Quantum Optimal Transport with Quantum Channels

Abstract

We propose a new generalization to quantum states of the Wasserstein distance, which is a fundamental distance between probability distributions given by the minimization of a transport cost. Our proposal is the first where the transport plans between quantum states are in natural correspondence with quantum channels, such that the transport can be interpreted as a physical operation on the system. Our main result is the proof of a modified triangle inequality for our transport distance. We also prove that the distance between a quantum state and itself is intimately connected with the Wigner-Yanase metric on the manifold of quantum states. We then specialize to quantum Gaussian systems, which provide the mathematical model for the electromagnetic radiation in the quantum regime. We prove that the noiseless quantum Gaussian attenuators and amplifiers are the optimal transport plans between thermal quantum Gaussian states, and that our distance recovers the classical Wasserstein distance in the semiclassical limit.

Appendices

Appendix A. Continuity of the Quantum Wasserstein Distance

Compactness of Quantum Plans

Recall that a sequence $\{ A_n \}_{n=0}^\infty \subset {\mathcal {B}}({\mathcal {H}})$ weakly converges towards $A \in {\mathcal {B}}({\mathcal {H}})$ if, for every $|\phi \rangle , |\psi \rangle \in {\mathcal {H}}$,

$$\begin{aligned} \lim _{n \rightarrow \infty } \langle \phi | A_n| \psi \rangle = \langle \phi | A | \psi \rangle . \end{aligned}$$

(136)

Lemma 4

Given $\{\rho _n\}_{n =0 }^\infty \subseteq {\mathcal {S}}({\mathcal {H}})$, there exists $\rho \in {\mathcal {T}}_1({\mathcal {H}})$, $\rho \ge 0$, $\mathrm {Tr}_{{\mathcal {H}}}\left[ \rho \right] \le 1$ and a subsequence $\{\rho _{n_k}\}_{k=0}^\infty$ weakly converging towards $\rho$ . If $\mathrm {Tr}_{{\mathcal {H}}}\left[ \rho \right] =1$ then convergence holds in the trace norm.

Proof

The first statement is a consequence of Banach–Alaoglu theorem and lower semicontinuity of the trace, see also the proof of [35, Theorem 11.2]. The second statement is [35, Lemma 11.1]. $\square$

Because of the result above, we may consider weak convergence or equivalently in the trace norm for a sequence of quantum states $\{\rho _n\}_{n =0}^\infty$, provided that the limit $\rho$ is a quantum state. Hence, we often omit to specify which type of convergence we consider on quantum states in the results below.

Proposition 8

Let $\rho$, $\sigma \in {\mathcal {S}}({\mathcal {H}})$ be quantum states and let $\{\Pi _n\}_{n=0}^\infty \subset {\mathcal {S}}({\mathcal {H}}\otimes {\mathcal {H}}^*)$ be a sequence of quantum couplings $\Pi _n \in {\mathcal {C}}(\rho _n, \sigma _n)$ such that $\lim _n \rho _n = \rho$, $\lim _n \sigma _n = \sigma$. Then, there exists a subsequence $\{ \Pi _{n_k} \}_{k=0}^\infty$ converging towards $\Pi \in {\mathcal {C}}(\rho , \sigma )$.

Proof

Existence of a weak limit $\Pi = \lim _{k} \Pi _{n_k} \in {\mathcal {T}}_1({\mathcal {H}})$, $\Pi \ge 0$, follows from Proposition 4. To argue that $\mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*} [\Pi ] = 1$, let P, $Q \in {\mathcal {B}}({\mathcal {H}})$ be projectors with $P+Q = {\mathbb {I}}_{{\mathcal {H}}}$ and P with finite rank. We have

$$\begin{aligned} \mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*}[ \Pi ]&\ge \mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*}[ (P \otimes P^T) \Pi (P \otimes P^T)]\nonumber \\&= \lim _{n\rightarrow \infty } \mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*}[ (P \otimes P^T) \Pi _n (P \otimes P^T)], \end{aligned}$$

(137)

where the limit holds because $P\otimes P^T$ has finite rank. Then,

$$\begin{aligned}&\mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*}[ (P \otimes P^T) \Pi _n (P \otimes P^T)] \nonumber \\&= \mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*} [\Pi _n] - \mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*}[ (P \otimes Q^T) \Pi _n (P \otimes Q^T)]\nonumber \\&\quad -\mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*}[ (Q \otimes {\mathbb {I}}_{{\mathcal {H}}^*}) \Pi _n (Q \otimes {\mathbb {I}}_{{\mathcal {H}}^*})]\nonumber \\&= 1 - \mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*}[ (P \otimes Q^T) \Pi _n (P \otimes Q^T)] -\mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*}[ (Q \otimes {\mathbb {I}}_{{\mathcal {H}}^*}) \Pi _n (Q \otimes {\mathbb {I}}_{{\mathcal {H}}^*})]. \end{aligned}$$

