$\displaystyle-\partial_{a}\left[\left(\partial^{a}H^{[\alpha,2]}+G^{[\alpha,1]}{}^{a}\right)f\right]+\frac{1}{2}\partial_{a}[D^{[\alpha,0]}{}^{ab}\partial_{b}f]$
(5.47)
$\displaystyle\begin{split}:=\,&\left(\partial_{a}H^{[\alpha,2]}+\operatorname{Im}\sum_{k}L^{[\alpha,0]}_{k}\partial_{a}M^{[\alpha,2]}_{k}{}^{*}\right)(\partial^{a}f)+\partial^{a}\left(f\operatorname{Im}\sum_{k}M^{[\alpha,1]}_{k}\partial_{a}M^{[\alpha,1]}_{k}{}^{*}\right)\\\
&\qquad+\frac{\hbar}{2}\partial_{a}\left[\operatorname{Re}\sum_{k}\big{(}\partial^{a}M^{[\alpha,1]}_{k}\big{)}\big{(}\partial^{b}M^{[\alpha,1]}_{k}{}^{*}\big{)}\partial_{b}f\right]\end{split}$
(5.48)

We collect these approximations in the following definition:

######  Definition 5.1 (Harmonic Approximation).

Given the classical Markovian dynamics

$\displaystyle\mathcal{L}_{\mathrm{c}}[f]=$
$\displaystyle-\partial_{a}[f(\partial^{a}H+G^{a})]+\frac{1}{2}\partial_{a}(D^{ab}\partial_{b}f)$
(5.49)

we define the classical harmonic approximation to the dynamics at $\alpha$ as

$\displaystyle\mathcal{L}^{(\alpha)}[f]:=$
$\displaystyle-\partial_{a}[f(\partial^{a}H^{[\alpha,2]}+G^{[\alpha,1]}{}^{a})]+\frac{1}{2}\partial_{a}(D^{[\alpha,0]ab}\partial_{b}f)$
(5.50)

where $E^{[\alpha,m]}$ denotes the $m$-th order Taylor approximation to the
phase space function $E$ at $\alpha$. Likewise, given quantum Markovian
dynamics

$\displaystyle\hat{\mathcal{L}}[\rho]$
$\displaystyle=-\frac{i}{\hbar}[\hat{H},\rho]+\frac{1}{\hbar}\sum_{k}\left(\hat{L}_{k}\rho\hat{L}_{k}^{\dagger}-\frac{1}{2}\\{\hat{L}_{k}^{\dagger}\hat{L}_{k},\rho\\}\right)$
(5.51)

we define the quantum harmonic approximation to the dynamics at $\alpha$ as

$\displaystyle\hat{\mathcal{L}}^{(\alpha)}[\rho]$
$\displaystyle:=-\frac{i}{\hbar}\left[\hat{H}^{[\alpha,2]}+\operatorname{Im}\sum_{k}\hat{L}^{[\alpha,0]}_{k}\hat{M}^{[\alpha,2]}_{k}{}^{\dagger},\rho\right]+\frac{1}{\hbar}\sum_{k}\left(\hat{M}^{[\alpha,1]}_{k}\rho\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}-\frac{1}{2}\left\\{\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\hat{M}^{[\alpha,1]}_{k},\rho\right\\}\right)$
(5.52)

where $M_{k}:=L_{k}-L^{[\alpha,0]}_{k}$ and
$\hat{E}^{[\alpha,m]}=\mathcal{W}_{\hbar}[E^{[\alpha,m]}]$. The respective
errors are denoted

$\displaystyle\delta\mathcal{L}^{(\alpha)}:=\mathcal{L}-\mathcal{L}^{(\alpha)},\qquad\qquad\delta\hat{\mathcal{L}}^{(\alpha)}:=\hat{\mathcal{L}}-\hat{\mathcal{L}}^{(\alpha)}$
(5.54)

Importantly, evolving $\tilde{\rho}$ with $\hat{\mathcal{L}}^{(\alpha)}$ is
equivalent to evolving its Wigner function
$W_{\rho}=\mathcal{W}_{\hbar}[\rho]$ with $\mathcal{L}^{(\alpha)}$:

######  Lemma 5.2 (Quantum-classical harmonic equivalence).

Consider the exact classical dynamics (5.49) corresponding (in the sense of
Definition 3.1) to the exact quantum dynamics (5.51) with $H$ and $L_{k}$
twice differentiable. Then their respective harmonic approximations
$\hat{\mathcal{L}}^{(\alpha)}$ and $\mathcal{L}^{(\alpha)}$ at any point
$\alpha$ are equivalent in the sense of being directly related by the Wigner
transform:

$\displaystyle\mathcal{L}^{(\alpha)}\circ\mathcal{W}_{\hbar}=\mathcal{W}_{\hbar}\circ\hat{\mathcal{L}}^{(\alpha)}$
(5.55)

###### Proof.

By Definition 3.1, $\mathcal{L}$ is the classical limiting dynamics
corresponding to $\hat{\mathcal{L}}$ when
$G^{a}(\alpha)=\operatorname{Im}\sum_{k}L_{k}\partial^{a}L_{k}^{*}$ and
$D^{ab}=\hbar\operatorname{Re}\sum_{k}(\partial^{a}L_{k})(\partial^{b}L_{k}^{*})$,
so

$\displaystyle G^{[\alpha,1]}{}^{a}$
$\displaystyle=\operatorname{Im}\sum_{k}(L^{[\alpha,0]}_{k}\partial^{a}M^{[\alpha,2]}_{k}{}^{*}+M^{[\alpha,1]}_{k}\partial^{a}M^{[\alpha,1]}_{k}{}^{*}),$
(5.56) $\displaystyle D^{[\alpha,0]ab}$
$\displaystyle=\hbar\operatorname{Re}\sum_{k}(\partial^{a}L^{[\alpha,1]}_{k})(\partial^{b}L^{[\alpha,1]}_{k}{}^{*})$
(5.57)

Then (5.55) can be checked through direct computation with the Moyal product
(4.16) using, for example,

$\displaystyle\begin{split}&\mathcal{W}_{\hbar}\left[\hat{M}^{[\alpha,1]}_{k}\rho\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}-\frac{1}{2}\left\\{\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\hat{M}^{[\alpha,1]}_{k},\rho\right\\}\right]\\\
&\qquad=-\partial_{a}\left[\operatorname{Im}\big{(}M^{[\alpha,1]}_{k}\partial^{a}M^{[\alpha,1]}_{k}{}^{*}\big{)}W_{\rho}\right]+\frac{\hbar}{2}\left[\operatorname{Re}\big{(}\partial^{a}M^{[\alpha,1]}_{k}\big{)}\big{(}\partial^{b}M^{[\alpha,1]}_{k}{}^{*}\big{)}\right](\partial_{a}\partial_{b}W_{\rho}).\end{split}$
(5.58)

∎

##  6 Proof of Theorem 3.1

In this section we give a detailed outline of the proof of Theorem 3.1, up to
lemmas that are deferred to later sections.

###  6.1 Defining the Gaussian mixture $\tilde{\rho}(t)$

In this section, we define an evolution

$\displaystyle\tilde{\rho}(t)=\int\\!\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(t)\hat{\tau}_{\alpha,\sigma},$
(6.1)

for some probability distribution $p_{\alpha,\sigma}(t)$ which we will
construct to satisfy

$\frac{\mathrm{d}}{\mathrm{d}t}\tilde{\rho}(t)=\iint\\!\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(t)\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}].$
(6.2)

Recall that $\hat{\mathcal{L}}^{(\alpha)}$ is the harmonic approximation to
the Lindbladian, defined in Section 5.4.

The double integral sign is used to emphasize that the integral is taken over
both phase space, $\mathbb{R}^{2d}$, and the space of all covariance matrices
for pure Gaussian states, i.e., positive semidefinite $\sigma$ where
$\sigma/(\hbar/2)$ is symplectic.

We now invoke our lemma from Section 5.3, restated here for convenience, about
the evolution of Gaussian states under harmonic dynamics: See 5.1 By
Definition 5.1 and Lemma 5.2, the dynamics $\hat{\mathcal{L}}^{(\alpha)}$ are
harmonic and characterized by quadratic Hamiltonian $H^{[\alpha,2]}$, linear
friction $G^{[\alpha,1]}$, and constant scaled diffusion
$\Delta^{[\alpha,0]}$. So it follows from Lemma 5.1 and the chain rule
that494949Note well that the appearances of $H$, etc., in Lemma 5.1 refer to
the parameters of some harmonic dynamics, while in the main body $H$ refers to
general dynamics with harmonic approximations $H^{[\alpha,2]}$, etc., to which
the theorem is applied.

$\displaystyle\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]=\partial_{t}\hat{\tau}_{\alpha,\sigma}=[(\partial
H^{[\alpha,2]}(\alpha)+G^{[\alpha,1]}(\alpha))\partial_{\alpha}+S^{[\alpha,0]}(\alpha,\sigma)\partial_{\sigma}]\hat{\tau}_{\alpha,\sigma}=[U(\alpha)\partial_{\alpha}+S(\alpha,\sigma)\partial_{\sigma}]\hat{\tau}_{\alpha,\sigma}$
(6.3)

for all $\alpha$, where we have introduced the abbreviations
$A\partial_{\alpha}\partial_{\alpha}:=A^{ab}\partial_{a}\partial_{b}=A^{ab}\partial^{2}/\partial\alpha^{a}\partial\alpha^{b}$
and $A\partial_{\sigma}:=A^{ab}\partial/\partial\sigma^{ab}$, and where

$\displaystyle S^{ab}(\alpha,\sigma):=$
$\displaystyle\,K(\alpha)^{a}_{\phantom{a}c}\sigma^{cb}+\sigma^{ac}K^{b}_{\phantom{a}c}(\alpha)+D^{ab}(\alpha)$
(6.4) $\displaystyle U^{a}(\alpha):=$
$\displaystyle\,\partial^{a}H(\alpha)+G^{a}(\alpha)$ (6.5) $\displaystyle
K^{a}_{\phantom{a}b}(\alpha):=$
$\displaystyle\,\partial^{a}U_{b}(\alpha)=F^{a}_{\phantom{a}b}(\alpha)+\Gamma^{a}_{\phantom{a}b}(\alpha)$
(6.6) $\displaystyle F^{a}_{\phantom{a}b}(\alpha)$
$\displaystyle:=\partial^{a}\partial_{b}H(\alpha)$ (6.7)
$\displaystyle\Gamma^{a}_{\phantom{a}b}(\alpha)$
$\displaystyle:=\partial^{a}G_{b}(\alpha)$ (6.8) $\displaystyle
D^{ab}(\alpha):=$
$\displaystyle\,\operatorname{Re}\sum_{k}(\partial^{a}L_{k})(\alpha)(\partial^{b}L_{k}^{*})(\alpha)$
(6.9) $\displaystyle G^{a}(\alpha):=$
$\displaystyle\,\operatorname{Im}\sum_{k}L_{k}(\alpha)\partial^{a}L_{k}^{*}(\alpha)$
(6.10)

(The third equality in (6.3) is because $E^{[\alpha,m]}(\alpha)=E(\alpha)$ for
all $m\geq 0$.) Then our desired condition (6.2) becomes

$\begin{split}\iint\\!\mathrm{d}\alpha\mathrm{d}\sigma\,\hat{\tau}_{\alpha,\sigma}\frac{\mathrm{d}}{\mathrm{d}t}p_{\alpha,\sigma}(t)=\iint\\!\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(t)\left[U(\alpha)\partial_{\alpha}+S(\alpha,\sigma)\partial_{\sigma}\right]\hat{\tau}_{\alpha,\sigma}.\end{split}$
(6.11)

We could integrate the right-hand side of (6.11) by parts in $\sigma$ and
$\alpha$ to obtain a transport equation for $p_{\alpha,\sigma}$, but we would
quickly lose control of the covariance matrix $\sigma$ which can be stretched
by the evolution. Instead, we observe that a component of the flow in the
$\sigma$ direction (toward increasing mixedness of the state) can also be
interpreted as diffusion in the $\alpha$ direction.505050When
$p_{\alpha,\sigma}$ has support only on a single value of the covariance
matrix $\sigma$, it is essentially the Glauber-Sudarshan P function, and it
has long been known that diffusive dynamics, which would increases the
mixedness of a single Gaussian state, can often be re-cast as diffusion in the
P function over pure states [99, 100, 101, 102, 103, 84]. What makes the
present approach distinct is that we are considering a more general
distribution $p_{\alpha,\sigma}$ supported on a large (but restricted) space
of pure-state $\sigma$. Increasing this allowed space to include $\sigma$
corresponding to mixed states may allow our main result to be generalized
further, but we defer that to future work. In particular, for Gaussian states
$\tau_{\alpha,\sigma}(\beta)=\exp[-(\beta-\alpha)^{\top}\sigma^{-1}(\beta-\alpha)/2]/((2\pi)^{d}\sqrt{\det\sigma})$,

$\displaystyle\partial_{\sigma}\tau_{\alpha,\sigma}=\frac{1}{2}\partial_{\alpha}\partial_{\alpha}\tau_{\alpha,\sigma},$
(6.12)

as reviewed in Appendix B.1. Therefore, for any decomposition
$S=S_{\mathrm{D}}+S_{0}$, we have

$\left[U\partial_{\alpha}+S\partial_{\sigma}\right]\hat{\tau}_{\alpha,\sigma}=\left[U\partial_{\alpha}+S_{0}\partial_{\sigma}+\frac{1}{2}S_{\mathrm{D}}\partial_{\alpha}\partial_{\alpha}\right]\hat{\tau}_{\alpha,\sigma}.$
(6.13)

Plugging this into (6.11) and integrating by parts, we see that (6.2) is
satisfied (with $p_{\alpha,\sigma}$ not necessarily non-negative) so long as
$p_{\alpha,\sigma}(t)$ solves

$\frac{\mathrm{d}}{\mathrm{d}t}p_{\alpha,\sigma}(t)=\left[-\partial_{\alpha}U(\alpha)-\partial_{\sigma}S_{0}(\alpha,\sigma)+\frac{1}{2}\partial_{\alpha}\partial_{\alpha}S_{\mathrm{D}}(\alpha,\sigma)\right]p_{\alpha,\sigma}(t)$
(6.14)

in brackets above. Note that the partial derivatives in (6.14) are understood
to act on everything to the right including $p_{\alpha,\sigma}$ outside the
brackets.

The main question remaining is the definition of $S_{0}$ and $S_{\mathrm{D}}$
such that $p_{\alpha,\sigma}(t)$ remains non-negative and supported on the
pure NTS states. This is deferred to Section 7, but we state here the main
condition.

######  Lemma 6.1 ($\mathrm{NTS}$-preservation condition).

Suppose that an initial probability distribution $p_{\alpha,\sigma}(0)$ is
supported on the set $\mathbb{R}^{2d}\times\mathcal{S}_{\mathrm{NTS}}(\mu)$
and that

$\displaystyle
S(\alpha,\sigma)=[F(\alpha)+\Gamma(\alpha)]\sigma+\sigma[F(\alpha)+\Gamma(\alpha)]^{\top}+D(\alpha)$
(6.15)

for matrix-valued function $F$, $\Gamma$, and $D$ that are Hamiltonian, skew-
Hamiltonian, and positive semidefinite, respectively, and which satisfy
$D/\hbar+i\Gamma\omega\geq 0$ (as is guaranteed for all Lindbladian dynamics).
Suppose also that $\mu$ satisfies

$\displaystyle 2\hbar\mu\zeta_{\mathrm{max}}+\mu^{2}\chi_{\mathrm{max}}<1$
(6.16)

where

$\displaystyle\zeta_{\mathrm{max}}\,$
$\displaystyle:=\sup_{\alpha}\|F(\alpha)\|\|D^{-1}(\alpha)\|,$ (6.17)
$\displaystyle\chi_{\mathrm{max}}\,$
$\displaystyle:=\sup_{\alpha}\|D(\alpha)\|\|D^{-1}(\alpha)\|,$ (6.18)

are the respective extremal ratios taken by $\|F\|$ and $\|D\|$ relative to
the minimum eigenvalue $\lambda_{\mathrm{min}}[D]=\|D^{-1}\|^{-1}$. (The
latter is just the maximum condition number taken by $D$ over phase space.) It
is also sufficient that $\Gamma=0$ and $\hbar\mu\zeta_{\mathrm{max}}<1$. Then
there exists a decomposition $S=S_{0}+S_{\mathrm{D}}$ such that when
$p_{\alpha,\sigma}(t)$ is evolved according to
$\partial_{t}p_{\alpha,\sigma}(t)=\mathchoice{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{5.00003pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\displaystyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{5.00003pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\textstyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{3.5pt}{2.72223pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptstyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{2.5pt}{1.94444pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptscriptstyle\mathcal{L}\hss$\crcr}}}\limits}^{(\alpha,\sigma)}[p_{\alpha,\sigma}(t)]$
where

$\displaystyle\mathchoice{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{5.00003pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\displaystyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{5.00003pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\textstyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{3.5pt}{2.72223pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptstyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{2.5pt}{1.94444pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptscriptstyle\mathcal{L}\hss$\crcr}}}\limits}^{(\alpha,\sigma)}[p_{\alpha,\sigma}]:=\left[-\partial_{\alpha}U(\alpha)-\partial_{\sigma}S_{0}(\alpha,\sigma)+\frac{1}{2}\partial_{\alpha}\partial_{\alpha}S_{\mathrm{D}}(\alpha,\sigma)\right][p_{\alpha,\sigma}]$
(6.27)

$p_{\alpha,\sigma}(t)$ remains non-negative and supported on
$\mathbb{R}^{2d}\times\mathcal{S}_{\mathrm{NTS}}(\mu)$ for all times $t\geq
0$.

