Quantum Relative Entropy, Data Processing, and Strong Subadditivity — Examples

Put $P_j=\mathrm{\mid }j⟩⟨j\mathrm{\mid }$ for $j=0,1$. Then $P_j=P_j^{†}=P_j^2$ and $P_0+P_1=I$ by [F3]. The map $\mathrm{\Delta }(X)={\sum }_j{P}_{j}X{P}_j$ is complex-linear. For any finite ancilla dimension $r$ and positive $Q$ on $H\otimes {\mathbb{C}}^r$, 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 $⟨Q(P_j\otimes I_r{)}^{†}z,(P_j\otimes I_r{)}^{†}z⟩\ge 0$. Thus [F4] gives complete positivity. Also [F6] gives $\mathrm{tr}(\mathrm{\Delta }(X))={\sum }_j\mathrm{tr}(P_{j}X{P}_j)=\mathrm{tr}(X(P_0+P_1))=\mathrm{tr}(X)$. Hence $\mathrm{\Delta }$ is a channel by [F5].

The state $\rho$ has eigenvalues $1,0$, while $\sigma$ has eigenvalues $3\mathrm{/}4,1\mathrm{/}4$; both are density operators by direct trace and positivity. Since $\rho$ is the rank-one projection onto $\mathrm{\mid }+⟩$, $\mathrm{tr}(\rho \ln \rho )=0$ under $0\ln 0=0$, and $⟨+\mathrm{\mid }\ln \sigma \mathrm{\mid }+⟩=\frac{1}{2}\ln (3\mathrm{/}4)+\frac{1}{2}\ln (1\mathrm{/}4)$. Therefore [F1] gives $D(\rho \mathrm{\parallel }\sigma )=\frac{1}{2}\ln (16\mathrm{/}3)$. Dephasing gives $\mathrm{\Delta }(\rho )=I\mathrm{/}2$ and $\mathrm{\Delta }(\sigma )=\sigma$; the diagonal trace formula gives $D(I\mathrm{/}2\mathrm{\parallel }\sigma )=\frac{1}{2}\ln ((1\mathrm{/}2)\mathrm{/}(3\mathrm{/}4))+\frac{1}{2}\ln ((1\mathrm{/}2)\mathrm{/}(1\mathrm{/}4))=\frac{1}{2}\ln (4\mathrm{/}3)$. Since $16\mathrm{/}3>4\mathrm{/}3$, 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.

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

By [F4], the vectors $g=\mathrm{\mid }\mathrm{G}\mathrm{H}\mathrm{Z}⟩$, $h=(\mathrm{\mid }000⟩-\mathrm{\mid }111⟩)\mathrm{/}\sqrt{2}$, and the six computational basis vectors other than $\mathrm{\mid }000⟩,\mathrm{\mid }111⟩$ form an orthonormal basis. Direct multiplication by ${\rho }_{ABC}=g{g}^{\ast }$ gives ${\rho }_{ABC}g=g$ and zero on the other seven basis vectors. Thus its spectrum is $(1,0,\dots ,0)$; it is non-negative by [F8] and has trace one, so it is a state by [F2]. Consequently $S({\rho }_{ABC})=-1\ln 1=0$ by [F1] and [F7].

Expanding ${\rho }_{ABC}$ gives the four terms $\frac{1}{2}\mathrm{\mid }000⟩⟨000\mathrm{\mid }$, $\frac{1}{2}\mathrm{\mid }000⟩⟨111\mathrm{\mid }$, $\frac{1}{2}\mathrm{\mid }111⟩⟨000\mathrm{\mid }$, and $\frac{1}{2}\mathrm{\mid }111⟩⟨111\mathrm{\mid }$. For any traced register, the first- or last-factor formula [F3], after the wire permutation [F9] when needed, kills both off-diagonal terms because $\mathrm{tr}(\mathrm{\mid }0⟩⟨1\mathrm{\mid })=\mathrm{tr}(\mathrm{\mid }1⟩⟨0\mathrm{\mid })=0$; the resulting two-register marginal is $\frac{1}{2}\mathrm{\mid }00⟩⟨00\mathrm{\mid }+\frac{1}{2}\mathrm{\mid }11⟩⟨11\mathrm{\mid }$. Tracing out two registers similarly gives $I\mathrm{/}2$ for each one-register marginal. These diagonal matrices have positive eigenvalues $1\mathrm{/}2,1\mathrm{/}2$, with two additional zero eigenvalues for each two-register marginal. Across each one-versus-two split, the displayed vector itself is a Schmidt expansion of the form [F5], with coefficients $1\mathrm{/}\sqrt{2},1\mathrm{/}\sqrt{2}$.

For each reduced state in step 1.2, [F1] gives entropy $-2(\frac{1}{2}\ln \frac{1}{2})=\ln 2$. Hence the left side of the claimed SSA instance is $2\ln 2$, and the right side is $\ln 2+0$. Since $2>1$ and $\ln 2>0$ by strict monotonicity in [F7], the inequality is strict and equals the stated comparison.