(138)

We have the inequality

$$\begin{aligned} \mathrm {Tr}_{{\mathcal {H}}}[(P\otimes Q^T) \Pi _n (P\otimes Q^T)] \le \mathrm {Tr}_{{\mathcal {H}}}[({\mathbb {I}}_{{\mathcal {H}}} \otimes Q^T) \Pi _n ( {\mathbb {I}}_{{\mathcal {H}}}\otimes Q^T)] \le Q^T \rho _n^T Q^T. \end{aligned}$$

(139)

Taking the partial trace with respect to ${\mathcal {H}}^*$,

$$\begin{aligned} \limsup _{n\rightarrow \infty } \mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*}[ (P \otimes Q^T) \Pi _n (P \otimes Q^T)]&\le \limsup _{n\rightarrow \infty } \mathrm {Tr}_{{\mathcal {H}}^*}[ Q^T \rho ^T_n Q^T]\nonumber \\&= \mathrm {Tr}_{{\mathcal {H}}^*}[Q^T\rho ^TQ^T] = \mathrm {Tr}_{{\mathcal {H}}}[Q\rho Q]. \end{aligned}$$

(140)

Similarly, $\limsup _n \mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*}[ (Q \otimes {\mathbb {I}}_{{\mathcal {H}}^*}) \Pi _n (Q \otimes {\mathbb {I}}_{{\mathcal {H}}^*})] \le \mathrm {Tr}_{{\mathcal {H}}}[ Q \sigma Q]$. It follows that

$$\begin{aligned} \mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*}[ \Pi ] \ge 1- \mathrm {Tr}_{{\mathcal {H}}}[ Q \rho Q]-\mathrm {Tr}_{{\mathcal {H}}}[ Q \sigma Q]. \end{aligned}$$

(141)

Letting $P \rightarrow {\mathbb {I}}_{{\mathcal {H}}}$ weakly, both $Q\rho Q$ and $Q \sigma Q$ converge to 0 in the trace norm, hence the thesis. $\square$

A similar argument gives the following result.

Lemma 5

Given quantum states $\rho$, $\sigma \in {\mathcal {S}}({\mathcal {H}})$ let $\{ P_n \}_{n=0}^\infty$, $\{ Q_n\}_{n=0}^\infty \subset {\mathcal {B}}({\mathcal {H}})$ be such that $0 \le P_n \le {\mathbb {I}}_{{\mathcal {H}}}$, $0 \le Q_n \le {\mathbb {I}}_{{\mathcal {H}}}$, and $\lim _n P_n\rho P_n = \rho$, $\lim _n Q_n \sigma Q_n= \sigma$. Then, for any $\Pi \in {\mathcal {C}}(\rho , \sigma )$,

$$\begin{aligned} \lim _{n\rightarrow \infty } \left( Q_n \otimes P^T_n\right) \Pi \left( Q_n \otimes P^T_n\right) = \Pi \end{aligned}$$

(142)

in the trace norm.

Proof

The assumption $\lim _n P_n \rho P_n = \rho$ implies that $\{P_n\}_{n=0}^\infty$ weakly converges to the identity operator on $\mathrm {supp}\, \rho$. By a density argument, it is sufficient to prove that, for any $|\phi \rangle$ eigenvector for $\rho$, $\rho |\phi \rangle = p | \phi \rangle$, with $p>0$, one has $\lim _n P_n |\phi \rangle = |\phi \rangle$. Since $p |\phi \rangle \langle \phi | \le \rho$, we have indeed

$$\begin{aligned}&\Vert ({\mathbb {I}}_{{\mathcal {H}}} - P_n) |\phi \rangle \Vert ^2 = \mathrm {Tr} [ ({\mathbb {I}}_{{\mathcal {H}}} - P_n) |\phi \rangle \langle \phi | ({\mathbb {I}}_{{\mathcal {H}}} - P_n)] \nonumber \\&\le p^{-1} \mathrm {Tr}[({\mathbb {I}}_{{\mathcal {H}}} - P_n) \rho ({\mathbb {I}}_{{\mathcal {H}}} - P_n)] \rightarrow 0. \end{aligned}$$

(143)

Similarly, we have that $\{Q_n\}_{n=0}^\infty$ weakly converges to the identity operator on $\mathrm {supp}\, \sigma$. We deduce that $\{ Q_n \otimes P_n^T \}_{n=0}^\infty$ weakly converges to the identity operator on $\mathrm {supp}\, \sigma \otimes \mathrm {supp}\, \rho ^T$, which contains $\mathrm {supp}\, \Pi$. Hence, (142) holds in the weak sense.

