Quantifying the irreversibility of channels

Abstract

In contrast to unitary evolutions, which are reversible, generic quantum processes (operations and quantum channels) are often irreversible. However, the degree of irreversibility is different for different channels, and it is desirable to have a quantitative characterization of irreversibility. In this paper, by exploiting the channel–state duality implemented by the Jamiołkowski–Choi isomorphism, we quantify the irreversibility of channels via entropy of the Jamiołkowski–Choi states of the corresponding channels and compare it with the notions of entanglement fidelity and entropy exchange. General properties of a reasonable measure of irreversibility are discussed from an intuitive perspective, and entropic measures of irreversibility are introduced. Several relations between irreversibility, entanglement fidelity, the degree of nonunitality, and decorrelating power are established. Some measures of irreversibility for a variety of prototypical channels are evaluated explicitly, revealing some information-theoretic aspects of the structure of channels from the perspective of irreversibility.

Appendix: Proofs of Propositions 1–5

Below, we give the detailed proofs of Propositions 1–5. We further discuss an alternative measure of irreversibility in terms of the Tsallis entropy, which is easier to compute than the irreversibility based on the von Neumann entropy. We make a comparative study of this quantity and that introduced in the main text.

Appendix A: Proof of Proposition 1

1. By the properties of von Neumann entropy, we have $0\le S(J_{\mathcal E})\le\ln d^2$, $S(J_{\mathcal E})=0$ if and only if $J_{\mathcal E}$ is a pure state, and $S(J_{\mathcal E})=\ln d^2$ if and only if $J_{\mathcal E}=\frac{1}{d^2}\mathbf 1\otimes\mathbf 1$ is a maximally mixed state on $H\otimes H$. Consequently,

$$0\le S(\mathcal E)=\frac{1}{2}S(J_\mathcal E)\le\ln d.$$

It is clear that $S(\mathcal E)=0$ if and only if $J_{\mathcal E}$ is a pure state, and $S(\mathcal E)=\ln d$ if and only if $J_{\mathcal E}=\mathbf 1\otimes\mathbf 1/d^2$ is a maximally mixed state on $H\otimes H$. Next, we show that $J_{\mathcal E}$ is a pure state if and only if $\mathcal E$ is a unitary channel, and $J_{\mathcal E}=\frac{1}{d^2}\mathbf 1\otimes\mathbf 1$ is equivalent to $\mathcal E$ being the completely depolarizing channel. For the first equivalence, we suppose that $\mathcal E(\rho)=U\rho U^\dagger$ is a unitary channel; then

$$J_{\mathcal E}=|\Phi^{+}_U\rangle\langle\Phi^{+}_U|,$$

where $|\Phi^{+}_U\rangle=\frac{1}{\sqrt d}\sum_{i}|i\rangle\otimes U|i\rangle$ is a pure state, with $S(\mathcal E)=0$. Conversely, we suppose that

$$J_{\mathcal E}=\frac{1}d\sum_{i,j}|i\rangle\langle j|\otimes\mathcal E(|i\rangle\langle j|)$$

is a pure state; then $J_{\mathcal E}$ is a rank-1 operator, which makes $\mathcal E(|i\rangle\langle j|)$ an operator with rank not greater than 1 for any $i$ and $j$. Because $\mathcal E(|i\rangle\langle i|)$ are quantum states, $\operatorname{rank}(\mathcal E(|i\rangle\langle i|))=1$ for any $i=1,2,\ldots,d$. If there is $\mathcal E(|i\rangle\langle j|)=0$ for some $i\neq j$, we have $\operatorname{rank}(J_\mathcal E)\ge\operatorname{rank}(\mathcal E(|i\rangle\langle i|))+\operatorname{rank}(\mathcal E(|j\rangle\langle j|))=2$, which contradicts the rank condition $\operatorname{rank}(J_{\mathcal E})=1$. Thus, $\mathcal E(|i\rangle\langle j|)$ is a rank-1 operator for any $i$ and $j$. Combining $\operatorname{rank}(J_{\mathcal E})=1$ and $\operatorname{rank}(\mathcal E(|i\rangle\langle i|))=1$ for any $i$, we see that $\mathcal E(|i\rangle\langle j|)$ can be written as

$$\mathcal E(|i\rangle\langle j|)=|\phi_i\rangle\langle\phi_j|,\qquad i,j=1,2,\ldots,d,$$

