and hence

$$Q_f(x, \xi) = Q_f(x, \xi)^* \geq 0, \quad \text{for a.e. } (x, \xi) \in \mathbf{R}^d \times \mathbf{R}^d.$$

Moreover

$$\mathrm{trace}_{\mathfrak{S}}(Q(x, \xi) + \mathbf{1}_{\mathcal{N}}(x, \xi)|u\rangle\langle u|) = f(x, \xi) + \mathbf{1}_{\mathcal{N}}(x, \xi),$$

so that

$$\mathrm{trace}_{\mathfrak{S}}(Q_f(x, \xi)) = 1, \quad \text{for a.e. } (x, \xi) \in \mathbf{R}^d \times \mathbf{R}^d.$$

Finally

$$f(x, \xi)Q_f(x, \xi) = \frac{f(x, \xi)Q(x, \xi)}{f(x, \xi) + \mathbf{1}_{\mathcal{N}}(x, \xi)} = Q(x, \xi)$$

for a.e. $(x, \xi) \in \mathbf{R}^d \times \mathbf{R}^d$. Indeed, whenever $f(x, \xi) > 0$, one has $\mathbf{1}_{\mathcal{N}}(x, \xi) = 0$ and the claimed equality is obvious. On the other hand, since

$$Q(x, \xi) = Q(x, \xi)^* \geq 0 \quad \text{and} \quad \mathrm{trace}_{\mathfrak{S}}(Q(x, \xi)) = f(x, \xi)$$

for a.e. $(x, \xi) \in \mathbf{R}^d \times \mathbf{R}^d$, it follows that

$$f(x, \xi) = 0 \implies Q(x, \xi) = 0 = f(x, \xi)Q_f(x, \xi).$$

□

With Lemma 19 at our disposal, the proof of Theorem 16 in the case where the intermediate point is a classical density follows the proof of Theorem 7.3 in [63].

**Proof of Theorem 16: the case of a classical intermediate density.** Consider for example the case where both $\rho_1$ and $\rho_3 \in \mathcal{D}_2(\mathfrak{S})$, and assume that $\rho_2 = f(y, \eta)dyd\eta \in \mathcal{P}_2(\mathbf{R}^d \times \mathbf{R}^d)$. Choose couplings $Q^1 \in \mathcal{C}(\rho_1, f)$ while $Q^3 \in \mathcal{C}(f, \rho_3)$. Call $Q_f^3$ the disintegration of $Q^3$ with respect to $f$ as in Lemma 19. Set

$$T(y, \eta) := Q^1(y, \eta) \otimes Q_f^3(y, \eta).$$

By construction

$$T(y, \eta) = T(y, \eta)^* \geq 0,$$

and

$$\mathrm{trace}_1(T(y, \eta)) = f(y, \eta)Q_f^3(y, \eta) = Q^3(y, \eta),$$

$$\mathrm{trace}_3(T(y, \eta)) = Q^1(y, \eta) \mathrm{trace}_{\mathfrak{S}}(Q_f^3(y, \eta)) = Q^1(y, \eta).$$

In particular

$$\int_{\mathbf{R}^{2d}} T(y, \eta)dyd\eta := Q \in \mathcal{C}(\rho_1, \rho_3).$$

By the second inequality in Lemma 17

$$\begin{aligned}
\mathfrak{D}(\rho_1, \rho_3)^2 &\leq \mathrm{trace}_{\mathfrak{S} \otimes 2}(Q^{\frac{1}{2}} C_h(x, \hbar \nabla_x, z, \hbar \nabla_z) Q^{\frac{1}{2}}) \\
&= \int_{\mathbf{R}^{2d}} \mathrm{trace}_{\mathfrak{S} \otimes 2} (T(y, \eta)^{\frac{1}{2}} C_h(x, \hbar \nabla_x, z, \hbar \nabla_z) T(y, \eta)^{\frac{1}{2}}) dyd\eta \\
&\leq (1 + \alpha) \int_{\mathbf{R}^{2d}} \mathrm{trace}_1 \left( \mathrm{trace}_3(T(y, \eta)^{\frac{1}{2}} c_h(x, \hbar \nabla_x, y, \eta) T(y, \eta)^{\frac{1}{2}}) \right) dyd\eta \\
&\quad + (1 + \frac{1}{\alpha}) \int_{\mathbf{R}^{2d}} \mathrm{trace}_3 \left( \mathrm{trace}_1(T(y, \eta)^{\frac{1}{2}} c_h(x, \hbar \nabla_x, y, \eta) T(y, \eta)^{\frac{1}{2}}) \right) dyd\eta \\
&\leq (1 + \alpha) \int_{\mathbf{R}^{2d}} \mathrm{trace}_{\mathfrak{S}}(Q^1(y, \eta)^{\frac{1}{2}} c_h(x, \hbar \nabla_x, y, \eta) Q^1(y, \eta)^{\frac{1}{2}}) dyd\eta \\
&\quad + (1 + \frac{1}{\alpha}) \int_{\mathbf{R}^{2d}} \mathrm{trace}_{\mathfrak{S}}(Q_f^3(y, \eta)^{\frac{1}{2}} c_h(x, \hbar \nabla_x, y, \eta) Q_f^3(y, \eta)^{\frac{1}{2}}) f(y, \eta) dyd\eta.
\end{aligned}$$
Minimizing the last right-hand side in $Q^1 \in \mathcal{C}(\rho_1, \rho_2)$ and in $Q^3 \in \mathcal{C}(\rho_2, \rho_3)$ leads to the inequality

$$ \mathfrak{D}(\rho_1, \rho_3)^2 \leq (1 + \alpha)\mathfrak{D}(\rho_1, \rho_2)^2 + (1 + \frac{1}{\alpha})\mathfrak{D}(\rho_2, \rho_3)^2, $$

and we conclude as in the rank-1 case. □

**Quiz 28.**

(1) Complete the missing details in the proof of Theorem 16 in the case where $\rho_1, \rho_3 \in \mathcal{D}_2(\mathfrak{H})$ and $\rho_2 = f(y, \eta)dyd\eta$. In particular, prove the identity

$$ \operatorname{trace}_{\mathfrak{H}\otimes 2}(Q^{\frac{1}{2}} C_h(x, \hbar \nabla_x, z, \hbar \nabla_z) Q^{\frac{1}{2}}) \\ = \int_{\mathbb{R}^{2d}} \operatorname{trace}_{\mathfrak{H}\otimes 2}(T(y, \eta)^{\frac{1}{2}} C_h(x, \hbar \nabla_x, z, \hbar \nabla_z) T(y, \eta)^{\frac{1}{2}}) dyd\eta. $$

(2) Write the proof of Theorem 16 in the missing cases.

**3.2. Applications of the Restricted Triangle Inequalities.** Before going further in our discussion of the triangle inequality for the pseudometric $\mathfrak{D}$, we shall present two easy applications of the (restricted) triangle inequalities already established in the previous section.

