On the Convergence of Time Splitting Methods for Quantum Dynamics in the Semiclassical Regime

Abstract

By using the pseudo-metric introduced in Golse and Paul (Arch Ration Mech Anal 223:57–94, 2017), which is an analogue of the Wasserstein distance of exponent 2 between a quantum density operator and a classical (phase-space) density, we prove that the convergence of time splitting algorithms for the von Neumann equation of quantum dynamics is uniform in the Planck constant $\hbar$. We obtain explicit uniform in $\hbar$ error estimates for the first-order Lie–Trotter, and the second-order Strang splitting methods.

Appendices

Appendix A: Proof of Theorem 2.7

Let $Q^{in}\in {\mathcal {C}}(f^{in},R^{in})$. Set

$$\begin{aligned} Q(t,X,\varXi ):=U(t)^*Q^{in}\circ \varPhi _{-t}(X,\varXi )U(t) \end{aligned}$$

for all $t\in {\mathbf {R}}$ and a.e. $(x,\xi )\in {\mathbf {R}}^d\times {\mathbf {R}}^d$, and

$$\begin{aligned} {\mathcal {E}}(t):=\iint _{{\mathbf {R}}^{2d}}\hbox {trace}_{{\mathfrak {H}}}(Q(t,X,\varXi )^{1/2}c(X,\varXi ,y,\hbar D_y)Q(t,X,\varXi )^{1/2})\mathrm{d}X\mathrm{d}\varXi . \end{aligned}$$

Since $\varPhi _t$ leaves the phase space volume element $\mathrm{d}x\mathrm{d}\xi$ invariant

$$\begin{aligned} {\mathcal {E}}(t)=\iint _{{\mathbf {R}}^{2d}}\hbox {trace}_{{\mathfrak {H}}}(\sqrt{Q^{in}(x,\xi )} U(t)c(\varPhi _t(x,\xi ),y,\hbar D_y)U(t)^*\sqrt{Q^{in}(x,\xi )})\mathrm{d}x\xi . \end{aligned}$$

By construction, $Q(t,\cdot ,\cdot )\in {\mathcal {C}}(f(t,\cdot ,\cdot ),R(t))$. Indeed, for a.e. $(X,\varXi )\in {\mathbf {R}}^d$,

$$\begin{aligned} 0\le Q^{in}(\varPhi _{-t}(X,\varXi ))=Q^{in}(\varPhi _{-t}(X,\varXi ))^*\in {\mathcal {L}}({\mathfrak {H}}) \end{aligned}$$

so that $Q(t,X,\varXi )\in {\mathcal {L}}({\mathfrak {H}})$ satisfies

$$\begin{aligned} \begin{aligned} Q(t,X,\varXi )=&U(t)^*Q^{in}(\varPhi _{-t}(X,\varXi ))U(t) \\ =&(U(t)^*Q^{in}(\varPhi _{-t}(X,\varXi ))U(t))=Q(t,X,\varXi )^*\ge 0. \end{aligned} \end{aligned}$$

Besides

$$\begin{aligned} \hbox {trace}_{\mathfrak {H}}(Q(t,X,\varXi ))=\hbox {trace}_{\mathfrak {H}}(Q^{in}(\varPhi _{-t}(X,\varXi ))) =f^{in}(\varPhi _{-t}(X,\varXi ))=f(t,X,\varXi ) \end{aligned}$$

while

$$\begin{aligned} \begin{aligned}&\iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}Q(t,X,\varXi )\mathrm{d}x\mathrm{d}\varXi =U(t)^*\left( \iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}Q^{in}(\varPhi _{-t}(X,\varXi ))\mathrm{d}x\mathrm{d}\varXi \right) U(t)\\&\quad =U(t)^*\left( \iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}Q^{in}(x,\xi )\mathrm{d}x\mathrm{d}\xi \right) U(t) =U(t)^*R^{in}U(t)=R(t). \end{aligned} \end{aligned}$$

In particular

$$\begin{aligned} {\mathcal {E}}(t)\ge E_{\hbar }(f(t),R(t)),\quad \hbox { for each }t\ge 0. \end{aligned}$$

Let $e_j(x,\xi ,\cdot )$ for $j\in {\mathbf {N}}$ be a ${\mathfrak {H}}$-complete orthonormal system of eigenvectors of $Q^{in}(x,\xi )$ for a.e. $x,\xi \in {\mathbf {R}}^d$. Hence

$$\begin{aligned} \begin{aligned}&\hbox {trace}_{{\mathfrak {H}}}(\sqrt{Q^{in}(x,\xi )}U(t)c(\varPhi _t(x,\xi ),y,\hbar D_y)U(t)^*\sqrt{Q^{in}(x,\xi )})\\&\quad =\sum _{j\in {\mathbf {N}}}\rho _j(x,\xi )\langle U(t)^*e_j(x,\xi )|c(\varPhi _t(x,\xi ),y,\hbar D_y)|U(t)^*e_j(x,\xi )\rangle \end{aligned} \end{aligned}$$

where $\rho _j(x,\xi )$ is the eigenvalue of $Q^{in}(x,\xi )$ defined by

$$\begin{aligned} Q^{in}(x,\xi )e_j(x,\xi )=\rho _j(x,\xi )e_j(x,\xi ),\quad \hbox { for a.e. }(x,\xi )\in {\mathbf {R}}^d\times {\mathbf {R}}^d. \end{aligned}$$