We prove next that

$$\begin{aligned} \lim _{n\rightarrow \infty } \mathrm {Tr}[ (Q_n \otimes P^T_n) \Pi (Q_n \otimes P^T_n)] = 1. \end{aligned}$$

(144)

We write

$$\begin{aligned}&\mathrm {Tr}[ (Q_n \otimes P^T_n) \Pi (Q_n \otimes P^T_n)] \nonumber \\&= 1-\mathrm {Tr}[ \left( Q_n \otimes ({\mathbb {I}}_{{\mathcal {H}}^*} - P_n^T)\right) \, \Pi \, \left( Q_n \otimes ( {\mathbb {I}}_{{\mathcal {H}}^*} - P_n^T\right) ]\nonumber \\&\quad -\mathrm {Tr}[ \left( ( {\mathbb {I}}_{{\mathcal {H}}} -Q_n) \otimes {\mathbb {I}}_{{\mathcal {H}}^*} \right) \Pi \left( ({\mathbb {I}}_{{\mathcal {H}}} -Q_n) \otimes {\mathbb {I}}_{{\mathcal {H}}^*} \right) ]. \end{aligned}$$

(145)

If we take the partial trace with respect to ${\mathcal {H}}$, then

$$\begin{aligned}&\mathrm {Tr}_{{\mathcal {H}}} [ \left( Q_n \otimes ({\mathbb {I}}_{{\mathcal {H}}^*} - P_n^T)\right) \, \Pi \, \left( Q_n \otimes ( {\mathbb {I}}_{{\mathcal {H}}^*} - P_n^T\right) ]\nonumber \\&\le ({\mathbb {I}}_{{\mathcal {H}}^*} - P_n^T) \mathrm {Tr}_{{\mathcal {H}}} [\Pi ] ({\mathbb {I}}_{{\mathcal {H}}^*} - P_n^T) = (\rho -P_n \rho P_n)^T, \end{aligned}$$

(146)

so that

$$\begin{aligned} \lim _{n\rightarrow \infty } \mathrm {Tr}[ \left( Q_n \otimes ({\mathbb {I}}_{{\mathcal {H}}^*} - P_n^T)\right) \, \Pi \, \left( Q_n \otimes ( {\mathbb {I}}_{{\mathcal {H}}^*} - P_n^T\right) ] =0. \end{aligned}$$

(147)

Arguing similarly, taking the partial trace with respect to ${\mathcal {H}}^*$,

$$\begin{aligned} \lim _{n\rightarrow \infty } \mathrm {Tr}\left[ \left( ( {\mathbb {I}}_{{\mathcal {H}}} -Q_n) \otimes {\mathbb {I}}_{{\mathcal {H}}^*} \right) \Pi \left( ({\mathbb {I}}_{{\mathcal {H}}} -Q_n) \otimes {\mathbb {I}}_{{\mathcal {H}}^*}\right) \right]&\le \lim _{n\rightarrow \infty } \mathrm {Tr}_{{\mathcal {H}}}\left[ \sigma - Q_n \sigma Q_n\right] \nonumber \\&=0, \end{aligned}$$

(148)

and (144) follows.

To conclude that (142) holds in the trace norm, define for n sufficiently large the quantum state

$$\begin{aligned} \Pi _n = (Q_n \otimes P^T_n) \Pi (Q_n \otimes P^T_n) \bigg / \mathrm {Tr}\left[ (Q_n \otimes P^T_n) \Pi (Q_n \otimes P^T_n)\right] , \end{aligned}$$

(149)

so that $\lim _n \Pi _n = \Pi$ weakly, hence in the trace norm. By (144) this implies convergence in the trace norm also in (142). $\square$

Energy

We prove in this subsection some results that are needed to deal with the energy of quantum states with respect to unbounded operators.

Lemma 6

Let ${\mathcal {H}}_A$, ${\mathcal {H}}_B$ be Hilbert spaces and let A be a self-adjoint operator on ${\mathcal {H}}_A$. Then, there exists a sequence of bounded self-adjoint operators $\{A_n\}_{n=0}^\infty \subset {\mathcal {B}}({\mathcal {H}})$ such that, for every for every quantum state $\rho \in {\mathcal {S}}({\mathcal {H}}_A\otimes {\mathcal {H}}_B)$, $E_{A_n \otimes {\mathbb {I}}_B}(\rho )$ increases towards $E_{A \otimes {\mathbb {I}}_B}(\rho )$.

Proof

It is sufficient to consider the case of a pure state $\rho = |\psi \rangle \langle \psi |$, because the general case follows by Remark 6 and the monotone convergence theorem for series.

