Amortized channel divergence for asymptotic quantum channel discrimination

Abstract

It is well known that for the discrimination of classical and quantum channels in the finite, non-asymptotic regime, adaptive strategies can give an advantage over non-adaptive strategies. However, Hayashi (IEEE Trans Inf Theory 55(8):3807–3820, 2009. arXiv:0804.0686) showed that in the asymptotic regime, the exponential error rate for the discrimination of classical channels is not improved in the adaptive setting. We extend this result in several ways. First, we establish the strong Stein’s lemma for classical–quantum channels by showing that asymptotically the exponential error rate for classical–quantum channel discrimination is not improved by adaptive strategies. Second, we recover many other classes of channels for which adaptive strategies do not lead to an asymptotic advantage. Third, we give various converse bounds on the power of adaptive protocols for general asymptotic quantum channel discrimination. Intriguingly, it remains open whether adaptive protocols can improve the exponential error rate for quantum channel discrimination in the asymmetric Stein setting. Our proofs are based on the concept of amortized distinguishability of quantum channels, which we analyse using data-processing inequalities.

Appendices

Appendix A: Amortization does not increase the Hilbert $\alpha$-channel divergences

In this appendix, we prove that the Hilbert $\alpha$-divergence from [64, Section III] obeys a data-processed triangle inequality, and as a consequence, channel divergences based on it do not increase under amortization. We also remark how other metrics based on quantum fidelity obey a data-processed triangle inequality, and so their corresponding channel divergences do not increase under amortization.

The Hilbert $\alpha$-divergence of states $\rho$ and $\sigma$ is defined for $\alpha \ge 1$ as [64, Section III]

$$\begin{aligned} H_{\alpha }(\rho \Vert \sigma )&:=\frac{\alpha }{\alpha -1} \log \sup {}_{\alpha }(\rho /\sigma ) \end{aligned}$$

(A1)

$$\begin{aligned} \sup {}_{\alpha }(\rho /\sigma )&:=\sup _{\alpha ^{-1}I\le \varLambda \le I} \frac{\mathrm{Tr}[\varLambda \rho ]}{\mathrm{Tr}[\varLambda \sigma ]}. \end{aligned}$$

(A2)

It is known that [64, Theorem 1]

$$\begin{aligned} \lim _{\alpha \rightarrow 1}H_{\alpha }(\rho \Vert \sigma )&=\frac{1}{2\ln 2} \left\| \rho -\sigma \right\| _{1}, \end{aligned}$$

(A3)

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }H_{\alpha }(\rho \Vert \sigma )&=D_{\max } (\rho \Vert \sigma ). \end{aligned}$$

(A4)

Here we prove that this quantity obeys a data-processed triangle inequality for all $\alpha \ge 1$.

Lemma 42

(Data-processed triangle inequality) Let ${\mathcal {P}}_{A\rightarrow B}$ be a positive trace-preserving map, and let $\rho _{A},\sigma _{A}\in S(A)$ and $\omega _{B}\in S(B)$. Then the following inequality holds for all $\alpha \ge 1$:

$$\begin{aligned} H_{\alpha }({\mathcal {P}}_{A\rightarrow B}(\rho _{A})\Vert \omega _{B}) \le H_{\alpha }(\rho _{A}\Vert \sigma _{A})+H_{\alpha }({\mathcal {P}}_{A \rightarrow B}(\sigma _{A})\Vert \omega _{B}). \end{aligned}$$

(A5)

Proof

For $\alpha =1$, we have that $\lim _{\alpha \rightarrow 1}H_{\alpha }(\rho \Vert \sigma )=\frac{1}{2\ln 2}\left\| \rho -\sigma \right\| _{1}$, as recalled above. The statement then follows from the usual triangle inequality:

$$\begin{aligned} \left\| {\mathcal {P}}_{A\rightarrow B}(\rho _{A})-\omega _{B}\right\| _{1}&\le \left\| {\mathcal {P}}_{A\rightarrow B}(\rho _{A}) -{\mathcal {P}}_{A\rightarrow B}(\sigma _{A})\right\| _{1} +\left\| {\mathcal {P}}_{A\rightarrow B}(\sigma _{A}) -\omega _{B}\right\| _{1} \end{aligned}$$

(A6)

$$\begin{aligned}&\le \left\| \rho _{A}-\sigma _{A}\right\| _{1} +\left\| {\mathcal {P}}_{A\rightarrow B}(\sigma _{A}) -\omega _{B}\right\| _{1}, \end{aligned}$$

(A7)

and the fact that trace distance is monotone with respect to positive, trace-preserving maps.

