Subsystem Entropy

Subsystem entropy is the von Neumann entropy of a reduced density operator. For a bipartite state $\rho_{AB}$,

$$\rho_A = \operatorname{Tr}_B\rho_{AB}, \qquad \rho_B = \operatorname{Tr}_A\rho_{AB},$$

and the subsystem entropies are

$$S_A = S(\rho_A), \qquad S_B = S(\rho_B),$$

where

$$S(\rho) = -\operatorname{Tr}(\rho\log_2\rho).$$

The same formula supports several interpretations. If the joint state $\rho_{AB}$ is pure, $S_A=S_B$ measures bipartite entanglement across the $A$ versus $B$ split. If the joint state is mixed, $S_A$ and $S_B$ are local mixedness measures; they can include classical uncertainty, local noise, environmental entanglement, and ordinary correlation. They are not automatically entanglement measures. For the first density-operator introduction to $S(\rho)$, see Entropy Overview.

Helpful background. Reduced density operators identifies the subsystem state; spectral decomposition supplies the eigenvalues used to calculate its entropy.

Von Neumann entropy from eigenvalues

Let $\rho$ be a density operator on a finite-dimensional Hilbert space. If its spectral decomposition is

$$\rho = \sum_\alpha p_\alpha \lvert\alpha\rangle\langle\alpha\rvert,$$

then

$$S(\rho) = -\sum_\alpha p_\alpha\log_2 p_\alpha,$$

with the convention

$$0\log_20=0.$$

The logarithm base fixes the units:

• $\log_2$ gives entropy in bits;
• $\ln$ gives entropy in nats.

This page uses base $2$ throughout. In thermodynamics, $k_{\mathrm B}\ln2$ converts this dimensionless entropy in bits to physical entropy units when the density operator describes the thermodynamic state.

Von Neumann entropy depends only on the eigenvalues of $\rho$. It is unchanged by a unitary change of basis:

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

It satisfies

$$S(\rho)\ge0.$$

For a $d$-dimensional density operator,

$$S(\rho)\le \log_2 d,$$

with equality for the maximally mixed state $\rho=I/d$.

Entropy is not the uncertainty of an arbitrary measurement

A pure density operator has spectrum $(1,0,\ldots)$ and entropy zero. A maximally mixed state $I/d$ has $d$ equal eigenvalues and entropy $\log_2d$.

The probabilities in a chosen measurement basis are not generally the eigenvalues. For $|+\rangle$, the von Neumann entropy is zero, but a $Z$ measurement gives two equally likely outcomes and one bit of Shannon entropy. For a complete rank-one projective measurement, outcome entropy is at least $S(\rho)$, with equality in an eigenbasis. Arbitrary POVMs need not satisfy that simple comparison; a one-outcome POVM always has zero outcome entropy.

For a qubit with eigenvalues $p$ and $1-p$, define

$$H_2(p)=-p\log_2p-(1-p)\log_2(1-p).$$

It vanishes at $p=0,1$ and reaches one bit at $p=1/2$.

As a worked example, consider

$$\rho=\frac12|0\rangle\langle0|+\frac12|+\rangle\langle+| =\begin{pmatrix}3/4&1/4\\1/4&1/4\end{pmatrix}.$$

Its determinant is $1/8$, so its eigenvalues are

$$\lambda_\pm=\frac{1\pm1/\sqrt2}{2}.$$

Consequently $S(\rho)=H_2(\lambda_+)\approx0.601$ bits. The preparation uses a fair classical label with entropy one bit, while a $Z$ measurement gives entropy $H_2(3/4)\approx0.811$ bits. These three numbers answer different questions. Nonorthogonal ensemble members are not perfectly distinguishable, so the preparation-label entropy is not intrinsic to the density operator.

Entropy of a Subsystem

For a joint state $\rho_{AB}$, the entropy of subsystem $A$ is

$$S_A = -\operatorname{Tr}_A(\rho_A\log_2\rho_A), \qquad \rho_A=\operatorname{Tr}_B\rho_{AB}.$$

This number is a property of the local state $\rho_A$. It tells how mixed the local density operator is. It does not, by itself, tell why $\rho_A$ is mixed.