If $\phi \equiv \phi (y)\in C^\infty _c({\mathbf {R}}^d)$, the map

$$\begin{aligned} t\mapsto \langle U(t)^*\phi |c(\varPhi _t(x,\xi ),y,\hbar D_y)|U(t)^*\phi \rangle \end{aligned}$$

is of class $C^1$ on ${\mathbf {R}}$, and one has

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\langle U(t)^*\phi |c(\varPhi _t(x,\xi ),y,\hbar D_y)|U(t)^*\phi \rangle \\&\quad =\bigg \langle \frac{i}{\hbar }{\mathcal {H}}U(t)^*\phi \Big |c(\varPhi _t(x,\xi ),y,\hbar D_y)\Big |U(t)^*\phi \bigg \rangle \\&\qquad +\bigg \langle U(t)^*\phi \Big |c(\varPhi _t(x,\xi ),y,\hbar D_y)\Big |\frac{i}{\hbar }{\mathcal {H}}U(t)^*\phi \bigg \rangle \\&\qquad +\langle U(t)^*\phi |\{H(\varPhi _t(x,\xi )),c(\varPhi _t(x,\xi ),y,\hbar D_y)\}|U(t)^*\phi \rangle . \end{aligned} \end{aligned}$$

In other words

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\langle U(t)^*\phi |c(\varPhi _t(x,\xi ),y,\hbar D_y)|U(t)^*\phi \rangle \\&\quad =\bigg \langle U(t)^*\phi \Big |\frac{i}{\hbar }[{\mathcal {H}},c(\varPhi _t(x,\xi ),y,\hbar D_y)]\Big |U(t)^*\phi \bigg \rangle \\&\qquad +\langle U(t)^*\phi |\{H(\varPhi _t(x,\xi )),c(\varPhi _t(x,\xi ),y,\hbar D_y)\}|U(t)^*\phi \rangle . \end{aligned} \end{aligned}$$

A straightforward computation shows that

$$\begin{aligned} \begin{aligned}&\{H_{\lambda }(\varPhi _t(x,\xi )),c(\varPhi _t(x,\xi ),y,\hbar D_y)\}+\frac{i}{\hbar }[{\mathcal {H}}_{\lambda },c_{\lambda }(\varPhi _t(x,\xi ),y,\hbar D_y)]\\&\quad ={\lambda }\sum _{k=1}^d\left( (X_k-y_k)(\varXi _k-\hbar D_{y_k})+(\varXi _k-\hbar D_{y_k})(X_k-y_k)\right) \\&\qquad -\sum _{k=1}^d\left( (\partial _kV(X)-\partial _kV(y))(\varXi _k-\hbar D_{y_k})+(\varXi _k-\hbar D_{y_k})(\partial _kV(X)-\partial _kV(y))\right) \\&\quad \le {\lambda }\sum _{k=1}^d(|X_k-y_k|^2+|\varXi _k-\hbar D_{y_k}|^2)+\sum _{k=1}^d\left( |\partial _kV(X)-\partial _kV(y)|^2+|\varXi _k-\hbar D_{y_k}|^2\right) \\&\quad \le {\lambda }\sum _{k=1}^d(|X_k-y_k|^2+|\varXi _k-\hbar D_{y_k}|^2)\\&\qquad +\max (1,\hbox {Lip}({\nabla }V)^2)\sum _{k=1}^d\left( |X_k-y|^2+|\varXi _k-\hbar D_{y_k}|^2\right) \\&\quad \le \left( {\lambda }+\max (1,\hbox {Lip}({\nabla }V)^2)\right) c(X,\varXi ,y,\hbar D_y). \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} \begin{aligned}&\langle U(t)^*\phi |c(\varPhi _t(x,\xi ),y,\hbar D_y)|U(t)^*\phi \rangle \le \langle \phi |c(x,\xi ,y,\hbar D_y)|\phi \rangle \\&\quad +\left( {\lambda }+\max (1,\hbox {Lip}({\nabla }V)^2)\right) \int _0^t\langle U(s)^*\phi |c(\varPhi _s(x,\xi ),y,\hbar D_y)|U(s)^*\phi \rangle \mathrm{d}s \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}&\langle U(t)^*\phi |c(\varPhi _t(x,\xi ),y,\hbar D_y)|U(t)^*\phi \rangle \\&\quad \le \langle \phi |c(x,\xi ,y,\hbar D_y)|\phi \rangle \exp \left( \left( {\lambda }+\max (1,\hbox {Lip}({\nabla }V)^2)\right) t\right) \end{aligned} \end{aligned}$$

for each $\phi \in C^\infty _c({\mathbf {R}}^d)$. By density of $C^\infty _c({\mathbf {R}}^d)$ in the form domain of $c(x,\xi ,y,\hbar D_y)$

$$\begin{aligned} \begin{aligned} 0&\le \langle U(t)^*e_j(x,\xi )|c(\varPhi _t(x,\xi ),y,\hbar D_y)|U(t)^*e_j(x,\xi )\rangle \\&\le \langle e_j(x,\xi )|c_{\lambda }(x,\xi ,y,\hbar D_y)|e_j(x,\xi )\rangle \exp \left( \left( {\lambda }+\max (1,\hbox {Lip}({\nabla }V)^2)\right) t\right) \end{aligned} \end{aligned}$$