For $\theta \in [a,\pi \mathrm{/}4]$, ${\rho }_{{\sin }^2\theta }=\mathrm{diag}({\cos }^2\theta ,{\sin }^2\theta )$ is nonnegative and has trace one by [F11], so is a state by [F5]. Since $a>0$ by [F15], [F8] and [F12] make sine positive and strictly increasing on $[a,\pi \mathrm{/}4]$; squaring preserves strict order there, so the angle parametrization is injective and $T_a$ is well-defined. Its output angle lies in $[0,\pi \mathrm{/}4-a]$, so its output is also a density operator.

The family ${\mathcal{S}}_a$ is the line segment $\{{\rho }_p:{\sin }^{2}a\le p\le 1\mathrm{/}2\}$. If $T_a$ were affine on this segment, interpolation between its endpoint states would make its output's $\mathrm{\mid }1⟩$ diagonal entry affine in $p$. The same is necessary for the restriction of a quantum channel, by complex-linearity. Thus there would be constants $A,B$ with ${\sin }^2(\theta -a)=A+B{\sin }^2\theta$ for every $\theta \in [a,\pi \mathrm{/}4]$. Differentiating twice on the open interval and using [F10]–[F13] gives $\cos (2\theta -2a)=B\cos (2\theta )$. By [F14], cosine is continuous, so taking $\theta \uparrow \pi \mathrm{/}4$ yields $\sin (2a)=0$, because $\cos (\pi \mathrm{/}2-2a)=\sin (2a)$ and $\cos (\pi \mathrm{/}2)=0$. But $0<2a=\pi \mathrm{/}6<\pi \mathrm{/}2$ by [F15], so [F8]–[F9] give $\sin (2a)>0$, a contradiction. Hence $T_a$ is not affine on the family, and no quantum channel agrees with it there.

For $\theta ,\varphi \in [a,\pi \mathrm{/}4]$, the square roots in [F6] are diagonal with entries $(\cos \theta ,\sin \theta )$ and $(\cos \varphi ,\sin \varphi )$. Their product is positive diagonal, so by [F2] and [F4], $F({\rho }_{{\sin }^2\theta },{\rho }_{{\sin }^2\varphi })=\cos \theta \cos \varphi +\sin \theta \sin \varphi =\cos (\theta -\varphi )\mathrm{.}$ The cosine addition formula [F10] justifies the last equality. Applying $T_a$ subtracts $a$ from both angles, leaving their difference unchanged; root fidelity is therefore preserved for every pair.

For diagonal qubit states, [F3] and [F4] give $d_{\mathrm{t}\mathrm{r}}({\rho }_p,{\rho }_q)=\mathrm{\mid }p-q\mathrm{\mid }$. Take $\theta \ge \varphi$ in $[a,\pi \mathrm{/}4]$. The addition formulas and Pythagorean identity [F10]–[F11] give ${\sin }^{2}u-{\sin }^{2}v=\sin (u+v)\sin (u-v)$. Therefore ${\sin }^2\theta -{\sin }^2\varphi =\sin (\theta +\varphi )\sin (\theta -\varphi ),$ ${\sin }^2(\theta -a)-{\sin }^2(\varphi -a)=\sin (\theta +\varphi -2a)\sin (\theta -\varphi )\mathrm{.}$ Here $0\le \theta +\varphi -2a\le \theta +\varphi \le \pi \mathrm{/}2$, so [F8]–[F12] make the first sine factor in the second line no larger than the corresponding nonnegative factor in the first line; the other factor is the same and nonnegative. Thus the output coordinate difference is at most the input difference, proving trace-distance contraction for every pair.

Take $\theta =\pi \mathrm{/}4$ and $\varphi =a$. By strict increase of sine from $\sin 0=0$ to $\sin (\pi \mathrm{/}2)=1$ in [F8], [F9], and [F12], every angle strictly between zero and $\pi \mathrm{/}2$ gives $0<{\sin }^2\theta <1$; thus both input diagonal states have full support and their relative entropy is finite by [F1] and [F7], since the positive spectral entries have finite natural logarithms. The outputs are ${\rho }_{{\sin }^2(\pi \mathrm{/}4-a)}$ and ${\rho }_0=\mathrm{\mid }0⟩⟨0\mathrm{\mid }$. Since $0<\pi \mathrm{/}4-a<\pi \mathrm{/}2$ by [F15], the first output has full support, while the second has one-dimensional support; support inclusion fails and [F1] gives output divergence $+\mathrm{\infty }$.

Independently, the root fidelity of $\mathrm{\mid }0⟩⟨0\mathrm{\mid }$ and $\mathrm{\mid }+⟩⟨+\mathrm{\mid }$ is $1\mathrm{/}\sqrt{2}$: the product of their positive square roots is $(1\mathrm{/}\sqrt{2})\mathrm{\mid }0⟩⟨+\mathrm{\mid }$, whose only nonzero singular value is $1\mathrm{/}\sqrt{2}$. Their difference is $\frac{1}{2}\left(\begin{array}{cc}1& -1\\ -1& -1\end{array}\right)$, whose square is $I\mathrm{/}2$, so both singular values are $1\mathrm{/}\sqrt{2}$ and [F3]–[F4] give trace distance $1\mathrm{/}\sqrt{2}$. The first support is not contained in the second, so [F1] gives infinite relative entropy.