Thus, under the assumptions of Theorem 3.1, we have defined a trajectory
$\tilde{\rho}(t)$ that is at all times a mixture (6.1) of Gaussian states with
covariances matrices from the set $\mathcal{S}_{\mathrm{NTS}}(\mu)$. We now
turn to proving the bounds (3.21) and (3.20)

###  6.2 Bounding $\|\rho-\tilde{\rho}\|_{\mathrm{Tr}}$

First observe what may be understood as a version of Duhamel’s principle,

$\displaystyle\tilde{\rho}(t)-\rho(t)$
$\displaystyle=\int_{0}^{t}\mathrm{d}s\,e^{(t-s)\hat{\mathcal{L}}}\left(\partial_{s}-\hat{\mathcal{L}}\right)[\tilde{\rho}(s)]$
(6.28)

because the anti-derivative of the integrand (with respect to $s$) is
$e^{(t-s)\hat{\mathcal{L}}}[\tilde{\rho}(s)]$ (and $\tilde{\rho}(0)=\rho(0)$).
Then

$\displaystyle\left\lVert\tilde{\rho}(t)-\rho(t)\right\rVert_{\mathrm{Tr}}$
$\displaystyle=\left\lVert\int_{0}^{t}\mathrm{d}s\,e^{(t-s)\hat{\mathcal{L}}}\left(\partial_{s}-\hat{\mathcal{L}}\right)[\tilde{\rho}(s)]\right\rVert_{\mathrm{Tr}}$
(6.29) $\displaystyle\leq\int_{0}^{t}\mathrm{d}s\,\left\lVert
e^{(t-s)\hat{\mathcal{L}}}\left(\partial_{s}-\hat{\mathcal{L}}\right)[\tilde{\rho}(s)]\right\rVert_{\mathrm{Tr}}$
(6.30)
$\displaystyle\leq\int_{0}^{t}\mathrm{d}s\,\left\lVert\left(\partial_{s}-\hat{\mathcal{L}}\right)[\tilde{\rho}(s)]\right\rVert_{\mathrm{Tr}}$
(6.31)
$\displaystyle=\int_{0}^{t}\mathrm{d}s\,\left\lVert\int\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(s)\left(\hat{\mathcal{L}}^{(\alpha)}-\hat{\mathcal{L}}\right)[\hat{\tau}_{\alpha,\sigma}]\right\rVert_{\mathrm{Tr}}$
(6.32)
$\displaystyle=\int_{0}^{t}\mathrm{d}s\,\left\lVert\int\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(s)\delta\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]\right\rVert_{\mathrm{Tr}}$
(6.33)
$\displaystyle\leq\int_{0}^{t}\mathrm{d}s\,\int\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(s)\left\lVert\delta\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]\right\rVert_{\mathrm{Tr}}$
(6.34)
$\displaystyle\leq\max_{\alpha}\max_{\sigma\in\mathcal{S}_{\mathrm{NTS}}(\mu)}\left\lVert\delta\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]\right\rVert_{\mathrm{Tr}}\int_{0}^{t}\mathrm{d}s\,\int\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(t)$
(6.35) $\displaystyle\leq
t\max_{\alpha}\max_{\sigma\in\mathcal{S}_{\mathrm{NTS}}(\mu)}\left\lVert\delta\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]\right\rVert_{\mathrm{Tr}}$
(6.36)

where (6.31) follows from the fact that $e^{(t-s)\hat{\mathcal{L}}}$ is a CP
map and so cannot increase the trace norm, (6.32) follows from dynamics (6.2)
for $\tilde{\rho}$, in (6.33) we have used
$\hat{\mathcal{L}}=\hat{\mathcal{L}}^{(\alpha)}+\delta\hat{\mathcal{L}}^{(\alpha)}$,
in (6.35) we have used the fact that $p_{\alpha,\sigma}$ is supported on
$\mathcal{S}_{\mathrm{NTS}}(\mu)$ which follows from Lemma 6.1 and in (6.36)
we have used that $p_{\alpha,\sigma}$ is a probability distribution so that
$\int p_{\alpha,\sigma}=1$. We will now make use of the following lemma on the
error introduced. by the harmonic approximation, proved in Section 8.2:

######  Lemma 6.2 (Error in harmonic approximation to quantum dynamics).

The error $\delta\mathcal{L}^{(\alpha)}$ in the local harmonic approximation
to the quantum dynamics acting on coherent state $\tau_{\alpha,\sigma}$
satisfies

$\displaystyle\left\lVert\delta\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]\right\rVert_{\mathrm{Tr}}\leq$
$\displaystyle\,C_{d}\frac{\|\sigma\|^{3/2}}{\hbar}\left(B^{\mathrm{anh}}_{\mathrm{q}}[H,L_{k},\hbar]+B^{\mathrm{anh}}_{\mathrm{q}^{\prime}}[L_{k},\hbar,\|\sigma\|^{1/2}]\right)$
(6.37)

where $B^{\mathrm{anh}}_{\mathrm{q}}[H,L_{k},\hbar]$ and
$B^{\mathrm{anh}}_{\mathrm{q}^{\prime}}[L_{k},\hbar,\nu]$ are defined in
(3.18) and (3.19).

Applying Lemma 6.2 gives

$\displaystyle\left\lVert\delta\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]\right\rVert_{\mathrm{Tr}}\,$
$\displaystyle\leq\frac{\|\sigma\|^{3/2}}{\hbar}\left(B^{\mathrm{anh}}_{\mathrm{q}}[H,L_{k},\hbar]+B^{\mathrm{anh}}_{\mathrm{q}^{\prime}}[L_{k},\hbar,\|\sigma\|^{1/2}]\right)$
(6.38)
$\displaystyle\leq\hbar^{1/2}g^{3/2}\left(B^{\mathrm{anh}}_{\mathrm{q}}[H,L_{k},\hbar]+B^{\mathrm{anh}}_{\mathrm{q}^{\prime}}[L_{k},\hbar,\|\sigma\|^{1/2}]\right)$
(6.39)

for all $\alpha$ and $\sigma\in\mathcal{S}_{\mathrm{NTS}}(g/2)$, recalling
$\sigma\in\mathcal{S}_{\mathrm{NTS}}(\mu)$ satisfy
$\left\lVert\sigma\right\rVert\leq(2\mu)^{-1}\hbar$ and that
$B^{\mathrm{anh}}_{\mathrm{q}^{\prime}}[L_{k},\hbar,\nu]$ is a monotonically
increasing function of $\nu$, so (6.36) implies

$\displaystyle\left\lVert\tilde{\rho}(t)-\rho(t)\right\rVert_{\mathrm{Tr}}\leq$
$\displaystyle\,t\hbar^{1/2}(2\mu)^{-3/2}\left(B^{\mathrm{anh}}_{\mathrm{q}}[H,L_{k},\hbar]+B^{\mathrm{anh}}_{\mathrm{q}^{\prime}}[L_{k},\hbar,(2\mu)^{-1/2}\hbar^{1/2}]\right)$
(6.40)

###  6.3 Bounding $\|\mathcal{W}_{\hbar}[\tilde{\rho}]-f\|_{\mathrm{L^{1}}}$

The calculation is similar to the that of the previous subsection. Again we
observe a version of Duhamel’s principle,

$\displaystyle\mathcal{W}_{\hbar}[\tilde{\rho}(t)]-f(t)$
$\displaystyle=\int_{0}^{t}\mathrm{d}s\,e^{(t-s)\mathcal{L}}\left(\partial_{s}-\mathcal{L}\right)[\mathcal{W}_{\hbar}[\tilde{\rho}](s)]$
(6.41)

which follows for a similar reason as (6.28). Then we have

$\displaystyle\left\lVert\mathcal{W}_{\hbar}[\tilde{\rho}(t)]-f(t)\right\rVert$
$\displaystyle=\left\lVert\int_{0}^{t}\mathrm{d}s\,e^{(t-s)\mathcal{L}}\left(\partial_{s}-\mathcal{L}\right)[\tilde{\rho}(s)]\right\rVert_{L^{1}}$
(6.42) $\displaystyle\leq\int_{0}^{t}\mathrm{d}s\,\left\lVert
e^{(t-s)\mathcal{L}}\left(\partial_{s}-\mathcal{L}\right)[\tilde{\rho}(s)]\right\rVert_{L^{1}}$
(6.43)
$\displaystyle\leq\int_{0}^{t}\mathrm{d}s\,\left\lVert\left(\partial_{s}-\mathcal{L}\right)[\tilde{\rho}(s)]\right\rVert_{L^{1}}$
(6.44)
$\displaystyle=\int_{0}^{t}\mathrm{d}s\,\left\lVert\int\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(s)\left(\mathcal{L}^{(\alpha)}-\mathcal{L}\right)[\tau_{\alpha,\sigma}]\right\rVert_{L^{1}}$
(6.45)
$\displaystyle=\int_{0}^{t}\mathrm{d}s\,\left\lVert\int\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(s)\delta\mathcal{L}^{(\alpha)}[\tau_{\alpha,\sigma}]\right\rVert_{L^{1}}$
(6.46)
$\displaystyle\leq\int_{0}^{t}\mathrm{d}s\,\int\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(s)\left\lVert\delta\mathcal{L}^{(\alpha)}[\tau_{\alpha,\sigma}]\right\rVert_{L^{1}}$
(6.47)
$\displaystyle\leq\max_{\alpha}\max_{\sigma\in\mathcal{S}_{\mathrm{NTS}}(\mu)}\left\lVert\mathcal{L}^{(\alpha)}[\tau_{\alpha,\sigma}]\right\rVert_{L^{1}}\int_{0}^{t}\mathrm{d}s\,\int\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(t)$
(6.48) $\displaystyle\leq
t\max_{\alpha}\max_{\sigma\in\mathcal{S}_{\mathrm{NTS}}(\mu)}\left\lVert\delta\mathcal{L}^{(\alpha)}[\tau_{\alpha,\sigma}]\right\rVert_{L^{1}}$
(6.49)

where (6.44) follows from the fact that $e^{(t-s)\mathcal{L}}$ (flow and
diffusion) does not increase $L^{1}$ norm, (6.45) follows from
$\mathcal{W}_{\hbar}[\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]]=\mathcal{L}^{(\alpha)}[\tau_{\alpha,\sigma}]$
(by Lemma 5.2), (6.46) follows from
$\hat{\mathcal{L}}=\hat{\mathcal{L}}^{(\alpha)}+\delta\hat{\mathcal{L}}^{(\alpha)}$,
in (6.48) we used that $p_{\alpha,\sigma}$ is supported on
$\mathcal{S}_{\mathrm{NTS}}(\mu)$ (by Lemma 6.1), and (6.49) follows from
$\int\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(t)=1$. We will now
make use of the following bound on the error in the classical dynamics
introduced in the harmonic approximation, proved in Section 8:

######  Lemma 6.3 (Error in harmonic approximation to classical dynamics).

The error $\delta\mathcal{L}^{(\alpha)}:=\mathcal{L}-\mathcal{L}^{(\alpha)}$
in the local harmonic approximation to the classical dynamics acting on
coherent state $\tau_{\alpha,\sigma}$ satisfies

$\left\lVert\delta\mathcal{L}^{(\alpha)}[\tau_{\alpha,\sigma}]\right\rVert_{L^{1}}\leq
14d^{\frac{3}{2}}\frac{1}{\hbar}\left\lVert\sigma\right\rVert^{\frac{3}{2}}B^{\mathrm{anh}}_{\mathrm{c}}[H,L_{k}]$
(6.50)

with anharmonicity factor

$\displaystyle
B^{\mathrm{anh}}_{\mathrm{c}}[H,L_{k}]:=\left(\left|H\right|_{C^{3}}+\left|G\right|_{C^{2}}+\left|\Delta\right|_{C^{1}}\right).$
(6.51)

depending only on the classical Hamiltonian and Lindblad functions through
$G^{a}=\operatorname{Im}\sum_{k}L_{k}\partial^{a}L_{k}^{*}$ and
$\Delta^{ab}=\operatorname{Re}\sum_{k}(\partial^{a}L_{k})(\partial^{b}L_{k}^{*})$.

By Lemma 6.3 we have

$\displaystyle\|\delta\mathcal{L}^{(\alpha)}[\tau_{\alpha,\sigma}]\|_{L^{1}}\leq
14d^{\frac{3}{2}}\frac{1}{\hbar}\left\lVert\sigma\right\rVert^{\frac{3}{2}}B^{\mathrm{anh}}_{\mathrm{c}}[H,L_{k}]\leq
14d^{\frac{3}{2}}\hbar^{1/2}(2\mu)^{-3/2}B^{\mathrm{anh}}_{\mathrm{c}}[H,L_{k}]$
(6.52)

for all $\alpha$ and $\sigma\in\mathcal{S}_{\mathrm{NTS}}(g/2)$, recalling
$\sigma\in\mathcal{S}_{\mathrm{NTS}}(\mu)$ satisfy
$\left\lVert\sigma\right\rVert\leq(2\mu)^{-1}\hbar$, so (6.49) implies

$\displaystyle\left\lVert\mathcal{W}_{\hbar}[\tilde{\rho}(t)]-f(t)\right\rVert_{L^{1}}\leq
14d^{\frac{3}{2}}\hbar^{1/2}(2\mu)^{-3/2}tB^{\mathrm{anh}}_{\mathrm{c}}[H,L_{k}]$
(6.53)

###  6.4 Concluding the proof

We construct $\tilde{\rho}(t)$ as in Section 6.1. We need to choose our class
of NTS states such that the condition sufficiently strong diffusion in Lemma
6.1 is satisfied:

$\displaystyle\begin{cases}2\hbar\mu\zeta_{\mathrm{max}}+\mu^{2}\chi_{\mathrm{max}}<1&\text{if}\
G^{a}\neq 0\,\,\text{(frictionful)}\\\
\hbar\mu\zeta_{\mathrm{max}}<1&\text{if}\
G^{a}=0\,\,\text{(frictionless)}\end{cases}$ (6.54)

where we recall

$\displaystyle\zeta_{\mathrm{max}}\,$
$\displaystyle:=\sup_{\alpha}\|F(\alpha)\|\|D^{-1}(\alpha)\|=\hbar^{-1}\left(\inf_{\alpha}\frac{\lambda_{\mathrm{min}}[\Delta(\alpha)]}{\lambda_{\mathrm{max}}[\nabla^{2}H(\alpha)]}\right)^{-1},$
(6.55) $\displaystyle\chi_{\mathrm{max}}\,$
$\displaystyle:=\sup_{\alpha}\|D(\alpha)\|\|D^{-1}(\alpha)\|=\left(\inf_{\alpha}\frac{\lambda_{\mathrm{min}}[D(\alpha)]}{\lambda_{\mathrm{max}}[D(\alpha)]}\right)^{-1}.$
(6.56)

Therefore we pick NTS states characterized
$\sigma\in\mathcal{S}_{\mathrm{NTS}}(\mu=g/2)$, where we recall: See 3.2 In
other words, we choose

$\displaystyle\mu=\frac{g}{2}=\begin{cases}\min\left\\{(4\hbar\zeta_{\mathrm{max}})^{-1},\chi_{\mathrm{max}}^{-1/2}\right\\},&\text{if}\
G^{a}\neq 0\,\,\text{(frictionful)}\\\
\min\left\\{(2\hbar\zeta_{\mathrm{max}})^{-1},1\right\\},&\text{if}\
G^{a}=0\,\,\text{(frictionless)}\end{cases}$ (6.57)

This choice ensures (6.54), so we have by Lemma 6.1 that the evolution of
$\tilde{\rho}$ given by (6.14) preserves the property that
$p_{\alpha,\sigma}(t)$ is always supported on
$\mathbb{R}^{2d}\times\mathcal{S}_{\mathrm{NTS}}(g/2)$. Then by (6.40) we
conclude

$\displaystyle\left\lVert\tilde{\rho}(t)-\rho(t)\right\rVert_{\mathrm{Tr}}\leq$
$\displaystyle\,t\,\hbar^{\frac{1}{2}}g^{-\frac{3}{2}}\left(B^{\mathrm{anh}}_{\mathrm{q}}[H,L_{k},\hbar]+B^{\mathrm{anh}}_{\mathrm{q}^{\prime}}[L_{k},\hbar,\sqrt{\hbar/g}]\right)$
(6.58)

Likewise by (6.53) we conclude

$\displaystyle\left\lVert\mathcal{W}_{\hbar}[\tilde{\rho}(t)]-f(t)\right\rVert_{L^{1}}\leq
14d^{\frac{3}{2}}t\,\hbar^{\frac{1}{2}}g^{-\frac{3}{2}}B^{\mathrm{anh}}_{\mathrm{c}}[H,L_{k}]$
(6.59)

This concludes the proof of Theorem 3.1, our main result. The proof depended
on lemmas concerning the preservation of the NTS condition (Lemma 6.1, proven
in Section 7) and the size of the error from the classical and quantum
harmonic approximations (Lemma 6.2 and Lemma 6.3, proven in Section 8).

##  7 NTS Preservation

In this section we prove Lemma 6.1, which we now restate. See 6.1

The proof of Lemma 6.1 relies primarily on Lemma 7.1, a statement about just
linear algebra which we use to define the decomposition
$S=S_{0}+S_{\mathrm{D}}$. To state our decomposition we recall from Section
4.2 that a matrix $A$ is defined to be symplectic, Hamiltonian, or skew-
Hamiltonian when it satisfies the respective conditions $A^{\top}\omega
A=\omega$, $A^{\top}=-\omega^{\top}A\omega$, $A^{\top}=\omega^{\top}A\omega$.
When $A$ is symmetric, symplectic, and invertible (as is true for the
covariance matrix for all pure Gaussian states) it therefore satisfies
$\omega^{\top}A\omega=A^{-1}$.

######  Lemma 7.1 (Decomposition of covariance dynamics).

Suppose $F$ is a Hamiltonian matrix ($F^{\top}=-\omega^{\top}F\omega$),
$\Delta+i\Gamma\omega\geq 0$ is a positive semidefinite matrix with real and
imaginary parts $\Delta$ and $\Gamma\omega$ satisfying $\Delta\geq
c_{\Delta}\mathds{1}$, and $S$ is the function

$S(\sigma):=(F+\Gamma)\sigma+\sigma(F+\Gamma)^{\top}+\hbar\Delta$ (7.1)

on positive definite matrices $\sigma$. Suppose moreover that $\mu\in(0,1]$
obeys

$\displaystyle c_{\Delta}>2\mu\|F\|+\mu^{2}\|\Delta\|.$ (7.2)

or, alternatively, that $\Gamma=0$ and $\mu$ satisfies the weaker condition
$c_{\Delta}>\mu\|F\|$. Then there exists a decomposition
$S=S_{0}+S_{\mathrm{D}}$ satisfying

  * •

“Diffusion positivity”: $S_{\mathrm{D}}(\sigma)$ is positive semidefinite
whenever $\sigma\in\mathcal{S}_{\mathrm{NTS}}(\mu)$;

  * •

“Purity preservation”: $S_{0}(\sigma)$ is symmetric and
$\sigma^{-1}S_{0}(\sigma)$ is Hamiltonian; and

  * •

“NTS preservation”: $v^{\top}S_{0}(\sigma)v>0$ whenever $v$ is an eigenvector
of $\sigma$ with eigenvalue $\lambda\leq(\hbar/2)\mu$.

###### Proof.

In the following proof, we will let an overline denote division by $\hbar/2$,
so $\bar{\sigma}=\sigma/(\hbar/2)$. We work with these normalized quantities
because $\bar{\sigma}=\omega^{\top}\bar{\sigma}^{-1}\omega>0$ is symplectic
exactly when $\sigma$ is the covariance matrix of a pure Gaussian state.
Likewise, $\sigma\in\mathcal{S}_{\mathrm{NTS}}(\mu)$ implies
$\mu\mathds{1}_{2d}\leq\bar{\sigma}\leq\mu^{-1}\mathds{1}_{2d}$.