where $\{|\phi_i\rangle\colon i=1,2,\ldots,d\}$ is a set of pure states. In what follows, we show that $\{|\phi_i\rangle:i=1,2,\ldots,d\}$ constitutes an orthonormal basis. By

$$\mathcal E(|i\rangle\langle i|)=\sum_k E_k|i\rangle\langle i|E_k^\dagger=|\phi_i\rangle\langle\phi_i|,$$

we have $E_k|i\rangle=c_{k i}|\phi_i\rangle$ for some complex numbers $c_{ki}$. Thus,

$$\mathcal E(|i\rangle\langle j|)=\biggl(\sum_k c_{ki}\bar c_{kj}\biggr)|\phi_i\rangle\langle\phi_j|=|\phi_i\rangle\langle\phi_j|,$$

which implies that $\sum_k c_{k i}\bar c_{k j}=1$ for any $i$, $j$. The orthogonality of $\{|\phi_i\rangle\colon i=1,2,\ldots,d\}$ follows from

$$\langle i|j\rangle=\Big\langle i\bigg|\sum_k E_k^\dagger E_k\bigg|j\Big\rangle= \biggl(\sum_k\bar c_{ki}c_{kj}\biggr)\langle\phi_i|\phi_j\rangle= \langle\phi_i|\phi_j\rangle,\qquad i,j=1,2,\ldots,d.$$

Thus, $\mathcal E$ maps the orthonormal basis $\{|i\rangle\colon i=1,2,\ldots,d\}$ to the orthonormal basis $\{|\phi_i\rangle\colon i=1,2,\ldots,d\}$. Because $S(\mathcal E)$ is independent of the choice of the orthonormal basis $\{|i\rangle\colon i=1,2,\ldots,d\}$, $\mathcal E$ maps any orthonormal basis to an orthonormal basis, which implies that $\mathcal E$ is a unitary channel.

To prove the second equivalence, we suppose that $\mathcal E$ is the completely depolarizing channel. Then

$$J_{\mathcal E}=\frac{1}{d}\sum_{i,j}|i\rangle\langle j|\otimes\mathcal E(|i\rangle\langle j|)=\frac{1}{d^2}\mathbf 1\otimes\mathbf 1$$

is the maximally mixed state on $H\otimes H$. Conversely, if $J_{\mathcal E}=\mathbf 1\otimes\mathbf 1/d^2$ is the maximally mixed state, then we conclude from

$$J_{\mathcal E}=\frac{1}d\sum_{i,j}|i\rangle\langle j|\otimes\mathcal E(|i\rangle\langle j|)= \frac{1}d\sum_{i}|i\rangle\langle i|\otimes\mathcal E(|i\rangle\langle i|)+ \frac{1}d\sum_{i\neq j}|i\rangle\langle j|\otimes\mathcal E(|i\rangle\langle j|)$$

that $\mathcal E(|i\rangle\langle j|)=0$ for any $i\neq j$ and $\mathcal E(|i\rangle\langle i|)=\mathbf 1/d$ for any $i$. Therefore,

$$\mathcal E(\rho)=\mathcal E\biggl(\sum_{i,j}\langle i|\rho|j\rangle|i\rangle\langle j|\biggr)= \sum_{i,j}\langle i|\rho|j\rangle\mathcal E(|i\rangle\langle j|)=\frac{1}d \mathbf 1$$

for any state $\rho$, i.e., $\mathcal E$ is the completely depolarizing channel.

2. Direct calculation shows that

$$J_{p_1\mathcal E_1+p_2\mathcal E_2}=p_1J_{\mathcal E_1}+p_2J_{\mathcal E_2}.$$

By the concavity of von Neumann entropy, we have

$$\begin{aligned} \, S(p_1\mathcal E_1+p_2\mathcal E_2)&=\frac{1}{2}S(J_{p_1\mathcal E_1+p_2\mathcal E_2})=\frac{1}{2}S(p_1J_{\mathcal E_1}+p_2J_{\mathcal E_2})\ge \\ &\ge\frac{1}{2}\bigl(p_1S(J_{\mathcal E_1})+p_2S(J_{\mathcal E_2})\bigr)=p_1S(\mathcal E_1)+p_2S(\mathcal E_2). \end{aligned}$$

