matrix it acts on, and the idea of a "state operator" all trace to one person, working out how to make quantum mechanics and thermodynamics coexist.

Origins · 1 minJohn von Neumann and how this was discovered

John von Neumann

1903–1957 · mathematical foundations of quantum mechanics and quantum statistical mechanics

John von Neumann built the modern mathematical scaffolding of quantum theory, and the entropy that bears his name was a byproduct of a deeper project: putting quantum statistical mechanics on rigorous footing. In a 1927 paper and then in his 1932 treatise Mathematische Grundlagen der Quantenmechanik, he introduced the density operator $\rho$ to describe both statistical mixtures and subsystems of larger pure states, and defined $S=-\mathrm{Tr}(\rho\log\rho)$ (with a Boltzmann constant in front) as the thermodynamic entropy of a quantum ensemble. His motivation was a thought experiment about extracting work from a gas of quantum particles: he needed an entropy that respected the indistinguishability of non-orthogonal states, and the trace formula was the unique answer consistent with the second law. He proved its basic properties — non-negativity, unitary invariance, monotonicity under measurement — decades before anyone connected it to information. That connection waited for Shannon's 1948 coding theory and for Schumacher's 1995 proof that von Neumann's quantity is the exact qubit compression rate. The name "entropy" that von Neumann reportedly suggested to Shannon "because no one knows what entropy really is" he had, twenty years earlier, already pinned to the quantum operator that now anchors a field.

History

1927

— von Neumann introduces the density operator and defines S = −Tr(ρ log ρ)

1932

— his Mathematical Foundations of Quantum Mechanics systematizes quantum entropy

1970

— Araki and Lieb prove the triangle inequality for marginal entropies

1973

— Lieb and Ruskai prove strong subadditivity

1995

— Schumacher gives S(ρ) its operational meaning as the qubit compression rate

Going further

Texts. Nielsen & Chuang, Quantum Computation and Quantum Information, Ch. 11 (Thm 11.7 concavity, Thm 11.8 subadditivity, §11.3.4 Araki–Lieb, Thm 11.15 strong subadditivity) · Preskill, Lecture Notes on Quantum Information, Ch. 10 · Watrous, The Theory of Quantum Information, Ch. 5 · Wilde, Quantum Information Theory, Ch. 11 (Klein's inequality, Thm 11.8.1) · Kitaev, Shen & Vyalyi, Classical and Quantum Computation · Mermin, Quantum Computer Science. The three boxed derivations reduce each inequality to the non-negativity of relative entropy (Klein's inequality) and to the Schmidt decomposition. Key numerical anchors: (1) $S(I/d)=\log d$ and $S=0$ for any pure state; (2) the qubit value $2-\tfrac34\log_2 3=0.8113$ bit; (3) the Bell-state values $S(\rho_A)=1$ , $S(\rho_{AB})=0$ ; (4) the partially entangled marginal $S=-0.8\log_2 0.8-0.2\log_2 0.2=0.7219$ bit; (5) the Holevo example eigenvalues $\tfrac12(1\pm1/\sqrt2)$ giving $\chi=0.601$ bit. Strong subadditivity is the Lieb–Ruskai (1973) theorem; Araki–Lieb is the Araki–Lieb (1970) result.

Prerequisites. Shannon Entropy and Classical Information Measures · Density Operators and the Spectral Theorem · Partial Trace and Reduced States. Leads to. Quantum Relative Entropy · Quantum Conditional Entropy and Mutual Information · Strong Subadditivity of Quantum Entropy · Schumacher Quantum Data Compression · The Holevo Bound.

Ask the tutor: Ask me why a Bell pair has zero joint entropy but one full bit in each half, or to compute the entanglement entropy of any state you name.

Is the von Neumann entropy the same as thermodynamic entropy?