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

$$\begin{aligned} \begin{aligned}&\hbox {trace}_{{\mathfrak {H}}}(\sqrt{Q^{in}(x,\xi )}U(t)c(\varPhi _t(x,\xi ),y,\hbar D_y)U(t)^*\sqrt{Q^{in}(x,\xi )})\\&\quad =\sum _{j\in {\mathbf {N}}}\rho _j(x,\xi )\langle U(t)^*e_j(x,\xi )|c(\varPhi _t(x,\xi ),y,\hbar D_y)|U(t)^*e_j(x,\xi )\rangle \\&\quad \le \exp \left( \left( {\lambda }+\max (1,\hbox {Lip}({\nabla }V)^2)\right) t\right) \sum _{j\in {\mathbf {N}}}\rho _j(x,\xi )\langle e_j(x,\xi )|c(x,\xi ,y,\hbar D_y)|e_j(x,\xi )\rangle \\&\quad =\exp \left( \left( {\lambda }+\max (1,\hbox {Lip}({\nabla }V)^2)\right) t\right) \hbox {trace}_{{\mathfrak {H}}}(\sqrt{Q^{in}(x,\xi )}c(x,\xi ,y,\hbar D_y)\sqrt{Q^{in}(x,\xi )}). \end{aligned} \end{aligned}$$

Integrating both sides of this inequality over ${\mathbf {R}}^d\times {\mathbf {R}}^d$ shows that

$$\begin{aligned} {\mathcal {E}}(t)\le {\mathcal {E}}(0)\exp \left( \left( {\lambda }+\max (1,\hbox {Lip}({\nabla }V)^2)\right) t\right) . \end{aligned}$$

Hence, for each $t\ge 0$ and each $Q^{in}\in {\mathcal {C}}(f,R)$, one has

$$\begin{aligned} E_{\hbar }(f(t),R(t))^2\le {\mathcal {E}}(0)\exp \left( \left( {\lambda }+\max (1,\hbox {Lip}({\nabla }V)^2)\right) t\right) . \end{aligned}$$

Minimizing the right hand side of this inequality as $Q^{in}$ runs through ${\mathcal {C}}(f^{in},R^{in})$ leads to the desired inequality.

Appendix B: Expressing Theorems 3.1, 3.2 in Terms of Density Operators

In this appendix, we express our main results, Theorems 3.1, 3.2 with their Corollaries 3.4 and 3.5, as upper bounds on some appropriate distance between the exact solution of the von Neumann equation and its time-splitting approximation.

Definition B.1

For all $R,S\in {\mathcal {D}}({\mathfrak {H}})$ and each integer $M\ge 0$, we set

$$\begin{aligned} d_M(R,S):=\sup _{\begin{array}{c} \max \limits _{\begin{array}{c} |{\alpha }|,|{\beta }| \le M \end{array}}\Vert {\mathcal {D}}^{\alpha }_{-i\hbar \nabla }{\mathcal {D}}^{\beta }_{x}F\Vert _{1}\le 1 \end{array}}|\hbox {trace}{(F(R-S))}|, \end{aligned}$$

where ${\mathcal {D}}_A=\tfrac{1}{i\hbar }[A,\cdot ]$ for each (possibly unbounded) self-adjoint operator A on ${\mathfrak {H}}$.

One might worry that the definition of $d_M$ involves ${\hbar }$ through the operator $-i\hbar {\nabla }$ and in the definition of ${\mathcal {D}}_A$. However, the correspondence principle in quantum mechanics stipulates that $\tfrac{i}{\hbar }[\cdot ,\cdot ]$ is the quantum analogue of the usual Poisson bracket $\{\cdot ,\cdot \}$, while $-i\hbar {\nabla }$ is the momentum operator, that is the quantum analogue of the momentum variable in the Hamiltonian formulation of classical mechanics. Therefore, both expressions $\tfrac{i}{\hbar }[\cdot ,\cdot ]$ and $-i\hbar {\nabla }$ should be thought of as being “of order $\hbar ^0$ ” in the semiclassical regime.

First we check that $d_M$ metrizes ${\mathcal {D}}({\mathfrak {H}})$. The (uninteresting) case $M=0$ corresponds to the distance associated to the operator norm: $d_0(R,S)=\Vert R-S\Vert$.

Lemma B.2

The function $d_M:\,{\mathcal {D}}({\mathfrak {H}})\times {\mathcal {D}}({\mathfrak {H}})\rightarrow [0,+\infty [$ is a distance.

Proof

That $d_M$ is symmetric and satisfies the triangle inequality is obvious by construction. The only thing to check is the separation property.

Let $f\equiv f(q,p)\in {\mathcal {S}}({\mathbf {R}}^d\times {\mathbf {R}}^d)$ and set $F:=\hbox {OP}^T_{\hbar }(f)$. Elementary computations show that

$$\begin{aligned} {[}x_j,\hbox {OP}^T_{\hbar }(f)]=\hbox {OP}^T_{\hbar }(-i{\hbar }\partial _pf),\quad {[}-i{\hbar }\partial _{x_j},\hbox {OP}^T_{\hbar }(f)]=\hbox {OP}^T_{\hbar }(i{\hbar }\partial _qf). \end{aligned}$$

Moreover, for each bounded Borel measure $\mu$ on ${\mathbf {R}}^d\times {\mathbf {R}}^d$, one has

$$\begin{aligned} \Vert \hbox {OP}^T_{\hbar }(\mu )\Vert _1\le \tfrac{1}{(2\pi {\hbar })^d}\iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}|\mu |(\mathrm{d}q\mathrm{d}p). \end{aligned}$$