The dynamics $S(\sigma)$ generate a very general positivity-preserving linear
dynamics for $\sigma$, and our goal is to break this up into a piece $S_{0}$
that additionally preserves the symplectic property (“purity preservation”)
and a remainder $S_{\mathrm{D}}$ that is equivalent to diffusion of the state
in phase space (“diffusion positivity”). The purity-preservation condition is
that $\sigma^{-1}S_{0}(\sigma)=\bar{\sigma}^{-1}\bar{S}_{0}(\sigma)$ is
Hamiltonian, which is equivalent to the form
$\bar{S}_{0}(\sigma)=\mathring{F}(\bar{\sigma})\bar{\sigma}+\bar{\sigma}\mathring{F}(\bar{\sigma})^{\top}$
for some Hamiltonian matrix $\mathring{F}(\bar{\sigma})$.
Intuitively,515151Indeed, if the overall dynamics are pure Hamiltonian (i.e.,
if all Lindblad terms are zero), then we could just make the choice
$\mathring{F}(\bar{\sigma})=F$. we want $\mathring{F}$ to include the
Hamiltonian part $F$ of the overall dynamics $S$ plus an extra piece
$Y(\bar{\sigma})$ that will fight against any squeezing that risks violating
the NTS-preservation condition. Therefore we look for
$\mathring{F}(\bar{\sigma})=F+Y(\bar{\sigma})$,

$\displaystyle\bar{S}_{0}(\sigma)$
$\displaystyle=[F+Y(\bar{\sigma})]\bar{\sigma}+\bar{\sigma}[F+Y(\bar{\sigma})]^{\top},$
(7.3) $\displaystyle\bar{S}_{\mathrm{D}}(\sigma)$
$\displaystyle=2\Delta+[\Gamma-Y(\bar{\sigma})]\bar{\sigma}+\bar{\sigma}[\Gamma-Y(\bar{\sigma})]^{\top}.$
(7.4)

In the frictionful case ($\Gamma\neq 0$) we make the ansatz

$\displaystyle
Y(\bar{\sigma})=\frac{1}{2}[{\Delta}\bar{\sigma}^{-1}-\bar{\sigma}\omega^{\top}{\Delta}\omega],$
(7.5)

which is the Hamiltonian part of ${\Delta}\bar{\sigma}^{-1}$. Because
$\bar{\sigma}$ is symplectic,
$\omega^{\top}\bar{\sigma}\omega=\bar{\sigma}^{-1}$, so $Y(\bar{\sigma})$ is
Hamiltonian by construction and hence preserves purity. Furthermore,

$\displaystyle\bar{S}_{\mathrm{D}}(\sigma)\,$
$\displaystyle={\Delta}+\bar{\sigma}\omega^{\top}{\Delta}\omega\bar{\sigma}+\Gamma\bar{\sigma}+\bar{\sigma}\Gamma^{\top}$
(7.6)
$\displaystyle=\frac{1}{2}\left[(1+i\omega\bar{\sigma})^{\dagger}({\Delta}+i\Gamma\omega)(1+i\omega\bar{\sigma})+\mathrm{tp.}\right]$
(7.7)

is a positive semidefinite because ${\Delta}+i\Gamma\omega$ is a positive
semidefinite matrix. (Above, “$\mathrm{tp.}$” denotes the transpose of the
preceding expression.) This ensures diffusion positivity. Finally, for $v$ an
eigenvalue of $\bar{\sigma}$ with eigenvalue $\bar{\lambda}\leq\mu$, we
consider

$\displaystyle v^{\top}\bar{S}_{0}(\sigma)v\,$
$\displaystyle=v^{\top}[F\bar{\sigma}+\bar{\sigma}F^{\top}+{\Delta}-\bar{\sigma}\omega^{\top}{\Delta}\omega\bar{\sigma}]v,$
(7.8)
$\displaystyle=2\bar{\lambda}v^{\top}Fv+v^{\top}{\Delta}v-\bar{\lambda}^{2}v^{\top}\omega^{\top}{\Delta}\omega
v,$ (7.9)
$\displaystyle\geq-2\bar{\lambda}\|F\|+c_{{\Delta}}-\bar{\lambda}^{2}\|{\Delta}\|$
(7.10)

This is guaranteed to be positive, and thus NTS preserving, when (7.2) holds
because $0<\bar{\lambda}\leq\mu\leq 1$.

Alternatively, in the frictionless case ($\Gamma=0$) we make the same ansatz
(7.5) except with $\Delta\to 2c_{\Delta}/(1-\mu^{2})$, i.e.,

$\displaystyle
Y(\bar{\sigma})=\left(\frac{c_{\Delta}}{\mu}\right)\frac{\bar{\sigma}^{-1}-\bar{\sigma}}{\mu^{-1}-\mu}=Y(\bar{\sigma})^{\top}.$
(7.11)

Again, $Y(\bar{\sigma})$ is a Hamiltonian matrix and so preserves purity.
Furthermore,

$\displaystyle\bar{S}_{\mathrm{D}}(\sigma)\,$
$\displaystyle=2\Delta-2c_{\Delta}\left(\frac{\bar{\sigma}}{\mu}\right)\left(\frac{\bar{\sigma}^{-1}-\bar{\sigma}}{\mu^{-1}-\mu}\right)$
(7.12) $\displaystyle\geq 2(\Delta-c_{\Delta}\mathds{1}_{2d})$ (7.13)
$\displaystyle\geq 0,$ (7.14)

ensuring diffusion positivity for $\sigma\in\mathcal{S}_{\mathrm{NTS}}(\mu)$
because $\mu\mathds{1}\leq\bar{\sigma}\leq\mu^{-1}\mathds{1}$. Lastly, with
$v$ again an eigenvector of $\bar{\sigma}$ with eigenvalue
$\bar{\lambda}\leq\mu\leq 1$, we have

$\displaystyle v^{\top}\bar{S}_{0}(\sigma)v\,$
$\displaystyle=v^{\top}[F\bar{\sigma}+\bar{\sigma}F^{\top}+2Y(\bar{\sigma})\bar{\sigma}]v,$
(7.15)
$\displaystyle=2\bar{\lambda}(v^{\top}Fv)+2c_{\Delta}\left(\frac{\bar{\lambda}}{\mu}\right)\frac{\bar{\lambda}^{-1}-\bar{\lambda}}{\mu^{-1}-\mu}$
(7.16) $\displaystyle\geq
2\bar{\lambda}\left(\frac{c_{\Delta}}{\mu}-\|F\|\right)$ (7.17)

This is guaranteed to be positive, and thus NTS preserving, when
$c_{\Delta}>\mu\|F\|$. ∎

Intuitively, Lemma 7.1 has established that the dynamics (7.1) for the
covariance matrix of our Gaussian states can always be reinterpreted as
diffusion of the center of the Gaussian plus Hamiltonian (i.e., purity-
preserving) dynamics that confine the covariance matrix to
$\mathcal{S}_{\mathrm{NTS}}(\mu)$. We will now make this precise.

###### Proof of Lemma 6.1.

First, observe that this Lemma’s assumed form for $S(\alpha,\sigma)$ and the
constraint on $\mu$ (given by (6.15) and (6.16), respectively) ensures that,
for any fixed $\alpha$, $S(\alpha,\sigma)$ satisfies the form for $S(\alpha)$
and the constraint for $\mu$ in Lemma 7.1 (given by (7.1) and (7.2),
respectively) when taking $\Delta=\hbar^{-1}D$ as expected. In other words, we
can apply Lemma 7.1 freely at all points $\alpha$ in phase space.

We will show that for any probability distribution $p_{\alpha,\sigma}$ evolved
by $\mathchoice{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{5.00003pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\displaystyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{5.00003pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\textstyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{3.5pt}{2.72223pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptstyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{2.5pt}{1.94444pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptscriptstyle\mathcal{L}\hss$\crcr}}}\limits}^{(\alpha,\sigma)}$,
the total probability mass of $\mathrm{NTS}$ states,

$m_{\mathrm{NTS}}(t):=\iint\mathrm{d}\alpha\mathrm{d}\sigma
p_{\alpha,\sigma}(t)I_{\mathrm{NTS}}(\sigma),$ (7.18)

is non-decreasing. Here $I_{\mathrm{NTS}}(\sigma)$, is the indicator function
enforcing the minimum-eigenvalue condition525252Note that everywhere we work
with covariance matrices of pure states (i.e., $\frac{\sigma}{\hbar/2}$
symplectic and positive definite), so it’s not necessary to enforce those
conditions with the indicator function. for the NTS covariance matrices,

$I_{\mathrm{NTS}}(\sigma):=\begin{cases}1,&\lambda_{\mathrm{min}}[\sigma]\geq(\hbar/2)\mu\\\
0,&\lambda_{\mathrm{min}}[\sigma]<(\hbar/2)\mu,\end{cases}$ (7.19)

which we observe can also be written
$I_{\mathrm{NTS}}(\sigma)=\Theta(\lambda_{\mathrm{min}}[\sigma]-(\hbar/2)\mu)$,
where $\Theta$ is the Heaviside step function. Therefore

$\displaystyle\partial_{\sigma}I_{\mathrm{NTS}}(\sigma)=\delta(\lambda_{\mathrm{min}}[\sigma]-(\hbar/2)\mu)\partial_{\sigma}\lambda_{\mathrm{min}}[\sigma].$
(7.20)

We use this to compute the time derivative of $m_{\mathrm{NTS}}$ by using
(6.27), the definition of $\mathchoice{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{5.00003pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\displaystyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{5.00003pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\textstyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{3.5pt}{2.72223pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptstyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{2.5pt}{1.94444pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptscriptstyle\mathcal{L}\hss$\crcr}}}\limits}^{(\alpha,\sigma)}$,
and applying Lemma 7.1:

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}m_{\mathrm{NTS}}(t)\,$
$\displaystyle=\iint\\!\mathrm{d}\alpha\mathrm{d}\sigma\,I_{\mathrm{NTS}}(\sigma)\mathchoice{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{5.00003pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\displaystyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{5.00003pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\textstyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{3.5pt}{2.72223pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptstyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{2.5pt}{1.94444pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptscriptstyle\mathcal{L}\hss$\crcr}}}\limits}^{(\alpha,\sigma)}[p_{\alpha,\sigma}(t)]$
(7.29)
$\displaystyle=\iint\\!\mathrm{d}\alpha\mathrm{d}\sigma\,I_{\mathrm{NTS}}(\sigma)\left[-\partial_{\alpha}U(\alpha)-\partial_{\sigma}S_{0}(\alpha,\sigma)+\frac{1}{2}\partial_{\alpha}\partial_{\alpha}S_{\mathrm{D}}(\alpha,\sigma)\right]p_{\alpha,\sigma}(t)$
(7.30)
$\displaystyle=\iint\\!\mathrm{d}\alpha\mathrm{d}\sigma\,I_{\mathrm{NTS}}(\sigma)\left[-\partial_{\sigma}S_{0}(\alpha,\sigma)\right]p_{\alpha,\sigma}(t)$
(7.31)
$\displaystyle=\iint\\!\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(t)S_{0}(\alpha,\sigma)\partial_{\sigma}I_{\mathrm{NTS}}(\sigma)$
(7.32)
$\displaystyle=\iint\\!\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(t)S_{0}(\alpha,\sigma)\delta(\lambda_{\mathrm{min}}[\sigma]-(\hbar/2)\mu)\partial_{\sigma}\lambda_{\mathrm{min}}[\sigma].$
(7.33)

In Eqs. (7.29) and (7.30) we have simply applied the definitions of
$m_{\mathrm{NTS}}$ and $\mathchoice{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{5.00003pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\displaystyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{5.00003pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\textstyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{3.5pt}{2.72223pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptstyle\mathcal{L}\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern
0.80002pt$\hss\leavevmode\resizebox{2.5pt}{1.94444pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptscriptstyle\mathcal{L}\hss$\crcr}}}\limits}^{(\alpha,\sigma)}$.
In Eqs. (7.31) and (7.32) we have integrated by parts in $\alpha$ (against the
constant function $1$) and in $\sigma$. Lastly in (7.33) we applied (7.20).

To conclude we need to show that
$S_{0}(\alpha,\sigma)\partial_{\sigma}\lambda_{\mathrm{min}}[\sigma]\geq 0$.
Note that

$S_{0}(\alpha,\sigma)\partial_{\sigma}\lambda_{\mathrm{min}}[\sigma]=\left.\frac{\mathrm{d}}{\mathrm{d}t}\lambda_{\mathrm{min}}[\sigma+tS_{0}(\alpha,\sigma)]\right|_{t=0}=v^{\top}S_{0}(\alpha,\sigma)v$
(7.34)

when $\sigma v=\lambda_{\mathrm{min}}[\sigma]v$ for unit eigenvector $v$.
Using the “NTS preservation condition” of Lemma 7.1, it follows that
$v^{\top}S_{0}(\alpha,\sigma)v\geq 0$ when
$\lambda_{\mathrm{min}}[\sigma]=(\hbar/2)\mu$. Therefore
$S_{0}(\alpha,\sigma)\partial_{\sigma}\lambda_{\mathrm{min}}[\sigma]\geq 0$,
so it follows that

$\frac{\mathrm{d}}{\mathrm{d}t}m_{\mathrm{NTS}}(t)\geq 0.$ (7.35)

We note that $p_{\alpha,\sigma}(t)\geq 0$ is guaranteed by the “diffusion
positivity” condition in Lemma 7.1 ($S_{\mathrm{D}}(\alpha,\sigma)\geq 0$) and
that $\int\\!\mathrm{d}\alpha\mathrm{d}\sigma\,p_{\alpha,\sigma}(t)=1$ is
conserved, so we have $m_{\mathrm{NTS}}(t)=1$ for $t\geq 0$. Combining this
with the “purity preservation condition” in Lemma 7.1, we conclude that
$p_{\alpha,\sigma}(t)\geq 0$ is supported on the set
$\mathbb{R}^{2d}\times\mathcal{S}_{\mathrm{NTS}}(\mu)$ for all times $t\geq
0$.

∎

##  8 Harmonic approximation error

In this section we prove Lemma 6.3 about the error in the harmonic
approximation to the classical dynamics and Lemma 6.2 about the error in the
harmonic approximation to the quantum dynamics. In both cases we find that the
instantaneous error scales as
$\frac{1}{\hbar}\left\lVert\sigma\right\rVert^{\frac{3}{2}}$, where $\sigma$
is the covariance matrix of the pure state on which the dynamics act.

###  8.1 Classical case

Consider the classical state
$\tau_{\alpha,\sigma}(\alpha+\beta)=\exp(-\beta^{a}\sigma^{-1}_{ab}\beta^{b}/2)/((2\pi)^{d}\sqrt{\det\sigma})$,
a Gaussian probability distribution over phase space centered on $\alpha$
that’s equal to the Wigner function of the quantum state
$\hat{\tau}_{\alpha,\sigma}=|\alpha,\sigma\rangle\langle\alpha,\sigma|$. We
want to bound the error

$\displaystyle\left\lVert(\mathcal{L}-\mathcal{L}^{(\alpha)})[\tau_{\alpha,\sigma}]\right\rVert_{L^{1}}=\left\lVert\delta\mathcal{L}^{(\alpha)}[\tau_{\alpha,\sigma}]\right\rVert_{L^{1}}$
(8.1)

due to approximating the true classical dynamics, generated by

$\displaystyle\mathcal{L}[f]=$
$\displaystyle-\left(\partial^{a}H\right)(\partial_{a}f)-\partial_{a}\left[f\operatorname{Im}\sum_{k}L_{k}\partial^{a}L_{k}^{*}\right]+\frac{\hbar}{2}\partial_{a}\left[(\partial_{b}f)\operatorname{Re}\sum_{k}(\partial^{a}L_{k})(\partial^{b}L_{k}^{*})\right]$
(8.2)
$\displaystyle-\left(\partial^{a}H\right)(\partial_{a}f)-\partial_{a}\left[fG{}^{a}\right]+\frac{\hbar}{2}\partial_{a}\left[(\partial_{b}f)\Delta^{ab}\right]$
(8.3)

acting on the state $\tau_{\alpha,\sigma}$, with the linearization
$\mathcal{L}^{(\alpha)}$ given by (see Section 5.4)

$\displaystyle\mathcal{L}^{(\alpha)}[f]=$
$\displaystyle-\partial_{a}\left[f\left(\partial^{a}H^{[\alpha,2]}+G^{[\alpha,1]}{}^{a}\right)\right]+\frac{\hbar}{2}\Delta^{[\alpha,0]ab}\partial_{a}\partial_{b}f$
(8.4) $\displaystyle=$
$\displaystyle-\partial_{a}\left[f\left(\partial^{a}H^{[\alpha,2]}+\operatorname{Im}\sum_{k}\left(L_{k}(\alpha)\partial^{a}M^{[\alpha,2]}_{k}{}^{*}+M^{[\alpha,1]}_{k}\partial^{a}M^{[\alpha,1]}_{k}{}^{*}\right)\right)\right]$
(8.5)
$\displaystyle\qquad+\frac{\hbar}{2}\left[\operatorname{Re}\sum_{k}(\partial^{a}M^{[\alpha,1]}_{k})(\partial^{b}M^{[\alpha,1]}_{k}{}^{*})\right](\partial_{a}\partial_{b}f)$
(8.6)

where we recall $M_{k}=L_{k}-L_{k}(\alpha)$,
$G_{a}=\operatorname{Im}\sum_{k}L_{k}\partial_{a}L_{k}^{*}$, and
$\Delta^{ab}=\operatorname{Re}\sum_{k}(\partial^{a}M_{k})^{*}(\partial^{b}M_{k})$.
(As described earlier, $E^{[\alpha,m]}$ denotes the $m$-th order Taylor
approximation to $E$ at $\alpha$.) The main result of this section will be to
prove Lemma 6.3, which we now re-state:

See 6.3

###### Proof.

In the proof below, because we are after an explicit constant in (6.51), we
compute the Gaussian integrals explicitly.