3. Let $U$ and $V$ be any unitary operators. Then

$$\begin{aligned} \, J_{\mathcal E_U \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E_V}&= \frac{1}{d}\sum_{i,j}|i\rangle\langle j|\otimes U\mathcal E( V|i\rangle\langle j|V^\dagger)U^\dagger= \\ &=(V^\dagger\otimes U)\biggl(\frac{1}{d}\sum_{i,j}V|i\rangle\langle j|V^\dagger\otimes\mathcal E(V|i\rangle\langle j|V^\dagger)\biggr) (V^\dagger\otimes U)^\dagger. \end{aligned}$$

Thus

$$S(\mathcal E_U \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E_V)= \frac{1}{2}S(J_{\mathcal E_U \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E_V})= \frac{1}{2}S \biggl(\frac{1}{d}\sum_{i,j}V|i\rangle\langle j|V^\dagger\otimes\mathcal E(V|i\rangle\langle j|V^\dagger)\biggr)=\frac{1}{2}S(J_{\mathcal E})=S(\mathcal E)$$

whence item 3 follows.

4. Let $H^a$ and $H$ have the respective dimensions $d_a$ and $d$, and orthonormal bases $\{|\mu\rangle\}$ and $\{|i\rangle\}$. For the channel $\mathcal I^a\otimes\mathcal E$ on the composite system Hilbert space $H^a\otimes H$, we have

$$\begin{aligned} \, d_adJ_{\mathcal I^a\otimes\mathcal E}&= \sum_{\mu,\nu,i,j}|\mu\rangle\langle\nu|\otimes|i\rangle\langle j|\otimes|\mu\rangle \langle\nu|\otimes\mathcal E(|i\rangle\langle j|)= \\ &=\sum_{\mu,\nu,i,j}(\mathbf 1^a\otimes F_{}\otimes\mathbf 1)(|\mu\rangle\langle\nu|\otimes|\mu\rangle \langle\nu|\otimes|i\rangle\langle j|\otimes\mathcal E(|i\rangle\langle j|)) (\mathbf 1^a\otimes F_{}^\dagger\otimes\mathbf 1)= \\ &=d_ad(\mathbf 1^a\otimes F_{}\otimes\mathbf 1)(|\Phi^{+}_a\rangle\langle\Phi^{+}_a|\otimes J_{\mathcal E})(\mathbf 1^a\otimes F^\dagger_{}\otimes\mathbf 1), \end{aligned}$$

where $F=\sum_{\mu i}|i\rangle\langle\mu|\otimes|\mu\rangle\langle i|$ is the swap operator on the composite system Hilbert space $H^a\otimes H$ and $|\Phi^{+}_a\rangle=\frac{1}{\sqrt{d_a}}\sum_\mu|\mu\rangle\otimes|\mu\rangle$. Consequently,

$$\begin{aligned} \, S(\mathcal I^a\otimes\mathcal E)=\frac{1}{2}S(J_{\mathcal I^a\otimes\mathcal E})&= \frac{1}{2}S\bigl((\mathbf 1^a\otimes F\otimes\mathbf 1)(|\Phi_a^{+}\rangle\langle\Phi_a^{+}|\otimes J_{\mathcal E})(\mathbf 1^a\otimes F^\dagger_{}\otimes\mathbf 1)\bigr)= \\ &=\frac{1}{2}S(J_{\mathcal E}\otimes|\Phi_a^{+}\rangle\langle\Phi_a^{+}|)=\frac{1}{2} S(J_{\mathcal E})=S(\mathcal E), \end{aligned}$$

whence item 4 follows.

5. Let $\mathcal F$ be a unital channel on a $d$-dimensional system satisfying $\mathcal F(\mathbf 1)=\mathbf 1$. By the definition of Jamiołkowski–Choi states, we have $J_{\mathcal F \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E}=\mathcal I\otimes\mathcal F(J_{\mathcal E})$. It is obvious that $\mathcal I\otimes{\mathcal F}$ is also a unital channel. In view of the monotonicity of von Neumann entropy for a unital channel, we obtain

$$S(\mathcal F \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E)= \frac{1}{2}S(J_{\mathcal F \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E})=\frac{1}{2}S(\mathcal I\otimes\mathcal F(J_{\mathcal E}))\ge \frac{1}{2}S(J_{\mathcal E})=S(\mathcal E).$$

