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.
Quantum Information Theory · Chapter 03
Which information can survive processing, and how does entropy quantify its loss?
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.
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 \(D(\operatorname{diag}(0.8,0.2)\|I/2)\).
The binary entropy is \(h_2(0.2)\approx0.721928\), so the relative entropy is \(1-h_2(0.2)\approx0.278072\) bits.
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 \(I(A:C|B)\).
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.
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?
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.
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
What is the entropy of a pure qutrit?
Inspect the spectrum (1,0,0).
It is zero bits. Measurement outcomes can still be random in another basis.
What is the maximum entropy of two qubits?
Use their joint dimension.
The dimension is four, so the maximum is log₂4=2 bits, attained by I/4.
Compute mutual information for a product state.
Use additivity of entropy on tensor products.
S(AB)=S(A)+S(B), hence I(A:B)=0.
Compute D(|0⟩⟨0| ∥ I/2) and the reverse.
Check supports before taking logarithms.
The first is one bit. The reverse is infinite because I/2 has support outside the one-dimensional support of |0⟩.
Compute S(A|B) for a product of two maximally mixed qubits.
The joint entropy is two.
S(A|B)=2−1=1 bit. Without correlation, conditioning on B changes nothing.
Show the mutual-information chain rule.
Expand the entropy definitions.
\(I(A:BC)-I(A:B)=S(BC)-S(ABC)-S(B)+S(AB)=I(A:C|B)\).
For a pure tripartite state, simplify I(A:C|B).
Use equality of complementary entropies.
\(S(AB)=S(C), S(BC)=S(A), S(B)=S(AC)\), so \(I(A:C|B)=I(A:C)\).
Is mutual information an entanglement measure?
Try the shared classical bit.
No. A separable mixture of |00⟩ and |11⟩ has mutual information one bit. Mutual information counts total correlations.
Does data processing say entropy always increases?
Consider a channel replacing every input with |0⟩.
No. Relative entropy contracts when both arguments undergo the same channel. An individual state’s entropy may increase or decrease.
Explain why relative entropy is additive on independent copies.
Expand log of a tensor product.
\(\log(\rho\otimes\rho)=\log\rho\otimes I+I\otimes\log\rho\) on the support. Trace factorization gives \(D(\rho^{\otimes n}\|\sigma^{\otimes n})=nD(\rho\|\sigma)\).