For a system in thermal equilibrium, yes, up to Boltzmann's constant. The Gibbs state $\rho=e^{-\beta H}/Z$ maximizes $S(\rho)$ at fixed mean energy, and its von Neumann entropy equals the thermodynamic entropy $S_\text{thermo}=k_B\ln 2\cdot S(\rho)$ (with $S$ in bits). This is no coincidence: von Neumann built the quantity to be the entropy of a quantum ensemble, and the maximum-entropy principle that selects thermal states is literally the maximization of $S$ under constraints. Away from equilibrium the identification loosens — $S(\rho)$ of a closed system is constant under unitary evolution while thermodynamic entropy appears to rise, a tension resolved by using coarse-grained or subsystem entropies rather than the global one.

Why does S depend only on the eigenvalues and not on how I prepared the state?

Because $S(\rho)=-\mathrm{Tr}(\rho\log\rho)$ is a trace of a function of $\rho$ , and a trace is invariant under any change of basis. Two ensembles that yield the same density operator are physically indistinguishable — no measurement, however clever, can tell them apart — so it would be inconsistent for a physical quantity to depend on which one you "really" used. The eigenvalues encode the mutually exclusive alternatives the state genuinely contains; the labels you attach during preparation may be more numerous (carrying more Shannon entropy), but the extra structure is not recoverable, so $S$ ignores it. This is exactly why the Holevo $\chi$ can be strictly below the mixing entropy.

How can a whole system have zero entropy while its parts do not?

This is the defining quantum surprise, and it is entirely due to entanglement. If $|\psi\rangle_{AB}$ is pure, the joint state is perfectly known, so $S(\rho_{AB})=0$ ; yet tracing out $B$ leaves $A$ mixed whenever the two are entangled, so $S(\rho_A)>0$ . Classically this is impossible, because $H(X,Y)\ge H(X)$ . The resolution is that the "missing" certainty is not local information at all — it lives in the correlations between $A$ and $B$ . Knowing the whole means knowing how $A$ and $B$ relate, not what each is doing on its own. The von Neumann entropy of a part therefore measures entanglement with the rest of the world, not intrinsic randomness.

What is the difference between von Neumann entropy and Rényi entropies?

They are members of one family, $S_\alpha(\rho)=\tfrac{1}{1-\alpha}\log\mathrm{Tr}\,\rho^\alpha$ , and the von Neumann entropy is the $\alpha\to1$ limit. The other orders trade operational sharpness for measurability: $S_2=-\log\mathrm{Tr}\,\rho^2$ (collision entropy) is fixed by the purity and can be estimated by interfering two copies of a state, and $S_\infty=-\log\lambda_{\max}$ (min-entropy) governs randomness extraction. But only $S_1$ has the additivity and tight source-coding meaning that make it the right rate for i.i.d. compression and channel capacity — so people measure an accessible $S_\alpha$ and use it to bound $S$ , since $S_\alpha$ is non-increasing in $\alpha$ .

Can the von Neumann entropy decrease?

Under closed-system (unitary) evolution, never — it is exactly conserved, because unitaries preserve the spectrum. It can decrease only through processes that are not unitary on the system alone: a measurement whose outcome you record, or an interaction with a cold environment that carries entropy away (the basis of cooling and error correction). Even then the total entropy of system-plus-environment does not fall — you are simply moving it across the cut. This is why erasing a qubit, lowering its entropy by up to $1$ bit, must dump at least $k_BT\ln2$ of heat into the surroundings: Landauer's principle. Entropy is conserved globally and only shuffled locally.

Why is strong subadditivity considered so important?

Because almost every quantitative theorem in quantum information rests on it, and because it was genuinely hard to prove. Strong subadditivity, $S(\rho_{ABC})+S(\rho_B)\le S(\rho_{AB})+S(\rho_{BC})$ , is equivalent to the statement that quantum relative entropy cannot increase under a channel — the data-processing inequality. From that one fact flow the monotonicity of mutual information, the Holevo bound, and the converse theorems that establish channel capacities. Classically the analogue is nearly trivial, but quantum mechanically it resisted proof until Lieb and Ruskai used deep operator-convexity results in 1973. It is, in a real sense, the single mathematical fact that makes quantum Shannon theory hold together.