6. Let $\mathcal E^a$ and $\mathcal E^b$ be channels on systems $a$ and $b$, with orthonormal bases $\{|\mu\rangle\}$ and $\{|i\rangle\}$. Because

$$J_{\mathcal E^a\otimes\mathcal E^b}= \frac{1}{d_ad_b}\sum_{\mu\nu ij}|\mu\rangle\langle\nu|\otimes|i\rangle\langle j|\otimes\mathcal E^a(|\mu\rangle\langle\nu|)\otimes\mathcal E^b(|i\rangle\langle j|),$$

we have

$$\begin{aligned} \, (\mathbf 1^a\otimes F_{ab}^\dagger\otimes\mathbf 1^b)&J_{\mathcal E^a\otimes\mathcal E^b}(\mathbf 1^a\otimes F_{ab}\otimes\mathbf 1^b)= \\ &=\frac{1}{d_ad_b}\sum_{\mu,\nu,i,j}|\mu\rangle\langle\nu|\otimes\mathcal E^a(|\mu\rangle\langle\nu|)\otimes |i\rangle\langle j|\otimes\mathcal E^b(| i\rangle\langle j|)=J_{\mathcal E^a}\otimes J_{\mathcal E^b}, \end{aligned}$$

where $F_{ab}=\sum_{\mu i}|\mu\rangle\langle i|\otimes|i\rangle\langle\mu|$ is the swap operator on the system Hilbert space $H^a\otimes H^b$. Consequently,

$$\begin{aligned} \, S(\mathcal E^a\otimes\mathcal E^b)=\frac{1}{2}S(J_{\mathcal E^a\otimes\mathcal E^b})&= \frac{1}{2}S \bigl((\mathbf 1^a\otimes F_{ab}^\dagger\otimes\mathbf 1^b) J_{\mathcal E^a\otimes\mathcal E^b}(\mathbf 1^a\otimes F_{ab}^\dagger\otimes\mathbf 1^b)\bigr)= \\ &=\frac{1}{2}S(J_{\mathcal E^a}\otimes J_{\mathcal E^b})= \frac{1}{2}S(J_{\mathcal E^a})+\frac{1}{2}S(J_{\mathcal E^b})= S(\mathcal E^a)+S(\mathcal E^b), \end{aligned}$$

whence item 6 follows.

Appendix B: Proof of Proposition 2

For any random unitary channel $\mathcal E_{\mathrm{ru}}$, we have

$$J_{\mathcal E_{\mathrm{ru}}}= \sum_k p_k\biggl(\frac{1}{d}\sum_{i,j}|i\rangle\langle j|\otimes U_k|i\rangle\langle j|U_k^\dagger\biggr)= \sum_k p_k J_{\mathcal E_{U_k}},$$

whence

$$0\le S(J_{\mathcal E_{\mathrm{ru}}})= S\biggl(\sum_k p_k J_{\mathcal E_{U_k}}\biggr)\le H(\{p_k\})+\sum_k p_k S(J_{\mathcal E_{U_k}})=H(\{p_k\}),$$

which implies that

$$0\le S(\mathcal E_{\mathrm{ru}})\le\frac{1}{2}H(\{p_k\}).$$

By the properties of von Neumann entropy, $S(J_{\mathcal E_{\mathrm{ru}}})=0$ if and only if $\mathcal E_{\mathrm{ru}}$ is actually a unitary channel and $S(J_{\mathcal E_{\mathrm{ru}}})=H(\{p_k\})$ if and only if the states $J_{\mathcal E_{U_k}}$ have support on orthogonal subspaces, i.e., $\operatorname{tr} (J_{\mathcal E_{U_k}}J_{\mathcal E_{U_l}})=0$ for any $k\neq l$. Direct manipulation of the definition of a Jamiołkowski–Choi state shows that

$$\operatorname{tr} (J_{\mathcal E_{U_k}}J_{\mathcal E_{U_l}})=\frac{1}{d^2}| \operatorname{tr} U_k^\dagger U_l|^2.$$

Therefore, $S(J_{\mathcal E_{\mathrm{ru}}})=H(\{p_k\})$ if and only if $\operatorname{tr} (U_k^\dagger U_l)=0$ for any $k\neq l$.