We start by observing the following identity for the derivative of the
Gaussian state $\tau_{\alpha,\sigma}$:

$\displaystyle\begin{split}(\partial_{c}\tau_{\alpha,\sigma})(\alpha+\beta)&=\frac{\partial}{\partial\beta^{c}}\frac{\exp\left[-\beta^{a}\sigma^{-1}_{ab}\beta^{b}/2\right]}{(2\pi)^{d}\sqrt{\det\sigma}}\\\
&=-\sigma^{-1}_{cd}\beta^{d}\frac{\exp\left[-\beta^{a}\sigma^{-1}_{ab}\beta^{b}/2\right]}{(2\pi)^{d}\sqrt{\det\sigma}}\\\
&=-m_{c}\tau_{\alpha,\sigma}(\alpha+\beta),\\\ \end{split}$ (8.7)

where
$m^{a}:=(\sigma^{-1})^{a}_{\phantom{a}b}\beta^{b}=\omega^{ac}\sigma^{-1}_{cb}\beta^{b}$.
Then expanding $\mathcal{L}[f]-\mathcal{L}^{(\alpha)}[f]$ using the
expressions in (8.2) and (8.4) with $f=\tau_{\alpha,\sigma}$, and using the
triangle inequality,

$\displaystyle\begin{split}&\left\lVert\delta\mathcal{L}^{(\alpha)}[\tau_{\alpha,\sigma}]\right\rVert_{L^{1}}\leq\left\lVert\tau_{\alpha,\sigma}m^{a}(\partial_{a}\delta
H^{[\alpha,2]})\right\rVert_{L^{1}}+\left\lVert\tau_{\alpha,\sigma}m^{a}\delta
G^{[\alpha,1]}_{a}\right\rVert_{L^{1}}+\left\lVert\tau_{\alpha,\sigma}\partial^{a}\delta
G^{[\alpha,1]}_{a}\right\rVert_{L^{1}}\\\
&\qquad\qquad\qquad\qquad\qquad+\frac{\hbar}{2}\left\lVert\tau_{\alpha,\sigma}m_{a}m_{b}\delta\Delta^{[\alpha,0]ab}\right\rVert_{L^{1}}\end{split}$
(8.8)

where $\delta E^{[\alpha,m]}=E-E^{[\alpha,m]}$ denotes error on the $m$-th
order Taylor approximation to $E$.

Starting with the first term on the right-hand side, we use Taylor’s theorem
on the derivative of $\delta H^{(\alpha)}$:

$\displaystyle\begin{split}(\partial_{d}\delta
H^{[\alpha,2]})(\alpha+\beta)&=(\partial_{d}H)(\alpha+\beta)-(\partial_{d}H^{[\alpha,2]})(\alpha+\beta)\\\
&=\frac{1}{2!}\beta^{a}\beta^{b}(\partial_{a}\partial_{b}\partial_{d}H)(\alpha+z\beta)\end{split}$
(8.9)

for some ($\beta$-dependent) choice of $z\in[0,1]$. We can bound this
derivative

$\displaystyle\begin{split}\left|m^{a}\partial_{a}\delta
H^{(\alpha)}(\alpha)\right|^{2}=\frac{1}{4}\left|m^{a}\beta^{b}\beta^{c}(\partial_{a}\partial_{b}\partial_{c}H)(\alpha)\right|^{2}\leq\frac{1}{4}|m|^{2}|\beta|^{4}\left|H\right|_{C^{3}}^{2}.\end{split}$
(8.10)

using the $C^{3}$ seminorm defined in (4.7),

$\displaystyle\begin{split}\left|H\right|_{C^{3}}:=\sup_{\alpha}\left\|\nabla^{3}H(\alpha)\right\|_{\mathrm{gop}}=\sup_{\alpha}\sup_{\|\beta_{i}\|=1}|\beta_{1}^{a}\beta_{2}^{b}\beta_{3}^{c}\partial_{a}\partial_{b}\partial_{c}H(\alpha)|\end{split}$
(8.11)

which gives a global upper bound on the third derivatives of the classical
Hamiltonian function. Likewise norms like

$\displaystyle|m|^{2}=m^{a}\mathds{1}_{ab}m^{b}=\beta^{a}(\sigma^{-1})_{a}^{\phantom{a}c}\mathds{1}_{cd}(\sigma^{-1})^{d}_{\phantom{a}b}\beta^{b}=\beta^{a}\sigma^{-2}_{ab}\beta^{b}=(\beta^{\top}\sigma^{-2}\beta)$
(8.12)

are computed not with the symplectic form but with the Euclidean inner
product.535353Note that, physically, this inner product depends on our choice
of units. See Appendix A.

We can then perform the Gaussian integral

$\displaystyle\left\lVert\tau_{\alpha,\sigma}|m||\beta|^{2}\right\rVert_{L^{1}}^{2}$
$\displaystyle\leq\left\lVert\tau_{\alpha,\sigma}^{1/2}\right\rVert_{L^{2}}^{2}\left\lVert\tau_{\alpha,\sigma}^{1/2}|m||\beta|^{2}\right\rVert_{L^{2}}^{2}$
(8.13)
$\displaystyle=\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)\left(\beta^{\top}\sigma^{-2}\beta\right)|\beta|^{4}$
(8.14)
$\displaystyle=\operatorname{Tr}[\sigma^{-1}](\operatorname{Tr}\sigma)^{2}+2\operatorname{Tr}[\sigma^{-1}]\operatorname{Tr}[\sigma^{2}]+4\operatorname{Tr}[\sigma]\operatorname{Tr}[\sigma^{0}]+8\operatorname{Tr}[\sigma]$
(8.15)
$\displaystyle\leq\hbar^{-2}[2^{5}(d+1)d^{2}]\|\sigma\|^{3}+[2^{4}(d+1)d]\|\sigma\|$
(8.16) $\displaystyle\leq\hbar^{-2}[2^{5}(d+2)(d+1)d]\|\sigma\|^{3}$ (8.17)
$\displaystyle\leq\hbar^{-2}[2^{6}3d^{3}]\|\sigma\|^{3}$ (8.18)

where in the first line we have used Cauchy-Schwartz inequality,545454More
specifically, the Cauchy-Schwartz is $|(v,w)|^{2}\leq(v,v)(w,w)$ and we choose
$v=\sqrt{\tau_{\alpha,\sigma}}$ and $w=\sqrt{\tau_{\alpha,\sigma}}|\beta|$ so
$\|vw\|^{2}_{L^{1}}=[\int\\!\mathrm{d}\beta\,|v(\beta)w(\beta)|]^{2}=|(v,w)|^{2}\leq(v,v)(w,w)=\int\\!\mathrm{d}\beta\,|w(\beta)|^{2}$
because $(v,v)=\int\\!\mathrm{d}\beta\,\tau_{\alpha,\sigma}(\beta)=1$. and in
the last two lines we have used $d\geq 1$ and the fact that, by the
uncertainty principle, $\|\sigma\|\geq\hbar/2$ because $\sigma$ is the
covariance matrix of a pure Gaussian state. (The Gaussian integrals we use in
this section are recalled in Appendix B.2.) Thus we can bound the first term
on the right-hand side of (LABEL:eq:classical-bound-decomp):

$\displaystyle\left\lVert\tau_{\alpha,\sigma}m^{a}(\partial_{a}\delta
H^{[\alpha,2]})\right\rVert_{L^{1}}$
$\displaystyle\leq\hbar^{-1}\left|H\right|_{C^{3}}[2^{4}3d^{3}]^{1/2}\|\sigma\|^{3/2}$
(8.19)

Now we bound the rest of the terms in (LABEL:eq:classical-bound-decomp) in
basically the same way. Using Taylor’s theorem again, we have

$\displaystyle\begin{split}\delta
G^{[\alpha,1]}_{a}(\alpha+\beta)&=\frac{1}{2!}\beta^{b}\beta^{c}\partial_{b}\partial_{c}G_{a}(\alpha+z\beta)\\\
\partial^{a}\delta
G^{[\alpha,1]}_{a}(\alpha+\beta)&=\delta(\partial^{a}G_{a})^{[\alpha,0]}(\alpha+\beta)=\beta^{c}\partial_{c}\partial^{a}G_{a}(\alpha+z\beta)\\\
\delta\Delta^{[\alpha,0]ab}(\alpha+\beta)&=\beta^{c}\partial_{c}\Delta^{ab}(\alpha+z\beta)\end{split}$
(8.20)

where on each line $z\in[0,1]$ depends separately on $\beta$ on that line.
Then applying the $C^{k}$ seminorms as before

$\displaystyle\begin{split}|m^{a}\delta
G^{[\alpha,1]}_{a}(\alpha+\beta)|&\leq|m||\beta|^{2}\left|G\right|_{C^{2}}/2!\\\
|\partial^{a}\delta
G^{[\alpha,1]}_{a}(\alpha+\beta)|&\leq|\beta|\left|G\right|_{C^{2}}\\\
|m_{a}m_{b}\delta\Delta^{[\alpha,0]ab}(\alpha+\beta)|&\leq|m|^{2}|\beta|\left|\Delta\right|_{C^{1}}\end{split}$
(8.21)

where analogously to $\left|H\right|_{C^{3}}$ in (8.11) we defined

$\displaystyle\left|G\right|_{C^{2}}:=$
$\displaystyle\sup_{\alpha}\left\|\nabla^{2}G(\alpha)\right\|_{\mathrm{gop}}=\sup_{\alpha}\sup_{\|\beta_{i}\|=1}|\beta_{1}^{a}\beta_{2}^{b}\beta_{3}^{c}\partial_{a}\partial_{b}G_{c}|$
(8.22)
$\displaystyle\begin{split}=&\sup_{\alpha}\sup_{\|\beta_{i}\|=1}\Bigg{|}\beta_{1}^{a}\beta_{2}^{b}\beta_{3}^{c}\operatorname{Im}\sum_{k}\Big{[}(\partial_{a}\partial_{b}L_{k})^{*}(\alpha)(\partial_{c}L_{k})(\alpha)+2(\partial_{a}L_{k})^{*}(\partial_{b}\partial_{c}L_{k})(\alpha)\\\
&\hskip
220.00034pt+(L_{k})^{*}(\alpha)(\partial_{a}\partial_{b}\partial_{c}L_{k})(\alpha)\Big{]}\Bigg{|},\end{split}$
(8.23) $\displaystyle\left|\Delta\right|_{C^{1}}:=$
$\displaystyle\sup_{\alpha}\left\|\nabla^{1}\Delta(\alpha)\right\|_{\mathrm{gop}}=\sup_{\alpha}\sup_{\|\beta_{i}\|=1}|\beta_{1}^{a}\beta_{2}^{b}\beta_{3}^{c}\partial_{a}\Delta_{bc}|$
(8.24) $\displaystyle=$
$\displaystyle\sup_{\alpha}\sup_{\|\beta_{i}\|=1}\left|\beta_{1}^{a}\beta_{2}^{b}\beta_{3}^{c}\operatorname{Re}\sum_{k}\Big{[}(\partial_{a}\partial_{b}L_{k})^{*}(\partial_{c}L_{k})+(\partial_{b}L_{k})^{*}(\alpha)(\partial_{a}\partial_{c}L_{k})(\alpha)\Big{]}\right|.$
(8.25)

As mentioned in Section 4.3, the $C^{k}$ norm of a rank-$r$ tensor $E$ is the
generalized operator norm of the rank-$(k+r)$ tensor $\nabla^{k}E$ of partial
derivatives:
$\left|E\right|_{C^{k}}=\sup_{\alpha}\sup_{\|\beta_{j}\|=1}\left|\beta_{1}^{a_{1}}\cdots\beta_{k+r}^{a_{k+r}}\partial_{a_{1}}\cdots\partial_{a_{k}}E_{a_{k+1}\cdots
a_{k+r}}(\alpha)\right|$.

We again use Cauchy-Schwartz (see footnote 54) to get

$\displaystyle\begin{split}\left\lVert\tau_{\alpha,\sigma}|\beta|\right\rVert_{L^{1}}^{2}&\leq\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)|\beta|^{2}=\operatorname{Tr}[\sigma]\leq(2d)\|\sigma\|\leq\hbar^{-2}(8d)\|\sigma\|^{3}\\\
\left\lVert\tau_{\alpha,\sigma}|m|^{2}|\beta|\right\rVert_{L^{1}}^{2}&\leq\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)|m|^{4}|\beta|^{2}\\\
&=(\operatorname{Tr}[\sigma^{-1}])^{2}\operatorname{Tr}[\sigma]+4\operatorname{Tr}[\sigma^{-1}]\operatorname{Tr}[\sigma^{0}]+\operatorname{Tr}[\sigma]\operatorname{Tr}[\sigma^{-2}]+8\operatorname{Tr}[\sigma^{-1}]\\\
&\leq\hbar^{-2}[2^{4}(2d^{2}+5d+4)d]\|\sigma\|^{3}\\\
&\leq\hbar^{-2}[2^{4}11d^{3}]\|\sigma\|^{3}\end{split}$ (8.26)

where the relevant Gaussian integrals are recalled in Appendix B.2.

Pulling this all together we can now bound the error (LABEL:eq:classical-
bound-decomp) on the classical harmonic approximation for a Gaussian state:

$\displaystyle\begin{split}\left\lVert\delta\mathcal{L}^{(\alpha)}[\tau_{\alpha,\sigma}]\right\rVert_{L^{1}}&\leq\frac{\|\sigma\|^{3/2}}{\hbar}\Big{[}(\left|H\right|_{C^{3}}+\left|G\right|_{C^{2}})\left[2^{4}3d^{3}\right]^{1/2}+\left|G\right|_{C^{2}}\left[8d\right]^{1/2}+(\hbar/2)\left|\Delta\right|_{C^{1}}\left[2^{4}11d^{3}\right]^{1/2}\Big{]}\\\
&\leq\frac{\|\sigma\|^{3/2}}{\hbar}B^{\mathrm{anh}}_{\mathrm{c}}[H,L_{k}]\end{split}$
(8.27)

where

$\displaystyle
B^{\mathrm{anh}}_{\mathrm{c}}[H,L_{k}]=14d^{3/2}\left(\left|H\right|_{C^{3}}+\left|G\right|_{C^{2}}+\left|\Delta\right|_{C^{1}}\right).$
(8.28)

is a measure of the anharmonicity of the classical Hamiltonian and Lindblad
functions $H$ and $L_{k}$ (and in particular does not depent on $\hbar$ or
$\sigma$). It may seem unusual that $B^{\mathrm{anh}}_{\mathrm{c}}$ depends on
$\left|G\right|_{C^{2}}$, which in turn can diverge if the Lindblad functions
$L_{k}$ becomes arbitrarily large without its third derivative vanishing (see
the last term in (8.23)). However, this may be expected due to the fact that
the overall dynamics are _not_ invariant under $L_{k}(\alpha)\to
L_{k}(\alpha)+L_{0}$ (in contrast to the case of the Hamiltonian shift
$H(\alpha)\to H(\alpha)+H_{0}$, which does preserve the dynamics).

That concludes the proof of Lemma 6.3. ∎

###  8.2 Quantum case

We recall from (5.42) that, for any $\alpha$, we can express the exact quantum
dynamics with $\hat{M}_{k}:=\hat{L}_{k}-L_{k}(\alpha)$ as

$\displaystyle\hat{\mathcal{L}}[\rho]$
$\displaystyle=-\frac{i}{\hbar}\left[\hat{H}+\operatorname{Im}\sum_{k}L_{k}(\alpha)\hat{M}_{k}^{\dagger},\rho\right]+\frac{1}{\hbar}\sum_{k}\left(\hat{M}_{k}\rho\hat{M}_{k}^{\dagger}-\frac{1}{2}\left\\{\hat{M}_{k}^{\dagger}\hat{M}_{k},\rho\right\\}\right)$
(8.29)

We emphasize that although we express $\hat{\mathcal{L}}$ above in terms of
$\alpha$, the object is independent of $\alpha$. As discussed in Section 5.4,
the linearized dynamics
$\hat{\mathcal{L}}^{(\alpha)}=\hat{\mathcal{L}}^{[\alpha,2]}$ at $\alpha$,
which of course _do_ depend on $\alpha$, are

$\displaystyle\hat{\mathcal{L}}^{(\alpha)}[\rho]$
$\displaystyle=-\frac{i}{\hbar}\left[\hat{H}^{[\alpha,2]}+\operatorname{Im}\sum_{k}L_{k}(\alpha)\hat{M}^{[\alpha,2]}_{k}{}^{\dagger},\rho\right]+\frac{1}{\hbar}\sum_{k}\left(\hat{M}^{[\alpha,1]}_{k}\rho\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}-\frac{1}{2}\left\\{\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\hat{M}^{[\alpha,1]}_{k},\rho\right\\}\right)$
(8.30)

We want a global (independent of $\alpha$) bound on the error

$\displaystyle\left\lVert(\hat{\mathcal{L}}-\hat{\mathcal{L}}^{(\alpha)})[\hat{\tau}_{\alpha,\sigma}]\right\rVert_{\mathrm{Tr}}=\left\lVert\delta\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]\right\rVert_{\mathrm{Tr}}.$
(8.31)

for the Gaussian state
$\hat{\tau}_{\alpha,\sigma}=|\alpha,\sigma\rangle\langle\alpha,\sigma|$ with
covariance matrix $\sigma$ located at $\alpha$. We can now re-state the lemma
concerning the harmonic approximation which we prove in this subsection:

See 6.2

Our proof of Lemma 6.2 is more more involved than the classical case (Lemma
6.3) in the previous subsection. We use many variations of the same basic
trick, and we expect (for reasons related to the above discussion) that a more
abstract understanding of how Lindbladians and Liouvillians are Taylor
approximated would make tighter, simpler bounds possible.

In the proof of Lemma 6.2 we will not keep track of the constant $C_{d}$,
choosing instead to use the notation $A{\,\lesssim\,}B$ to mean that $A\leq
CB$ for some constant $C$ depending only on dimension. The implicit constant
can change from line to line.

