advanced27 min studyLesson 424 in Physics

Mutual information and relative entropy

Quantify total correlations with relative entropy and derive the direction and assumptions of quantum data processing.

On this page

What you’ll understand

  • Compute mutual information from three consistent entropy spectra.
  • Relate mutual information to relative entropy from a product of marginals.
  • Apply data processing to local channels without equating correlation with entanglement.

Correlations mean that knowing a joint state gives information absent from a product of its local descriptions. Quantum mutual information measures this total correlation. It includes classical correlation and entanglement, so a positive value alone does not certify either resource separately. Relative entropy supplies a second description: compare the actual state with the product state that has the same marginals. Both viewpoints use base-two logarithms and information units of bits throughout this lesson.

Compare states with an asymmetric divergence

Quantum relative entropy compares a state rho with a reference sigma using the trace of rho times the difference of their logarithms. Its finite value requires the support of rho to lie inside the support of sigma. Otherwise it is positive infinity: the reference assigns zero support to an event the actual state permits. Zero eigenvalues outside the support cause no contribution. Diagonal states reduce to the classical sum of probability times the logarithm of a probability ratio.

The divergence is nonnegative and vanishes exactly when the normalized states agree. It is not symmetric and is not a distance metric, so there is no general triangle inequality to invoke. Its operational significance concerns distinguishability in repeated asymmetric hypothesis tests, with carefully specified error criteria. For elementary calculations, first check supports and a common eigenbasis. If the matrices do not commute, separately diagonalize them to define their logarithms; taking logarithms of individual matrix entries is invalid.

Drel(ρ∥σ)=Tr⁡ρ(log⁡2ρ−log⁡2σ)≥0D_{\rm rel}(\rho\|\sigma)=\operatorname{Tr}\rho(\log_2\rho-\log_2\sigma)\ge0
The divergence is in bits for normalized finite-dimensional density operators. It is infinite when support containment fails; D_rel is distinct from half-trace-norm distance D.↔ Scroll wide equations horizontally.

A product reference exposes total correlation

For declared factors A and B, take the joint state and compare it with the tensor product of its partial traces. Expanding the logarithm of a tensor product gives the sum of the local logarithms on their respective factors. The partial-trace identity then turns the two local expectation terms into local entropies. This proves that mutual information equals the sum of the marginal entropies minus the joint entropy and is nonnegative.

A fair classical mixture of zero-zero and one-one has local entropies one bit and joint entropy one bit, hence mutual information one bit. A Bell pair has the same marginals but joint entropy zero, hence mutual information two bits. The latter does not mean that measuring one qubit releases two ordinary classical bits. Quantum mutual information counts total correlation before a measurement and can exceed the classical bound set by either marginal entropy.

I(A:B)ρ=S(ρA)+S(ρB)−S(ρAB)=Drel(ρAB∥ρA⊗ρB)I(A:B)_\rho=S(\rho_A)+S(\rho_B)-S(\rho_{AB})=D_{\rm rel}(\rho_{AB}\|\rho_A\otimes\rho_B)
The marginal states are obtained by tracing the opposite tensor factor. I is in bits, equals zero exactly for a product state, and does not by itself measure entanglement.↔ Scroll wide equations horizontally.

Local loss cannot create total correlation

The data-processing theorem says that applying the same completely positive, trace-preserving map to the two arguments cannot increase relative entropy. Applying a channel locally on B therefore cannot increase mutual information between A and the channel output. The comparison product remains the product of the new marginals because the channel is local. A joint operation acting on both factors can create correlations, so the word local is essential to this conclusion.

Dephasing a Bell pair in the computational basis removes its off-diagonal terms and leaves the classically correlated mixture. Its mutual information drops from two bits to one, while its local states remain unchanged. Conditional entropy, defined as joint minus conditioning-system entropy, correspondingly changes from negative one bit to zero. Negative quantum conditional entropy is allowed. Postselecting on an announced outcome defines a different conditional task; a selected branch need not satisfy the unconditioned comparison without accounting for its classical record.

Erase coherence without changing marginals

Start from a Bell pair on A:B. Completely dephase B in the zero–one basis. Compute mutual information before and after, and both conditional entropies S(A|B).

  1. Before dephasing the joint state is pure, so S(AB)=0, while both marginal states are I/2 and each has entropy one bit. Therefore I(A:B)=1+1−0=2 bits.

  2. Dephasing B removes the two Bell off-diagonal entries. The remaining joint state has two eigenvalues one half and two zeros, giving S(AB)=1 bit; both marginals remain I/2.

  3. After dephasing I(A:B)=1+1−1=1 bit. Conditional entropy S(A|B)=S(AB)−S(B) changes from −1 to 0 bits, consistent with losing quantum coherence through a local channel.

The result

Mutual information falls from 2 to 1 bit. Identical marginal states do not imply unchanged joint correlations.

Explore the model

Interactive lab

Interactive lab · markov

A classical channel forgets its initial bit

Reset and set p=0.5 and q=0.5. Before stepping the chain starts in state zero. Predict that one Step, a single transition, makes the state-one probability 0.5 and all later probabilities remain 0.5. Compare a fresh Reset with both controls p=0 and q=0, which retains state zero. This illustrates classical information loss only: the lab has one fixed initial distribution and no joint-state, mutual-information, dephasing, or quantum-channel readout.

A classical channel forgets its initial bitExact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA map that replaces either input by the same distribution cannot preserve information about that input.

Check your understanding

Which operation is guaranteed not to increase I(A:B) for every input state?

Try it yourself

For ρ=diag⁡(3/4,1/4)\rho=\operatorname{diag}(3/4,1/4) and σ=I/2\sigma=I/2, compute Drel(ρ∥σ)D_{\rm rel}(\rho\|\sigma) and the reverse divergence. Are they equal?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use probability ratios separately in each direction and keep logarithms in base two.

Compare with the solution

The forward divergence is (3/4)log⁡2(3/2)+(1/4)log⁡2(1/2)=1−h2(1/4)≈0.188722(3/4)\log_2(3/2)+(1/4)\log_2(1/2)=1-h_2(1/4)\approx0.188722 bits. The reverse is (1/2)log⁡2(2/3)+(1/2)log⁡22=(1/2)log⁡2(4/3)≈0.207519(1/2)\log_2(2/3)+(1/2)\log_2 2=(1/2)\log_2(4/3)\approx0.207519 bits. Both supports are full and both answers are nonnegative, but they differ. Reversing a reference comparison is therefore a substantive change, even for commuting states.

Take it with you

  • Mutual information compares a joint state with the product of its marginals.
  • Classical correlation can give positive mutual information without entanglement.
  • Relative entropy and mutual information contract under the relevant deterministic channels.

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