Appendix C: Proof of Proposition 3

To prove Proposition 3, we first recall Pinsker’s inequality, which states that [59]

$$ S(\rho|\sigma)\ge 2T^2(\rho,\sigma)$$

(15)

for any states $\rho$ and $\sigma$, where $S(\rho|\sigma)= \operatorname{tr} \rho (\ln\rho-\ln\sigma)$ is the quantum relative entropy, and

$$T(\rho,\sigma)=\frac{1}{2} \operatorname{tr} |\rho-\sigma|$$

is the trace distance between states $\rho$ and $\sigma$ with $|A|=\sqrt{A^\dagger A}$ for any operator $A$.

By Pinsker’s inequality (15) and the triangle inequality for the trace distance, we have

$$S(J_\mathcal E|J_{\mathrm{cde}})\ge 2T^2(J_\mathcal E,J_{\mathrm{cde}})\ge 2\bigl(T(J_\mathcal E,J_{\mathcal I})-T(J_{\mathcal I},J_{\mathrm{cde}})\bigr)^2= 2\biggl(1-\frac{1}{d^2}-T(J_\mathcal E,J_{\mathcal I})\biggr)^{\!2},$$

where the equality holds by virtue of $T(J_{\mathcal I},J_{\mathrm{cde}})=1-1/d^2$. By the well-known inequality relating fidelity and trace distance [11]

$$T(\rho,\sigma)\le\sqrt{1-F(\rho,\sigma)},$$

we further have

$$\begin{aligned} \, \sqrt{S(J_\mathcal E|J_{\mathrm{cde}})}&\ge\sqrt{2}\biggl(1-\frac1{d^2}-T(J_\mathcal E,J_{\mathcal I})\biggr)\ge \\ &\ge\sqrt{2}\biggl(1-\frac{1}{d^2}-\sqrt{1-F(J_\mathcal E,J_{\mathcal I})}\biggr)= \sqrt{2}\biggl(1-\frac{1}{d^2}-\sqrt{1-F(\mathcal E)}\biggr), \end{aligned}$$

whence the desired inequality follows.

Appendix D: Proof of Proposition 4

By Eqs. (9) and (12), we have

$$\begin{aligned} \, D(\mathcal E)&=I(J_{\mathcal I})-I(J_{\mathcal E})= 2S(\mathbf 1/d)-\bigl( S(\mathbf 1/d) +S(\mathcal E(\mathbf 1/d))- S(J_\mathcal E)\bigr)= \\ &=\ln d-S(\mathcal E(\mathbf 1/d))+S(J_\mathcal E)=N(\mathcal E)+2S(\mathcal E). \end{aligned}$$

Appendix E: Proof of Proposition 5

1. By Eq. (13), we have $D(\mathcal E)\ge0$ and the equality holds if and only if $S(J_\mathcal E)=0$ and ${S(\mathcal E(\mathbf 1/d)|\mathbf 1/d)\!=\!0}$, which implies that $\mathcal E$ is a unitary channel. For the upper bound $D(\mathcal E)\le 2\ln d$, noting that

$$\begin{aligned} \, D(\mathcal E)&=S(J_\mathcal E)+S(\mathcal E(\mathbf 1/d)|\mathbf 1/d)= S(J_\mathcal E)+S(\mathbf 1/d)-S(\mathcal E(\mathbf 1/d))= \\ &=2S(\mathbf 1/d)-S(J_\mathcal E|\mathbf 1/d\otimes\mathcal E(\mathbf 1/d))\le 2S(\mathbf 1/d)=2\ln d, \end{aligned}$$

we have $D(\mathcal E)\le 2\ln d$ and $D(\mathcal E)=2\ln d$ if and only if $J_\mathcal E=\mathbf 1/d\otimes\mathcal E(\mathbf 1/d)$, which implies that $\mathcal E(\rho)=\mathcal E(\mathbf 1/d)$ for any state $\rho$.