To prove the inequality for $\alpha >1$, let $\varLambda _{B}$ be an arbitrary operator such that $\alpha ^{-1}I_{B}\le \varLambda _{B}\le I_{B}$. The map ${\mathcal {P}} _{A\rightarrow B}^{\dag }$ is positive and unital because ${\mathcal {P}} _{A\rightarrow B}$ is positive and trace preserving by assumption. Then $\alpha ^{-1} I_{A}\le {\mathcal {P}}_{A\rightarrow B}^{\dag }(\varLambda _{B})\le I_{A}$ and

$$\begin{aligned} \frac{\mathrm{Tr}[\varLambda _{B}{\mathcal {P}}_{A\rightarrow B}(\rho _{A})]}{\mathrm{Tr}[\varLambda _{B}\omega _{B}]}&=\frac{\mathrm{Tr}\left[ {\mathcal {P}}_{A\rightarrow B}^{\dag } (\varLambda _{B})(\rho _{A})\right] }{\mathrm{Tr}[\varLambda _{B}\omega _{B}]} \end{aligned}$$

(A8)

$$\begin{aligned}&=\frac{\mathrm{Tr}\left[ {\mathcal {P}}_{A\rightarrow B}^{\dag }(\varLambda _{B}) (\rho _{A})\right] }{\mathrm{Tr}\left[ {\mathcal {P}}_{A\rightarrow B}^{\dag } (\varLambda _{B})(\sigma _{A})\right] }\frac{\mathrm{Tr}\left[ {\mathcal {P}}_{A\rightarrow B}^{\dag } (\varLambda _{B})(\sigma _{A})\right] }{\mathrm{Tr}[\varLambda _{B}\omega _{B}]} \end{aligned}$$

(A9)

$$\begin{aligned}&=\frac{\mathrm{Tr}\left[ {\mathcal {P}}_{A\rightarrow B}^{\dag }(\varLambda _{B}) (\rho _{A})\right] }{\mathrm{Tr}\left[ {\mathcal {P}}_{A\rightarrow B}^{\dag } (\varLambda _{B})(\sigma _{A})\right] }\frac{\mathrm{Tr}[\varLambda _{B}{\mathcal {P}}_{A\rightarrow B} (\sigma _{A})]}{\mathrm{Tr}[\varLambda _{B}\omega _{B}]} \end{aligned}$$

(A10)

$$\begin{aligned}&\le \left( \sup _{\alpha ^{-1}I_{A}\le \varGamma _{A}\le I_{A}}\frac{\mathrm{Tr}[\varGamma _{A} (\rho _{A})]}{\mathrm{Tr}[\varGamma _{A} (\sigma _{A})]}\right) \nonumber \\&\qquad \cdot \left( \sup _{\alpha ^{-1}I_{B}\le \varLambda _{B}\le I_{B}}\frac{\mathrm{Tr}[\varLambda _{B}{\mathcal {P}}_{A\rightarrow B} (\sigma _{A})]}{\mathrm{Tr}[\varLambda _{B}\omega _{B}]}\right) \end{aligned}$$

(A11)

$$\begin{aligned}&=\sup {}_{\alpha }(\rho _{A}/\sigma _{A})\cdot \sup {}_{\alpha } ({\mathcal {P}}_{A\rightarrow B}(\sigma _{A})/\omega _{B}). \end{aligned}$$

(A12)

Since the inequality holds for all $\varLambda _{B}$ such that $\alpha ^{-1}I_{B}\le \varLambda _{B}\le I_{B}$, we conclude that

$$\begin{aligned} \sup {}_{\alpha }({\mathcal {P}}_{A\rightarrow B}(\rho _{A})/\omega _{B}) \le \sup {}_{\alpha }(\rho _{A}/\sigma _{A})\cdot \sup {}_{\alpha } ({\mathcal {P}}_{A\rightarrow B}(\sigma _{A})/\omega _{B}). \end{aligned}$$

(A13)

Finally, we take a logarithm and multiply by $\alpha /\left( \alpha -1\right)$ to conclude the statement of the lemma. $\square$

By the same proof that we gave for the max-relative entropy in Proposition 10 and using the fact that the Hilbert $\alpha$-divergence is strongly faithful [64, Theorem 1(i)], we conclude that there is an amortization collapse for the Hilbert $\alpha$-divergence of quantum channels. As special case, we conclude that the diamond norm of the difference of two channels does not increase under amortization.