(To see this, split $\mu$ into its positive and negative parts as $\mu ^+-\mu ^-$, and observe that

$$\begin{aligned} \Vert \hbox {OP}^T_{\hbar }(\mu ^\pm )\Vert _1=\hbox {trace}_{\mathfrak {H}}(\hbox {OP}^T_{\hbar }(\mu ^\pm )) =\tfrac{1}{(2\pi {\hbar })^d}\iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}\mu ^\pm (\mathrm{d}q\mathrm{d}p), \end{aligned}$$

together with the fact that $\mu ^\pm \ge 0$ implies that $\hbox {OP}^T_{\hbar }(\mu ^\pm )\ge 0$, while $|\mu |=\mu ^++\mu ^-$.) Therefore

$$\begin{aligned} \begin{aligned} d_M(R,S)&\ge \sup _{\begin{array}{c} f\in {\mathcal {S}}({\mathbf {R}}^d\times {\mathbf {R}}^d)\\ \max \limits _{\begin{array}{c} |{\alpha }|,|{\beta }| \le M \end{array}}\Vert \partial ^{\alpha }_{\xi }\partial ^{\beta }_{x}f\Vert _{L^\infty }\le 1 \end{array}}|\hbox {trace}_{\mathfrak {H}}(\hbox {OP}^T_{\hbar }(f)(R-S))|\\&=\sup _{\begin{array}{c} f\in {\mathcal {S}}({\mathbf {R}}^d\times {\mathbf {R}}^d)\\ \max \limits _{\begin{array}{c} |{\alpha }|,|{\beta }| \le M \end{array}}\Vert \partial ^{\alpha }_{\xi }\partial ^{\beta }_{x}f\Vert _{L^\infty }\le 1 \end{array}}\iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}f(q,p)({{\tilde{W}}}_{\hbar }(R)-\tilde{W}_{\hbar }(S))(q,p)\mathrm{d}q\mathrm{d}p. \end{aligned} \end{aligned}$$

(24)

Therefore $d_M(R,S)=0$ implies that ${{\tilde{W}}}_{\hbar }(R)=\tilde{W}_{\hbar }(S)$ in ${\mathcal {S}}'({\mathbf {R}}^d\times {\mathbf {R}}^d)$ and therefore pointwise on ${\mathbf {R}}^d\times {\mathbf {R}}^d$ (since ${{\tilde{W}}}_{\hbar }(R),{{\tilde{W}}}_{\hbar }(S)$ are analytic functions on phase space). Since any density operator on ${\mathfrak {H}}$ is uniquely determined by its Husimi transform (see Remark 2.3 on p. 64 in [10]), this implies in turn that $R=S$. $\square$

Next we formulate our main results, i.e. Theorems 3.1, 3.2 with their Corollaries 3.4 and 3.5 in terms of the distance $d_M$ between the exact solution of the von Neumann equation and its time-splitting approximation. Here is the result for the simple splitting method.

Theorem B.3

(Simple splitting) Under the same assumptions and with the same constants as in Theorem 3.1, one has

$$\begin{aligned} d_{[d/2]+2}(R^n,R(n{\varDelta }t))\le C_T{\varDelta }t+2\sqrt{d{\hbar }}\left( D_d+\exp \left( \tfrac{1}{2}T(1+\max (1,\hbox {Lip}({\nabla }V)^2))\right) \right) , \end{aligned}$$

where the constant $D_d>0$ depends only on the dimension d and is defined in (25).

Under the same assumptions and with the same constants as in Corollary 3.4,

$$\begin{aligned} d_{[d/2]+2}(R^n,R(n{\varDelta }t))\le C'[T,V,\mu ^{in}]{\varDelta }t^{1/3}, \end{aligned}$$

where $C'[T,V,\mu ^{in}]$ is defined in (27).

As for the Strang splitting method, one has the following convergence estimate.

Theorem B.4

(Strang splitting) Under the same assumptions and with the same constants as in Theorem 3.2, one has

$$\begin{aligned} d_{[d/2]+2}(R^n,R(n{\varDelta }t))\le D_T{\varDelta }t^2+2\sqrt{d{\hbar }}\left( D_d+\exp \left( \tfrac{1}{2}T(1+\max (1,\hbox {Lip}({\nabla }V)^2))\right) \right) , \end{aligned}$$

where the constant $D_d>0$ depends only on the dimension d and is defined in (25).

Under the same assumptions and with the same constants as in Corollary 3.5,

$$\begin{aligned} d_{[d/2]+2}(R^n,R(n{\varDelta }t))\le D'[T,V,\mu ^{in}]{\varDelta }t^{2/3}, \end{aligned}$$

where $D'[T,V,\mu ^{in}]$ is defined in (28).

These four results are obvious or straightforward corollaries of Theorems 3.1 and 3.2, and of Corollaries 3.4 and 3.5, and of the following proposition.

Proposition B.5

Let $R,S\in {\mathcal {D}}^2({\mathfrak {H}})$, with Wigner transforms $W_{\hbar }(R),W_{\hbar }(S)$ and Husimi transforms $\tilde{W}_{\hbar }(R),{{\tilde{W}}}_{\hbar }(S)$. For each integer $M\ge 0$, define

$$\begin{aligned} \begin{aligned}&\delta _M(W_{\hbar }(R),W_{\hbar }(S))\\&\quad :=\sup _{\begin{array}{c} f\in L^2({\mathbf {R}}^{2d})\\ \max \limits _{|{\alpha }|,|{\beta }|\le M}\Vert \partial ^{\alpha }_x\partial ^{\beta }_\xi f\Vert _{L^\infty }\le 1 \end{array}}\left| \int (W_{\hbar }(R)(x,\xi )-W_{\hbar }(S)(x,\xi ))f(x,\xi )\mathrm{d}x\mathrm{d}\xi \right| . \end{aligned} \end{aligned}$$