2. Direct calculations show that

$$J_{p_1\mathcal E_1+p_2\mathcal E_2}=p_1J_{\mathcal E_1}+p_2J_{\mathcal E_2}.$$

By the joint convexity of relative entropy, we have

$$\begin{aligned} \, I(J_{p_1\mathcal E_1+p_2\mathcal E_2})&=S(J_{p_1\mathcal E_1+p_2\mathcal E_2}|\mathbf 1/d\otimes(p_1\mathcal E_1+p_2\mathcal E_2)(\mathbf 1/d))= \\ &=S(p_1J_{\mathcal E_1}+p_2J_{\mathcal E_2}|p_1\mathbf 1/d\otimes\mathcal E_1(\mathbf 1/d)+p_2\mathbf 1/d\otimes\mathcal E_2(\mathbf 1/d))\le \\ &\le p_1S(J_{\mathcal E_1}|\mathbf 1/d\otimes\mathcal E_1(\mathbf 1/d))+p_2S(J_{\mathcal E_2}|\mathbf 1/d\otimes\mathcal E_2(\mathbf 1/d))= \\ &=p_1I(J_{\mathcal E_1}) +p_2I(J_{\mathcal E_2}), \end{aligned}$$

whence item 2 follows.

3. By the unitary invariance of von Neumann entropy, we have

$$\begin{aligned} \, I(J_{\mathcal E_U \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E_V})&= S(\mathbf 1/d)+S(\mathcal E_U \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E_V(\mathbf 1/d))- S(\mathcal I\otimes\mathcal E_U \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E_V(|\Phi^{+}\rangle\langle\Phi^{+}|))= \\ &=S(\mathbf 1/d)+S(\mathcal E(\mathbf 1/d))-S(\mathcal I\otimes\mathcal E(|\Phi^{+}\rangle\langle\Phi^{+}|))=I(J_{\mathcal E}) \end{aligned}$$

for any unitary operators $U$ and $V$, which naturally implies the desired property.

4. Let $H^a$ and $H$ have respective dimensions $d_a$ and $d$, and orthonormal bases $\{|\mu\rangle\}$ and $\{|i\rangle\}$. For the channel $\mathcal I^a\otimes\mathcal E$ on the composite system Hilbert space $H^a\otimes H$, we have

$$\begin{aligned} \, d_a dJ_{\mathcal I^a\otimes\mathcal E}&= \sum_{\mu\nu ij}|\mu\rangle\langle\nu|\otimes|i\rangle\langle j|\otimes|\mu\rangle\langle\nu|\otimes\mathcal E (|i\rangle\langle j|)= \\ &=\sum_{\mu\nu ij}(\mathbf 1^a\otimes F_{ab}\otimes\mathbf 1) (|\mu\rangle\langle\nu|\otimes|\mu\rangle\langle\nu|\otimes|i\rangle\langle j|\otimes\mathcal E(|i\rangle\langle j|)) (\mathbf 1^a\otimes F^\dagger_{ab}\otimes\mathbf 1)= \\ &=d_a d(\mathbf 1^a\otimes F_{ab}\otimes\mathbf 1) (|\Phi^{+}_a\rangle\langle\Phi^{+}_a|\otimes J_{\mathcal E}) (\mathbf 1^a\otimes F^\dagger_{ab}\otimes\mathbf 1) \end{aligned}$$

where $F_{ab}=\sum_{\mu i}|i\rangle\langle\mu|\otimes|\mu\rangle\langle i|$ is the swap operator on the composite system $H^a\otimes H$ and $|\Phi^{+}_a\rangle=\frac{1}{\sqrt{d_a}}\sum_\mu|\mu\rangle\otimes|\mu\rangle$. Consequently,

$$\begin{aligned} \, S(J_{\mathcal I^a\otimes\mathcal E})&= S\bigl((\mathbf 1^a\otimes F_{ab}\otimes\mathbf 1) (|\Phi_a^{+}\rangle\langle\Phi_a^{+}|\otimes J_{\mathcal E}) (\mathbf 1^a\otimes F^\dagger_{ab}\otimes\mathbf 1)\bigr)= \\ &=S(J_{\mathcal E}\otimes|\Phi_a^{+}\rangle\langle\Phi_a^{+}|)=S(J_{\mathcal E}) \end{aligned}$$

and

$$\begin{aligned} \, S(\mathbf 1^a/d_a\otimes\mathcal E(\mathbf 1/d)|\mathbf 1^a/{d_a}\otimes\mathbf 1/{d})&= S(\mathbf 1^a/d_a)+S(\mathbf 1/d)-S(\mathbf 1^a/d_a\otimes\mathcal E(\mathbf 1/d))= \\ &=S(\mathbf 1^a/d_a)+S(\mathbf 1/d)-S(\mathbf 1^a/d_a)-S(\mathcal E(\mathbf 1/d))= \\ &=S(\mathbf 1/d)-S(\mathcal E(\mathbf 1/d))=S(\mathcal E(\mathbf 1/d)|\mathbf 1/d), \end{aligned}$$

whence item 4 follows.