Proposition 43

Let ${\mathcal {N}}_{A\rightarrow B},{\mathcal {M}}_{A\rightarrow B}\in {\mathcal {Q}}(A\rightarrow B)$. Then for all $\alpha \ge 1$, we have the following amortization collapse:

$$\begin{aligned} H_{\alpha }^{{\mathcal {A}}}({\mathcal {N}}\Vert {\mathcal {M}}) =H_{\alpha }({\mathcal {N}}\Vert {\mathcal {M}}). \end{aligned}$$

(A14)

We can establish related results for the c-distance and the Bures distance of quantum states, both of which are based on the quantum fidelity. For states $\rho$ and $\sigma$, the c-distance [91,92,93,94] and Bures distance [95] are, respectively, defined as

$$\begin{aligned} c(\rho ,\sigma ) :=\sqrt{1-F(\rho ,\sigma )}, \qquad B(\rho \Vert \sigma ) :=\sqrt{2\left( 1-\sqrt{F(\rho ,\sigma )}\right) }. \end{aligned}$$

(A15)

(In the above and what follows, we use the notation $c(\rho ,\sigma )$ for c-distance in order to differentiate this quantity from the Chernoff divergence $C(\rho \Vert \sigma )$.) The same proof as in (A6)–(A7), along with the fact that the quantum fidelity is monotone with respect to positive, trace-preserving maps [7, Corollary A.5], implies that the following data-processed triangle inequalities hold:

Lemma 44

(Data-processed triangle inequalities) Let ${\mathcal {P}}_{A\rightarrow B}$ be a positive trace-preserving map, and let $\rho _{A},\sigma _{A}\in S(A)$ and $\omega _{B}\in S(B)$. Then the following inequalities hold:

$$\begin{aligned} c({\mathcal {P}}_{A\rightarrow B}(\rho _{A}),\omega _{B})&\le c(\rho _{A},\sigma _{A})+c({\mathcal {P}}_{A\rightarrow B} (\sigma _{A}),\omega _{B}), \end{aligned}$$

(A16)

$$\begin{aligned} B({\mathcal {P}}_{A\rightarrow B}(\rho _{A})\Vert \omega _{B})&\le B(\rho _{A}\Vert \sigma _{A})+B({\mathcal {P}}_{A\rightarrow B} (\sigma _{A})\Vert \omega _{B}). \end{aligned}$$

(A17)

By the same reasoning as above, we then conclude that the induced channel divergences do not increase under amortization:

Proposition 45

Let ${\mathcal {N}}_{A\rightarrow B},{\mathcal {M}}_{A\rightarrow B}\in {\mathcal {Q}}(A\rightarrow B)$. Then we have the following amortization collapses:

$$\begin{aligned} c^{{\mathcal {A}}}({\mathcal {N}},{\mathcal {M}})=c({\mathcal {N}},{\mathcal {M}}), \qquad B^{{\mathcal {A}}}({\mathcal {N}}\Vert {\mathcal {M}}) =B({\mathcal {N}}\Vert {\mathcal {M}}). \end{aligned}$$

(A18)

Appendix B: Generalized Fuchs–van-de-Graaf inequality

A well-known inequality in quantum information theory is the following Fuchs-van-de-Graaf inequality [96]:

$$\begin{aligned} \frac{1}{2}\left\| \rho -\sigma \right\| _{1}\le \sqrt{1-F(\rho ,\sigma )}, \end{aligned}$$

(B1)

which holds for density operators $\rho$ and $\sigma$, and $F(\rho ,\sigma ):=\Vert \sqrt{\rho }\sqrt{\sigma } \Vert _{1}^{2}$. The following lemma, proved in [70, Supplementary Lemma 3], generalizes this relation to the case of positive semi-definite operators A and B, and it also represents a tighter bound than that given in [6, Theorem 7]. (Note that [6, Theorem 7] generalizes one of the inequalities in [97, Equation (1)] to positive semi-definite operators.) The proof of Lemma 46 that we give below is very similar to the proof of Theorem 7 of [6], but it features a minor change in the reasoning. The proof is also different from that given in [70, Supplementary Lemma 3].