###### Proof of Lemma 6.2.

We have

$\displaystyle\begin{split}\delta\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]&=\hat{\mathcal{L}}[\hat{\tau}_{\alpha,\sigma}]-\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]\\\
&=\frac{-i}{\hbar}\left[\delta\hat{H}^{(\alpha)}+\operatorname{Im}\sum_{k}L_{k}(\alpha)\delta\hat{M}^{[\alpha,2]}_{k},|\alpha,\sigma\rangle\langle\alpha,\sigma|\right]\\\
&\qquad+\frac{1}{\hbar}\sum_{k}\Bigg{[}\delta\hat{M}^{[\alpha,1]}_{k}\mathinner{|{\alpha,\sigma}\rangle}\mathinner{\langle{\alpha,\sigma}|}\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}+\hat{M}^{[\alpha,1]}_{k}\mathinner{|{\alpha,\sigma}\rangle}\mathinner{\langle{\alpha,\sigma}|}\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\\\
&\qquad\qquad\qquad\qquad+\delta\hat{M}^{[\alpha,1]}_{k}\mathinner{|{\alpha,\sigma}\rangle}\mathinner{\langle{\alpha,\sigma}|}\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}-\frac{1}{2}\left\\{\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\hat{M}^{[\alpha,1]}_{k},\mathinner{|{\alpha,\sigma}\rangle}\mathinner{\langle{\alpha,\sigma}|}\right\\}\\\
&\qquad\qquad\qquad\qquad-\frac{1}{2}\left\\{\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\delta\hat{M}^{[\alpha,1]}_{k},\mathinner{|{\alpha,\sigma}\rangle}\mathinner{\langle{\alpha,\sigma}|}\right\\}-\frac{1}{2}\left\\{\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\delta\hat{M}^{[\alpha,1]}_{k},\mathinner{|{\alpha,\sigma}\rangle}\mathinner{\langle{\alpha,\sigma}|}\right\\}\Bigg{]}\end{split}$
(8.32)

For any state vectors $\mathinner{|{\psi}\rangle}$ and
$\mathinner{|{\phi}\rangle}$, we have
$\left\lVert\mathinner{|{\psi}\rangle}\mathinner{\langle{\phi}|}\right\rVert_{\mathrm{Tr}}^{2}=\operatorname{Tr}[(\mathinner{|{\psi}\rangle}\\!\langle\phi|\phi\rangle\\!\mathinner{\langle{\psi}|})^{1/2}]^{2}=\langle\psi|\psi\rangle\langle\phi|\phi\rangle=\|\psi\|^{2}\|\phi\|^{2}$,
where the unlabeled norms denote Hilbert space norm. Therefore,555555For a
tighter bound, but only by a constant, we can compute more exactly. Note
$\left\lVert|v\rangle\langle w|-|w\rangle\langle v|\right\rVert=\left\lVert
v\right\rVert\left\lVert w\right\rVert\sqrt{1-|\langle
v|w\rangle|^{2}/(\left\lVert v\right\rVert\left\lVert w\right\rVert)}$, so
$\left\lVert[\hat{H},\hat{\tau}_{\alpha,\sigma}]\right\rVert=\left\lVert\hat{H}|\alpha\rangle\langle\alpha|-|\alpha\rangle\langle\alpha|\hat{H}\right\rVert=\left\lVert\hat{H}|\alpha\rangle\right\rVert\sqrt{1-\langle\alpha|\hat{H}|\alpha\rangle^{2}/\langle\alpha|\hat{H}^{2}|\alpha\rangle}=\sqrt{\langle\alpha|\hat{H}^{2}|\alpha\rangle-\langle\alpha|\hat{H}|\alpha\rangle^{2}}=\sqrt{\textrm{Var}(\hat{H})_{|\alpha\rangle}}$.
And also $\left\lVert|v\rangle\langle w|-|w\rangle\langle
v|\right\rVert_{1}=2\left\lVert|v\rangle\langle w|-|w\rangle\langle
v|\right\rVert$.

$\displaystyle\begin{split}\left\lVert\delta\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]\right\rVert_{\mathrm{Tr}}\leq&\frac{2}{\hbar}\left\lVert\delta\hat{H}^{[\alpha,2]}|\alpha,\sigma\rangle\right\rVert+\frac{2}{\hbar}\sum_{k}L_{k}(\alpha)\left\lVert\delta\hat{M}^{[\alpha,2]}_{k}|\alpha,\sigma\rangle\right\rVert\\\
&\quad+\frac{1}{\hbar}\sum_{k}\Bigg{[}2\left\lVert\delta\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert\left\lVert\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert+\left\lVert\delta\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert^{2}+\left\lVert\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert\\\
&\qquad\qquad\qquad+\left\lVert\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\delta\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert+\left\lVert\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\delta\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert\Bigg{]}\end{split}$
(8.33)

These terms all become small in the classical limit for essentially the same
reason: a Gaussian state of scale $\hbar$ cannot easily “see” the third order
corrections to the harmonic approximation. However, since our current
techniques are not strong enough show this rigorously in one fell swoop, we
will bound each of these terms individually. To simplify the bounds we will
use the quantities $\mathcal{Q}^{q,r}_{\hbar}[E]$ and
$\mathcal{N}^{q,r}_{\hbar;s,\nu}[E]$ defined in (3.16) and (3.17) and recalled
below for convenience:

$\displaystyle\mathcal{Q}^{q,r}_{\hbar}[E]$
$\displaystyle:=\sum_{j=q}^{r}\hbar^{(j-q)/2}\left|E\right|_{C^{j}}$
$\displaystyle\mathcal{N}^{q,r}_{\hbar;s,\nu}[E](\alpha)$
$\displaystyle:=\sum_{j=q}^{r}\hbar^{(j-q)/2}\sup_{\beta}\frac{\left\|\nabla^{j}E(\alpha+\beta)\right\|_{\mathrm{gop}}}{(1+\nu^{-1}|\beta|)^{s}}.$

With this notation we will show that we can bound the terms of (8.33) by

$\displaystyle\|\delta\hat{H}^{[\alpha,2]}|\alpha,\sigma\rangle\|$
$\displaystyle{\,\lesssim\,}\|\sigma\|^{3/2}\mathcal{Q}^{3,2d+4}_{\hbar}[H]$
(8.34)
$\displaystyle|L_{k}(\alpha)|\left\lVert\delta\hat{M}^{[\alpha,2]}_{k}|\alpha,\sigma\rangle\right\rVert$
$\displaystyle{\,\lesssim\,}\|\sigma\|^{3/2}\sup_{\beta}|L_{k}(\beta)|\mathcal{N}^{3,2d+6}_{\hbar;1,\|\sigma\|^{1/2}}[L_{k}](\beta)$
(8.35)
$\displaystyle\left\lVert\delta\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert$
$\displaystyle{\,\lesssim\,}\|\sigma\|\mathcal{Q}^{2,2d+3}_{\hbar}[L_{k}]$
(8.36)
$\displaystyle\left\lVert\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert$
$\displaystyle{\,\lesssim\,}\|\sigma\|^{1/2}\mathcal{Q}^{1,1}[L_{k}]$ (8.37)
$\displaystyle\begin{split}\left\lVert\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert&{\,\lesssim\,}\|\sigma\|^{3/2}\mathcal{Q}^{1,1}[L_{k}]\mathcal{Q}^{2,2d+3}_{\hbar}[L_{k}]\end{split}$
(8.38)
$\displaystyle\begin{split}\left\lVert\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\delta\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert&{\,\lesssim\,}\|\sigma\|^{3/2}\mathcal{Q}^{1,1}[L_{k}]\mathcal{Q}^{2d+3}_{\hbar}[L_{k}]\end{split}$
(8.39)
$\displaystyle\begin{split}\left\lVert\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\delta\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert&{\,\lesssim\,}\|\sigma\|^{2}\big{(}\mathcal{Q}^{2,4d+6}_{\hbar}[L_{k}]\big{)}^{2}\end{split}$
(8.40)

Assuming these bounds, the error (8.33) on the quantum harmonic approximation
for a Gaussian state becomes

$\displaystyle\begin{split}\left\lVert\delta\hat{\mathcal{L}}^{(\alpha)}[\hat{\tau}_{\alpha,\sigma}]\right\rVert_{\mathrm{Tr}}&{\,\lesssim\,}\frac{\|\sigma\|^{3/2}}{\hbar}\Bigg{[}\mathcal{Q}^{3,2d+4}_{0;\hbar}[H]+\sum_{k}\bigg{(}\mathcal{Q}^{1,1}[L_{k}]\mathcal{Q}^{2,2d+3}_{\hbar}[L_{k}]+\sup_{\beta}|L_{k}(\beta)|\mathcal{N}^{3,2d+6}_{\hbar;1,\|\sigma\|^{1/2}}[L_{k}](\beta)\\\
&\hskip
260.0004pt+(\mathcal{Q}^{2,4d+6}_{\hbar}[L_{k}])^{2}\|\sigma\|^{1/2}\bigg{)}\Bigg{]}\\\
&{\,\lesssim\,}\frac{\|\sigma\|^{3/2}}{\hbar}\left(B^{\mathrm{anh}}_{\mathrm{q}}[H,L_{k},\hbar]+B^{\mathrm{anh}}_{\mathrm{q}^{\prime}}[L_{k},\hbar,\|\sigma\|^{1/2}]\right)\end{split}$
(8.41)

for the anharmonicity factors $B^{\mathrm{anh}}_{\mathrm{q}}[H,L_{k},\hbar]$
and $B^{\mathrm{anh}}_{\mathrm{q}^{\prime}}[L_{k},\hbar,\nu]$ defined by
(3.18) and (3.19).

Thus, to complete the proof of Lemma 6.2, all we have to do is demonstrate the
bounds (8.34–8.40). To do this we will use the identity 565656That the
expectation value of an operator can be computed by integrating its Weyl
symbol against the Wigner function is generally taken as a defining feature of
the Weyl transformation, and it can be confirmed by using the integral
definition of the Weyl transformation in footnote 1.

$\|\hat{E}\mathinner{|{\alpha,\sigma}\rangle}\|^{2}=\operatorname{Tr}[\hat{E}^{2}\hat{\tau}_{\alpha,\sigma}]=\int(E\star
E)(\beta)\tau_{\alpha,\sigma}(\beta)\mathrm{d}\beta,$ (8.42)

for any symbol $E$ with Weyl quantization
$\hat{E}=\operatorname{Op}_{\hbar}[E]$, where $\star$ is the Moyal product. To
bound this Moyal product in turn we use the following proposition, which is
technically involved and is proved separately in Section 9.

######  Proposition 8.1 (Moyal Product Bound).

Let $\delta E^{[\alpha,m]}=E-E^{[\alpha,m]}$ be the remainder to the $m$-th
Taylor approximation
$E^{[\alpha,m]}(\alpha+\beta)=\sum_{k=0}^{m}\beta^{a_{1}}\cdots\beta^{a_{k}}(\partial_{a_{1}}\cdots\partial_{a_{k}}E)(\alpha)/k!$
at $\alpha$ of some function $E$ over $2d$-dimensional phase space. Then, for
any non-negative integer $s$ and phase-space length scale $\nu>0$,

$\begin{split}|\delta E^{[\alpha,m]}{}^{*}\star\delta
E^{[\alpha,m]}(\alpha+\beta)|&\leq(C_{0}(d+m))^{2(2d+2+m)}(1+\nu^{-2s}|\beta|^{2s})\\\
&\qquad(|\beta|^{2m+2}+\hbar^{m+1})[\mathcal{N}_{\hbar;s,\nu}^{m+1,2d+2+m+s}[E](\alpha)]^{2}\end{split}$
(8.43)

where $\mathcal{N}_{\hbar;s,\nu}^{q,r}[E](\alpha)$, defined in (3.17), is an
upper bound on the $q$-th through $r$-th derivatives of $E$ near $\alpha$,
weighted by an $s$-th order polynomial decay in the distance $\beta$ from
$\alpha$. Likewise for non-negative integer $s$ we have

$\displaystyle\begin{split}|E^{[\alpha,m]}{}^{*}\star E^{[\alpha,m]}\star
E^{[\alpha,m]}{}^{*}\star
E^{[\alpha,m]}(\alpha+\beta)|&\leq(C_{0}(d+m))^{4(2d+2+m)}(1+\nu^{-4s}|\beta|^{4s})\\\
&\qquad(|\beta|^{4m+4}+\hbar^{2(m+1)})[\mathcal{N}_{\hbar;2s,\nu}^{m+1,2(2d+2+m+s)}[E](\alpha)]^{4}\end{split}$
(8.44)

for the thrice iterated Moyal product. Above $C_{0}$ is a large absolute
constant which we have not computed.

We will predominantly make use of Proposition 8.1 in the special case of
$s=0$, in which case the right hand side simplifies to

$\displaystyle|\delta E^{[\alpha,m]}{}^{*}\star\delta
E^{[\alpha,m]}(\alpha+\beta)|{\,\lesssim\,}(|\beta|^{2(m+1)}+\hbar^{m+1})(\mathcal{Q}^{m+1,2d+m+2}_{\hbar}[E])^{2}$
(8.45)

and

$\displaystyle\begin{split}&|E^{[\alpha,m]}{}^{*}\star E^{[\alpha,m]}\star
E^{[\alpha,m]}{}^{*}\star
E^{[\alpha,m]}(\alpha+\beta)|{\,\lesssim\,}(|\beta|^{4(m+1)}+\hbar^{2(m+1)})(\mathcal{Q}^{m+1,2(2d+m+2)}_{\hbar}[E])^{4}.\end{split}$
(8.46)

We break up the rest of the proof into parts corresponding to Eqs.
(8.34–8.40), and we apply Proposition 8.1 in all but one. For the purposes of
this proof, we introduce the shorthand $\ell_{a}=\partial_{a}L_{k}(\alpha)$
since we consider just one Lindblad operator at a time and $\alpha$ is just a
fixed point we are expanding around.

Also, because we are not carefully computing the constants involved as we did
in the classical case, we will mostly simply use the following bound for the
Gaussian integral:

$\int\mathop{}\\!\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)|\beta|^{2k}{\,\lesssim\,}\|\sigma\|^{k}.$
(8.47)

Proof of (8.34), (8.36), and (8.40):

Using (8.42),

$\displaystyle\begin{split}\left\lVert\delta\hat{H}^{[\alpha,2]}|\alpha,\sigma\rangle\right\rVert^{2}&=\langle\alpha,\sigma|(\delta\hat{H}^{(\alpha)})^{2}|\alpha,\sigma\rangle=\operatorname{Tr}\left[\hat{\tau}_{\alpha,\sigma}(\delta\hat{H}^{[\alpha,2]})^{2}\right]\\\
&=\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)\left(\delta
H^{[\alpha,2]}\star\delta H^{[\alpha,2]}\right)(\alpha+\beta)\end{split}$
(8.48)

where
$\tau_{\alpha,\sigma}(\alpha+\beta)=\exp(-\beta^{a}\sigma^{-1}_{ab}\beta^{b}/2)/(2\pi\sqrt{\det\sigma})$,
a positive-valued function on phase space, is the Wigner function of the pure
Gaussian state $|\alpha,\sigma\rangle$. (Note that $\delta\hat{H}^{(\alpha)}$
on the left-hand side is an operator while $\delta H^{(\alpha)}$ on the right-
hand side is just a classical scalar function of the phase-space location
$\beta$.) Applying Proposition 8.1 with $m=2$ gives

$\displaystyle\begin{split}\left\lVert\delta\hat{H}^{[\alpha,2]}|\alpha,\sigma\rangle\right\rVert^{2}&{\,\lesssim\,}(\mathcal{Q}^{3,2d+4}_{\hbar}[E])^{2}\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)(|\beta|^{6}+\hbar^{3})\\\
&{\,\lesssim\,}(\mathcal{Q}^{3,2d+4}_{\hbar}[H])^{2}\|\sigma\|^{3}.\end{split}$
(8.49)

The proof of (8.36) follows the same strategy, first applying (8.42), then
Proposition 8.1 with $m=1$, then performing the Gaussian integral, and finally
applying elementary inequalities:

$\displaystyle\left\lVert\delta\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert^{2}$
$\displaystyle=\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)\left(\delta
M^{[\alpha,1]}_{k}\star\delta M^{[\alpha,1]}_{k}\right)(\alpha+\beta)$ (8.50)
$\displaystyle{\,\lesssim\,}(\mathcal{Q}^{2,2d+3}_{\hbar}[L_{k}])^{2}\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)(|\beta|^{4}+\hbar^{2})$
(8.51)
$\displaystyle{\,\lesssim\,}(\mathcal{Q}^{2,2d+3}_{\hbar}[L_{k}])^{2}\|\sigma\|^{2}$
(8.52)

For (8.40), we deploy the triple-Moyal-product part of Proposition 8.1 with
$m=1$, giving

$\displaystyle\left\lVert\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\delta\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert^{2}$
$\displaystyle=\langle\alpha,\sigma|\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\delta\hat{M}^{[\alpha,1]}_{k}\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\delta\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle$
(8.53)
$\displaystyle=\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)\left(\delta
M^{[\alpha,1]}_{k}{}^{*}\star\delta M^{[\alpha,1]}_{k}\star\delta
M^{[\alpha,1]}_{k}{}^{*}\star\delta M^{[\alpha,1]}_{k}\right)(\alpha+\beta)$
(8.54)
$\displaystyle{\,\lesssim\,}(\mathcal{Q}^{2,4d+6}_{\hbar}[L_{k}])^{4}\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)\left(\hbar^{4}+|\beta|^{8}\right)$
(8.55)
$\displaystyle{\,\lesssim\,}(\mathcal{Q}^{2,4d+6}_{\hbar}[L_{k}])^{4}\|\sigma\|^{4}$
(8.56)

Proof of (8.35):

The presence of leading term $L_{k}(\alpha)$ on the left-hand side of (8.35)
introduces a complication for our goal of bounding that side with a constant
independent of $\alpha$. In particular, we want our bound to hold in the
special case of linear Lindblad operators,
$L_{k}(\alpha)=\ell_{k;a}\alpha^{a}$, so we cannot bound $|L_{k}(\alpha)|$ and
$\|\delta\hat{M}^{[\alpha,2]}_{k}|\alpha,\sigma\rangle\|$ separately. It is
for this term, and this term only, that we will use the form of Proposition
8.1 with $s=1$ rather than $s=0$,

$\displaystyle\left\lVert\delta\hat{M}^{[\alpha,2]}_{k}|\alpha,\sigma\rangle\right\rVert^{2}$
$\displaystyle=\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)\left(\delta
M^{[\alpha,2]}_{k}{}^{*}\star\delta M^{[\alpha,2]}_{k}\right)(\alpha+\beta)$
(8.57)
$\displaystyle{\,\lesssim\,}(\mathcal{N}^{3,2d+6}_{\hbar;1,\nu}[L_{k}](\alpha))^{2}\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)(|\beta|^{6}+\hbar^{3}+\nu^{-2}|\beta|^{8}+\nu^{-2}\hbar^{3}|\beta|^{2})$
(8.58)
$\displaystyle{\,\lesssim\,}(\mathcal{N}^{3,2d+6}_{\hbar;1,\nu}[L_{k}](\alpha))^{2}\left[\|\sigma\|^{3}+\nu^{-2}\|\sigma\|^{4}\right]$
(8.59)

Then

$\displaystyle|L_{k}(\alpha)|\left\lVert\delta\hat{M}^{[\alpha,2]}_{k}|\alpha,\sigma\rangle\right\rVert$
$\displaystyle{\,\lesssim\,}|L_{k}(\alpha)|\mathcal{N}^{3,2d+6}_{\hbar;1,\nu}[L_{k}](\alpha)\left[\|\sigma\|^{3}+\nu^{-2}\|\sigma\|^{4}\right]^{1/2}$
(8.60)
$\displaystyle{\,\lesssim\,}|L_{k}(\alpha)|\mathcal{N}^{3,2d+6}_{\hbar;1,\nu}[L_{k}](\alpha)(\alpha)\left[\|\sigma\|^{3/2}+\nu^{-1}\|\sigma\|^{2}\right]$
(8.61)

Choosing $\nu=\|\sigma\|^{1/2}$ we obtain (8.35).