**3.2.1. Definition of $\mathfrak{D}$ on $\mathcal{P}_2(\mathbb{R}^d \times \mathbb{R}^d) \times \mathcal{D}_2(\mathfrak{H})$.** Our first application is of a quite fundamental nature, since it completes our definition of $\mathfrak{D}$ in Lecture I. So far we have defined $\mathfrak{D}(\mu, R) = \mathfrak{D}(R, \mu)$ for $\mu \in \mathcal{P}_2(\mathbb{R}^d \times \mathbb{R}^d)$ and $R \in \mathcal{D}_2(\mathfrak{H})$ only when $\mu = f(x, \xi)dxd\xi$, with $f$ a probability density on $\mathbb{R}^d \times \mathbb{R}^d$ — i.e. only when $\mu$ is absolutely continuous with respect to the phase-space Lebesgue measure, by the Radon-Nikodym theorem.

**Theorem 20.** For each $R \in \mathcal{D}_2(\mathfrak{H})$, the map $f \mapsto \mathfrak{D}(f, R)$, defined for all probability density $f$ with finite second order moments on $\mathbb{R}^d \times \mathbb{R}^d$ has a unique extension to $\mathcal{P}_2(\mathbb{R}^d \times \mathbb{R}^d)$ satisfying

$$ |\mathfrak{D}(\mu, R) - \mathfrak{D}(\nu, R)| \leq \mathcal{W}_2(f, g), \quad \mu, \nu \in \mathcal{P}_2(\mathbb{R}^d \times \mathbb{R}^d). $$

See section 3 in [40].

**Remark.** This extension of $\mathfrak{D}$ obviously satisfies the restricted triangle inequality

$$ \mathfrak{D}(\rho_1, \rho_3) \leq \mathfrak{D}(\rho_1, \rho_2) + \mathfrak{D}(\rho_2, \rho_3) $$

for all $\rho_1, \rho_2, \rho_3 \in \mathfrak{D}$ provided that $\rho_2 \in \mathcal{P}_2(\mathbb{R}^d \times \mathbb{R}^d)$ or if one of the $\rho_j$s is a rank-1 density operator on $\mathfrak{H} = L^2(\mathbb{R}^d)$.

**Proof.** For all $f, g$ probability densities with finite 2nd order moments on $\mathbb{R}^d \times \mathbb{R}^d$, one has the triangle inequality

$$ \mathfrak{D}(f, R) \leq \mathfrak{D}(f, g) + \mathfrak{D}(g, R) $$

by Theorem 16, so that

$$ \mathfrak{D}(f, R) - \mathfrak{D}(g, R) \leq \mathfrak{D}(f, g) = \mathcal{W}_2(f, g). $$

Exchanging $f$ and $g$ in the inequality above implies that

$$ |\mathfrak{D}(f, R) - \mathfrak{D}(g, R)| \leq \mathcal{W}_2(f, g). $$

Thus the function $f \mapsto \mathfrak{D}(f, R)$ is Lipschitz-continuous for the metric $\mathcal{W}_2$. It has therefore a unique Lipschitz-continuous extension to $\mathcal{P}_2(\mathbb{R}^d \times \mathbb{R}^d)$ by the density argument recalled in Lemma 21 below. □
**Lemma 21.** Let $\mu \in \mathcal{P}_2(\mathbf{R}^n)$ and let $\chi_\epsilon(x) = \chi(x/\epsilon)/\epsilon^n$ be an even $C^\infty$ mollifier with support in $B_\epsilon(0)$. Then $f_\epsilon := \chi_\epsilon \star \mu$ is a $C^\infty$ probability density on $\mathbf{R}^n$ and

$$W_2(f_\epsilon, \mu) \to 0 \quad \text{as } \epsilon \to 0.$$

This is Lemma 3.2 in [40].

*Proof.* For all $\phi \in C_0(\mathbf{R}^n)$, one has

$$\int_{\mathbf{R}^n} \phi(x) \mu(dx) - \int_{\mathbf{R}^n} f_\epsilon(x) \phi(x) dx = \int_{\mathbf{R}^n} (\phi(x) - \chi_\epsilon \star \phi(x)) \mu(dx)$$

since $\chi_\epsilon$ is even, so that

$$\left| \int_{\mathbf{R}^n} \phi(x) \mu(dx) - \int_{\mathbf{R}^n} f_\epsilon(x) \phi(x) dx \right| \le \|\phi - \phi \star \chi_\epsilon\|_{L^\infty(\mathbf{R}^n)} \to 0.$$

Hence $f_\epsilon \to \mu$ weakly in $\mathcal{P}(\mathbf{R}^n)$ as $\epsilon \to 0$.

It remains to establish the tightness property (see the properties of $W_2$ recalled in Lecture I). Since $\chi_\epsilon$ is even,

$$\int_{\mathbf{R}^n} \mathbf{1}_{|x|>R} |x|^2 \chi_\epsilon \star \mu(x) dx = \int_{\mathbf{R}^n} \chi_\epsilon \star (\mathbf{1}_{|x|>R} |x|^2) \mu(dx).$$

On the other hand, for all $\epsilon \in (0, 1)$

$$\begin{aligned} \chi_\epsilon \star (\mathbf{1}_{|x|>R} |x|^2) &\le \mathbf{1}_{|x|+1 \ge R} \int_{\mathbf{R}^n} |x - \epsilon y|^2 \chi(y) dy \\ &\le 2 \mathbf{1}_{|x|+1 \ge R} \left( |x|^2 + \epsilon^2 \underbrace{\int_{\mathbf{R}^n} |y|^2 \chi(y) dy}_{\le 1} \right) \end{aligned}$$

Hence

$$\sup_{0<\epsilon<1} \int_{\mathbf{R}^n} \mathbf{1}_{|x|>R} |x|^2 \chi_\epsilon \star \mu(x) dx \le 2 \int_{\mathbf{R}^n} \mathbf{1}_{|x|+1>R} (|x|^2 + 1) \mu(dx) \to 0$$

as $R \to \infty$, by dominated convergence.

Therefore, by Theorem 7.12 of [63]

$$W_2(\chi_\epsilon \star \mu, \mu) \to 0 \quad \text{as } \epsilon \to 0.$$

□

### 3.2.2. $W_2$ is the Classical Limit of $\mathfrak{D}$.
Our next application of the restricted triangle inequalities presented in Theorem 16 can be thought of as a confirmation of the geometric picture proposed in Lecture I.

We recall the idea of considering $\mathcal{P}_2(\mathbf{R}^d \times \mathbf{R}^d)$ as a limit set, or boundary set, of $\mathcal{D}_2(\mathfrak{H})$ in $\mathfrak{D}$. The next result completes this picture by showing that $W_2$ is the limiting metric deduced from the pseudometric $\mathfrak{D}$ on $\mathcal{D}(\mathfrak{H})$.