Lemma 46

([70]) For positive semi-definite, trace class operators A and B acting on a separable Hilbert space, we have that

$$\begin{aligned} \left\| A-B\right\| _{1}^{2}+4\left\| A^{1/2}B^{1/2}\right\| _{1}^{2} \le \big (\mathrm{Tr}[A+B]\big )^2. \end{aligned}$$

(B2)

Proof

For convenience, we give a complete proof and follow the proof of Theorem 7 of [6] quite closely. Consider two general operators P and Q, and define their sum and difference as $S=P+Q$ and $D=P-Q$. Then $P=\left( S+D\right) /2$ and $Q=\left( S-D\right) /2$. Consider that

$$\begin{aligned} PP^{\dag }-QQ^{\dag }&=\frac{1}{4}\left( \left( S+D\right) \left( S+D\right) ^{\dag }-\left( S-D\right) \left( S-D\right) ^{\dag }\right) \end{aligned}$$

(B3)

$$\begin{aligned}&=\frac{1}{2}\left( SD^{\dag }+DS^{\dag }\right) . \end{aligned}$$

(B4)

Then, we have

$$\begin{aligned} \left\| PP^{\dag }-QQ^{\dag }\right\| _{1}&=\frac{1}{2} \left\| SD^{\dag }+DS^{\dag }\right\| _{1} \end{aligned}$$

(B5)

$$\begin{aligned}&\le \frac{1}{2}\left( \left\| SD^{\dag }\right\| _{1} +\left\| DS^{\dag }\right\| _{1}\right) \end{aligned}$$

(B6)

$$\begin{aligned}&=\left\| SD^{\dag }\right\| _{1} \end{aligned}$$

(B7)

$$\begin{aligned}&\le \left\| S\right\| _{2}\left\| D\right\| _{2}. \end{aligned}$$

(B8)

Now pick $P=A^{1/2}U$ and $Q=B^{1/2}$, where U is an arbitrary unitary. Then $S,D=A^{1/2}U\pm B^{1/2}$, and we find that

$$\begin{aligned} \left\| A-B\right\| _{1}\le \left\| A^{1/2}U+B^{1/2}\right\| _{2} \left\| A^{1/2}U-B^{1/2}\right\| _{2}. \end{aligned}$$

(B9)

Squaring this gives

$$\begin{aligned}&\left\| A-B\right\| _{1}^{2}\nonumber \\&\quad \le \left\| A^{1/2}U+B^{1/2}\right\| _{2}^{2} \left\| A^{1/2}U-B^{1/2}\right\| _{2}^{2} \end{aligned}$$

(B10)

$$\begin{aligned}&\quad =\mathrm{Tr}\Big [\left( A^{1/2}U+B^{1/2}\right) ^{\dag }\left( A^{1/2}U+B^{1/2}\right) \Big ] \cdot \mathrm{Tr}\Big [\left( A^{1/2}U-B^{1/2}\right) ^{\dag }\left( A^{1/2}U-B^{1/2}\right) \Big ] \end{aligned}$$

(B11)

$$\begin{aligned}&\quad =\mathrm{Tr}\Big [A+B+B^{1/2}A^{1/2}U+U^{\dag }A^{1/2}B^{1/2} \Big ]\cdot \mathrm{Tr}\Big [A+B-B^{1/2}A^{1/2}U-U^{\dag }A^{1/2}B^{1/2}\Big ] \end{aligned}$$

(B12)

$$\begin{aligned}&\quad =\left( \mathrm{Tr}[A+B]+2\mathrm{Re}\Big \{\mathrm{Tr}\Big [B^{1/2}A^{1/2}\Big ]\Big \}\right) \Big (\mathrm{Tr}[A+B]-2\mathrm{Re}\Big \{\mathrm{Tr}\Big [B^{1/2}A^{1/2}U\big ]\Big \}\Big ) \end{aligned}$$

(B13)

$$\begin{aligned}&\quad =\left( \mathrm{Tr}[A+B]\right) ^{2}-4\left( \mathrm{Re}\Big \{\mathrm{Tr} \Big [B^{1/2}A^{1/2}U\Big ]\Big \}\right) ^{2}. \end{aligned}$$

(B14)

Note that the unitary U in the above is arbitrary. So we can finally pick the unitary U to be the operator from the polar decomposition of $B^{1/2}A^{1/2}$ as

$$\begin{aligned} B^{1/2}A^{1/2}U=\sqrt{B^{1/2}AB^{1/2}}. \end{aligned}$$

(B15)

Then, we get

$$\begin{aligned} \left\| A-B\right\| _{1}^{2}&\le \left( \mathrm{Tr}[A+B]\right) ^{2} -4\left( \mathrm{Re}\Big \{\mathrm{Tr}\Big [\sqrt{B^{1/2}AB^{1/2}}\Big ]\Big \}\right) ^{2} \end{aligned}$$

(B16)

$$\begin{aligned}&=\left( \mathrm{Tr}[A+B]\right) ^{2}-4\left\| A^{1/2}B^{1/2}\right\| _{1}^{2} \end{aligned}$$

(B17)

and the proof is concluded. $\square$