The reason matters. The same local state

$$\rho_A = \frac12 I$$

can arise from:

• a qubit entangled with another qubit in a pure Bell state;
• a separable mixture with a shared classical label;
• a product state $\rho_A\otimes\rho_B$ with no correlation at all;
• an open-system state correlated with an environment outside the model.

Therefore subsystem entropy must be interpreted together with the joint state, the subsystem split, and the physical preparation.

Pure Bipartite States

For a pure bipartite state,

$$\rho_{AB} = \lvert\Psi\rangle\langle\Psi\rvert_{AB},$$

the Schmidt decomposition gives

$$\lvert\Psi\rangle = \sum_r s_r \lvert u_r\rangle_A \lvert v_r\rangle_B, \qquad \sum_r s_r^2=1.$$

The reduced states are

$$\rho_A = \sum_r s_r^2 \lvert u_r\rangle\langle u_r\rvert_A, \qquad \rho_B = \sum_r s_r^2 \lvert v_r\rangle\langle v_r\rvert_B.$$

Thus $\rho_A$ and $\rho_B$ have the same nonzero eigenvalues. If

$$p_r=s_r^2,$$

then

$$S_A = S_B = -\sum_r p_r\log_2 p_r.$$

For a pure bipartite state, this common value is the entanglement entropy. Entanglement entropy develops its properties and pure-state interpretation.

In this special pure-state setting:

• $S_A=0$ exactly when $\lvert\Psi\rangle$ is a product state;
• $S_A>0$ exactly when $\lvert\Psi\rangle$ is entangled across the chosen split;
• $S_A=S_B$ because the nonzero reduced-state spectra coincide;
• the maximum possible value is $\log_2 d$, where $d=\min(\dim\mathcal H_A,\dim\mathcal H_B)$.

Three states with identical local entropy

The Bell projector $\rho_{\mathrm{Bell}}=|\Phi^+\rangle\langle\Phi^+|$, the correlated mixture $\rho_{\mathrm{cc}}=(|00\rangle\langle00|+|11\rangle\langle11|)/2$, and the uncorrelated state $I_4/4$ all have $\rho_A=\rho_B=I/2$. Thus each has $S_A=S_B=1$ bit, but their global entropy differs.

Define mutual information $I(A:B)=S_A+S_B-S_{AB}$ and quantum conditional entropy $S(A|B)=S_{AB}-S_B$. In bits:

Joint state	$S_{AB}$	$I(A:B)$	$S(A\mid B)$
Bell projector	$0$	$2$	$-1$
Correlated mixture	$1$	$1$	$0$
Uncorrelated $I_4/4$	$2$	$0$	$1$

For the Bell state, the local entropy is entanglement entropy. For the correlated mixture, it reflects a shared classical label. For $I_4/4$, the same local entropy occurs without any correlation. Local entropy alone cannot distinguish these cases.

Negative quantum conditional entropy has no classical analogue. It certifies entanglement, but some entangled mixed states have nonnegative conditional entropy. It is a difference of state entropies, not the ordinary Shannon uncertainty of a measurement outcome conditioned on another recorded outcome.

Mixed Joint States

When $\rho_{AB}$ is mixed, the subsystem entropies need not be equal. For example,

$$\rho_{AB} = \lvert0\rangle\langle0\rvert_A \otimes \frac12 I_B$$

has

$$S_A=0, \qquad S_B=1$$

in bits.

Even when $S_A=S_B$, the equality does not imply that $\rho_{AB}$ is pure or entangled. The classically correlated state $\rho_{\mathrm{cc}}$ above has equal subsystem entropies and no entanglement.

For mixed joint states, several entropy combinations become useful:

$$I(A:B) = S_A+S_B-S_{AB}$$

measures total correlation, both classical and quantum. Conditional entropy,

$$S(A\vert B) = S_{AB}-S_B,$$

is another useful quantity, especially in quantum information. These quantities are related to entanglement in important ways, but none of them is simply “the entanglement” of an arbitrary mixed state.