**Theorem 22.** Let $R_\hbar, S_\hbar \in \mathcal{D}_2(\mathfrak{H})$ and $\mu, \nu \in \mathcal{P}_2(\mathbf{R}^d \times \mathbf{R}^d)$. Assume that $\mu, \nu$ are the classical limits of $R_\hbar, S_\hbar$ respectively, i.e.

$$\mathfrak{D}(\mu, R_\hbar) + \mathfrak{D}(\nu, S_\hbar) \to 0 \quad \text{as } \hbar \to 0.$$

Then

$$\lim_{\hbar \to 0} \mathfrak{D}(R_\hbar, S_\hbar) = \mathfrak{D}(\mu, \nu).$$

This is Theorem C (see also Theorem 5.5) in [40].
<figure><img src="image_12.png" /><figcaption>FIGURE 7. The Wasserstein distance $W_2$ is the classical limit of the pseudometric $\mathfrak{D}$ as $\hbar \to 0$.</figcaption></figure>

*Proof.* By the restricted triangle inequality

$$ \mathfrak{D}(R_h, S_h) \le \mathfrak{D}(R_h, \mu) + \mathfrak{D}(\mu, \nu) + \mathfrak{D}(\nu, S_h), $$

so that

$$ \overline{\lim}_{\hbar \to 0} \mathfrak{D}(R_h, S_h) \le \mathfrak{D}(\mu, \nu) = W_2(\mu, \nu). $$

On the other hand, by Theorem 1 (2) of Lecture I,

$$ \mathfrak{D}(R_h, \mu)^2 \ge W_2(\mathcal{H}[R_h], \mu)^2 - d\hbar \implies \lim_{\hbar \to 0} W_2(\mathcal{H}[R_h], \mu) = 0, $$

$$ \mathfrak{D}(S_h, \nu)^2 \ge W_2(\mathcal{H}[S_h], \nu)^2 - d\hbar \implies \lim_{\hbar \to 0} W_2(\mathcal{H}[S_h], \nu) = 0. $$

Hence

$$ \mathfrak{D}(R_h, S_h)^2 \ge W_2(\mathcal{H}[R_h], \mathcal{H}[S_h])^2 - 2d\hbar, $$

implies that

$$ \overline{\lim}_{\hbar \to 0} \mathfrak{D}(R_h, S_h) \ge \lim_{\hbar \to 0} W_2(\mathcal{H}[R_h], \mathcal{H}[S_h]) = W_2(\mu, \nu). $$

(This part of the argument does not appeal to the restricted triangle inequality, and had been already established in Theorem 2.3 (2) of [37], albeit in a slightly different form). Summarizing

$$ W_2(\mu, \nu) \le \overline{\lim}_{\hbar \to 0} \mathfrak{D}(R_h, S_h) \le \overline{\lim}_{\hbar \to 0} \mathfrak{D}(R_h, S_h) \le W_2(\mu, \nu). $$

□

**3.3. Quantum Kantorovich Duality.** In the classical setting, Kantorovich duality has several important consequences. The Knott-Smith and Brenier theorems, giving invaluable information on the structure of optimal couplings between two (Borel) probability measures on $\mathbb{R}^n$ are among the most important applications of duality — at least in the case of the Wasserstein distance $W_2$.

It is therefore natural to seek extensions of Kantorovich duality to the quantum setting. However, the examples studied at the end of Lecture I (in the section showing that “quantum optimal transport is cheaper”) suggest that the structure of quantum optimal couplings might differ significantly from that of classical optimal couplings.

First we consider the case of $\mathfrak{D}(f, R)$, where $f$ is a probability density on $\mathbb{R}^d \times \mathbb{R}^d$ with finite 2nd order moments and $R \in \mathcal{D}_2(\mathfrak{S})$.
Define the set $\mathfrak{k}$ of test Kantorovich potentials, as follows:

$$ \mathfrak{k} := \{(a, B) : a \in C_b(\mathbf{R}^d \times \mathbf{R}^d) \text{ and } B = B^* \in \mathcal{L}(\mathfrak{h}) \text{ s.t. } a(x, \xi) I_{\mathfrak{h}} + B \le c_h(x, \xi)\}. $$

The operator inequality in the definition of $\mathfrak{k}$ means that

$$ a(x, \xi) \|\phi\|_{\mathfrak{h}}^2 + \langle \phi | B | \phi \rangle \le \langle \phi | c_h(x, \xi) | \phi \rangle, \quad x, \xi \in \mathbf{R}^d, $$

for all $\phi \in H^1(\mathbf{R}^d) \cap L^2(\mathbf{R}^d; |y|^2 dy)$.

**Theorem 23.** Let $f$ be a probability density on $\mathbf{R}^d \times \mathbf{R}^d$ with finite second order moments and let $R \in \mathcal{D}_2(\mathfrak{h})$. Then

$$ \mathfrak{d}(f, R)^2 = \sup_{(a,B) \in \mathfrak{k}} \left( \int_{\mathbf{R}^{2d}} a(x, \xi) f(x, \xi) dxd\xi + \text{trace}_{\mathfrak{h}}(BR) \right). $$

The proof of Theorem 23 is somewhat involved technically, and we shall not repeat it in these notes. We refer instead the interested reader to Theorem 4.1 and section 4 of [40].

However, it is a good idea to keep in mind the core argument in that proof, which is based on convex duality, exactly as in the classical setting.

Set $\mathcal{E} := C_b(\mathbf{R}^{2d}; \mathcal{L}(\mathfrak{h}))$ with

$$ \|T\|_{\mathcal{E}} := \sup_{x, \xi \in \mathbf{R}^d} \|T(x, \xi)\|_{\mathcal{L}(\mathfrak{h})}. $$

Define

$$ G(T) := \begin{cases} 0 & \text{if } T(x, \xi) = T(x, \xi)^* \ge -c_h(x, \xi), \\ + \infty & \text{otherwise,} \end{cases} $$

and

$$ H(T) := \begin{cases} \int_{\mathbf{R}^{2d}} \tilde{a} f(x, \xi) dxd\xi + \text{trace}_{\mathfrak{h}}(\tilde{B}R) & \text{if } \begin{cases} T(x, \xi) = T(x, \xi)^* \\ = \tilde{a}(x, \xi) I_{\mathfrak{h}} + \tilde{B}, \end{cases} \\ + \infty & \text{otherwise.} \end{cases} $$

Theorem 23 follows from the Fenchel-Rockafellar duality formula

$$ \inf_{T \in \mathcal{E}} (G(T) + H(T)) = \max_{\Lambda \in \mathcal{E}'} (-G^*(-\Lambda) - H^*(\Lambda)). $$

The role of the functional $G$ in this formulation is obviously to penalize the inequality constraint in the definition of $\mathfrak{k}$. (Sometimes, in convex analysis, the functional equal to 0 if a constraint is satisfied, and to $+\infty$ otherwise is referred to as the “indicator function” of that constraint. This is not to be confused with the classical notion of indicator function of a set, equal to one on the set, and to zero outside.)

Next we consider the case of $\mathfrak{d}(R, S)$ where $R, S \in \mathcal{D}_2(\mathfrak{h})$. Here again, there is an analogue of the classical Kantorovich duality theorem.

Define the set $\mathfrak{R}$ of test Kantorovich potentials as follows

$$ \mathfrak{R} := \{(A, B) : A = A^* \text{ and } B = B^* \in \mathcal{L}(\mathfrak{h}) \text{ s.t. } A \otimes I_{\mathfrak{h}} + I_{\mathfrak{h}} \otimes B \le C_h\}. $$

The operator inequality means that for all $\Phi \equiv \Phi(x, y) \in \mathfrak{h} \otimes \mathfrak{h}$ such that

$$ (\nabla_x - \nabla_y)\Phi \in L^2(\mathbf{R}^d \times \mathbf{R}^d) \quad \text{and} \quad \Phi \in L^2(\mathbf{R}^d \times \mathbf{R}^d; |x-y|^2 dxdy), $$

it holds

$$ \langle \Phi | A \otimes I_{\mathfrak{h}} + I_{\mathfrak{h}} \otimes B | \Phi \rangle \le \langle \Phi | C_h | \Phi \rangle. $$
**Theorem 24.** For all $R, S \in \mathcal{D}_2(\mathfrak{H})$, one has

$$ \mathfrak{D}(R, S)^2 = \sup_{(A,B) \in \mathcal{R}} \mathrm{trace}_{\mathfrak{H}}(AR + BS). $$

Here again, the proof of Theorem 24 is too technically involved to be of immediate interest for these lecture notes. We refer the interested reader to [20] for a detailed exposition of duality in the quantum-to-quantum setting, together with various applications thereof.