Define by functional calculus [66, Theorem VIII.5] $A_n = A \chi _{(-n, n)}(A)$, where $\chi _{(-n,n)}(x) = 1$ if $x \in (-n,n)$, $\chi _{(-n,n)}(x) = 0$ otherwise. Let us argue without the Hilbert space ${\mathcal {H}}_B$ first (i.e., ${\mathcal {H}}_B = {\mathbb {C}}$). Then,

$$\begin{aligned} E_{A_n}(\rho ) = \left\| A_n|\psi \rangle \right\| ^2 = \int _{-n}^n \lambda ^2 \mathrm {d} \langle \psi | P_\lambda | \psi \rangle , \end{aligned}$$

(150)

where $P_\lambda = \chi _{(-\infty , \lambda ]}(A)$ denotes the resolution of the identity associated to A. The expression is increasing with respect to n and $\sup _{n} \left\| A_n|\psi \rangle \right\| ^2 < \infty$ if and only if $\psi$ belongs to the domain of A, because, defining $|\psi _n\rangle = \chi _{(-n,n)}(A) |\psi \rangle$, the sequence $\{ |\psi _n\rangle \}_{k=0}^\infty \subset {\mathcal {H}}$ converges towards $|\psi \rangle$ with $A |\psi _n\rangle$ bounded and A is a closed operator. In this case, we also have $\lim _n A_n |\psi \rangle = A |\psi \rangle$ by [66, Theorem VIII.5, c)], hence the conclusion in this case. The thesis with the additional Hilbert space ${\mathcal {H}}_B$ follows from repeating the argument by noticing that

$$\begin{aligned} A_n \otimes {\mathbb {I}}_{B} = (A \otimes {\mathbb {I}}_B) \chi _{(-n,n)}(A \otimes {\mathbb {I}}_B). \end{aligned}$$

(151)

$\square$

The following result shows that $E_A(\rho )$ does not depend on the ensemble that generates $\rho$.

Proposition 9

If $\rho = \sum _{k=0}^\infty |\phi _k\rangle \langle \phi _k|$ with $\{|\phi _k\rangle \}_{k=0}^\infty \subset {\mathcal {H}}$, then

$$\begin{aligned} E_A(\rho ) = \sum _{k=0}^\infty \left\| A|\phi _k\rangle \right\| ^2, \end{aligned}$$

(152)

where we let $\left\| A|\phi \rangle \right\| ^2 =\infty$ if $| \phi \rangle$ does not belong to the domain of A.

Analogously, let $\mu$ be a probability measure on ${\mathcal {H}}$, and let

$$\begin{aligned} \rho = \int _{\mathcal {H}}|\psi \rangle \langle \psi |\,\mathrm {d}\mu (\psi )\,. \end{aligned}$$

(153)

Then,

$$\begin{aligned} E_A(\rho ) = \int _{\mathcal {H}}\left\| A|\psi \rangle \right\| ^2\mathrm {d}\mu (\psi )\,. \end{aligned}$$

(154)

Proof

If we define $|{{\hat{\phi }}}_k\rangle = |\phi _k\rangle / \Vert |\phi _k\rangle \Vert$, then the thesis can be rewritten as

$$\begin{aligned} E_A(\rho ) = \sum _{k=0}^\infty \Vert |\phi _k \rangle \Vert ^2 E_A(|\hat{\phi }_k \rangle \langle {{\hat{\phi }}}_k | ), \end{aligned}$$

(155)

hence, by Lemma 6 and the monotone convergence theorem for series, it is sufficient to prove the result for $A \in {\mathcal {B}}({\mathcal {H}})$.

Arguing as in [34, Theorem 2.6] (noticing that the proof holds also in the infinite dimensional case), there exists a unitary map $U: \ell ^2({\mathbb {N}}) \rightarrow \ell ^2({\mathbb {N}})$, $U^\dagger U = {\mathbb {I}}_{\ell ({\mathbb {N}})}$, i.e., $(u_{ij})_{i,j=0}^\infty$ with $\sum _{k=0}^\infty u_{ki}^* u_{kj} = \delta _{ij}$, such that

$$\begin{aligned} |\phi _k\rangle = \sum _{i=0}^\infty u_{ki} \sqrt{p_i} | \psi _i\rangle . \end{aligned}$$

(156)

Then,

$$\begin{aligned}&\sum _{k=0}^\infty \left\| A|\phi _k\rangle \right\| ^2 = \sum _{k=0}^\infty \langle \phi _k| A^2 | \phi _k\rangle = \sum _{k=0}^\infty \sum _{i,j=0}^\infty \sqrt{p_i p_j} u_{ki}^* u_{kj} \langle \psi _i| A^2 | \psi _j\rangle \nonumber \\&= \sum _{i,j=0}^\infty \sqrt{p_i p_j}\langle \psi _i| A^2 | \psi _j\rangle \left( \sum _{k=0}^\infty u_{ki}^* u_{kj}\right) = \sum _{i=0}^\infty p_i \langle \psi _i| A^2 | \psi _i\rangle = E_A(\rho ), \end{aligned}$$

(157)

and the claim (152) follows.

