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Qubit dephasing decreases quantum relative entropy

Statement

On the qubit space with computational basis $∣ 0 ⟩, ∣ 1 ⟩$ , define the dephasing channel $Δ (X) = ∣ 0 ⟩ ⟨ 0 ∣ X ∣ 0 ⟩ ⟨ 0 ∣ + ∣ 1 ⟩ ⟨ 1 ∣ X ∣ 1 ⟩ ⟨ 1 ∣ .$ For $∣ + ⟩ = (∣ 0 ⟩ + ∣ 1 ⟩) / \sqrt{2}$ and $ρ = ∣ + ⟩ ⟨ + ∣$ , $σ = diag (3 / 4, 1 / 4)$ , one has $D (ρ ∥ σ) = \frac{1}{2} \ln (16 / 3) > \frac{1}{2} \ln (4 / 3) = D (Δ (ρ) ∥ Δ (σ)) .$ For the orthogonal pure states $∣ + ⟩, ∣ - ⟩ = (∣ 0 ⟩ - ∣ 1 ⟩) / \sqrt{2}$ , the input relative entropy is $+ \infty$ , whereas both dephased states equal $I / 2$ and their relative entropy is $0$ .

Facts & Assumptions

Given: The finite qubit register with ordered orthonormal basis $∣ 0 ⟩, ∣ 1 ⟩$ and the operators in the Statement.

[F1]

Finite-dimensional quantum relative entropy: the finite branch is $D (A ∥ B) = tr (A \ln A) - tr (A \ln B)$ , and support failure gives $+ \infty$ .

[F2]

Density operator: a state is nonnegative with trace one.

[F3]

Finite registers and their ordered computational bases: $∣ 0 ⟩, ∣ 1 ⟩$ is an orthonormal basis of the two-dimensional qubit register.

[F4]

Completely positive maps: complete positivity means every finite ancillary amplification sends positive operators to positive operators.

[F5]

Finite-dimensional quantum channel: a quantum channel is complex-linear, completely positive, and trace-preserving.

[F6]

For $A \in M_{m \times n} (F)$ and $B \in M_{n \times m} (F)$ , $tr (A B) = tr (B A)$ : for compatible finite-dimensional operators, $tr (A B) = tr (B A)$ .

[F7]

The natural logarithm as the inverse of the exponential function and Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: $\ln$ is the natural logarithm on positive numbers, is strictly increasing, and satisfies the product and quotient laws.

[F8]

A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S: a positive semidefinite operator is diagonalizable with nonnegative eigenvalues; rank-one orthogonal projections have eigenvalues $0, 1$ .

[F9]

Channel data processing for finite quantum relative entropy: a finite quantum channel does not increase quantum relative entropy, including its support-sensitive extended value.

Diagram convention

The one wire is the qubit operator input/output, and the rectangular box denotes the linear map $Δ$ defined above. This is only a map schematic; the channel and entropy claims are proved by the finite matrix calculations below.

\begin{center} \begin{tikzpicture}[>=stealth, every node/.style={font=\small}] \draw[->] (0,0) -- (1.2,0) node[midway,above] { $X$ }; \node[draw,rounded corners,minimum width=1.5cm,minimum height=.75cm] (deph) at (2,0) { $Δ$ }; \draw[->] (deph.east) -- (3.3,0) node[midway,above] { $Δ (X)$ }; \end{tikzpicture} \end{center}

Proof

1.1givenF3F4F5F6algebra

Put $P_{j} = ∣ j ⟩ ⟨ j ∣$ for $j = 0, 1$ . Then $P_{j} = P_{j}^{†} = P_{j}^{2}$ and $P_{0} + P_{1} = I$ by [F3]. The map $Δ (X) = \sum_{j} P_{j} X P_{j}$ is complex-linear. For any finite ancilla dimension $r$ and positive $Q$ on $H \otimes C^{r}$ , its amplification is $\sum_{j} (P_{j} \otimes I_{r}) Q (P_{j} \otimes I_{r})^{†}$ , a sum of positive operators since each summand has quadratic form $⟨ Q (P_{j} \otimes I_{r})^{†} z, (P_{j} \otimes I_{r})^{†} z ⟩ \geq 0$ . Thus [F4] gives complete positivity. Also [F6] gives $tr (Δ (X)) = \sum_{j} tr (P_{j} X P_{j}) = tr (X (P_{0} + P_{1})) = tr (X)$ . Hence $Δ$ is a channel by [F5].

2.1givenF1F2F3F7F8F9step 1.1algebra

The state $ρ$ has eigenvalues $1, 0$ , while $σ$ has eigenvalues $3 / 4, 1 / 4$ ; both are density operators by direct trace and positivity. Since $ρ$ is the rank-one projection onto $∣ + ⟩$ , $tr (ρ \ln ρ) = 0$ under $0 \ln 0 = 0$ , and $⟨ + ∣ \ln σ ∣ + ⟩ = \frac{1}{2} \ln (3 / 4) + \frac{1}{2} \ln (1 / 4)$ . Therefore [F1] gives $D (ρ ∥ σ) = \frac{1}{2} \ln (16 / 3)$ . Dephasing gives $Δ (ρ) = I / 2$ and $Δ (σ) = σ$ ; the diagonal trace formula gives $D (I / 2 ∥ σ) = \frac{1}{2} \ln ((1 / 2) / (3 / 4)) + \frac{1}{2} \ln ((1 / 2) / (1 / 4)) = \frac{1}{2} \ln (4 / 3)$ . Since $16 / 3 > 4 / 3$ , strict monotonicity of the logarithm in [F7] proves the strict decrease; this explicit calculation is consistent with [F9] applied to the channel in step 1.1.

3.1givenF1F3algebra∎

The supports of $∣ + ⟩ ⟨ + ∣$ and $∣ - ⟩ ⟨ - ∣$ are distinct one-dimensional subspaces, so the first is not contained in the second and [F1] gives input divergence $+ \infty$ . Their diagonal entries are both $(1 / 2, 1 / 2)$ , so $Δ (∣ + ⟩ ⟨ + ∣) = Δ (∣ - ⟩ ⟨ - ∣) = I / 2$ ; consequently the output relative entropy is $D (I / 2 ∥ I / 2) = 0$ by [F1].

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Sources

John Watrous, The Theory of Quantum Information (2018), Chapter 5 (standard reference, not scraped)
Mark M. Wilde, From Classical to Quantum Shannon Theory, second-edition prepublication v8 (2019) (standard reference, not scraped)