Although we have chosen to avoid reproducing the proofs of Theorems 23 and 24, we believe that both results are of key importance in the study of the pseudometric $\mathfrak{D}$, and in our approach of quantum optimal transport. As a matter of fact, the end of Lecture III will be focussed on two applications of Theorem 23, which we believe are of some importance.

**3.4. Generalized Triangle Inequalities.** In this section, we return to the question of the triangle inequality for $\mathfrak{D}$ for $\rho_1, \rho_2, \rho_3 \in \mathcal{D}$, and consider the case where the intermediate point $\rho_2$ is a density operator and none of the points $\rho_1, \rho_2, \rho_3$ are rank-1 density operators.

In that case, we do not know of any analogue of Lemma 19, and at the time of this writing we do not know how to “glue” $Q_{12} \in \mathcal{C}(\rho_1, \rho_2)$ and $Q_{23} \in \mathcal{C}(\rho_2, \rho_3)$ along $\rho_2$ as in Lemma 7.6 in [63], to mimic the proof of Theorem 7.3 in [63] in the classical setting.

At the beginning of this lecture, we have proved “restricted” triangle inequalities for $\mathfrak{D}$, i.e. the triangle inequality under additional assumptions on $\rho_1, \rho_2, \rho_3$ — specifically if one of the points $\rho_1, \rho_2, \rho_3$ is a rank-1 density operator, or if the intermediate point $\rho_2$ is an element of $\mathcal{P}_2(\mathbf{R}^d \times \mathbf{R}^d)$.

In the sequel, we shall prove a “generalized triangle inequality”, which holds for all $\rho_1, \rho_2, \rho_3 \in \mathcal{D}$, but includes a correction term of order $\sqrt{\hbar}$ in the right-hand side. **Theorem 25.** For all $\rho_1, \rho_2, \rho_3 \in \mathcal{D}$, one has

$$ \mathfrak{D}(\rho_1, \rho_3) < \mathfrak{D}(\rho_1, \rho_2) + \mathfrak{D}(\rho_2, \rho_3) + \sqrt{d\hbar}. $$

In particular

$$ \mathfrak{D}(\rho_1, \rho_3) < \mathfrak{D}(\rho_1, \rho_2) + \mathfrak{D}(\rho_2, \rho_3) + \frac{1}{\sqrt{2}} \mathfrak{D}(\rho_2, \rho_2). $$

**Remark.** One should compare this result with the De Palma-Trevisan generalized triangle inequality for their distance — which is reminiscent of our generalization of the Wasserstein distance $W_2$ to $\mathcal{D}_2(\mathfrak{H})$ in [37] (the restriction of $\mathfrak{D}$ to $\mathcal{D}_2(\mathfrak{H}) \times \mathcal{D}_2(\mathfrak{H}))$. They arrive at the inequality

$$ d_{dPT}(R, T) \le d_{dPT}(R, S) + d_{dPT}(R, T) + d_{dPT}(S, S) $$

for all $R, S, T \in \mathcal{D}_2(\mathfrak{H})$ for $d_{dPT}$ defined in formula (38) of Definition 8, on p. 3208 in [26]: see formula (51) in Theorem 2 of [26] on p. 3210. Notice however that $d_{dPT}$ differs from the restriction of $\mathfrak{D}$ to $\mathcal{D}_2(\mathfrak{H}) \times \mathcal{D}_2(\mathfrak{H})$ — in particular the definition of couplings in [26] uses the notion of quantum channel and is very different from the definition used in these lectures, which comes from [37].

**Quiz 29.** Can one slightly modify the definition of $\mathfrak{D}$ so as to obtain a functional $\tilde{\mathfrak{D}}$ defined on $\mathcal{D} \times \mathcal{D}$ that is symmetric and satisfies the (genuine) triangle inequality? Same question for the DePalma-Trevisan functional $d_{dPT}$. (Of course, the functional $\tilde{\mathfrak{D}}$ so obtained satisfies $\tilde{\mathfrak{D}} > \mathfrak{D} > 0$ and therefore is not a bona fide metric
either, since $\tilde{\mathfrak{d}}(\rho, \rho) > 0$ for each $\rho \in \mathcal{D}$. The functional $\tilde{\mathfrak{d}}$ obtained in this exercise is therefore of limited interest.)

The proof of Theorem 25 is very different from the proof of the triangle inequality for the original Wasserstein metric $W_2$ — although it uses at some point the restricted triangle inequality in Theorem 16, whose proof is modelled on the proof of Theorem 7.3 in [63].

A key step in the proof of Theorem 25 is the following lemma, which can be seen as a consequence of duality for the classical-to-quantum pseudometric presented in Theorem 23.

**Lemma 26.** For each $R, S \in \mathcal{D}_2(\mathfrak{h})$, one has

$$ \mathfrak{d}(R, S)^2 \geq \mathfrak{d}(R, \mathcal{H}[S])^2 - d\hbar. $$

Taking this lemma for granted, we give a quick proof of Theorem 25.

**Proof of Theorem 25.** Using $\mathcal{H}[\rho_2]$ as intermediate point, the restricted triangle inequality implies that

$$ \mathfrak{d}(\rho_1, \rho_3) \leq \mathfrak{d}(\rho_1, \mathcal{H}[\rho_2]) + \mathfrak{d}(\mathcal{H}[\rho_2], \rho_3). $$

Then, Lemma 26 implies that

$$ \mathfrak{d}(\rho_1, \mathcal{H}[\rho_2]) \leq \sqrt{\mathfrak{d}(\rho_1, \rho_2)^2 + d\hbar} < \mathfrak{d}(\rho_1, \rho_2) + \frac{1}{2}\sqrt{d\hbar}, $$

$$ \mathfrak{d}(\mathcal{H}[\rho_2], \rho_3) \leq \sqrt{\mathfrak{d}(\rho_2, \rho_3)^2 + d\hbar} < \mathfrak{d}(\rho_2, \rho_3) + \frac{1}{2}\sqrt{d\hbar}. $$

The second inequalities above result from the following elementary observation

$$ X > Y > 0 \implies \sqrt{X^2 + Y^2} \leq X + \frac{1}{2}Y, $$

whose proof is left to the reader as an (easy) exercise.

With the restricted triangle inequality above (Theorem 16), this implies the first generalized triangle inequality in Theorem 25.