The integral version (154) of the claim can be proved along the same lines employing Beppo Levi’s monotone convergence theorem [67, Theorem 2.8.2] and Fubini’s theorem [67, Theorem 3.4.4]. $\square$

Proposition 10

$E_{A}$ is lower semicontinuous, i.e., if $\{\rho _n\}_{n=0}^\infty \subset {\mathcal {S}}({\mathcal {H}})$ converge towards $\rho \in {\mathcal {S}}({\mathcal {H}})$, then $E_A(\rho ) \le \liminf _n E_A(\rho _n)$.

Proof

If A is bounded then, by Remark 7, $E_A(\rho ) = \mathrm {Tr}_{{\mathcal {H}}} \left[ A \, \rho \, A\right]$, which is continuous with respect to the trace norm. The general case follows by Lemma 6, since $E_A(\rho ) = \sup _{n} E_{A_n}(\rho )$ and a pointwise supremum of continuous functions is lower semicontinuous. $\square$

Proposition 11

Let A, B be, respectively, self-adjoint operators on Hilbert spaces ${\mathcal {H}}_A$, ${\mathcal {H}}_B$ and let $\Pi \in {\mathcal {S}}({\mathcal {H}}_A \otimes {\mathcal {H}}_B)$ with $\rho = \mathrm {Tr}_{A}[\Pi ]$, $\sigma = \mathrm {Tr}_B[\Pi ]$. Then,

$$\begin{aligned} E_{A \otimes {\mathbb {I}}_{B}} (\Pi ) = E_{A}(\sigma ), \quad E_{{\mathbb {I}}_A \otimes B} (\Pi ) = E_{B}(\sigma ). \end{aligned}$$

(158)

If both energies above are finite, then

$$\begin{aligned} E_{A \otimes {\mathbb {I}}_{B} + {\mathbb {I}}_A \otimes B} (\Pi ) + E_{A \otimes {\mathbb {I}}_{B} - {\mathbb {I}}_A \otimes B} (\Pi )= 2 E_{A}(\rho ) + 2 E_{B}(\sigma ). \end{aligned}$$

(159)

Proof

We prove the first identity in (158), the second one being similar. By Lemma 6, it is sufficient to argue when A is bounded. Then, by Remark 7,

$$\begin{aligned}&E_{A \otimes {\mathbb {I}}_B}(\Pi ) = \mathrm {Tr}_{AB} [ (A \otimes {\mathbb {I}}_B) \Pi (A \otimes {\mathbb {I}}_B)] \nonumber \\&\quad = \mathrm {Tr}_{A} [ A \mathrm {Tr}_B[ \Pi ] A ] = \mathrm {Tr}_{A}[A \sigma A] = E_{A}(\sigma ). \end{aligned}$$

(160)

To prove (159), by Remark 6, is it sufficient to consider the case of a pure state $\Pi = |\psi \rangle \langle \psi |$, so that it reads

$$\begin{aligned}&\Vert \left( A \otimes {\mathbb {I}}_{B} + {\mathbb {I}}_A \otimes B \right) | \psi \rangle \Vert ^2 + \Vert \left( A \otimes {\mathbb {I}}_{B} - {\mathbb {I}}_A \otimes B \right) | \psi \rangle \Vert ^2\nonumber \\&= 2 \Vert \left( A \otimes {\mathbb {I}}_{B} \right) | \psi \rangle \Vert ^2 + 2 \Vert \left( {\mathbb {I}}_A \otimes B \right) | \psi \rangle \Vert ^2 \end{aligned}$$

(161)

assuming that both terms in the right-hand side are finite, i.e., $|\psi \rangle$ belongs to the domain of $A\otimes {\mathbb {I}}_B$ and ${\mathbb {I}}_A \otimes B$. But then

$$\begin{aligned} \left( A \otimes {\mathbb {I}}_{B} + {\mathbb {I}}_A \otimes B \right) | \psi \rangle = \left( A \otimes {\mathbb {I}}_{B} \right) | \psi \rangle + \left( {\mathbb {I}}_A \otimes B \right) | \psi \rangle , \end{aligned}$$

(162)

$$\begin{aligned} \left( A \otimes {\mathbb {I}}_{B} - {\mathbb {I}}_A \otimes B \right) | \psi \rangle = \left( A \otimes {\mathbb {I}}_{B} \right) | \psi \rangle - \left( {\mathbb {I}}_A \otimes B \right) | \psi \rangle , \end{aligned}$$

(163)

hence the thesis by straightforward computation. $\square$

Convergence of the Quantum Wasserstein Distance

As in Sect. 3, we fix quadratures $\{R_1,\,\ldots ,\,R_N\}$, i.e., self-adjoint operators on ${\mathcal {H}}$. Then, the cost associated with a quantum plan $\Pi \in {\mathcal {C}}(\rho , \sigma )$ (Definition 5) reads

$$\begin{aligned} C(\Pi ) = \sum _{i=1}^N E_{R_i\otimes {\mathbb {I}}_{{\mathcal {H}}^*} - {\mathbb {I}}_{{\mathcal {H}}}\otimes R_i^T}(\Pi ) \end{aligned}$$

(164)

and a quantum state $\rho \in {\mathcal {S}}({\mathcal {H}})$ has finite energy (Definition 6) if

$$\begin{aligned} \sum _{i=1}^N E_{R_i}(\rho ) <\infty \,. \end{aligned}$$

(165)

We are in a position to give a proof of Proposition 3.