Proof of (8.37):

For this we do not actually need Proposition 8.1 because
$M^{[\alpha,1]}_{k}(\beta)=\ell_{a}(\beta^{a}-\alpha^{a})$ is just linear so,
by the explicit Moyal product (4.17),

$\displaystyle M^{[\alpha,1]}_{k}{}^{*}\star M^{[\alpha,1]}_{k}(\alpha+\beta)$
$\displaystyle=\left|M^{[\alpha,1]}_{k}(\alpha+\beta)\right|^{2}+\frac{i\hbar}{2}(\partial_{a}M^{[\alpha,1]}_{k}{}^{*})(\alpha+\beta)(\partial^{a}M^{[\alpha,1]}_{k})(\alpha+\beta)$
(8.62)
$\displaystyle=|\ell_{a}\beta^{a}|^{2}+\frac{i\hbar}{2}\ell_{a}^{*}\ell^{a}$
(8.63) $\displaystyle\leq|\ell|^{2}(|\beta|^{2}+\hbar/2)$ (8.64)
$\displaystyle\leq(\mathcal{Q}^{1,1}[L_{k}])^{2}(|\beta|^{2}+\hbar/2)$ (8.65)

where we have used the Cauchy-Schwartz inequality and $|\ell|=|\partial
L_{k}(\alpha)|\leq
C^{1,1}_{L_{k};0}=\sup_{\beta}[|\partial_{\mathrm{x}}L_{k}(\beta)|+|\partial_{\mathrm{p}}L_{k}(\beta)|]$.
Therefore

$\displaystyle\left\lVert\hat{M}^{[\alpha,1]}_{k}|\alpha,\sigma\rangle\right\rVert^{2}$
$\displaystyle=\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\beta)\left(M^{[\alpha,1]}_{k}{}^{*}\star
M^{[\alpha,1]}_{k}\right)(\alpha+\beta)$ (8.66)
$\displaystyle\leq(\mathcal{Q}^{1,1}[L_{k}])^{2}\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)(|\beta|^{2}+\hbar/2)$
(8.67)
$\displaystyle=(\mathcal{Q}^{1,1}[L_{k}])^{2}(\operatorname{Tr}[\sigma]+\hbar/2)$
(8.68) $\displaystyle{\,\lesssim\,}(\mathcal{Q}^{1,1}[L_{k}])^{2}\|\sigma\|$
(8.69)

as desired.

Proof of (8.38) and (8.39):

We note that the left hand sides of (8.39) and (8.38) are related by

$\displaystyle\|\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\delta\hat{M}^{[\alpha,1]}_{k}\mathinner{|{\alpha,\sigma}\rangle}\\!\|\leq\|\delta\hat{M}^{[\alpha,1]}_{k}\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\mathinner{|{\alpha,\sigma}\rangle}\\!\|+\|\big{[}\hat{M}^{[\alpha,1]}_{k}{}^{\dagger},\delta\hat{M}^{[\alpha,1]}_{k}\big{]}\mathinner{|{\alpha,\sigma}\rangle}\\!\|.$
(8.70)

Lets start by expanding the first part of the right-hand side:

$\|\delta\hat{M}^{[\alpha,1]}_{k}\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\mathinner{|{\alpha,\sigma}\rangle}\\!\|^{2}=\int\left(\delta
M^{[\alpha,1]}_{k}{}^{*}\star\delta
M^{[\alpha,1]}_{k}\right)(\alpha+\beta)\left(M^{[\alpha,1]}_{k}{}^{*}\star\tau_{\alpha,\sigma}\star
M^{[\alpha,1]}_{k}\right)(\alpha+\beta)\mathop{}\\!\mathrm{d}\beta,$ (8.71)

The first term inside the integrand of (8.71) can be bounded with Proposition
8.1.

$\left(\delta M^{[\alpha,1]}_{k}{}^{*}\star\delta
M^{[\alpha,1]}_{k}\right)(\alpha+\beta){\,\lesssim\,}(\mathcal{Q}^{2,2d+3}_{\hbar}[L_{k}])^{2}(|\beta|^{4}+\hbar^{2}).$
(8.72)

For the second term in the integrand of (8.71), we note that since
$M^{[\alpha,1]}_{k}$ is linear,
$M^{[\alpha,1]}_{k}{}^{*}\star\tau_{\alpha,\sigma}$ can be computed explicitly
with the Moyal product (4.17) as

$M^{[\alpha,1]}_{k}{}^{*}\star\tau_{\alpha,\sigma}(\alpha+\beta)=\ell_{a}^{*}\beta^{a}\tau_{\alpha,\sigma}(\alpha+\beta)+\frac{i\hbar}{2}\ell_{a}^{*}\partial^{a}\tau_{\alpha,\sigma}(\alpha+\beta).$
(8.73)

Now we recall (8.7),

$\displaystyle(\partial_{c}\tau_{\alpha,\sigma})(\alpha+\beta)$
$\displaystyle=-m_{c}\tau_{\alpha,\sigma}(\alpha+\beta)$ (8.74)

where
$m^{a}:=(\sigma^{-1})^{a}_{\phantom{a}b}\beta^{b}=\omega^{ac}\sigma^{-1}_{cb}\beta^{b}$,
so

$\displaystyle
M^{[\alpha,1]}_{k}{}^{*}\star\tau_{\alpha,\sigma}(\alpha+\beta)$
$\displaystyle=\ell_{a}^{*}\left[\beta^{a}-\frac{i\hbar}{2}m^{a}\right]\tau_{\alpha,\sigma}(\alpha+\beta)$
(8.75)
$\displaystyle=\ell_{a}^{*}\left[\delta^{a}_{\phantom{a}b}-\frac{i\hbar}{2}(\sigma^{-1})^{a}_{\phantom{a}b}\right]\beta^{b}\tau_{\alpha,\sigma}(\alpha+\beta).$
(8.76)

Then we can apply the Moyal product with $M^{[\alpha,1]}_{k}$ on the right to
get

$\displaystyle\begin{split}&M^{[\alpha,1]}_{k}{}^{*}\star\tau_{\alpha,\sigma}\star
M^{[\alpha,1]}_{k}(\alpha+\beta)\\\
&\qquad=\tau_{\alpha,\sigma}(\alpha+\beta)\left[|\ell_{a}\beta^{a}|^{2}-\frac{i\hbar}{2}(\ell_{a}^{*}(\sigma^{-1})^{a}_{\phantom{a}b}\beta^{b})(\ell_{c}\beta^{c})\right]-\frac{i\hbar}{2}\ell_{c}\left[\partial^{c}(M^{[\alpha,1]}_{k}{}^{*}\star\tau_{\alpha,\sigma})\right](\alpha+\beta).\end{split}$
(8.77)

We can compute

$\displaystyle\left[\partial_{c}(M^{[\alpha,1]}_{k}{}^{*}\star\tau_{\alpha,\sigma})\right](\alpha+\beta)=\left[\ell_{c}^{*}-\frac{i\hbar}{2}\ell_{a}^{*}(\sigma^{-1})^{a}_{\phantom{a}c}-\left(\ell_{a}^{*}\beta^{a}-\frac{i\hbar}{2}\ell_{a}^{*}(\sigma^{-1})^{a}_{\phantom{a}b}\beta^{b}\right)\sigma^{-1}_{cd}\beta^{d}\right]\tau_{\alpha,\sigma}(\alpha+\beta)$
(8.78)

so now inserting this into (8.77) we have

$\begin{split}M^{[\alpha,1]}_{k}{}^{*}\star\tau_{\alpha,\sigma}\star
M^{[\alpha,1]}_{k}(\alpha+\beta)&=\tau_{\alpha,\sigma}(\alpha+\beta)\Big{[}|\ell_{a}\beta^{a}|^{2}+i\frac{\hbar}{2}\big{(}\ell^{c}\ell^{*}_{c}-\ell^{*}_{a}\beta^{a}\ell^{c}\sigma^{-1}_{cb}\beta^{b}+\ell^{a}{}^{*}\sigma^{-1}_{ab}\beta^{b}\ell_{c}\beta^{c}\big{)}\\\
&\qquad\qquad+\frac{\hbar^{2}}{4}(\ell^{a}{}^{*}\sigma^{-1}_{ab}\beta^{b}\ell^{c}\sigma^{-1}_{cd}\beta^{d}-\ell^{c}{}^{*}(\sigma^{-1})_{ca}\ell^{a})\Big{]}\\\
&=\tau_{\alpha,\sigma}(\alpha+\beta)\Big{[}|\ell_{a}\beta^{a}|^{2}+\hbar\operatorname{Im}\left(\ell^{*}_{a}\beta^{a}\ell^{c}\sigma^{-1}_{cb}\beta^{b}-\ell^{c}\ell^{*}_{c}/2\right)\\\
&\qquad\qquad+\frac{\hbar^{2}}{4}\left(|\ell^{a}\sigma^{-1}_{ab}\beta^{b}|^{2}-\ell^{c}{}^{*}\sigma^{-1}_{ca}\ell^{a}\right)\Big{]}\\\
&=\tau_{\alpha,\sigma}(\alpha+\beta)\left[{\ell^{a}}\left(\frac{i\hbar}{2}\omega_{ab}+\frac{\hbar^{2}}{4}\sigma^{-1}_{ab}\right)\ell^{b}{}^{*}+\left|\ell^{a}\left(\omega_{ab}-\frac{i\hbar}{2}\sigma^{-1}_{ab}\right)\beta^{a}\right|^{2}\right]\\\
&=\tau_{\alpha,\sigma}(\alpha+\beta)\left[v_{0}+w_{a}^{*}w_{b}\beta^{a}\beta^{b}\right]\end{split}$
(8.79)

where

$\displaystyle w_{b}:=$
$\displaystyle\ell^{a}\left(\omega_{ab}-\frac{i\hbar}{2}\sigma^{-1}_{ab}\right)\in\mathbb{C}^{2d},$
(8.80) $\displaystyle v_{0}:=$
$\displaystyle\frac{i\hbar}{2}{\ell^{a}}\left(\omega_{ab}-\frac{i\hbar}{2}\sigma^{-1}_{ab}\right)\ell^{b}{}^{*}=\frac{i\hbar}{2}w_{b}\ell^{b}{}^{*}\in\mathbb{R}$
(8.81)

which obey

$\displaystyle|v_{0}|\leq$
$\displaystyle\,(\hbar/2)|\ell|^{2}\left(1+(\hbar/2)\|\sigma^{-1}\|\right)\leq(\mathcal{Q}^{1,1}[L_{k}])^{2}\left(\hbar/2+\|\sigma\|\right)\leq
2(\mathcal{Q}^{1,1}[L_{k}])^{2}\|\sigma\|$ (8.82)
$\displaystyle(w^{\dagger}\sigma^{n}w)=$ $\displaystyle
w_{a}^{*}(\sigma^{n})^{ab}w_{b}=(\ell^{\dagger}\sigma^{n}\ell)+\frac{\hbar^{2}}{4}(\ell^{\dagger}\sigma^{n-2}\ell)\leq(\mathcal{Q}^{1,1}[L_{k}])^{2}\left(\|\sigma^{n}\|+\hbar^{2}\|\sigma^{n-2}\|/4\right)\leq
2(\mathcal{Q}^{1,1}[L_{k}])^{2}\|\sigma^{n}\|$ (8.83)

If we insert (8.79) and (8.72) into (8.71) we get

$\displaystyle\begin{split}\|\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\hat{M}^{[\alpha,1]}_{k}\mathinner{|{\alpha,\sigma}\rangle}\|^{2}&\leq[C^{2,2d+3}_{L_{k};0;\hbar}]^{2}\int\tau_{\alpha,\sigma}(\alpha+\beta)\left[\hbar^{2}v_{0}+\hbar^{2}w_{a}^{*}w_{b}\beta^{a}\beta^{b}+v_{0}|\beta|^{4}+w_{a}^{*}w_{b}\beta^{a}\beta^{b}|\beta|^{4}\right]\mathop{}\\!\mathrm{d}\beta\end{split}$
(8.84)

and then perform the Gaussian integral

$\displaystyle\begin{split}\frac{\|\delta\hat{M}^{[\alpha,1]}_{k}{}^{\dagger}\hat{M}^{[\alpha,1]}_{k}\mathinner{|{\alpha,\sigma}\rangle}\|^{2}}{(\mathcal{Q}^{2,2d+3}_{\hbar}[L_{k}])^{2}}&\leq\int\tau_{\alpha,\sigma}(\alpha+\beta)\left[\hbar^{2}v_{0}+\hbar^{2}w_{a}^{*}w_{b}\beta^{a}\beta^{b}+v_{0}|\beta|^{4}+w_{a}^{*}w_{b}\beta^{a}\beta^{b}|\beta|^{4}\right]\mathop{}\\!\mathrm{d}\beta\\\
&=v_{0}\hbar^{2}+\hbar^{2}(w^{\dagger}\sigma
w)+v_{0}\left((\operatorname{Tr}\sigma)^{2}+2\operatorname{Tr}[\sigma^{2}]\right)\\\
&\qquad\qquad+\left((w^{\dagger}\sigma
w)(\operatorname{Tr}\sigma)^{2}+2(w^{\dagger}\sigma
w)\operatorname{Tr}[\sigma^{2}]+4\operatorname{Tr}[\sigma](w^{\dagger}\sigma^{2}w)+8(w^{\dagger}\sigma^{3}w)\right)\\\
&=(\mathcal{Q}^{1,1}[L_{k}])^{2}4\left[\hbar^{2}\|\sigma\|+8(d+1)^{2}\|\sigma\|^{3}\right]\\\
&{\,\lesssim\,}(\mathcal{Q}^{1,1}[L_{k}])^{2}\|\sigma\|^{3}.\end{split}$
(8.85)

which proves (8.38).

To prove (8.39), we need to handle the commutator in (8.70). Note that
$\hat{M}^{[\alpha,1]}_{k}=\ell_{a}(\hat{r}^{a}-\alpha^{a})$ is linear in the
phase space variables $\hat{r}=(\hat{x},\hat{p})$, so the Wigner transform of
(i.e., symbol for) the commutator
$[\hat{M}^{[\alpha,1]}_{k}{}^{\dagger},\delta\hat{M}^{[\alpha,1]}_{k}]$ can be
computed directly with the Moyal product (4.17) to be
$i\hbar\ell_{a}^{*}\partial^{a}\delta M^{[\alpha,1]}_{k}$. This is just the
remainder from the zeroth order Taylor approximation to the function
$i\hbar\ell_{a}^{*}\partial^{a}L_{k}=i\hbar\ell_{a}^{*}\partial^{a}M_{k}$:

$\displaystyle\ell_{a}^{*}\partial^{a}\delta
M^{[\alpha,1]}_{k}=\ell_{a}^{*}\partial^{a}\delta
L^{[\alpha,1]}_{k}=\delta(\ell_{a}^{*}\partial^{a}L_{k})^{[\alpha,0]},$ (8.86)

so that it satisfies the prerequisites of Proposition 8.1 with $m=0$, which we
can apply to get

$\displaystyle\left\lVert[\hat{M}^{[\alpha,1]}_{k}{}^{\dagger},\delta\hat{M}^{[\alpha,1]}_{k}]|\alpha,\sigma\rangle\right\rVert^{2}$
$\displaystyle=\hbar^{2}\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)\left(\delta(\ell_{a}^{*}\partial^{a}L_{k})^{[\alpha,0]}{}^{*}\star\delta(\ell_{a}^{*}\partial^{a}L_{k})^{[\alpha,0]}\right)(\alpha+\beta)$
(8.87)
$\displaystyle{\,\lesssim\,}\hbar^{2}\left(\mathcal{Q}^{1,2d+2}_{\hbar}[\ell_{a}^{*}\partial^{a}L_{k}]\right)^{2}\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)(|\beta|^{2}+\hbar^{1})$
(8.88)
$\displaystyle{\,\lesssim\,}\hbar^{2}(\mathcal{Q}^{1,1}[L_{k}]\mathcal{Q}^{2,2d+3}_{\hbar}[L_{k}])^{2}\left(\operatorname{Tr}[\sigma]+\hbar^{1}\right)$
(8.89)
$\displaystyle{\,\lesssim\,}\hbar^{2}(\mathcal{Q}^{1,1}[L_{k}]\mathcal{Q}^{2,2d+3}_{\hbar}[L_{k}])^{2}\|\sigma\|$
(8.90)
$\displaystyle{\,\lesssim\,}(\mathcal{Q}^{1,1}[L_{k}]\mathcal{Q}^{2,2d+3}_{\hbar}[L_{k}])^{2}\|\sigma\|^{3}$
(8.91)

where we have used

$\displaystyle\mathcal{Q}^{q,r}_{\hbar}[\ell_{a}^{*}\partial_{a}L_{k}]\leq|\ell|\mathcal{Q}^{q+1,r+1}_{\hbar}[L_{k}]\leq\mathcal{Q}^{1,1}[L_{k}]\mathcal{Q}^{q+1,r+1}_{\hbar}[L_{k}]$
(8.92)

Taking the square roots of (8.85) and (8.91) and inserting into (8.70) gives
(8.39).

Having now demonstrated all the bounds (8.34)–(8.39) with the help of
Proposition 8.1, Lemma 6.2 is proved. ∎

The only remaining task to complete the demonstration of our main result is to
justify Proposition 8.1, which is addressed in the next section.

##  9 Moyal product bound

In this section575757Note that, due to the regrettably finite size of
alphabets, in this section we have re-used variables previously defined for
other purposes elsewhere in the paper. This section should be considered self-
contained. we prove Proposition 8.1, whose statement we now recall:

See 8.1

Note that in Proposition 8.1 we do keep track of the dimensional dependence of
the constant, so we will no longer use the ${\,\lesssim\,}$ notation in this
section. We will however make use of the notation that $C$ is an unspecified
constant whose value may change from line to line.

We use the following integral formulation of the Moyal product:

$E\star G(\alpha)=(2\pi\hbar)^{-d}\int
e^{i\beta_{a}\gamma^{a}/(2\hbar)}E(\alpha+\beta/2)G(\alpha+\gamma/2)\mathop{}\\!\mathrm{d}\beta\mathop{}\\!\mathrm{d}\gamma.$
(9.1)

The proof of Proposition 8.1 splits into two main parts. First, in Section 9.1
we state Lemma 9.1 giving a bound for $F\star F(\alpha)$ in terms of a
convolution of derivatives of $F$. Then in Section 9.2 we show how Lemma 9.1
implies Proposition 8.1. The proof of Lemma 9.1 is deferred to Section 9.3.