To get the second inequality, observe that

$$ \rho_2 \in \mathcal{D}_2(\mathfrak{h}) \implies \mathfrak{d}(\rho_2, \rho_3) \geq \sqrt{2d\hbar}. $$

□

**Remark.** In fact, we have proved the slightly more precise inequality

$$ \mathfrak{d}(\rho_1, \rho_3) \leq \sqrt{\mathfrak{d}(\rho_1, \rho_2)^2 + d\hbar} + \sqrt{\mathfrak{d}(\rho_2, \rho_3)^2 + d\hbar}. $$

*Proof of Lemma 26.* For all $a \in C_b(\mathbb{R}^d \times \mathbb{R}^d)$ and all $B = B^* \in \mathcal{L}(\mathfrak{h})$ satisfying

$$ a(x, \xi)I_{\mathfrak{h}} + B \leq c_{\hbar}(x, \xi), $$

one applies the Toeplitz map to both sides of the inequality above in the variables $x, \xi$, to find

$$ \begin{aligned} \mathcal{T}[a] \otimes I_{\mathfrak{h}} + (2\pi\hbar)^d I_{\mathfrak{h}} \otimes B &\leq (2\pi\hbar)^d \int |q, p\rangle\langle q, p|c_{\hbar}(q, p)dqdp \\ &\leq (2\pi\hbar)^d (C_{\hbar} + d\hbar I_{\mathfrak{h}\otimes\mathfrak{h}}) \end{aligned} $$
(see the basic properties of the Toeplitz map and Quiz 8 in Lecture I). Thus, for all $T \in C(R, S)$, one has

$$ (2\pi\hbar)^d \left( \operatorname{trace}_{\mathfrak{S}\otimes\mathfrak{S}} \left( T^{\frac{1}{2}} C_h T^{\frac{1}{2}} \right) + d\hbar \right) \\ \geq \operatorname{trace}_{\mathfrak{S}\otimes\mathfrak{S}} \left( T^{\frac{1}{2}} (\mathcal{T}[a] \otimes I_{\mathfrak{S}} + (2\pi\hbar)^d I_{\mathfrak{S}} \otimes B) T^{\frac{1}{2}} \right) \\ = \operatorname{trace}_{\mathfrak{S}\otimes\mathfrak{S}} \left( T(\mathcal{T}[a] \otimes I_{\mathfrak{S}} + (2\pi\hbar)^d I_{\mathfrak{S}} \otimes B) \right) \\ = \operatorname{trace}_{\mathfrak{S}} (RT[a]) + (2\pi\hbar)^d \operatorname{trace}_{\mathfrak{S}} (SB). $$

Transforming $\operatorname{trace}_{\mathfrak{S}}(RT[a])$ into an integral involving the functions $a$ and $\mathcal{H}[R]$, i.e. (see Lecture I, formula (4) on Husimi transforms)

$$ \operatorname{trace}_{\mathfrak{S}}(RT[a]) = (2\pi\hbar)^d \int_{\mathbb{R}^{2d}} \mathcal{H}[R](q,p)a(q,p)dqdp, $$

we arrive at the formula

$$ (2\pi\hbar)^d \left( \operatorname{trace}_{\mathfrak{S}\otimes\mathfrak{S}} \left( T^{\frac{1}{2}} C_h T^{\frac{1}{2}} \right) + d\hbar \right) \\ \geq (2\pi\hbar)^d \left( \int_{\mathbb{R}^{2d}} \mathcal{H}[R](q,p)a(q,p)dqdp + \operatorname{trace}_{\mathfrak{S}}(SB) \right). $$

Maximizing the right-hand side above in $a \in C_b(\mathbb{R}^d \times \mathbb{R}^d)$ and $B = B^* \in \mathcal{L}(\mathfrak{S})$ s.t.

$$ a(x, \xi) I_{\mathfrak{S}} + B \leq c_{\hbar}(x, \xi), $$

and applying the duality formula in Theorem 16 shows that

$$ (2\pi\hbar)^d \left( \operatorname{trace}_{\mathfrak{S}\otimes\mathfrak{S}} \left( T^{\frac{1}{2}} C_h T^{\frac{1}{2}} \right) + d\hbar \right) \geq (2\pi\hbar)^d \mathcal{D}(\mathcal{H}[R], S)^2 $$

i.e.

$$ \operatorname{trace}_{\mathfrak{S}\otimes\mathfrak{S}} (T^{\frac{1}{2}} C_h T^{\frac{1}{2}}) \geq \mathcal{D}(\mathcal{H}[R], S)^2 - d. $$

Minimizing the left-hand side of the inequality above in $T \in C(R, S)$ leads to the desired inequality. $\square$

**Quiz 30.** Use Lemma 26 to recover the following result (already known as statement (2) in Theorem 1 of Lecture I)

$$ \mathcal{D}(R, S)^2 \geq \mathcal{D}(\mathcal{H}[R], \mathcal{H}[S])^2 - 2d\hbar. $$

**Remarks.**

(1) If you include the proof of the duality formula in Theorem 16, this is the longest and most difficult proof of the inequality above... On the other hand, Lemma 26 is a (much) stronger statement — it is the key to the generalized triangle inequality. That its proof is more involved than the proof of the inequality (2) in Theorem 1 is only natural.

(2) Summarizing, in order to prove the triangle inequality for $\mathcal{D}$ when the intermediate point is not a classical density and none of the density operators involved are rank-1 projections, you

(i) first use the exact (restricted) triangle inequality from Theorem 16

$$ \mathcal{D}(\rho_1, \rho_3) \leq \mathcal{D}(\rho_1, \mathcal{H}[\rho_2]) + \mathcal{D}(\mathcal{H}[\rho_2], \rho_3), $$

(ii) and then pay the price for replacing $\rho_2$ with its Husimi function

$$ \mathcal{D}(\rho_1, \rho_3) \leq \sqrt{\mathcal{D}(\rho_1, \rho_2)^2 + d\hbar} + \sqrt{\mathcal{D}(\rho_2, \rho_3)^2 + d\hbar} $$
which is the result in Lemma 26, based on the Kantorovich duality for the classical-to-quantum distance (Theorem 23). The end of the proof is Kindergarten analysis.

The reason for the detour through $\mathcal{H}[\rho_2]$ instead of $\rho_2$ is due to the fact that question (5) in the following quiz is answered in the negative$^{14}$.

Before working on this exercise, it is a good idea to review the proofs of Theorem 7.3 (the triangle inequality for $W_p$) and of Lemma 7.6 (i.e. the disintegration and the glueing of couplings) in [63].

**Quiz 31.** Pick $\rho_1, \rho_2, \rho_3 \in \mathcal{D}_2(\mathfrak{H})$, all of them or rank $\ge 2$ — otherwise, there is nothing to prove. Pick $R_{12}$ and $R_{23}$ to be optimal couplings of $\rho_1, \rho_2$ and $\rho_2, \rho_3$ (recall briefly why such couplings exist).

(1) Assume there exists $T \in \mathcal{D}(\mathfrak{H} \otimes \mathfrak{H} \otimes \mathfrak{H})$ such that

$$ \mathrm{trace}_1(T) = R_{23} \quad \text{and} \quad \mathrm{trace}_3(T) = R_{12}. $$

Prove that

$$ \mathfrak{D}(\rho_1, \rho_3) \le \mathfrak{D}(\rho_1, \rho_2) + \mathfrak{D}(\rho_2, \rho_3). $$

(Hint: observe that $\mathrm{trace}_2(T) \in \mathcal{C}(\rho_1, \rho_3)$.)

Therefore, proving the triangle inequality boils down to proving the existence of such a $T$. The classical analogue of this is precisely the content of Lemma 7.6 in Villani's book [63].