Then

$$\begin{aligned} d_{[d/2]+2}(R,S)\le \delta _{[d/2]+2}(W_{\hbar }(R),W_{\hbar }(S))\le \hbox {dist}_{{\mathrm{MK}},2}(\widetilde{W}_{\hbar }(R),{\widetilde{W}}_{\hbar }(S))+C_d\sqrt{\hbar } \end{aligned}$$

where $C_d>0$ depends only on the dimension d and is given by (29). Moreover

$$\begin{aligned} \delta _{[d/2]+1}(W_{\hbar }(R),W_{\hbar }(S))\le r_{d} \Vert R-S\Vert _1. \end{aligned}$$

Indeed, the first inequality in Proposition B.5 together with Theorem 3.1 (resp. Theorem 3.2) immediately implies the first inequality in Theorem B.3 (resp. Theorem B.4), with

$$\begin{aligned} D_d:=1+\frac{C_d}{2\sqrt{d}}. \end{aligned}$$

(25)

For the uniform bounds in Theorems B.3 and B.4, we proceed as follows: first, Proposition B.5 implies that

$$\begin{aligned} \begin{aligned}&d_{[d/2]+2}(R_{\hbar }^n,R_{\hbar }(n{\varDelta }t))\\&\quad \le \min (\Vert R_{\hbar }^n-R_{\hbar }(n{\varDelta }t)\Vert _1,\hbox {dist}_{{\mathrm{MK}},2}(\tilde{W}_{\hbar }(R_{\hbar }^n),{{\tilde{W}}}_{\hbar }(R_{\hbar }(n{\varDelta }t)))+C_d\sqrt{\hbar }). \end{aligned} \end{aligned}$$

(26)

Now, in the case of the simple splitting method, we bound $\Vert R_{\hbar }^n-R_{\hbar }(n{\varDelta }t)\Vert _1$ as in (18), and $\hbox {dist}_{{\mathrm{MK}},2}(\tilde{W}_{\hbar }(R_{\hbar }^n),{{\tilde{W}}}_{\hbar }(R_{\hbar }(n{\varDelta }t)))$ as in Theorem 3.1. This gives the inequality

$$\begin{aligned} \begin{aligned}&d_{[d/2]+2}(R_{\hbar }^n,R_{\hbar }(n{\varDelta }t))\\&\quad \le \min \Bigg (4\frac{{\varDelta }t}{\hbar }M(V)\left( M(V)t^2+\sqrt{2}\hbar +d+\int _{{\mathbf {R}}^{2d}}|p|\mu ^{in}(\mathrm{d}q\mathrm{d}p)\right) ,\\&\qquad C_T{\varDelta }t+2\sqrt{d{\hbar }}\left( D_d+\exp \left( \tfrac{1}{2}T\left( 1+\max \left( 1,\hbox {Lip}({\nabla }V)^2\right) \right) \right) \right) \Bigg ). \end{aligned} \end{aligned}$$

This implies the second inequality in Theorem B.3 by the same argument as in the proof of Corollary 3.4, with

$$\begin{aligned} \begin{aligned}&C'[T,V,\mu ^{in}]:=\max \Bigg (4\sqrt{2}M(V),C_T,4M(V)\left( M(V)T^2+d +\int _{{\mathbf {R}}^{2d}}|p|\mu ^{in}(\mathrm{d}q\mathrm{d}p)\right) ,\\&\quad 2\sqrt{d}\left( D_d+\exp \left( \tfrac{1}{2}T(1+\max (1,\hbox {Lip}({\nabla }V)^2))\right) \right) \Bigg ). \end{aligned} \end{aligned}$$

(27)

The case of the Strang splitting method is treated similarly: on the right hand side of (26), the first argument in the max, i.e. $\Vert R_{\hbar }^n-R_{\hbar }(n{\varDelta }t)\Vert _1$ is bounded by $M'[T,V,\mu ^{in}]{\varDelta }t^2/{\hbar }$ as in the proof of Corollary 3.5, while $\hbox {dist}_{{\mathrm{MK}},2}(\tilde{W}_{\hbar }(R_{\hbar }^n),{{\tilde{W}}}_{\hbar }(R_{\hbar }(n{\varDelta }t)))$, that is the second argument in the max, is bounded as in Theorem 3.2. Proceeding in this way, one arrives at

$$\begin{aligned} \begin{aligned}&d_{[d/2]+2}(R_{\hbar }^n,R_{\hbar }(n{\varDelta }t))\le \min \left( M'[T,V,\mu ^{in}]\frac{{\varDelta }t^2}{\hbar },D_T{\varDelta }t^2\right. \\&\qquad \left. +2\sqrt{d\hbar } \left( D_d+\exp \left( \tfrac{1}{2}t(1+\max (1,\hbox {Lip}({\nabla }V)^2))\right) \right) \right) \end{aligned} \end{aligned}$$

This implies the second inequality in Theorem B.4 by the same argument as in the proof of Corollary 3.5, with

$$\begin{aligned} \begin{aligned}&D'[T,V,\mu ^{in}]:=\max \Big (D_T,M'[T,V,\mu ^{in}],\\&\quad 2\sqrt{d}\left( D_d+\exp \left( \tfrac{1}{2}T(1+\max (1,\hbox {Lip}({\nabla }V)^2))\right) \right) \Big ). \end{aligned} \end{aligned}$$

(28)

Proof of Proposition B.5

The proof is split into several steps.