5. Let $\mathcal F$ be any channel on a system Hilbert space $H$. By the definition of Jamiołkowski–Choi states, we have $J_{\mathcal F \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E}=\mathcal I\otimes\mathcal F(J_{\mathcal E})$. In view of the monotonicity of relative entropy, item 5 follows from

$$\begin{aligned} \, I(J_{\mathcal F \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E})=I(\mathcal I\otimes\mathcal F(J_{\mathcal E}))&= S(\mathcal I\otimes\mathcal F(J_{\mathcal E})|\mathcal I\otimes\mathcal{F}(\mathbf 1/d\otimes\mathcal E(\mathbf 1/d)))\le \\ &\le S(J_{\mathcal E}|\mathbf 1/d\otimes\mathcal E(\mathbf 1/d))=I(J_{\mathcal E}). \end{aligned}$$

6. Let $\mathcal E^a$ and $\mathcal E^b$ be channels on the systems $a$ and $b$, with respective Hilbert spaces $H^a$ and $H^b$, and orthonormal bases $\{|\mu\rangle\}$ and $\{|i\rangle\}$. Because

$$J_{\mathcal E^a\otimes\mathcal E^b}=\frac{1}{d_ad_b}\sum_{\mu\nu ij} |\mu\rangle\langle\nu|\otimes|i\rangle\langle j|\otimes\mathcal E^a(|\mu\rangle\langle\nu|)\otimes\mathcal E^b(|i\rangle\langle j|),$$

we have

$$(\mathbf 1^a\otimes F^\dagger\otimes\mathbf 1^b) J_{\mathcal E^a\otimes\mathcal E^b}(\mathbf 1^a\otimes F\otimes\mathbf 1^b)=\frac{1}{d_ad_b}\sum_{\mu\nu ij} |\mu\rangle\langle\nu|\otimes\mathcal E^a(|\mu\rangle\langle\nu|)\otimes|i\rangle\langle j|\otimes\mathcal E^b(|i\rangle\langle j|)= J_{\mathcal E^a}\otimes J_{\mathcal E^b},$$

where $F_{ab}=\sum_{\mu i}|\mu\rangle\langle i|\otimes|i\rangle\langle\mu|$ is the swap operator on the system Hilbert space $H^a\otimes H^b$. Consequently,

$$S(J_{\mathcal E^a\otimes\mathcal E^b})= S\bigl((\mathbf 1^a\otimes F_{ab}\otimes\mathbf 1^b)J_{\mathcal E^a\otimes\mathcal E^b} (\mathbf 1^a\otimes F^\dagger_{ab}\otimes\mathbf 1^b)\bigr)=S(J_{\mathcal E^a}\otimes J_{\mathcal E^b})=S(J_{\mathcal E^a})+S(J_{\mathcal E^b}),$$

and

$$\begin{aligned} \, S(\mathcal E^a(\mathbf 1^a/d_a)\otimes\mathcal E^b&(\mathbf 1^b/d_b)|\mathbf 1^a/d_a\otimes\mathbf 1^b/d_b)=S(\mathbf 1^a/d_a\otimes\mathbf 1^b/d_b)-S(\mathcal E^a(\mathbf 1^a/d_a)\otimes\mathcal E^b(\mathbf 1^b/d_b))= \\ &=S(\mathbf 1^a/d_a)+S(\mathbf 1^b/d_b)-S(\mathcal E^a(\mathbf 1^a/d_a))-S(\mathcal E^b(\mathbf 1^b/d_b))= \\ &=S(\mathcal E^a(\mathbf 1^a/d_a)|\mathbf 1^a/d_a)+S(\mathcal E^b(\mathbf 1^b/d_b|\mathbf 1^b/d_b), \end{aligned}$$

whence item 6 follows.