Let us consider this problem in finite dimension: $\mathfrak{H} = \mathbf{C}^2$ — notice that 2 is the first interesting dimension, because if one of the densities $\rho_j$ for $j = 1, 2, 3$ has rank 1, the triangle inequality is already known (see Theorem 16 on the restricted triangle inequality).

(2) Let $R, R' \in M_2(\mathbf{C})$. Find a necessary and sufficient condition on $R, R'$ such that there exists $A, B, C \in M_2(\mathbf{C})$ for which the block-wise matrix

$$ T := \begin{pmatrix} A & B \\ B^* & C \end{pmatrix}, \quad A = A^*, C = C^*, $$

satisfies

$$ \tau'(T) := A + C = R \quad \text{and} \quad \tau(T) := \begin{pmatrix} \mathrm{trace}(A) & \mathrm{trace}(B) \\ \mathrm{trace}(B^*) & \mathrm{trace}(C) \end{pmatrix} = R'. $$

(3) Assume now that $R, R' \in M_2(M_2(\mathbf{C}))$. Find a necessary and sufficient condition on $R, R'$ such that there exists $A, B, C \in M_2(M_2(\mathbf{C}))$ for which the block-wise matrix

$$ T := \begin{pmatrix} A & B \\ B^* & C \end{pmatrix}, \quad A = A^*, C = C^*. $$

satisfies

$$ \tau'(T) = A + C = R \quad \text{and} \quad \begin{pmatrix} \mathrm{trace}_{M_2(\mathbf{C})}(A) & \mathrm{trace}_{M_2(\mathbf{C})}(B) \\ \mathrm{trace}_{M_2(\mathbf{C})}(B^*) & \mathrm{trace}_{M_2(\mathbf{C})}(C) \end{pmatrix} = R'. $$

<footnote>$^{14}$While preparing the final version of these lecture notes, I showed the problem to Prof. Denis Serre, who found a counterexample. At first sight, Serre's counterexample does not seem to suggest that the triangle inequality itself (without the extra $\sqrt{d\hbar}$ term on the right-hand side) should be wrong. However, it shows that the procedure of "glueing" classical couplings described in Lemma 7.6 of [63], which is key to proving the triangle inequality for $W_p$ with $1 < p < \infty$, does not have an analogue for general quantum couplings. This is a rather fundamental difference between classical and quantum optimal transport.</footnote>
The notation needs being explained. An element of $B \in M_2(M_2(C))$ is of the form

$$B = \begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{pmatrix} \quad \text{with } B_{kl} \in M_2(C).$$

Then

$$B^* := \left( \frac{\overline{B_{11}^T}}{B_{12}^T} \quad \frac{\overline{B_{21}^T}}{B_{22}^T} \right)$$

while

$$\mathrm{trace}_{M_2(C)}(B) := B_{11} + B_{22}.$$

(4) Explain how (3) is related to the problem of finding $T$ as in (1), in the case where $\rho_1, \rho_2, \rho_3 \in \mathcal{D}(C^2)$.

Now, there's the rub...

(5) Assuming that $R, R' \in \mathcal{D}(C^2)$, does (the) block-wise matrix (matrices) $T$ obtained in (3) satisfy $T = T^* \ge 0$?

**Remark.** A final observation on the generalized triangle inequality is in order. The presence of the additional term $\sqrt{d\hbar}$ in the right-hand side of the generalized triangle inequality may be related to the fact that some points in $\mathcal{D}$ have positive thickness, in the sense that $\mathcal{D}(R, R) \ge \sqrt{2d\hbar}$ for all $R \in \mathcal{D}_2(\mathfrak{H})$. However, the idea of relating this additional term to the “thickness” of the intermediate point $\rho_2$ in the triangle inequality could be misleading. In the first place, the genuine triangle inequality holds if $\rho_2$ is a rank-1 density operator, although $\mathcal{D}(\rho_2, \rho_2) \ge \sqrt{2d\hbar}$ in that case.

Besides, the extra term $\sqrt{d\hbar}$ on the right-hand side of the generalized triangle inequality seems more related to the method of proof of Theorem 25 (viz. the idea of replacing the intermediate point $\rho_2$ with its Husimi transform $\mathcal{H}[\rho_2]$) than to some intrinsic feature of $\mathcal{D}$.

Ultimately, it could be that question (5) in Quiz 31 is answered in the negative, and yet the genuine triangle inequality holds on $\mathcal{D}$ because there always exists an optimal coupling with a special structure for each pair of finite energy density operators on $\mathfrak{H}$, and this special structure acts in favor of the genuine triangle inequality.

This suggests investigating the structure of optimal couplings for the “pseudo-metric” $\mathcal{D}$ — however, the example discussed in Proposition 5 suggests that this is not an easy task.

**3.5. Towards Quantum Optimal Transport.** Until now, we have not said much about quantum optimal transport *per se*, although it is the topic of this school.

We shall conclude these lectures with some (partial) remarks in that direction. The material in this section mostly comes from [40].

We recall that the proofs of the Brenier, or of the Knott-Smith theorems use some form of Kantorovich duality to obtain some additional information on the structure of optimal couplings. That an optimal coupling for $W_2$ is supported in the graph of the subdifferential of some appropriate function comes from the specific structure of the inequality constraint in the Kantorovich dual formulation of the $W_2$ metric, and from the specific structure of the quadratic transport cost.

It is therefore natural to study the quantum analogue of Kantorovich duality in order to obtain some information on the structure of optimal couplings for $\mathcal{D}$. In the
sequel, we shall follow this approach with the classical-to-quantum duality formula in Theorem 23.

Our first task is to give a systematic procedure for constructing elements of the class $\mathfrak{k}$ of test Kantorovich potentials defined before the statement of Theorem 23.

Set $z := (x, \xi) \in \mathbf{R}^d \times \mathbf{R}^d$ and $Z := (y, -i\hbar\nabla_y)$, with the notation

$$z \cdot Z := x \cdot y - i\hbar\xi \cdot \nabla_y.$$

Thus $c_h(x, \xi) = |Z|^2 + |z|^2 I_{\mathfrak{H}} - 2z \cdot Z \ge d\hbar I_{\mathfrak{H}}$ and by Weyl's theorem (see Corollary 2 of Theorem XIII.14 in [55])

$$\tilde{B} \in \mathcal{L}(\mathfrak{H}) \implies c_h(z)^{-1} \tilde{B} \in \mathcal{K}(\mathfrak{H}) \implies \text{ess-spec}(c_h(z) - \tilde{B}) = \emptyset.$$

Assume that $\tilde{B} = \tilde{B}^*$ is such that $c_h(z) - \tilde{B}$ has nondegenerate ground state (i.e. with geometric multiplicity 1) for each $z \in \mathbf{R}^{2d}$. (For instance, choose for $\tilde{B}$ to be a bounded multiplication operator, and apply Theorem XIII.47 in [55]).