###  9.1 Main lemma

In the following lemma $\rho_{K}$ is the convolution kernel

$\rho_{K}(\alpha):=\hbar^{-d}(\hbar^{-1/2}|\alpha|+1)^{-K}.$

Note that when $K>2d$

$\displaystyle\int\rho_{K}(\alpha)|\alpha|^{j}\mathop{}\\!\mathrm{d}\alpha$
$\displaystyle=\hbar^{-d}\int(\hbar^{-1/2}|\alpha|+1)^{-K}|\alpha|^{j}\mathop{}\\!\mathrm{d}\alpha$
$\displaystyle=\hbar^{j/2}\int(|\alpha|+1)^{-K}|\alpha|^{j}\mathop{}\\!\mathrm{d}\alpha$
$\displaystyle=\hbar^{j/2}C_{d,K,j}$

Thus when $K>2d$ the convolution $\rho_{K}\ast E$ is well-defined:

$\rho_{K}\ast
E(\alpha):=\int\rho_{K}(\beta)E(\alpha-\beta)\mathop{}\\!\mathrm{d}\beta.$
(9.2)

######  Lemma 9.1.

Let $F,G\in C^{\infty}(\mathbb{R}^{2d})$ be smooth functions, and let
$K_{F},K_{G}\geq 0$ be nonnegative integers. Then

$\begin{split}|F\star
G(\alpha)|&\leq(CK)^{2K}C^{d}\left(\sum_{k=0}^{K_{F}}\big{(}\rho_{K_{G}}\ast\left\|\hbar^{k/2}\nabla^{k}F\right\|_{\mathrm{gop}}\big{)}(\alpha)\right)\left(\sum_{k=0}^{K_{G}}\big{(}\rho_{K_{F}}\ast\left\|\hbar^{k/2}\nabla^{k}G\right\|_{\mathrm{gop}}\big{)}(\alpha)\right).\end{split}$
(9.3)

where $C$ is an absolute constant, and $K=\max\\{K_{F},K_{G}\\}$.

######  Corollary 9.1 (Iterated Moyal bound).

Let $F\in C^{\infty}(\mathbb{R}^{2d})$ be a smooth function and let $K$ be a
nonnegative integer. Then

$|F\star F\star F\star
F(\alpha)|\leq(CK)^{4K}C^{d}\left[\rho_{K}\ast\left(\sum_{m=0}^{2K}\rho_{K}\ast\left\|\hbar^{m/2}\nabla^{m}F\right\|_{\mathrm{gop}}\right)^{2}\right]^{2}(\alpha)$
(9.4)

###### Proof using Lemma 9.1.

Using Lemma 9.1 we have

$|F\star F\star F\star
F(\alpha)|\leq(CK)^{2K}C^{d}\left(\sum_{k=0}^{K}(\rho_{K}\ast\left\|\hbar^{k/2}\nabla^{k}(F\star
F)\right\|_{\mathrm{gop}})(\alpha)\right)^{2}.$

To bound $\nabla^{k}(F\star F)$ we use the following product rule for the
partial derivative $\partial^{\vec{n}}$, defined as
$\partial_{1}^{n_{1}}\partial_{2}^{n_{2}}\cdots\partial_{2d}^{n_{2d}}$

$\partial^{\vec{n}}(F\star
F)=\sum_{\vec{m}\leq\vec{n}}\prod_{j=1}^{2d}\binom{n_{j}}{m_{j}}\partial^{\vec{m}}F\star(\partial^{\vec{n}-\vec{m}}F).$

The combinatorial factor $\prod_{j=1}^{2d}\binom{n_{j}}{m_{j}}$ is bounded by
$C^{K}$, so applying Lemma 9.1 to the terms in the right hand side we obtain

$\begin{split}&\hbar^{k/2}\left\|(\nabla^{j}F)\star(\nabla^{k-j}F)(\alpha)\right\|_{\mathrm{gop}}\\\
&\qquad\qquad\leq(CK)^{2K}C^{d}\left(\sum_{m=0}^{K}\rho_{K}\ast\left\|\hbar^{(m+j)/2}\nabla^{m+j}F\right\|_{\mathrm{gop}}(\alpha)\right)\left(\sum_{m=0}^{K}\rho_{K}\ast\left\|\hbar^{(m+k-j)/2}\nabla^{m+k-j}F\right\|_{\mathrm{gop}}(\alpha)\right)\\\
&\qquad\qquad\leq(CK)^{2K}C^{d}\left(\sum_{m=0}^{2K}\rho_{K}\ast\left\|\hbar^{m/2}\nabla^{m/2}F\right\|_{\mathrm{gop}}(\alpha)\right)^{2}.\end{split}$
(9.5)

∎

###  9.2 Proof of Proposition 8.1 from Lemma 9.1

First we prove a simple estimate for convolutions of functions against the
kernel $\rho_{K}$ in terms of the following weighted supremum

$M_{\hbar}^{q}[F](\alpha):=\sup_{\beta}(1+\hbar^{-1/2}|\beta-\alpha|)^{-q}|F(\beta)|.$
(9.6)

######  Lemma 9.2.

If $K\geq 2d+1$ then

$(\rho_{K}\ast F)(\alpha)\leq CM_{\hbar}^{K-2d-1}[F](\alpha)$ (9.7)

for some absolute constant $C$.

###### Proof.

Set $q=K-2d-1$. Then

$\begin{split}|\rho_{K}\ast
F(\alpha)|&\leq\hbar^{-d}\int(1+\hbar^{-1/2}|\beta-\alpha|)^{-K}|F(\beta)|\mathop{}\\!\mathrm{d}\beta\\\
&\leq\hbar^{-d}M_{\hbar}^{q}[F](\alpha)\int(1+\hbar^{-1/2}|\beta-\alpha|)^{2d-1}\mathop{}\\!\mathrm{d}\beta\\\
&\leq CM_{\hbar}^{q}[F](\alpha).\end{split}$ (9.8)

∎

We also note two quick facts about the weighted supremum. The first is the
weighted supremum of a monomial $m_{k}(\alpha)=|\alpha|^{k}$, using that
$(a+b)^{k}\leq 2^{k}(a^{k}+b^{k})$. For $q\geq k$ we have

$\begin{split}M_{\hbar}^{q}[m_{k}](\alpha)&=\sup_{\beta}(1+\hbar^{-1/2}|\beta-\alpha|)^{-q}|\beta|^{k}\\\
&\leq
2^{k}(|\alpha|^{k}+\sup_{\beta}(1+\hbar^{-1/2}|\beta-\alpha|)^{-q}|\alpha-\beta|^{k})\\\
&\leq 4^{k}(|\alpha|^{k}+\hbar^{k/2}),\end{split}$ (9.9)

where in the last step we used that $\sup_{t}(1+\hbar^{-1/2}t)^{-q}t^{k}\leq
2^{k}$ when $q\geq k$.

The second quick fact that we need is the following product rule for the
weighted supremum:

$M_{\hbar}^{q_{1}+q_{2}}[FG]\leq M_{\hbar}^{q_{1}}[F]M_{\hbar}^{q_{2}}[G].$
(9.10)

We will also use $M_{\hbar}^{q}[A]$ to refer to
$M_{\hbar}^{q}[\left\|A\right\|_{\mathrm{gop}}]$ when $A$ is a tensor-valued
or vector-valued quantity.

Finally, we introduce the weighted supremum at $\nu$-scale,

$M_{\nu}^{q}[F](\alpha):=\sup_{\beta}(1+\nu^{-1}|\beta-\alpha|)^{-q}|F(\beta)|.$
(9.11)

Now we are ready to prove Proposition 8.1.

###### Proof of Proposition 8.1 from Lemma 9.1.

The first step is to prove (8.43). First we need a bound for $\delta
E^{[\alpha,k]}$ that follows from the Taylor remainder formula,

$f(t)=\sum_{j=0}^{k_{0}}\frac{t^{j}}{j!}f^{(j)}(0)+\frac{1}{k_{0}!}\int_{0}^{t}(t-s)^{k_{0}}f^{(k_{0}+1)}(s)\mathop{}\\!\mathrm{d}s,$
(9.12)

which holds for functions in $C^{k_{0}+1}(\mathbb{R})$. Applying this with
$f(t)=E(t\alpha)$ and evaluating at $t=1$ we obtain

$E(\alpha)=E^{[0,k]}(\alpha)+\frac{1}{k!}\int_{0}^{1}(1-s)^{k}\alpha^{\otimes({k+1})}\cdot\nabla^{k+1}E(s\alpha)\mathop{}\\!\mathrm{d}s.$
(9.13)

Thus we obtain the formula

$\delta
E^{[0,k]}(\alpha)=\frac{1}{k!}\int_{0}^{1}(1-s)^{k}\alpha^{\otimes(k+1)}\cdot\nabla^{k+1}E(s\alpha)\mathop{}\\!\mathrm{d}s.$

Above $\alpha^{\otimes(k+1)}\nabla^{k+1}E$ is shorthand for
$\alpha^{a_{1}}\cdots\alpha^{a_{k}}\partial_{a_{1}}\cdots\partial_{a_{k}}E$
with an implicit sum over all indices.

Simply using the triangle inequality and estimating naively , this produces
the bound

$|\delta
E^{[0,k]}(\alpha)|\leq\sup_{|\beta|\leq|\alpha|}|\nabla^{k+1}E(\beta)||\alpha|^{k+1}\leq
M_{\nu}^{s}[\nabla^{k+1}E](0)(1+\nu^{-1}|\alpha|)^{s}|\alpha|^{k+1}.$

We also need bounds for the derivatives with respect to $\alpha$. When $0\leq
j\leq k$ we have

$\begin{split}\left\|\nabla^{j}\delta
E^{[0,k]}(\alpha)\right\|_{\mathrm{gop}}&\leq\sum_{j^{\prime}=0}^{j}\sup_{|\beta|\leq|\alpha|}\left\|\nabla^{k+1+j^{\prime}}E(\beta)\right\|_{\mathrm{gop}}\|\alpha\|^{k+1+j^{\prime}-j}\\\
&\leq\sum_{j^{\prime}=0}^{j}M_{\nu}^{s}[\nabla^{k+1+j^{\prime}}E](0)(1+\nu^{-1}|\alpha|)^{s}|\alpha|^{k+1+j^{\prime}-j}.\end{split}$
(9.14)

For higher order derivatives we note that $\nabla^{k+1}E^{[0,k]}=0$ and
therefore $\nabla^{k+1}\delta E^{[0,k]}=\nabla^{k+1}E$, so that for $j\geq
k+1$ we have

$\left\|\nabla^{j}\delta
E^{[0,k]}(\alpha)\right\|_{\mathrm{gop}}\leq\max_{|\beta|\leq|\alpha|}\left\|\nabla^{j}E(\beta)\right\|_{\mathrm{gop}}\leq
M_{\nu}^{s}[\nabla^{j}E](0)(1+\nu^{-1}|\alpha|)^{s}.$ (9.15)

We will estimate $\rho_{K}\ast|\hbar^{j/2}\nabla^{j}\delta E^{[0,k]}|(\alpha)$
in terms of the quantity

$Q_{k_{0},s}[E]:=\sum_{j^{\prime}=k_{0}+1}^{2d+k_{0}+s+2}\hbar^{(j^{\prime}-k-1)/2}M_{\nu}^{s}[\nabla^{j^{\prime}}E](0).$

We first work with $j\leq k$. We combine Combining (9.14) and Lemma 9.2 along
with (9.9) and the product rule (9.10) to see that for $K\geq 2d+k+s+1$ and
$j\leq k_{0}$ we have

$\begin{split}\rho_{K}\ast\left\|\hbar^{j/2}\nabla^{j}\delta
E^{[0,k]}\right\|_{\mathrm{gop}}(\alpha)&\leq
C\hbar^{j/2}M_{\hbar}^{K-2d-1}[\nabla^{j}\delta E^{[0,k]}](\alpha)\\\ &\leq
C\hbar^{j/2}\sum_{j^{\prime}=0}^{j}M_{\nu}^{s}[\nabla^{k+1+j^{\prime}}E](0)M_{\hbar}^{k+s+1}[(1+\nu^{-1}|\alpha|)^{s}|\alpha|^{k+1+j^{\prime}-j}]\\\
&\leq
C\hbar^{j/2}\sum_{j^{\prime}=0}^{j}M_{\nu}^{s}[\nabla^{k+1+j^{\prime}}E](0)M_{\hbar}^{s}[(1+\nu^{-1}|\alpha|)^{s}]M_{\hbar}^{k+1}[|\alpha|^{k+1+j^{\prime}-j}]\\\
&\leq
C(1+\nu^{-1}|\alpha|)^{s}\sum_{j^{\prime}=0}^{j}\hbar^{j/2}M_{\nu}^{s}[\nabla^{k+1+j^{\prime}}E](0)\left(|\alpha|^{k+1+j^{\prime}-j}+\hbar^{(k+1+j^{\prime}-j)/2}\right)\\\
&\leq
C(1+\nu^{-1}|\alpha|)^{s}Q_{k_{0},s}[E]\sum_{j^{\prime}=0}^{J}\left(\hbar^{(j-j^{\prime})/2}|\alpha|^{j-j^{\prime}}|\alpha|^{k+1}+\hbar^{(k+1)/2}\right)\end{split}$
(9.16)

for some constant $C$. In the final line, note that either
$\hbar^{1/2}>|\alpha|$, in which case the term $\hbar^{(k_{0}+1)/2}$
dominates, or else $\hbar^{1/2}<|\alpha|$, in which case
$(\hbar^{1/2}|\alpha|^{-1})^{(j-j^{\prime})}<1$ so that the first term in the
sum is bounded by $|\alpha|^{k+1}$. Therefore we can simplify the above bound
to

$\rho_{K}\ast|\hbar^{j/2}\nabla^{j}\delta E^{[0,k]}|(\alpha)\leq
CQ_{k_{0},s}[E](1+\nu^{-1}|\alpha|)^{s}(|\alpha|^{k+1}+\hbar^{(k+1)/2})$
(9.17)

For larger derivatives $j>k$, we simply use (9.15) to see that

$\rho_{K}\ast|\hbar^{j/2}\nabla^{j}\delta
E^{[0,k]}|(\alpha)\leq\hbar^{j/2}M_{\nu}^{s}[\nabla^{j}E](1+\nu^{-1}|\alpha|)^{s}\leq\hbar^{(k+1)/2}Q_{k_{0},s}[E](1+\nu^{-1}|\alpha|)^{s}.$

Taking $K=2d+k_{0}+s+2$ we have

$\sum_{j=0}^{K}(\rho_{K}\ast|\hbar^{j/2}\nabla^{j}\delta
E^{[0,k]}|)(\alpha)\leq
CQ_{k_{0},s}[E](1+\nu^{-1}|\alpha|)^{s})(|\alpha|^{k_{0}+1}+\hbar^{(k_{0}+1)/2}).$
(9.18)

The proof of (8.43) now follows from an application of Lemma 9.1.

To prove the bound on the triple Moyal product (8.44), we use the stronger
quantity

$Q^{\prime}_{k_{0},s}[E]:=\sum_{j=k_{0}+1}^{4d+2k_{0}+2s+4}\hbar^{(j-k_{0}-1)/2}M_{\nu}^{s}[\nabla^{j^{\prime}}E](0).$

Then taking (9.18) with $K^{\prime}=4d+2k_{0}+2s+4$, squaring both sides, and
convolving with $\rho_{K^{\prime}}$, we have

$\rho_{K^{\prime}}\ast\Bigg{(}\sum_{j=0}^{K^{\prime}}\rho_{K^{\prime}}\ast\left\|\hbar^{j/2}\nabla^{j}\delta
E^{[\alpha,k_{0}]}\right\|_{\mathrm{gop}}\Bigg{)}^{2}\leq
C(Q^{\prime}_{k_{0},s}[E])^{2}(1+\nu^{-2s}|\alpha|^{2s})(|\alpha|^{2k_{0}+2}+\hbar^{k_{0}+1}).$
(9.19)

Then (8.44) follows from Corollary 9.1. ∎

###  9.3 Proof of the Moyal product bound

###### Proof of Lemma 9.1.

The main idea is to use the following identity to integrate by parts in the
$\gamma$ variables in order to obtain decay in the $\beta$ variables:

$e^{i\beta_{a}\gamma^{a}/(2\hbar)}=-2i\hbar(\beta_{a})^{-1}\partial_{\gamma^{a}}e^{i\beta_{a}\gamma^{a}/(2\hbar)}.$
(9.20)

Symmetrically, we can integrate by parts in the $\beta$ variables to obtain
decay in the $\gamma$ variables.

To do this we introduce a partition of unity

$1=\chi_{0}(t)+\sum_{j=1}^{\infty}\chi_{j}(t),$

where $\chi_{0}\in C_{c}^{\infty}(\mathbb{R})$ is a smooth function supported
in $[-\hbar^{1/2},\hbar^{1/2}]$ and $\chi_{j}\in C_{c}^{\infty}(\mathbb{R})$
are supported in $\\{t\in\mathbb{R}\mid 2^{j-1}\hbar^{1/2}\leq|t|\leq
2^{j+2}\hbar^{1/2}\\}$. We choose this partition of unity so that it satisfies
the bounds

$\sup_{t}\left|\frac{\mathrm{d}^{k}}{\mathrm{d}t^{k}}\chi_{j}(t)\right|\leq
C_{k}2^{-jk}\hbar^{-k/2}.$ (9.21)

for $k\leq K:=\max\\{K_{F},K_{G}\\}$ and some constants $C_{k}$. It is
possible to take $C_{k}=(CK)^{k}$, as can be seen from constructing $\chi_{1}$
as an iterated convolution of the form $\chi=g^{\ast K}\ast\chi_{0}$ where
$\chi_{0}$ is an indicator function and $g$ is a $C^{1}$ function supported on
$[-1/K,1/K]$ and which integrates to $1$.

