Education / Quantum Information Theory

Quantum Information Theory · Chapter 03

Entropy and Data Processing

Which information can survive processing, and how does entropy quantify its loss?

In this chapter · 3 sections
  1. 3.1 Entropy and relative entropy
  2. 3.2 Correlations and conditioning
  3. 3.3 Many-copy testing
Reading guide · preparation and approach

Starting point: Chapters 1–2; eigenvalues and elementary logarithms.

Anchor example: A Bell pair and its classically dephased counterpart.

Method: Express information as relative entropy, then apply data processing.

Watch for: Quantum conditional entropy can be negative; entropy is not the uncertainty of every possible measurement.

This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.

3.1 Entropy and relative entropy

3.1.1 Spectra, support, and the reference state

The entropy of a state is \(S(\rho)=-\tr\rho\log_2\rho=-\sum_i\lambda_i\log_2\lambda_i\), with \(0\log0=0\). It measures the spread of the spectrum, not the spread of outcomes in an arbitrary basis. A pure qubit has zero entropy even if an incompatible measurement gives a uniformly random bit. The maximal entropy in dimension \(d\) is \(\log_2d\), reached by \(I/d\).

Relative entropy compares a state with a reference: \(D(\rho\|\sigma)=\tr\rho(\log_2\rho-\log_2\sigma)\) when \(\operatorname{supp}\rho\subseteq\operatorname{supp}\sigma\), and infinity otherwise. It is nonnegative but generally asymmetric and does not obey a triangle inequality. For diagonal states it becomes the classical expression \(\sum_i p_i\log_2(p_i/q_i)\). The support condition is operational: assigning zero probability to an event that actually occurs is a decisive modeling failure.

To evaluate a noncommuting example, diagonalize each matrix separately. The term \(\tr\rho\log\sigma\) depends on how the eigenvectors overlap; subtracting two lists of eigenvalues is not enough. For the special reference \(I/d\), however, \(\log_2(I/d)=-(\log_2d)I\), giving \(D(\rho\|I/d)=\log_2d-S(\rho)\). This identity measures deviation from maximal mixing and will later provide a resource monotone.

Worked check 3.1. Compute \(D(\operatorname{diag}(0.8,0.2)\|I/2)\).

Reasoned solution

The binary entropy is \(h_2(0.2)\approx0.721928\), so the relative entropy is \(1-h_2(0.2)\approx0.278072\) bits.

3.2 Correlations and conditioning

3.2.1 Negative conditional entropy is consistent

For a bipartite state, define \(I(A:B)=S(A)+S(B)-S(AB)\) and \(S(A|B)=S(AB)-S(B)\). Expanding the logarithm of \(\rho_A\otimes\rho_B\) shows \(I(A:B)=D(\rho_{AB}\|\rho_A\otimes\rho_B)\). Thus mutual information is nonnegative and vanishes precisely for product states. It counts total correlation, including classical correlation; it is not an entanglement measure.

A Bell pair has entropies \(S(A)=S(B)=1\), \(S(AB)=0\), giving \(I(A:B)=2\) and \(S(A|B)=-1\). Dephasing one half yields \((|00\rangle\langle00|+|11\rangle\langle11|)/2\), with joint entropy one. Its mutual information is one and conditional entropy zero. The local states have not changed, but coherent correlations have been lost.

The central theorem is data processing: for a channel \(\mathcal N\), \(D(\mathcal N(\rho)\|\mathcal N(\sigma))\le D(\rho\|\sigma)\). We state the full operator inequality without proof; its noncommuting proof is substantially deeper than the diagonal case. Applying it to a channel on one half proves that local processing cannot increase mutual information. Applying it to \(\rho_{ABC}\) and \(\rho_A\otimes\rho_{BC}\), then tracing out \(C\), gives strong subadditivity:

\[I(A:C|B)=S(AB)+S(BC)-S(B)-S(ABC)\ge0.\]

See Watrous, Chapter 5 for entropy inequalities and their proofs. The derivation of strong subadditivity here depends on the stated data-processing theorem, rather than proving that theorem.

Worked check 3.2. For three classical copies of one fair bit, compute \(I(A:C|B)\).

Reasoned solution

Every nonempty subset has entropy one, so \(I(A:C|B)=1+1-1-1=0\). A and C are correlated without conditioning, but B already supplies the common bit.