(a) Proof of the first right inequality. For each $f\in {\mathcal {S}}({\mathbf {R}}^d\times {\mathbf {R}}^d)$, write

$$\begin{aligned} \begin{aligned}&\iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}f(x,\xi )W_{\hbar }(R-S)(x,\xi )\mathrm{d}x\mathrm{d}\xi \\&\quad =\iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}f(x,\xi )\tilde{W}_{\hbar }(R-S)(x,\xi )\mathrm{d}x\mathrm{d}\xi \\&\qquad -\iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}g(x,\xi )W_{\hbar }(R-S)(x,\xi )\mathrm{d}x\mathrm{d}\xi , \end{aligned} \end{aligned}$$

where $g:=e^{{\hbar }{\varDelta }_{x,\xi }/4}f-f$. Since $R,S\in {\mathcal {L}}^1({\mathfrak {H}})\subset {\mathcal {L}}^2({\mathfrak {H}})$, and the Plancherel theorem implies that

$$\begin{aligned} \Vert W_{\hbar }(T)\Vert _{L^2({\mathbf {R}}^d\times {\mathbf {R}}^d)}^2=\frac{1}{(2\pi {\hbar })^d}\Vert T\Vert _2^2 \end{aligned}$$

for each Hilbert-Schmidt operator T, by using the formula expressing $\Vert T\Vert _2$ in terms of the integral kernel of T, all the integrals above are well defined. First

$$\begin{aligned} \left| \iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}f(x,\xi )\tilde{W}_{\hbar }(R-S)(x,\xi )\mathrm{d}x\mathrm{d}\xi \right| \le \hbox {Lip}(f)\hbox {dist}_{{\mathrm{MK}},2}({{\tilde{W}}}_{\hbar }(R),\tilde{W}_{\hbar }(S)) \end{aligned}$$

by using successively formulas (7.1) and formula (7.3) in chapter 7 of [20]. On the other hand

$$\begin{aligned} \iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}g(x,\xi )W_{\hbar }(R-S)(x,\xi )\mathrm{d}x\mathrm{d}\xi =\hbox {trace}_{\mathfrak {H}}(\hbox {OP}^W_{\hbar }(g)(R-S)) \end{aligned}$$

where $\hbox {OP}^W_{\hbar }(a)$ is the Weyl operator of symbol $a\equiv a(x,\xi )\in {\mathcal {S}}({\mathbf {R}}^d\times {\mathbf {R}}^d)$, whose integral kernel is the function

$$\begin{aligned} (x,y)\mapsto \tfrac{1}{(2\pi {\hbar })^d}\int _{{\mathbf {R}}^d}a\left( \tfrac{1}{2}(x+y), \zeta \right) e^{i\zeta \cdot (x-y)/{\hbar }}d\zeta . \end{aligned}$$

According to the Calderon–Vaillancourt theorem (see [3])

$$\begin{aligned} \Vert \hbox {OP}^W_{\hbar }(g)\Vert \le {\gamma }_d\max _{|{\alpha }|,|{\beta }|\le d/2+1}\Vert \partial _x^{\alpha }\partial _\xi ^{\beta }g\Vert _{L^\infty ({\mathbf {R}}^d\times {\mathbf {R}}^d)}. \end{aligned}$$

On the other hand

$$\begin{aligned} \begin{aligned}&\partial _x^{\alpha }\partial _\xi ^{\beta }g(x,\xi )=\left( (e^{{\hbar }{\varDelta }_{x,\xi }/4}-I)\partial _x^{\alpha }\partial _\xi ^{\beta }f\right) (x,\xi )\\&\quad =\tfrac{1}{\pi ^d}\iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}(\partial _x^{\alpha }\partial _\xi ^{\beta }f(x+{\hbar }^{1/2}z,\xi +{\hbar }^{1/2}\zeta )-\partial _x^{\alpha }\partial _\xi ^{\beta }f(x,\xi ))e^{-|z|^2-|\zeta |^2}\mathrm{d}z\mathrm{d}\zeta , \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}&\max _{|{\alpha }|,|{\beta }|\le d/2+1}\Vert \partial _x^{\alpha }\partial _\xi ^{\beta }g\Vert _{L^\infty ({\mathbf {R}}^d\times {\mathbf {R}}^d)}\\&\quad \le \sqrt{{\hbar }}\max _{|{\alpha }|,|{\beta }|\le d/2+2}\Vert \partial _x^{\alpha }\partial _\xi ^{\beta }f\Vert _{L^\infty ({\mathbf {R}}^d\times {\mathbf {R}}^d)}\tfrac{1}{\pi ^d}\iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d} (|z|+|\zeta |)e^{-|z|^2-|\zeta |^2}\mathrm{d}z\mathrm{d}\zeta \\&\quad \le d\pi ^{-1/2}\sqrt{{\hbar }}\max _{|{\alpha }|,|{\beta }|\le d/2+2}\Vert \partial _x^{\alpha }\partial _\xi ^{\beta }f\Vert _{L^\infty ({\mathbf {R}}^d\times {\mathbf {R}}^d)}. \end{aligned} \end{aligned}$$

Therefore

$$\begin{aligned} \begin{aligned}&\left| \iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}f(x,\xi )(W_{\hbar }(R-S))(x,\xi )\mathrm{d}x\mathrm{d}\xi \right| \le \hbox {Lip}(f)\hbox {dist}_{{\mathrm{MK}},2}(\tilde{W}_{\hbar }(R),{{\tilde{W}}}_{\hbar }(S))\\&\qquad +\tfrac{d}{\sqrt{\pi }}{\gamma }_d\sqrt{{\hbar }}\max _{|{\alpha }|,|{\beta }|\le d/2+2}\Vert \partial _x^{\alpha }\partial _\xi ^{\beta }f\Vert _{L^\infty ({\mathbf {R}}^d\times {\mathbf {R}}^d)}(\Vert R\Vert _1+\Vert S\Vert _1)\\&\quad \le \hbox {dist}_{{\mathrm{MK}},2}({{\tilde{W}}}_{\hbar }(R),\tilde{W}_{\hbar }(S))+\tfrac{2d}{\sqrt{\pi }}{\gamma }_d\sqrt{{\hbar }} \end{aligned} \end{aligned}$$

for all $f\in {\mathcal {S}}({\mathbf {R}}^d\times {\mathbf {R}}^d)$ such that

$$\begin{aligned} \max _{|{\alpha }|,|{\beta }|\le d/2+2}\Vert \partial _x^{\alpha }\partial _\xi ^{\beta }f\Vert _{L^\infty ({\mathbf {R}}^d\times {\mathbf {R}}^d)}\le 1. \end{aligned}$$