Appendix F: An alternative measure of irreversibility

We recall that the Tsallis $r$-entropy

$$S_r(\rho)=\frac {1- \operatorname{tr} \rho^r}{r-1},\qquad r\in\mathbb{R},$$

is a simple and significant quantity characterizing the mixedness of a state $\rho$ [60]. The case $r=1$ is understood as the limit $r\to 1$ and actually corresponds to the von Neumann entropy. If we take $\rho$ to be $J_\mathcal E$ of a channel $\mathcal E$, then we can regard

$$S_r(\mathcal E)=\frac{1}{2} S_r(J_\mathcal E)$$

as a measure of irreversibility of the channel $\mathcal E(\rho)=\sum_kE_k\rho E_k^\dagger$. For $r\in\mathbb{N}$, $S_r(\mathcal E)$ has the explicit form

$$S_r^{}(\mathcal E)=\frac{1}{2(r-1)} \biggl(1-\frac{1}{d^r}\sum_{k_1,\ldots,k_r} \operatorname{tr} (E_{k_1}^\dagger E_{k_2}^{}) \operatorname{tr} (E_{k_2}^\dagger E_{k_3}^{})\ldots \operatorname{tr} (E_{k_r}^\dagger E_{k_1}^{})\biggr).$$

In particular, the Tsallis 2-entropy $S_2(\rho)=1- \operatorname{tr} \rho^2$ is the linear entropy, and if we take $\rho$ to be $J_\mathcal E$ of a channel $\mathcal E$, then we can regard

$$S_2(\mathcal E)=\frac{1}{2} S_2(J_\mathcal E)$$

as a measure of irreversibility of the channel $\mathcal E$. It is interesting to note that this quantity can also be expressed as

$$S_2(\mathcal E)=\frac{1}{2}-\frac{1}{2d^2}\sum_{k}\|\mathcal E(X_k)\|^2$$

where $\{X_k\colon k=1,\ldots, d^2\}$ is any orthonormal basis of the operator space $L(H)$ of all observables (Hermitian operators) on $H$ with the Hilbert–Schmidt inner product $\langle A|B\rangle= \operatorname{tr} AB$. The quantity $S_2(\mathcal E)$ satisfies the following properties, which parallel those of $S(\mathcal E)$.

• 1. We have

  $$0\le S_2(\mathcal E)\le\frac{1}{2}-\frac{1}{2d^2},$$

  and $S_2(\mathcal E)=0$ if and only if $\mathcal E$ is a unitary channel, while $S_2(\mathcal E)$ attains the maximum value $(d^2-1)/2d^2$ if and only if $\mathcal E$ is the completely depolarizing channel $\mathcal E_{\mathrm{cde}}(\rho)=\mathbf 1/d$ for any state $\rho$.

• 2. $S_2(\mathcal E)$ is concave in $\mathcal E$, i.e.,

  $$S_2(p_1\mathcal E_1+p_2\mathcal E_2)\ge p_1 S_2(\mathcal E_1)+p_2 S_2(\mathcal E_2)$$

  for $p_1,p_2\ge 0$, $p_1+p_2=1$, and any channels $\mathcal E_1$ and $\mathcal E_2$.

• 3. $S_2(\,{\cdot}\,)$ is invariant under composition with unitary dynamics in the sense that

  $$S_2(\mathcal E_U \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E)=S_2(\mathcal E \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E_U)=S_2(\mathcal E)$$

  for any unitary channel $\mathcal E_U(\rho)=U\rho U^\dagger$ with $U$ being any unitary operator on the system Hilbert space.

• 4. $S_2(\,{\cdot}\,)$ is ancilla-independent in the sense that $S_2(\mathcal I^a\otimes\mathcal E)=S_2(\mathcal E)$, where $\mathcal I^a$ is the identity channel on any ancilla system $a$.

• 5. We have

  $$S_2(\mathcal E^a\otimes\mathcal E^b)=S_2(\mathcal E^a)+S_2(\mathcal E^b)-S_2(\mathcal E^a)S_2(\mathcal E^b),$$

  where $\mathcal E^a$ and $\mathcal E^b$ are channels on systems $a$ and $b$. This is a kind of nonextensitivity of the Tsallis entropy.

• 6. $S_2(\,{\cdot}\,)$ is monotonic in the sense that

  $$S_2(\mathcal F \mathbin{\stackrel{\scriptscriptstyle{\circ}}{{}_{\vphantom{.}}}} \mathcal E)\ge S_2(\mathcal E)$$

  for any unital channel $\mathcal F$.

The measure of irreversibility $S_2(\mathcal E)$ can be explicitly evaluated for various channels studied in Sec. 7. We list the results, together with those for $S(\mathcal E)$, in Table 1.