Define next

$$\tilde{a}(z) := \min \text{spec}(c_h(z) - \tilde{B}) = \inf_{\|\phi\|_{\mathfrak{H}}=1} \langle \phi | c_h(z) - \tilde{B} | \phi \rangle \\ \implies c_h(z) - \tilde{B} \ge \tilde{a}(z) I_{\mathfrak{H}}.$$

Besides, $z \mapsto \tilde{a}(z)$ is continuous (even real-analytic) by the Kato-Rellich theorem (Theorem XII.8 of [55]), and

$$\tilde{a}(z) \le \langle z | c_h(z) - \tilde{B} | z \rangle = d\hbar - \langle z | \tilde{B} | z \rangle \le d\hbar + \| \tilde{B} \|, \\ \tilde{a}(z) \ge d\hbar + \inf_{\|\phi\|_{\mathfrak{H}}=1} \langle \phi | - \tilde{B} | \phi \rangle \ge d\hbar - \| \tilde{B} \|.$$

Hence $\tilde{a} \in C_b(\mathbf{R}^d \times \mathbf{R}^d)$, and we have obtained in this way

$$(\tilde{a}, \tilde{B}) \in \mathfrak{k}.$$

The Kato-Rellich theorem recalled above also implies the existence of a continuous (even real-analytic) map

$$\mathbf{R}^d \times \mathbf{R}^d \ni z \mapsto \psi_z \in \mathfrak{H} \quad \text{s.t.} \quad \begin{cases} (c_h(z) - \tilde{B})\psi_z = \tilde{a}(z)\psi_z, \\ \text{and } \|\psi_z\|_{\mathfrak{H}} = 1, z \in \mathbf{R}^{2d}. \end{cases}$$

With this, we can define a notion of quantum optimal transport from $\mathcal{P}_2(\mathbf{R}^d \times \mathbf{R}^d)$ to $\mathcal{D}_2(\mathfrak{H})$.

**Theorem 27.** Let $\tilde{B} = \tilde{B}^*$ be such that $c_h(z) - \tilde{B}$ has nondegenerate ground state, set $\tilde{a}(z) := \min \text{spec}(c_h(z) - \tilde{B})$, and let $z \mapsto \psi_z$ be a continuous map from $\mathbf{R}^d \times \mathbf{R}^d$ to $\mathfrak{H}$ such that $\|\psi_z\|_{\mathfrak{H}} = 1$ and $\psi_z \in \text{Ker}(c_h(z) - \tilde{B} - \tilde{a}(z)I_{\mathfrak{H}})$.

Then, for each probability density $f$ with finite 2nd order moments, the map $z \mapsto f(z)|\psi_z\rangle\langle\psi_z|$ is an optimal coupling for the pseudometric $\mathfrak{d}$ between $f$ and the operator

$$\mathcal{T}^{\tilde{B}}[f] := \int_{\mathbf{R}^{2d}} f(z)|\psi_z\rangle\langle\psi_z|dz \in \mathcal{D}_2(\mathfrak{H}).$$

**Example.** Take for example $\tilde{B} = 0$; then, one easily checks that

$$\tilde{a}(z) = d\hbar, \quad \text{Ker}(c_h(z) - d\hbar I_{\mathfrak{H}}) = \mathbf{C}|z\rangle,$$

where $|z\rangle$ is the Schrödinger coherent state centered at $z := q + ip$, so that

$$\mathcal{T}^0[f] = \mathcal{T}[f]$$

is the Toeplitz operator of symbol $f$.
We already knew from Theorem 1 (1) in Lecture I that

$$ \mathcal{D}(f, \mathcal{T}[f]) = \sqrt{d\hbar} = \inf_{\mathcal{P}_2(\mathbb{R}^{2d}) \times \mathcal{D}_2(\mathfrak{S})} \mathcal{D}. $$

*Proof of Theorem 27.* Set $Q(z) := f(z)|\psi_z\rangle\langle\psi_z|$, so that $Q(z)^{\frac{1}{2}} = \sqrt{f(z)}|\psi_z\rangle\langle\psi_z|$, and

$$ \begin{aligned} \int_{\mathbb{R}^{2d}} \mathrm{trace}_{\mathfrak{S}} (Q(z)^{\frac{1}{2}} c_{\hbar}(z) Q(z)^{\frac{1}{2}}) dz &= \int_{\mathbb{R}^{2d}} \langle \psi_z | c_{\hbar}(z) | \psi_z \rangle f(z) dz \\ &= \int_{\mathbb{R}^{2d}} (\tilde{a}(z) \langle \psi_z | \psi_z \rangle + \langle \psi_z | B | \psi_z \rangle) f(z) dz \\ &= \int_{\mathbb{R}^{2d}} \tilde{a}(z) f(z) dz + \mathrm{trace}_{\mathfrak{S}} (\tilde{B} \mathcal{T}^{\tilde{B}}[f]). \end{aligned} $$

Since $(\tilde{a}, \tilde{B}) \in \mathfrak{k}$ and $Q \in \mathcal{C}(f, \mathcal{T}^{\tilde{B}}[f])$, this implies that

$$ \begin{aligned} & \int_{\mathbb{R}^{2d}} \mathrm{trace}_{\mathfrak{S}} (Q(z)^{\frac{1}{2}} c_{\hbar}(z) Q(z)^{\frac{1}{2}}) dz \\ = & \min_{T \in \mathcal{C}(f, \mathcal{T}^{\tilde{B}}[f])} \int_{\mathbb{R}^{2d}} \mathrm{trace}_{\mathfrak{S}} (T(z)^{\frac{1}{2}} c_{\hbar}(z) T(z)^{\frac{1}{2}}) dz = \mathcal{D} (f, \mathcal{T}^{\tilde{B}}[f])^2 . \end{aligned} $$

This also implies that

$$ \begin{aligned} & \int_{\mathbb{R}^{2d}} \tilde{a}(z) f(z) dz + \mathrm{trace}_{\mathfrak{S}} (\tilde{B} \mathcal{T}^{\tilde{B}}[f]) \\ = & \sup_{(a, B) \in \mathfrak{k}} \int_{\mathbb{R}^{2d}} a(z) f(z) dz + \mathrm{trace}_{\mathfrak{S}} (B \mathcal{T}^{\tilde{B}}[f]). \end{aligned} $$

Therefore, in this case, the sup is attained in $\mathfrak{k}$ (this is not true in general). □

## Remarks.

(1) Thus the optimal transport map for $\mathcal{D}$ between $\mathcal{P}_2(\mathbb{R}^d \times \mathbb{R}^d)$ and $\mathcal{D}_2(\mathfrak{S})$ can be thought of as a deformation of the Toeplitz quantization, at least when $\tilde{B}$ is such that $c_{\hbar}(z) - \tilde{B}$ has a ground state of geometric multiplicity 1.

(2) Notice that the starting point in Theorem 27 is the pair $(\tilde{a}, \tilde{B}) \in \mathfrak{k}$, and not the pair consisting of $f$ (the probability density) and the density operator which are the arguments of $\mathcal{D}$. This approach is vaguely reminiscent of the notion of geodesic in Riemannian geometry: the original definition of a geodesic curve on a Riemannian manifold is that of the shortest path between two points on the manifold. The calculus of variations shows that geodesics define local solutions of an ODE system set on the tangent, or cotangent bundle of the manifold. Conversely, solutions of this ODE system define local geodesic curves, which may differ from prescribing arbitrary end points and finding a shortest path between these points. There is a very loose analogy between the local theory of geodesics through the ODE system, and the definition of optimal couplings starting from elements of $\mathfrak{k}$ as explained in Theorem 27. In other words, the transport map $\mathcal{T}^{\tilde{B}}$ is independent of the choice of the endpoints — or more precisely the endpoints $f$ and $\mathcal{T}^{\tilde{B}}[f]$ follow from the optimal transport map, instead of the other way around.