By density of ${\mathcal {S}}({\mathbf {R}}^d\times {\mathbf {R}}^d)$ in $L^2({\mathbf {R}}^d\times {\mathbf {R}}^d)$, this implies the first right inequality with

$$\begin{aligned} C_d:=\tfrac{2d}{\sqrt{\pi }}{\gamma }_d \end{aligned}$$

(29)

where ${\gamma }_d$ is the constant that appears in the Calderon–Vaillancourt theorem (as stated in [3]).

(b) Proof of the first left inequality. Elementary computations show that

$$\begin{aligned} W_{\hbar }(T)(x,\xi )=\hbox {trace}_{\mathfrak {H}}(T{\mathcal {N}}(x,\xi )), \end{aligned}$$

where ${\mathcal {N}}(x,\xi )$ is the operator defined, for each $\phi \in {\mathfrak {H}}$, by the formula

$$\begin{aligned} ({\mathcal {N}}(x,\xi )\phi )(Y)=\frac{1}{(\pi {\hbar })^d}\phi (2x-Y)e^{-2i\xi \cdot (x-Y)/{\hbar }}. \end{aligned}$$

Moreover

$$\begin{aligned} {\mathcal {N}}(x,\xi )=\frac{1}{(\pi {\hbar })^d}e^{\frac{i}{{\hbar }}\xi \cdot Y}e^{-\frac{i}{{\hbar }}(-i{\hbar }x\cdot {\nabla }_Y)}Je^{\frac{i}{{\hbar }}(-i{\hbar }x\cdot {\nabla }_Y)}e^{-\frac{i}{{\hbar }}\xi \cdot Y} \end{aligned}$$

where $J\phi (Y):=\phi (-Y)$. Hence

$$\begin{aligned} \partial _{\xi _j}{\mathcal {N}}(x,\xi )=-\frac{1}{i{\hbar }}[Y_j,{\mathcal {N}}(x,\xi )], \quad \partial _{x_j}{\mathcal {N}}(x,\xi )=\frac{1}{i{\hbar }}[-i{\hbar }\partial _{Y_j},{\mathcal {N}}(x,\xi )],\quad j=1,\ldots ,d. \end{aligned}$$

Since ${\mathcal {D}}_{Y_j}$ and ${\mathcal {D}}_{-i{\hbar }\partial _{Y_k}}$ commute for all $j,k=1,\ldots ,d$ by the canonical commutation relations,

$$\begin{aligned} \partial _{x}^{\alpha }\partial _\xi ^{\beta }{\mathcal {N}}(x,\xi )=(-1)^{|{\beta }|}{\mathcal {D}}_Y^{\beta }{\mathcal {D}}_{-i{\hbar }\partial _Y}^{\alpha }{\mathcal {N}}(x,\xi ), \end{aligned}$$

and therefore

$$\begin{aligned} \begin{aligned} \partial _{x}^{\alpha }\partial _\xi ^{\beta }W_{\hbar }(T)(x,\xi )&=(-1)^{|{\beta }|}\hbox {trace}_{\mathfrak {H}}(T{\mathcal {D}}_Y^{\beta }{\mathcal {D}}_{-i{\hbar }\partial _Y}^{\alpha }{\mathcal {N}}(x,\xi ))\\&=(-1)^{|{\alpha }|}\hbox {trace}_{\mathfrak {H}}({\mathcal {N}}(x,\xi ){\mathcal {D}}_Y^{\beta }{\mathcal {D}}_{-i{\hbar }\partial _Y}^{\alpha }T)\\&=(-1)^{|{\alpha }|}W_{\hbar }({\mathcal {D}}_Y^{\beta }{\mathcal {D}}_{-i{\hbar }\partial _Y}^{\alpha }T)(x,\xi )\\&=(-1)^{|{\alpha }|}\hbox {trace}_{\mathfrak {H}}({\mathcal {N}}(x,\xi ){\mathcal {D}}_Y^{\beta }{\mathcal {D}}_{-i{\hbar }\partial _Y}^{\alpha }T) \end{aligned} \end{aligned}$$

for all ${\alpha },{\beta }\in {\mathbf {N}}^d$ and each T such that ${\mathcal {D}}_Y^{\beta }{\mathcal {D}}_{-i{\hbar }\partial _Y}^{\alpha }T\in {\mathcal {L}}^1({\mathfrak {H}})$. Hence

$$\begin{aligned} \Vert \partial _{x}^{\alpha }\partial _\xi ^{\beta }W_{\hbar }(T)\Vert _{L^\infty ({\mathbf {R}}^d\times {\mathbf {R}}^d)}\le \frac{1}{(\pi {\hbar })^d} \Vert {\mathcal {D}}_Y^{\beta }{\mathcal {D}}_{-i{\hbar }\partial _Y}^{\alpha }T\Vert _1 \end{aligned}$$

for all ${\alpha },{\beta }\in {\mathbf {N}}^d$ and each T such that ${\mathcal {D}}_Y^{\beta }{\mathcal {D}}_{-i{\hbar }\partial _Y}^{\alpha }T\in {\mathcal {L}}^1({\mathfrak {H}})$.