Proposition 3Let $\rho ,\,\sigma \in {\mathcal {S}}({\mathcal {H}})$ have finite energy. Then, any plan $\Pi \in {\mathcal {C}}(\rho ,\sigma )$ has finite cost.

Proof

For $i=1, \, \ldots , N$, we apply Proposition 11 with $A = R_i$, $B = R_i^T$ so that

$$\begin{aligned}&C(\Pi ) = \sum _{i=1}^N E_{R_i\otimes {\mathbb {I}}_{{\mathcal {H}}^*} - {\mathbb {I}}_{{\mathcal {H}}}\otimes R_i^T}(\Pi ) \nonumber \\&\quad = \sum _{i=1}^N\left( 2 E_{R_i}(\rho ) + 2 E_{R_i}(\sigma ) - E_{R_i\otimes {\mathbb {I}}_{{\mathcal {H}}^*} - {\mathbb {I}}_{{\mathcal {H}}}\otimes R_i^T}(\Pi )\right) < \infty , \end{aligned}$$

(166)

where we also used that $E_{R_i^T}(\rho ^T) = E_{R_i}(\rho )$. $\square$

Proposition 12

Let $\rho$, $\sigma \in {\mathcal {S}}({\mathcal {H}})$ be quantum states with finite energy and let $\{\Pi _n\}_{n=0}^\infty \subset {\mathcal {S}}({\mathcal {H}}\otimes {\mathcal {H}}^*)$ be a sequence of quantum couplings $\Pi _n \in {\mathcal {C}}(\rho _n, \sigma _n)$ converging towards $\Pi \in {\mathcal {C}}(\rho , \sigma )$. Then, $C(\Pi ) \le \liminf _{n} C(\Pi _n)$. If moreover $\lim _{n}E_{R_i}(\rho _n) = E_{R_i}(\rho )$ and $\lim _{n}E_{R_i}(\sigma _n) = E_{R_i}(\sigma )$ for $i =1, \ldots , N$, then $C(\Pi ) = \lim _n C(\Pi _n)$.

Proof

The first inequality follows from Proposition 10 applied to each term in (164). Assuming convergence of the energies of both marginals, it is sufficient to show that that $C(\Pi ) \ge \limsup _n C(\Pi _n)$. To this aim we use (166) and again Proposition 10 to argue that

$$\begin{aligned} -E_{\left( R_i\otimes {\mathbb {I}}_{{\mathcal {H}}^*} + {\mathbb {I}}_{{\mathcal {H}}}\otimes R_i^T\right) }(\Pi _n) \ge \limsup _{n \rightarrow \infty } - E_{\left( R_i\otimes {\mathbb {I}}_{{\mathcal {H}}^*} + {\mathbb {I}}_{{\mathcal {H}}}\otimes R_i^T\right) }(\Pi _n). \end{aligned}$$

(167)

$\square$

Proposition 13

Let $\rho$, $\sigma \in {\mathcal {S}}({\mathcal {H}})$ be quantum states with finite energy. Then, there exists $\Pi \in {\mathcal {C}}(\rho , \sigma )$ such that $C(\Pi ) = D(\rho , \sigma )^2$.

Proof

Let $\Pi _n \in {\mathcal {C}}(\rho , \sigma )$ be such that $\lim _n C(\Pi _n) = D(\rho , \sigma )^2$. By Lemma 8 we can assume, up to extracting a sub-sequence, that $\{\Pi _n\}_{n=0}^\infty$ converge towards $\Pi \in {\mathcal {C}}(\rho , \sigma )$. By Proposition 12, $D(\rho , \sigma )^2 \le C(\Pi ) \le \liminf _n C(\Pi _n) = D(\rho ,\sigma )^2$, hence $\Pi$ is optimal. $\square$

Theorem 9

Let $\rho$, $\sigma \in {\mathcal {S}}({\mathcal {H}})$ be quantum states with finite energy.