For finite-dimensional states, subadditivity and the Araki–Lieb inequality give

$$|S_A-S_B|\le S_{AB}\le S_A+S_B.$$

These are useful checks on a calculation. In infinite dimensions, a trace-class state can have infinite entropy; differences such as $\infty-\infty$ are undefined. Use the appropriate relative-entropy formulation or additional finiteness assumptions before applying the displayed differences.

Relation to Purification

Purification explains why the same entropy formula appears in different contexts. If $\rho_A$ has eigenvalues $p_k$, a spectral purification is

$$\lvert\Psi\rangle_{AR} = \sum_k \sqrt{p_k}\, \lvert k\rangle_A\lvert k\rangle_R.$$

Then

$$\operatorname{Tr}_R \bigl( \lvert\Psi\rangle\langle\Psi\rvert_{AR} \bigr) = \rho_A,$$

and the entanglement entropy of the pure state on $AR$ equals

$$S(\rho_A) = -\sum_k p_k\log_2 p_k.$$

This does not mean every use of $\rho_A$ requires a literal hidden reference system. It means every finite-dimensional mixed state can be represented as the marginal of a larger pure state if that representation is useful.

Common pitfalls

Use eigenvalues, not arbitrary diagonal entries. Basis probabilities can have greater entropy than the density operator because the measurement adds uncertainty relative to that basis.

The pure-state condition changes the interpretation. A positive marginal entropy quantifies entanglement only when the relevant joint state is pure.

Do not subtract divergent entropies. Infinite-dimensional states need additional care even when their trace and normalization are perfectly well defined.

References

• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, chapter 11.
• John Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chapter 5. Author-hosted text.
• Alfred Wehrl, “General properties of entropy,” Reviews of Modern Physics 50, 221–260 (1978). DOI.
• Mark M. Wilde, Quantum Information Theory, 2nd ed., Cambridge University Press, 2017, quantum entropy and its inequalities.

Exercises

1. For $|+\rangle$, compare the von Neumann entropy with the Shannon entropy of $Z$- and $X$-measurement outcomes.

Solution

The density operator is pure, so its eigenvalues are $1,0$ and $S=0$. A $Z$ measurement gives probabilities $1/2,1/2$, hence one bit. An $X$ measurement gives $+$ with certainty, hence zero bits. The same quantum state can have different measurement-outcome entropies.

2. Compute the entropy of the maximally mixed state $\rho_\ast=I/d$.

Solution

The eigenvalues are all $1/d$. Hence

$$S(\rho_\ast) = -\sum_{\alpha=1}^{d} \frac1d\log_2\frac1d = \log_2 d.$$

3. For the Bell state $\lvert\Phi^+\rangle$, compute $S_A$, $S_B$, and $S_{AB}$ in bits.

Solution

The Bell state is pure, so $S_{AB}=0$. Its one-qubit reductions are

$$\rho_A=\rho_B=\frac12 I.$$

Each has eigenvalues $1/2$ and $1/2$, so

$$S_A=S_B=1$$

bit.

4. Compare the Bell state with the classically correlated state $\rho_{\mathrm{cc}}$ using mutual information.

Solution

For the Bell state, $S_A=S_B=1$ and $S_{AB}=0$, so

$$I(A:B) = 1+1-0 = 2$$

bits.

For

$$\rho_{\mathrm{cc}} = \frac12\lvert00\rangle\langle00\rvert + \frac12\lvert11\rangle\langle11\rvert,$$

the nonzero joint eigenvalues are $1/2$ and $1/2$, so $S_{AB}=1$. Since $S_A=S_B=1$,

$$I(A:B) = 1+1-1 = 1$$

bit. The Bell state has pure-state entanglement; $\rho_{\mathrm{cc}}$ has classical correlation but no entanglement.

5. Give an example where $S_A\neq S_B$.

Solution

Take

$$\rho_{AB} = \lvert0\rangle\langle0\rvert_A \otimes \frac12 I_B.$$

Then $\rho_A=\lvert0\rangle\langle0\rvert$, so $S_A=0$, while $\rho_B=I/2$, so $S_B=1$ bit. The equality $S_A=S_B$ is not a general property of mixed joint states.