This raises the following question: in Brenier's theorem, the (classical) optimal transport map is the gradient of a convex function. Is there some analogous property in the quantum setting?
### Operator Legendre Duality and Quantum Optimal Transport.
If $(\tilde{a}, \tilde{B}) \in \mathfrak{k}$, one has

$$ \underbrace{|Z|^2 + |z|^2 I_{\mathfrak{H}} - 2z \cdot Z \ge \tilde{a}(z) I_{\mathfrak{H}} + \tilde{B} \quad \Leftrightarrow \quad a(z) + B \ge z \cdot Z,}_{=c_h(z)} $$

with

$$ a(z) := \frac{1}{2}(|z|^2 - \tilde{a}(z)), \quad B = \frac{1}{2}(|Z|^2 - \tilde{B}). $$

Besides, one has

$$ \mathrm{Dom}(c_h(z)) = \mathrm{Dom}(|Z|^2) = H^2(\mathbf{R}^d) \cap L^2(\mathbf{R}^d, |y|^4 dy) =: D. $$

After these preliminaries, we define a notion of Legendre transform of an (unbounded) operator on $\mathfrak{H}$.

**Definition.** Let $B$ satisfy $|Z|^2 - 2B \in \mathcal{L}(\mathfrak{H})$. The Legendre dual of $B$ is the convex function (upper envelope of affine functions)

$$ B^L(z) := \sup_{\phi \in D, \|\phi\|_{\mathfrak{H}}=1} (z \cdot \langle \phi | Z | \phi \rangle - \langle \phi | B | \phi \rangle). $$

Indeed, we recall that, if $T$ is an operator on $\mathfrak{H}$, one can think of $\psi \mapsto \langle \psi | T | \psi \rangle$ as the noncommutative analogue of the evaluation at $x \in \mathbf{R}^n$ of a real-valued function $f$ defined on $\mathbf{R}^n$. Therefore, the definition above is analogous to the usual definition of the Legendre(-Fenchel) transform

$$ \phi^*(\xi) = \sup_{x \in E} (\langle \xi, x \rangle_{E', E} - \phi(x)), \quad \xi \in E', $$

for all $\phi: E \to (-\infty, +\infty]$ where $E$ is a normed linear space on $\mathbf{R}$, and $\phi$ is not identically equal to $+\infty$, and $E'$ is the topological dual of $E$, i.e. the space of linear functionals on $E$ that are continuous for the norm topology.

**Theorem 28.** Let $\tilde{B} = \tilde{B}^*$ be such that $c_h(z) - \tilde{B}$ has nondegenerate ground state, set $\tilde{a}(z) := \min \mathrm{spec}(c_h(z) - \tilde{B})$, and let $z \mapsto \psi_z$ be a continuous map from $\mathbf{R}^d \times \mathbf{R}^d$ to $\mathfrak{H}$ such that $\|\psi_z\|_{\mathfrak{H}} = 1$ and $\psi_z \in \mathrm{Ker}(c_h(z) - \tilde{B} - \tilde{a}(z)I_{\mathfrak{H}})$.

(1) Setting

$$ a(z) := \frac{1}{2}(|z|^2 - \tilde{a}(z)), \quad B := \frac{1}{2}(|Z|^2 - \tilde{B}), $$

one has

$$ a = B^L. $$

(2) Besides

$$ \nabla a(z) = z - \nabla \tilde{a}(z) = \langle \psi_z | Z | \psi_z \rangle. $$

*Proof.* Statement (1) follows from the definition and the variational formula for the ground state. As for (2), differentiate in $z$ the identity

$$ B\psi_z - z \cdot Z\psi_z + a(z)\psi_z = 0, $$

and take the inner product with $\psi_z$ to get

$$ \underbrace{\langle \psi_z | B - z \cdot Z + a(z) | \psi_z \rangle}_{=0 \text{ since } B=B^*, Z=Z^*, a(z) \in \mathbf{R}} + \langle \psi_z | -Z + \nabla a(z) | \psi_z \rangle = 0. $$

□

### Remarks.

(1) In the Knott-Smith theorem recalled in Lecture I, optimal couplings for $W_2$ are supported in the graph of the subdifferential of a l.s.c. convex function, while, in the Brenier theorem, the optimal transport map is the gradient of a convex function. In both results, the function is obtained from an optimal Kantorovich potential by
the same transformation as $\tilde{a} \mapsto a$. Theorem 28 (2) is a partial analogue of this crucial piece of information, except that, in the quantum setting, density operators are not “functions of Z”.

(2) In classical optimal transport, there exist an optimal pair $(a, b)$ of Kantorovich potentials; they are l.s.c. proper convex functions and are Legendre duals of each other, so that $\nabla a \circ \nabla b = \text{Id}$; besides $a \in L_{\mu}^{1}$ and $b \in L_{\nu}^{1}$. If one tries to proceed by analogy, in the present case, one should define some notion of “quantum gradient” of the operator $B$.

One idea to do so is to use the phase space symplectic structure. For a smooth function $\alpha \equiv \alpha(x, \xi)$ on $\mathbf{R}^d \times \mathbf{R}^d$, one has

$$ \partial_{x_j} \alpha = \{\xi_j, \alpha\}, \quad \partial_{\xi_j} \alpha = -\{x_j, \alpha\}, \quad j = 1, \dots, d. $$

This suggests to define “quantum derivatives” as follows,

$$ \partial_{y_j}^Q B := \frac{i}{\hbar} [-i\hbar \partial_{y_j}, B], \quad \partial_{\eta_j}^Q B := -\frac{i}{\hbar} [y_j, B], \quad j = 1, \dots, d, $$

by using the correspondence principle and the analogy between commutator and Poisson bracket

$$ \frac{i}{\hbar}[\cdot, \cdot] \rightarrow \{\cdot, \cdot\} $$

recalled in Lecture II. Since

$$ B\psi_z = z \cdot Z\psi_z + a(z)\psi_z, \quad B = B^*, \quad Z = Z^* \text{ and } a(z) \in \mathbf{R}, $$

one easily checks that

$$ \begin{cases} x_j = \langle \psi_z | \partial_{y_j}^Q B | \psi_z \rangle \\ \xi_j = \langle \psi_z | \partial_{\eta_j}^Q B | \psi_z \rangle \end{cases} $$

This formula can be viewed as the inverse transform of Theorem 28 (2).

(3) Analogous ideas on a definition of an optimal transport “map” between elements of $\mathcal{D}_2(\mathfrak{h})$ can be found in [20]. Partial results analogous to Theorem 28 have been obtained there, but much remains to be done.

Following the proof of Theorems 27–28 suggests viewing the operator

$$ -\frac{1}{2}(|x|^2 - \hbar^2 \Delta_x - A) $$

as the “smallest eigenvalue” of the operator

$$ \frac{1}{2}(|y|^2 - \hbar^2 \Delta_y - B) - x \cdot y + \hbar^2 \nabla_x \cdot \nabla_y, $$

viewed as a “matrix” whose entries are operators in the $x$-variables. However, inequalities between operators do not define a total order relation, so that even the notion of ground state in this setting does not seem to make much sense.

New ideas on this problem are obviously needed.

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(F.G.) CMLS, ÉCOLE POLYTECHNIQUE, IP PARIS, 91128 PALAISEAU CEDEX, FRANCE
*Email address: francois.golse@polytechnique.edu*