• If $\{\rho _n\}_{n=0}^\infty$, $\{ \sigma _n\}_{n=0}^\infty \subset {\mathcal {S}}({\mathcal {H}})$ have finite energy and converge respectively towards $\rho$, $\sigma$, then

  $$\begin{aligned} D(\rho , \sigma ) \le \liminf _{n\rightarrow \infty } D(\rho _n, \sigma _n)\,. \end{aligned}$$

  (168)

• Let $\{ P_n \}_{n=0}^\infty$, $\{ Q_n\}_{n=0}^\infty \subset {\mathcal {B}}({\mathcal {H}})$ be such that $0 \le P_n \le {\mathbb {I}}_{{\mathcal {H}}}$, $0 \le Q_n \le {\mathbb {I}}_{{\mathcal {H}}}$, and $\lim _n P_n\rho P_n = \rho$, $\lim _{n} Q_n\sigma Q_n = \sigma$ in the trace norm. Define, for n sufficiently large,

  $$\begin{aligned} \rho _n = \frac{ P_n \rho P_n}{ \mathrm {Tr}[ P_n \rho P_n ]},\quad \sigma _n = \frac{ Q_n \sigma Q_n}{ \mathrm {Tr}[ Q_n \sigma Q_n]}, \end{aligned}$$

  (169)

  and assume that $\lim _{n}E_{R_i}(\rho _n) = E_{R_i}(\rho )$, $\lim _{n}E_{R_i}(\sigma _n) = E_{R_i}(\sigma )$ for $i =1, \ldots , N$. Then,

  $$\begin{aligned} D(\rho , \sigma ) = \lim _{n\rightarrow \infty } D(\rho _n, \sigma _n)\,. \end{aligned}$$

  (170)

Proof

By Proposition 13, choose $\Pi _n \in {\mathcal {C}}(\rho _n, \sigma _n)$ such that

$$\begin{aligned} C(\Pi _n) = D(\rho _n, \sigma _n)^2\,. \end{aligned}$$

(171)

By Lemma 8 and Proposition 12, we obtain $\Pi \in {\mathcal {C}}(\rho , \sigma )$ such that $C(\Pi ) \le \liminf _n C(\Pi _n)$, hence

$$\begin{aligned} D(\rho , \sigma )^2 \le C(\Pi ) \le \liminf _n D(\rho _n, \sigma _n)^2. \end{aligned}$$

(172)

To prove the second statement, let $\Pi \in {\mathcal {C}}(\rho , \sigma )$ be such that $C(\Pi ) = D(\rho , \sigma )^2$. Write

$$\begin{aligned} {\tilde{\Pi }}_n&= (Q_n \otimes P_n^T) \Pi (Q_n \otimes P_n^T),\nonumber \\ {\tilde{\rho }}_n^T&= \mathrm {Tr}_{{\mathcal {H}}}[{\tilde{\Pi }}_n] = P_n^T \mathrm {Tr}_{{\mathcal {H}}}[ (Q_n \otimes {\mathbb {I}}_{{\mathcal {H}}^*} ) \Pi (Q_n \otimes {\mathbb {I}}_{{\mathcal {H}}^*}) ] P_n^T,\nonumber \\ {\tilde{\sigma }}_n&= \mathrm {Tr}_{{\mathcal {H}}^*}[{\tilde{\Pi }}_n] = Q_n \mathrm {Tr}_{{\mathcal {H}}^*}[ ({\mathbb {I}}_{{\mathcal {H}}}\otimes P_n^T )\Pi ({\mathbb {I}}_{{\mathcal {H}}} \otimes P_n^T) ] Q_n,\nonumber \\ m_n&= \mathrm {Tr}_{{\mathcal {H}}\otimes {\mathcal {H}}^*} [{\tilde{\Pi }}_n] = \mathrm {Tr}_{{\mathcal {H}}^*}[{\tilde{\rho }}_n^T] = \mathrm {Tr}_{{\mathcal {H}}}[{\tilde{\sigma }}_n], \end{aligned}$$

(173)

and notice that ${\tilde{\rho }}_n^T \le P_n^T \rho ^T P_n^T = \rho _n^T$ and ${\tilde{\sigma }}_n \le Q_n \sigma Q_n = \sigma _n$. We define

$$\begin{aligned} \Pi _n = m_n {\tilde{\Pi }}_n + {\tilde{\sigma }}_n \otimes (\rho _n^T - {\tilde{\rho }}_n^T) + (\sigma _n - {\tilde{\sigma }}_n) \otimes {\tilde{\rho }}_n^T + (\sigma _n - {\tilde{\sigma }}_n) \otimes (\rho _n^T - {\tilde{\rho }}_n^T).\nonumber \\ \end{aligned}$$

(174)

It holds $\Pi _n \ge 0$,

$$\begin{aligned} \mathrm {Tr}_{\mathcal {H}}[\Pi _n] = m_n {\tilde{\rho }}_n^T + m_n ( \rho _n^T - {\tilde{\rho }}_n^T) + (1-m_n) {\tilde{\rho }}_n^T + (1-m_n)(\rho _n^T - {\tilde{\rho }}_n^T) = \rho _n^T,\nonumber \\ \end{aligned}$$

(175)

and similarly $\mathrm {Tr}_{\mathcal {H^*}}[\Pi _n] =\sigma _n$, so that $\Pi _n \in {\mathcal {C}}(\rho _n, \sigma _n)$.

