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

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

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.

Von Neumann Entropy

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

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

with the convention

$$0\log0=0.$$

The logarithm base fixes the units:

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

This page uses base $2$ for the two-qubit examples below.

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,

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

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

Pure and Mixed States

A pure state has density operator

$$\rho = \lvert\psi\rangle\langle\psi\rvert.$$

Its eigenvalues are one $1$ and the rest $0$, so

$$S(\rho) = -1\log1 - \sum_{\alpha>1}0\log0 = 0.$$

Conversely, a finite-dimensional density operator has zero von Neumann entropy exactly when it is pure.

For the maximally mixed state on a $d$-dimensional Hilbert space,

$$\rho_\ast = \frac1d I,$$

all eigenvalues are $1/d$, so

$$S(\rho_\ast) = -d\left(\frac1d\log\frac1d\right) = \log d.$$

For a qubit diagonal in the computational basis,

$$\rho = p\lvert0\rangle\langle0\rvert + (1-p)\lvert1\rangle\langle1\rvert,$$

the entropy in bits is the binary entropy

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

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

Entropy of a Subsystem

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

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

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

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

For a pure bipartite state, this common value is the entanglement entropy. The dedicated Entanglement Entropy page treats that pure-state entanglement measure as its canonical topic.

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 $\log d$, where $d=\min(\dim\mathcal H_A,\dim\mathcal H_B)$.

Three States with the Same Local Entropy

The following two-qubit examples all have maximally mixed one-qubit marginals, so

$$S_A=S_B=1 \quad \text{bit}.$$

Their joint states are nevertheless different.

First, the Bell state

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr)$$

is pure and entangled. Its total entropy is

$$S_{AB}=0,$$

and its mutual information is

$$I(A:B) = S_A+S_B-S_{AB} = 2 \quad \text{bits}.$$

The entropy $S_A=1$ bit is entanglement entropy in this case.

Second, the classically correlated separable state

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

has

$$S_A=S_B=1, \qquad S_{AB}=1, \qquad I(A:B)=1$$

in bits. The local entropy reflects a shared classical record, not entanglement.

Third, the product mixed state

$$\rho_{\mathrm{prod}} = \frac12 I_A\otimes\frac12 I_B$$

has

$$S_A=S_B=1, \qquad S_{AB}=2, \qquad I(A:B)=0$$

in bits. Here the same local entropy occurs with no correlation between $A$ and $B$.

These examples are the safest way to remember the warning: local entropy alone is not a correlation measure and is not a mixed-state entanglement measure.

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.

Relation to Purification

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

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

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 local shadow of a larger pure state if that representation is useful.

Common Mistakes

• Treating $S(\rho_A)>0$ as proof of entanglement when $\rho_{AB}$ is mixed.
• Forgetting that $S_A=S_B$ is guaranteed for pure bipartite joint states, not for arbitrary mixed states.
• Comparing numerical entropies without stating the logarithm base.
• Confusing total entropy $S_{AB}$ with subsystem entropy $S_A$ or $S_B$.
• Interpreting every mixed reduced state as ordinary ignorance about a pure local state.
• Forgetting the convention $0\log0=0$ when a density operator has zero eigenvalues.
• Ignoring the subsystem decomposition before assigning an entropy to a region, register, mode, or particle sector.

Cross-Links

• Reduced Density Operators
• Entropy Overview
• Entropy in Quantum Statistical Mechanics
• Thermal Entropy vs Entanglement Entropy
• Marginals and Correlations
• Purification
• Partial Trace
• Schmidt Decomposition
• Entanglement Entropy
• Renyi Entropies
• Mutual Information
• Bell States
• Classical Correlation versus Entanglement
• Pure Versus Mixed States
• Spectral Decomposition

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.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• M. M. Wilde, Quantum Information Theory, 2nd ed., Cambridge University Press, 2017.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018.

Exercises

1. Show that every pure density operator has zero von Neumann entropy.

Solution

For $\rho=\lvert\psi\rangle\langle\psi\rvert$, the spectrum is one eigenvalue equal to $1$ and all remaining eigenvalues equal to $0$. Therefore

$$S(\rho) = -1\log1 - \sum_{\alpha>1}0\log0 = 0.$$

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

Solution

The eigenvalues are all $1/d$. Hence

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

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.