Now, for each $R,S\in {\mathcal {D}}({\mathfrak {H}})$ and each $F\in {\mathcal {L}}^1({\mathfrak {H}})$, one has

$$\begin{aligned} \hbox {trace}_{\mathfrak {H}}(F^*(R-S))=(2\pi {\hbar })^d\iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d} \overline{W_{\hbar }(F)(x,\xi )}W_{\hbar }(R-S)(x,\xi )\mathrm{d}x\mathrm{d}\xi . \end{aligned}$$

Thus

$$\begin{aligned} \begin{aligned}&d_M(R,S)\le \sup _{\max _{|{\alpha }|,|{\beta }|\le M}\Vert {\mathcal {D}}_Y^{\beta }{\mathcal {D}}_{-i{\hbar }\partial _Y}^{\alpha }F\Vert _1\le 1}|\hbox {trace}_{\mathfrak {H}}(F(R-S))|\\&\quad \le \sup _{\max _{|{\alpha }|,|{\beta }|\le M}\Vert (\pi {\hbar })^d\partial _x^{\alpha }\partial _\xi ^{\beta }W_{\hbar }(F)\Vert _{L^\infty }\le 1}(2\pi {\hbar })^d\iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}W_{\hbar }(F)W_{\hbar }(R-S)(x,\xi )\mathrm{d}x\mathrm{d}\xi \\&\quad \le \sup _{\begin{array}{c} f\in L^2({\mathbf {R}}^d\times {\mathbf {R}}^d)\\ (\pi {\hbar })^d\max _{|{\alpha }|,|{\beta }|\le M}\Vert \partial _x^{\alpha }\partial _\xi ^{\beta }f\Vert _{L^\infty }\le 1 \end{array}}\left| \iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}(2\pi {\hbar })^dfW_{\hbar }(R-S)(x,\xi )\mathrm{d}x\mathrm{d}\xi \right| \\&\quad \le \sup _{\begin{array}{c} g\in L^2({\mathbf {R}}^d\times {\mathbf {R}}^d)\\ \max _{|{\alpha }|,|{\beta }|\le M}\Vert \partial _x^{\alpha }\partial _\xi ^{\beta }g\Vert _{L^\infty }\le 1 \end{array}}2^d\left| \iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}gW_{\hbar }(R-S)(x,\xi )\mathrm{d}x\mathrm{d}\xi \right| \\&\quad \le 2^d{\delta }_M(W_{\hbar }(R),W_{\hbar }(S)), \end{aligned} \end{aligned}$$

which implies the first left inequality.

(c) Proof of the second inequality. Proceeding as in step (a) above, one has, for each $g\in {\mathcal {S}}({\mathbf {R}}^d\times {\mathbf {R}}^d)$,

$$\begin{aligned} \begin{aligned} \left| \iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}gW_{\hbar }(R-S)(x,\xi )\mathrm{d}x\mathrm{d}\xi \right|&=|\hbox {trace}_{\mathfrak {H}}(\hbox {OP}^W_{\hbar }(g)(R-S))|\\&\le \Vert \hbox {OP}^W_{\hbar }(g)\Vert \Vert R-S\Vert _1\\&\le {\gamma }_d\max _{|{\alpha }|,|{\beta }|\le [d/2]+1}\Vert \partial _x^{\alpha }\partial _\xi ^{\beta }g\Vert _{L^\infty ({\mathbf {R}}^d\times {\mathbf {R}}^d)}\Vert R-S\Vert _1 \end{aligned} \end{aligned}$$

by the Calderon–Vaillancourt Theorem (see [3]). This implies the second inequality by density of ${\mathcal {S}}({\mathbf {R}}^d\times {\mathbf {R}}^d)$ in $L^2({\mathbf {R}}^d\times {\mathbf {R}}^d)$. $\square$

Remark B.6

Two remarks are in order after this proof.

(1) The same argument as in step (a) of the proof of Proposition B.5 implies that, for each probability density $\rho$ on ${\mathbf {R}}^d\times {\mathbf {R}}^d$ such that

$$\begin{aligned} \iint _{{\mathbf {R}}^d\times {\mathbf {R}}^d}(|x|^2+|\xi |^2)\rho (x,\xi )\mathrm{d}x\mathrm{d}\xi <\infty \end{aligned}$$

one has

$$\begin{aligned} \begin{aligned}&{\delta }_{[d/2]+2}(\rho ,W_{\hbar }(S))\le \hbox {dist}_{{\mathrm{MK}},2}(\rho ,\tilde{W}_{\hbar }(S))+\tfrac{1}{2}C_d\sqrt{{\hbar }}\\&\quad \le \sqrt{E_{\hbar }(\rho ,S)^2+d{\hbar }}+\tfrac{1}{2}C_d\sqrt{{\hbar }}\le E_{\hbar }(\rho ,S))+\left( \tfrac{1}{2}C_d+\sqrt{d}\right) \sqrt{{\hbar }}. \end{aligned} \end{aligned}$$

(The second inequality follows from Proposition 2.6 (b), the third being obvious.) While we do not use this bound here, it may be of independent interest.

(2) Of course, one can also use the first right inequality in Proposition B.5 to express Theorems B.3 and B.4 in terms of the distance $\delta _{[d/2]+2}(W_{\hbar }(R_{\hbar }^n),W_{\hbar }(R_{\hbar }(n{\varDelta }t)))$, instead of $\text { }d_{[d/2]+2}(R_{\hbar }^n,R_{\hbar }(n{\varDelta }t))$. We have chosen not to add these bounds in the statements of Theorems B.3 and B.4 for the sake of simplicity.