We claim that $\lim _n \Pi _n = \Pi$. Since $\lim _{n} P_n \rho P_n = \rho$ in the trace norm, $\lim _n\mathrm {Tr}[P_n \rho P_n] = 1$, so $\lim _n \rho _n =\rho$. Similarly, $\lim _n \sigma _n = \sigma$. Lemma 5 gives $\lim _n {\tilde{\Pi }}_n = \Pi$ in the trace norm and, by continuity of partial trace, $\lim _n {\tilde{\rho }}_n = \rho$, $\lim _n{\tilde{\sigma }}_n = \sigma$ and $\lim _n m_n = 1$, hence the claim is proved.

By Proposition 12, because the energies of the marginals converge, it follows that $C(\Pi ) = \lim _n C(\Pi _n)$. We conclude that

$$\begin{aligned} D(\rho , \sigma )^2 = C(\Pi ) \ge \limsup _{n\rightarrow \infty } D(\rho _n, \sigma _n)^2. \end{aligned}$$

(176)

$\square$

Remark 11

In Theorem 9, we may choose $P_n = \sum _{i=0}^n |\psi _i \rangle \langle \psi _i|$ where $\rho$ has the eigendecomposition $\rho = \sum _{k=1}^\infty p_i |\psi _i \rangle \langle \psi _i|$, and similarly for $Q_n$, by considering an eigendecomposition of $\sigma$. Indeed, by Lemma 9 it follows that $E_{R_i}(\rho _n)$ converge towards $E_{R_i}(\rho )$, and similarly for $\sigma$. We deduce that

$$\begin{aligned} \lim _{n\rightarrow \infty } D(\rho _n, \sigma _n) = D(\rho , \sigma )\,. \end{aligned}$$

(177)

Lemma 7

For any $X\in {\mathcal {T}}_2({\mathcal {H}})$ and any self-adjoint operator R on ${\mathcal {H}}$ such that $\left\| \left[ R,\,X\right] \right\| _2<\infty$, we have

$$\begin{aligned} \left\| \left[ R,\,X\right] \right\| _2^2&\ge \mathrm {Tr}_{{\mathcal {H}}}\left[ R\left( X^\dag \,X + X\,X^\dag \right) R - \sqrt{X^\dag \,X}\,R\,\sqrt{X^\dag \,X}\,R\right. \nonumber \\&\quad \left. - \sqrt{X\,X^\dag }\,R\,\sqrt{X\,X^\dag }\,R\right] \,. \end{aligned}$$

(178)

Proof

Let us consider the singular-value decomposition of X:

$$\begin{aligned} X = \sum _{i=0}^\infty x_i\,|\psi _i\rangle \langle \phi _i|\,,\qquad x_i\ge 0\,,\qquad \langle \psi _i|\psi _j\rangle = \langle \phi _i|\phi _j\rangle = \delta _{ij}\,, \end{aligned}$$

(179)

where the series converges in the Hilbert–Schmidt norm. We get

$$\begin{aligned}&\left\| \left[ R,\,X\right] \right\| _2^2 = \mathrm {Tr}_{{\mathcal {H}}}\left[ R\left( X^\dag \,X + X\,X^\dag \right) R - 2\,X^\dag \,R\,X\,R\right] \nonumber \\&\ge \mathrm {Tr}_{{\mathcal {H}}}\left[ R\left( X^\dag \,X + X\,X^\dag \right) R\right] - 2\left| \mathrm {Tr}_{{\mathcal {H}}}\left[ X^\dag \,R\,X\,R\right] \right| \nonumber \\&=\mathrm {Tr}_{{\mathcal {H}}}\left[ R\left( X^\dag \,X + X\,X^\dag \right) R\right] - 2\left| \sum _{i,\,j=0}^\infty x_i\,x_j\,\langle \psi _i|R|\psi _j\rangle \langle \phi _j|R|\phi _i\rangle \right| \nonumber \\&\ge \mathrm {Tr}_{{\mathcal {H}}}\left[ R\left( X^\dag \,X + X\,X^\dag \right) R\right] - \sum _{i,\,j=0}^\infty x_i\,x_j\left( \left| \langle \psi _i|R|\psi _j\rangle \right| ^2 + \left| \langle \phi _j|R|\phi _i\rangle \right| ^2\right) \nonumber \\&= \mathrm {Tr}_{{\mathcal {H}}}\left[ R\left( X^\dag \,X + X\,X^\dag \right) R - \sqrt{X^\dag \,X}\,R\,\sqrt{X^\dag \,X}\,R - \sqrt{X\,X^\dag }\,R\,\sqrt{X\,X^\dag }\,R\right] \,, \end{aligned}$$

(180)

and the claim follows. $\square$