3.3 Many-copy testing

3.3.1 Relative entropy as an error exponent

Return to the asymmetric test in Chapter 1, now with hypotheses \(\rho^{\otimes n}\) and \(\sigma^{\otimes n}\). Fix a tolerated miss probability \(0<\epsilon<1\), independently of \(n\), and minimize the false-acceptance probability \(\beta_{n,\epsilon}\). Quantum Stein’s lemma states that \(-\lim_{n\to\infty}n^{-1}\log_2\beta_{n,\epsilon}=D(\rho\|\sigma)\) for finite-dimensional states with finite relative entropy. Collective measurements are allowed. The Ogawa–Nagaoka strong-converse result completes the quantum Stein theorem. This is a theorem statement, not a proof; the asymptotic testing discussion in Preskill’s notes provides further context.

Why should a logarithm appear? For commuting states, probabilities of independent strings multiply. Their likelihood ratio is a product, and its logarithm is a sum. Under the first hypothesis, the average log likelihood approaches \(\sum_i p_i\log_2(p_i/q_i)\). This gives an intuition for the exponent; controlling error tails and noncommuting measurements needs additional arguments.

For the biased coin and fair coin above, the exponent is approximately 0.278072 bits per sample. That does not assert the exact finite-sample formula \(\beta=2^{-0.278072n}\); subleading terms and the allowed miss probability matter. Nor does it describe equal-prior minimum-error testing, where both errors are penalized symmetrically. Keeping these tasks separate prevents an entropy formula from being applied to the wrong experiment.

Worked check 3.3. Does the exponent alone determine the false-positive probability at n=20?

Reasoned solution

No. It specifies the limiting normalized logarithm. Finite-block bounds or an explicit test are needed at n=20; substituting the leading exponential is only an asymptotic estimate.

Summary

Questions and worked problems

These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.

1. Quick check §

What is the entropy of a pure qutrit?

Hint 1

Inspect the spectrum (1,0,0).

Worked solution

It is zero bits. Measurement outcomes can still be random in another basis.

Revisit the relevant explanation

2. Quick check §

What is the maximum entropy of two qubits?

Hint 1

Use their joint dimension.

Worked solution

The dimension is four, so the maximum is log₂4=2 bits, attained by I/4.

Revisit the relevant explanation

3. Quick check §

Compute mutual information for a product state.

Hint 1

Use additivity of entropy on tensor products.

Worked solution

S(AB)=S(A)+S(B), hence I(A:B)=0.

Revisit the relevant explanation

4. Worked problem §

Compute D(|0⟩⟨0| ∥ I/2) and the reverse.

Hint 1

Check supports before taking logarithms.

Worked solution

The first is one bit. The reverse is infinite because I/2 has support outside the one-dimensional support of |0⟩.

Revisit the relevant explanation

5. Worked problem §

Compute S(A|B) for a product of two maximally mixed qubits.

Hint 1

The joint entropy is two.

Worked solution

S(A|B)=2−1=1 bit. Without correlation, conditioning on B changes nothing.

Revisit the relevant explanation

6. Worked problem §

Show the mutual-information chain rule.

Hint 1

Expand the entropy definitions.

Worked solution

\(I(A:BC)-I(A:B)=S(BC)-S(ABC)-S(B)+S(AB)=I(A:C|B)\).

Revisit the relevant explanation

7. Worked problem §

For a pure tripartite state, simplify I(A:C|B).

Hint 1

Use equality of complementary entropies.

Worked solution

\(S(AB)=S(C), S(BC)=S(A), S(B)=S(AC)\), so \(I(A:C|B)=I(A:C)\).

Revisit the relevant explanation

8. Misconception §

Is mutual information an entanglement measure?

Hint 1

Try the shared classical bit.

Worked solution

No. A separable mixture of |00⟩ and |11⟩ has mutual information one bit. Mutual information counts total correlations.

Revisit the relevant explanation

9. Misconception §

Does data processing say entropy always increases?

Hint 1

Consider a channel replacing every input with |0⟩.

Worked solution

No. Relative entropy contracts when both arguments undergo the same channel. An individual state’s entropy may increase or decrease.

Revisit the relevant explanation

10. Extension §

Explain why relative entropy is additive on independent copies.

Hint 1

Expand log of a tensor product.

Worked solution

\(\log(\rho\otimes\rho)=\log\rho\otimes I+I\otimes\log\rho\) on the support. Trace factorization gives \(D(\rho^{\otimes n}\|\sigma^{\otimes n})=nD(\rho\|\sigma)\).

Revisit the relevant explanation

Search this book

Search the course.