For such $\chi_{j}$, we also have

$\sup_{t}\left|\frac{\mathrm{d}^{k}}{\mathrm{d}t^{k}}(t^{-a}\chi_{j}(t))\right|\leq
C_{a,k}2^{-j(k+a)}\hbar^{-(k+a)/2}.$ (9.22)

This follows from (9.21), the product rule, and the bound

$\sup_{2^{j-1}\hbar^{1/2}\leq|t|\leq
2^{j+2}\hbar^{1/2}}\left|\frac{d^{\ell}}{dt^{\ell}}t^{-a}\right|\leq
2^{a+\ell}(a+\ell)^{\ell}(2^{j}\hbar^{1/2})^{-a-\ell}.$

Then $C_{a,k}$ can be chosen to be

$\begin{split}C_{a,k}&\leq\sum_{j=0}^{k}\binom{k}{j}2^{a+j}(a+j)^{j}(CK)^{k-j}\leq
2^{a+k}(a+k+CK)^{k}.\end{split}$ (9.23)

Applying this partition of unity to each variable we we obtain the identity

$1=\prod_{a=1}^{2d}\left(\sum_{j=0}^{\infty}\chi_{j}(\beta_{a})\right)\prod_{b=1}^{2d}\left(\sum_{j=0}^{\infty}\chi_{j}(\gamma^{a})\right)=\sum_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\gamma}:[2d]\to{\mathbb{N}}}\prod_{a=1}^{2d}\chi_{j^{\beta}_{a}}(\beta_{a}),\prod_{a=1}^{2d}\chi_{j^{\gamma}_{a}}(\gamma^{a}).$
(9.24)

The latter sum is over pairs of tuples
${\mathbf{j}}^{\beta}=(j^{\beta}_{1},j^{\beta}_{2},\cdots,j^{\beta}_{2d})$,
${\mathbf{j}}^{\gamma}=(j^{\gamma}_{1},j^{\gamma}_{2},\cdots,j^{\gamma}_{2d})$.

We use this to split the moyal product $F\star G$ into terms indexed by
${\mathbf{j}}^{\beta}$ and ${\mathbf{j}}^{\gamma}$,

$F\star G=\sum_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\gamma}}(F\star
G)_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\gamma}},$ (9.25)

where

$(F\star
G)_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\alpha}}(\alpha):=(2\pi\hbar)^{-2d}\int
e^{i\beta_{a}\gamma^{a}/(2\hbar)}F(\alpha+\beta/2)G(\alpha+\gamma/2)\prod_{a=1}^{2d}\chi_{j^{\beta}_{a}}(\beta_{a})\prod_{a=1}^{2d}\chi_{j^{\gamma}_{a}}(\gamma_{a})\mathop{}\\!\mathrm{d}\beta_{a}\mathop{}\\!\mathrm{d}\gamma^{a}.$
(9.26)

We estimate the quantity $(F\star
G)_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\gamma}}$ differently depending on
whether ${\mathbf{j}}^{\beta}=0$ and/or ${\mathbf{j}}^{\gamma}=0$. We thus
split into four terms:

$F\star G=(F\star
G)_{{\mathbf{0}},{\mathbf{0}}}+\sum_{{\mathbf{j}}^{\beta}\not=0}(F\star
G)_{{\mathbf{j}}^{\beta},{\mathbf{0}}}+\sum_{{\mathbf{j}}^{\gamma}\not=0}(F\star
G)_{{\mathbf{0}},{\mathbf{j}}^{\gamma}}+\sum_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\gamma}}(F\star
G)_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\gamma}}.$

The first term, with ${\mathbf{j}}^{\beta}={\mathbf{j}}^{\gamma}={\mathbf{0}}$
being all zeros, can be bounded simply using the triangle inequality:

$\begin{split}|(F\star
G)_{{\mathbf{0}},{\mathbf{0}}}(\alpha)|&\leq\hbar^{-2d}\int|F(\alpha+\beta/2)||G(\alpha+\gamma/2)|\prod\chi_{0}(B_{a})\mathop{}\\!\mathrm{d}B\\\
&\leq\big{(}\hbar^{-d}\int_{|\alpha^{\prime}-\alpha|\leq\hbar^{1/2}}|F(\alpha^{\prime})|\mathop{}\\!\mathrm{d}\alpha^{\prime}\big{)}\big{(}\hbar^{-d}\int_{|\alpha^{\prime}-\alpha|\leq\hbar^{1/2}}|G(\alpha^{\prime})|\mathop{}\\!\mathrm{d}\alpha^{\prime}\big{)}\\\
&\leq
2^{K_{G}+K_{F}}(\rho_{K_{G}}\ast|F|(\alpha))(\rho_{K_{F}}\ast|G|(\alpha)).\end{split}$
(9.27)

In the last line the factor $2^{K_{G}+K_{F}}$ appears from the use of the fact
that $2^{K}\rho_{K}(\alpha)\geq 1$ when $|\alpha|\leq\hbar^{1/2}$.

For the remaining terms we integrate by parts using (9.20). We will assume for
this part that ${\mathbf{j}}^{\beta}\not={\mathbf{0}}$ and
${\mathbf{j}}^{\gamma}\not={\mathbf{0}}$ (this being the most technical case
to handle). Let $a_{0}=\operatorname*{argmax}{\mathbf{j}}^{\beta}$ and
$b_{0}=\operatorname*{argmax}{\mathbf{j}}^{\gamma}$ be the indices for which
$J^{\beta}:=j^{\beta}_{a_{0}}$ and $J^{\gamma}:=j^{\gamma}_{b_{0}}$ are
maximized.

We integrate by parts first $K_{G}$ times in the $\gamma^{a_{0}}$ variable to
obtain decay in $\beta_{a_{0}}$, obtaining

$\begin{split}(F\star
G)_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\alpha}}(\alpha)=(2\pi\hbar)^{-2d}(2i\hbar)^{K_{G}}&\int
e^{i\beta_{a}\gamma^{a}/(2\hbar)}\beta_{a_{0}}^{-K_{G}}F(\alpha+\beta/2)\partial_{\gamma^{a_{0}}}^{K_{G}}\big{(}\chi_{j^{\gamma}_{a_{0}}}(\gamma^{a_{0}})G(\alpha+\gamma/2)\big{)}\\\
&\qquad\qquad\times\prod_{a=1}^{2d}\chi_{j^{\beta}_{a}}(\beta_{a})\prod_{b\not=a_{0}}\chi_{j^{\gamma}_{b}}(\gamma^{b})\mathop{}\\!\mathrm{d}\beta_{a}\mathop{}\\!\mathrm{d}\gamma^{a}.\end{split}$
(9.28)

Next we integrate by parts $K_{F}$ times in $\beta_{b_{0}}$ to obtain decay in
$\gamma^{b_{0}}$, then use the product rule to split up the derivatives

$\begin{split}(F\star
G)_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\alpha}}(\alpha)&=(2\pi\hbar)^{-2d}(2i\hbar)^{K_{G}}(2i\hbar)^{K_{F}}\int
e^{i\beta_{a}\gamma^{a}/(2\hbar)}\partial_{\beta_{b_{0}}}^{K_{F}}\big{(}\beta_{a_{0}}^{-K_{G}}\chi_{j^{\beta}_{b_{0}}}(\beta_{b_{0}})F(\alpha+\beta/2)\big{)}\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times(\gamma^{b_{0}})^{-K_{F}}\partial_{\gamma^{a_{0}}}^{K_{G}}\big{(}\chi_{j^{\gamma}_{a_{0}}}(\gamma^{a_{0}})G(\alpha+\gamma/2)\big{)}\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times\prod_{a\not=b_{0}}\chi_{j^{\beta}_{a}}(\beta_{a})\prod_{b\not=a_{0}}\chi_{j^{\gamma}_{b}}(\gamma^{b})\mathop{}\\!\mathrm{d}\beta_{a}\mathop{}\\!\mathrm{d}\gamma^{a}\\\
&=(2\pi\hbar)^{-2d}(2i\hbar)^{K_{G}}(2i\hbar)^{K_{F}}\int
e^{i\beta_{a}\gamma^{a}/(2\hbar)}\sum_{k=0}^{K_{F}}\binom{K_{F}}{k}\partial_{\beta_{b_{0}}}^{K_{F}-k}\big{(}\beta_{a_{0}}^{-K_{G}}\chi_{j^{\beta}_{b_{0}}}(\beta_{b_{0}})\big{)}\partial_{\beta_{b_{0}}}^{k}F(\alpha+\beta/2)\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times(\gamma^{b_{0}})^{-K_{F}}\sum_{k^{\prime}=0}^{K_{G}}\binom{K_{G}}{k^{\prime}}\partial_{\gamma^{a_{0}}}^{K_{G}-k^{\prime}}\chi_{j^{\gamma}_{a_{0}}}(\gamma^{a_{0}})\partial_{\gamma^{a_{0}}}^{k^{\prime}}G(\alpha+\gamma/2)\big{)}\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times\prod_{a\not=b_{0}}\chi_{j^{\beta}_{a}}(\beta_{a})\prod_{b\not=a_{0}}\chi_{j^{\gamma}_{b}}(\gamma^{b})\mathop{}\\!\mathrm{d}\beta_{a}\mathop{}\\!\mathrm{d}\gamma^{a}.\end{split}$
(9.29)

In the quantity
$\partial_{\beta_{b_{0}}}^{K_{F}-k}(\beta_{a_{0}}^{-K_{G}}\chi_{j^{\beta}_{b_{0}}}(\beta_{b_{0}}))$
there are two cases to consider. If $a_{0}=b_{0}$ then we use (9.22), and if
$a_{0}\not=b_{0}$ then the derivative only falls on $\chi$ and we use (9.21).
Regardless we have the bound (using $\beta_{a_{0}}\geq
2^{J^{\beta}}\hbar^{1/2}$)

$|\partial_{\beta_{b_{0}}}^{K_{F}-k}(\beta_{a_{0}}^{-K_{G}}\chi_{j^{\beta}_{b_{0}}}(\beta_{b_{0}}))|\leq(CK)^{K_{F}-k}2^{-J^{\beta}K_{G}}\hbar^{-(K_{F}+K_{G}-k)/2}.$
(9.30)

Now we use the estimates (9.22) and (9.21) (using that $\gamma^{b_{0}}\sim
2^{J^{\gamma}}\hbar^{1/2}$ and $\beta_{a_{0}}\sim 2^{J^{\beta}}\hbar^{1/2}$)
to bound the derivatives hitting $\chi$, and then apply the triangle
inequality

$\begin{split}|(F\star
G)_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\alpha}}(\alpha)|&\leq
4^{K}\hbar^{K_{G}+K_{F}-2d}\int\sum_{k=0}^{K_{F}}\binom{K_{F}}{k}(CK)^{K_{F}-k}2^{-J^{\beta}K_{G}}\hbar^{-(K_{F}-K_{G}-k)/2}|\partial_{\beta_{b_{0}}}^{k}F(\alpha+\beta/2)|\\\
&\qquad\qquad\qquad\qquad\qquad\times 2^{-J^{\gamma}K_{F}}\hbar^{-K_{F}/2}\\\
&\qquad\qquad\qquad\qquad\qquad\times\sum_{k^{\prime}=0}^{K_{G}}\binom{K_{G}}{k^{\prime}}(CK)^{K_{G}-k^{\prime}}\hbar^{-(K_{G}-k^{\prime})/2}|\partial_{\gamma^{a_{0}}}^{k^{\prime}}G(\alpha+\gamma/2)\big{|}\\\
&\qquad\qquad\qquad\qquad\qquad\times\prod_{a}{\widetilde{\chi}}_{j^{\beta}_{a}}(\beta_{a})\prod_{b}{\widetilde{\chi}}_{j^{\gamma}_{b}}(\gamma^{b})\mathop{}\\!\mathrm{d}\beta_{a}\mathop{}\\!\mathrm{d}\gamma^{a}.\end{split}$
(9.31)

Above ${\widetilde{\chi}}_{a}$ is the indicator function for the support of
$\chi_{a}$. Collecting the constants, using $\binom{K}{k}\leq 2^{K}$ and
combining factors of $2$ and $\hbar$ we arrive at

$\begin{split}|(F\star
G)_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\alpha}}(\alpha)|&\leq(CK)^{2K}\sum_{k=0}^{K_{F}}\sum_{k^{\prime}=0}^{K_{G}}\hbar^{-2d}\int
2^{-J^{\beta}K_{G}}\hbar^{k/2}|\partial_{\beta_{b_{0}}}^{k}F(\alpha+\beta/2)|\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\times
2^{-J^{\gamma}K_{F}}\hbar^{k^{\prime}/2}|\partial_{\gamma^{a_{0}}}^{k^{\prime}}G(\alpha+\gamma/2)\big{|}\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\times\prod_{a}{\widetilde{\chi}}_{j^{\beta}_{a}}(\beta_{a})\prod_{b}{\widetilde{\chi}}_{j^{\gamma}_{b}}(\gamma^{b})\mathop{}\\!\mathrm{d}\beta_{a}\mathop{}\\!\mathrm{d}\gamma^{a}.\end{split}$
(9.32)

Since $J^{\beta}>j^{\beta}_{a}$ for any other index $a$ and on the support of
the integrand above $\beta_{a}\sim\hbar^{1/2}2^{j^{\beta}_{a}}$ (and similarly
for $\gamma$, it holds that

$2^{-KJ^{\beta}}\leq
C^{K}(\hbar^{-1/2}|\beta|+1)^{-K}\leq\hbar^{d}\rho_{K}(\beta)$

and similarly

$2^{-KJ^{\gamma}}\leq
C^{K}(\hbar^{-1/2}|\gamma|+1)^{-K}\leq\hbar^{d}\rho_{K}(\gamma).$

Recalling that $\rho_{K}(\beta)=\hbar^{-d}(\hbar^{1/2}|\beta|+1)^{-K}$, and
then noting that we can sum over all multi-indices ${\mathbf{j}}^{\beta}$ and
${\mathbf{j}}^{\gamma}$ using

$\sum_{{\mathbf{j}}^{\beta}}\prod_{a}{\widetilde{\chi}}_{j^{\beta}_{a}}(\beta_{a})\leq
4^{d}$

we can simplify the above bound to

$\begin{split}|(F\star
G)(\alpha)|&\leq(CK)^{2K}\hbar^{-2d}\sum_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\alpha}}\int
2^{-J^{\beta}K_{G}}\hbar^{k/2}|\partial_{\beta_{b_{0}}}^{k}F(\alpha+\beta/2)|\\\
&\qquad\qquad\qquad\qquad\times
2^{-J^{\gamma}K_{F}}\hbar^{k^{\prime}/2}|\partial^{k^{\prime}}_{\gamma^{a_{0}}}G(\alpha+\gamma/2)|\prod_{a}{\widetilde{\chi}}_{j^{\beta}_{a}}(\beta_{a})\prod_{a}{\widetilde{\chi}}_{j^{\gamma}_{b}}(\gamma^{b})\mathrm{d}\beta_{a}\mathrm{d}\gamma^{a}\\\
&\leq(CK)^{2K}\int\rho_{K_{G}}(\beta)\hbar^{k/2}(\sum_{k=0}^{K_{F}}\left\|\nabla^{k}F(\alpha+\beta/2)\right\|_{\mathrm{gop}})\\\
&\qquad\qquad\qquad\quad\times\rho_{K_{F}}(\gamma)\hbar^{k^{\prime}/2}(\sum_{k^{\prime}=0}^{K_{G}}\left\|\nabla^{k^{\prime}}G(\alpha+\gamma/2)\right\|_{\mathrm{gop}})\sum_{{\mathbf{j}}^{\beta},{\mathbf{j}}^{\alpha}}(\prod_{a}{\widetilde{\chi}}_{j^{\beta}_{a}}(\beta_{a})\prod_{b}{\widetilde{\chi}}_{j^{\gamma}_{b}}(\gamma^{b}))\mathrm{d}\beta_{a}\mathrm{d}\gamma^{a}\\\
&\leq(CK)^{2K}\int\rho_{K_{G}}(\beta)\hbar^{k/2}(\sum_{k=0}^{K_{F}}\left\|\nabla^{k}F(\alpha+\beta/2)\right\|_{\mathrm{gop}})\\\
&\qquad\qquad\qquad\quad\times\rho_{K_{F}}(\gamma)\hbar^{k^{\prime}/2}(\sum_{k^{\prime}=0}^{K_{G}}\left\|\nabla^{k^{\prime}}G(\alpha+\gamma/2)\right\|_{\mathrm{gop}})\mathrm{d}\beta_{a}\mathrm{d}\gamma^{a}\\\
&\leq(CK)^{2K}C^{d}\sum_{k=0}^{K_{F}}\sum_{k^{\prime}=0}^{K_{G}}\big{(}\rho_{K_{G}}\ast\left\|\hbar^{k/2}\nabla^{k}F\right\|_{\mathrm{gop}}(\alpha)\big{)}\big{(}\rho_{K_{F}}\ast\left\|\hbar^{k^{\prime}/2}\nabla^{k^{\prime}}G\right\|_{\mathrm{gop}}(\alpha)\big{)},\end{split}$
(9.33)

as desired. The remaining terms (with ${\mathbf{j}}^{\beta}\not=0$ and
${\mathbf{j}}^{\gamma}=0$ or vice versa) are handled similarly and bring
smaller constants. ∎

##  Appendix A Physical units, symplectic covariance, and a corollary

In this section we offer some informal discussion of symplectic symmetry and
the relationship to units. To illustrate this, we then define some preferred
choices of units and use them to state and prove Corollary A.1 of Theorem 3.1.
This corollary generalizes the main result from our companion paper [45] to
Hamiltonians not restricted to the form $\hat{H}=\hat{p}^{2}/2m+V(\hat{x})$ at
the expense of introducing the uncomputed constant $C_{d}$.

###  A.1 Symplectic transformations of the main result

For a symplectic matrix $\mathcal{Z}$
($\mathcal{Z}^{\top}\omega\mathcal{Z}=\omega$) representing a linear
symplectic transformation, we will use $\mathcal{Z}$ as a superscript on
scalar functions over phase space to denote the composition equivalent to the
change of coordinates associated with the matrix:
$E^{\mathcal{Z}}(\alpha):=(E\circ\mathcal{Z})(\alpha)=E(\mathcal{Z}\alpha)$.
The same notation is used for tensor functions, except we must additionally
transform the indices, e.g.,
$(E^{\mathcal{Z}})^{a}_{\phantom{a}b}(\alpha):=(\mathcal{Z}^{-1})^{a}_{\phantom{a}c}\mathcal{Z}^{d}_{\phantom{a}b}E^{c}_{\phantom{a}d}(\mathcal{Z}\alpha)$.

The general Fokker-Planck equation for a classical open system,

$\partial_{t}f=-\partial_{a}[f(\partial^{a}H+G^{a})]+\frac{1}{2}\partial_{a}(D^{ab}\partial_{b}f),$
(A.1)

and the Wigner representation of the Lindblad equation

$\displaystyle\partial_{t}W_{\rho}=$ $\displaystyle\,-\frac{i}{\hbar}(H\star
W_{\rho}-W_{\rho}\star H)+\frac{1}{\hbar}\sum_{k}\left(L_{k}\star
W_{\rho}\star L_{k}^{*}-\frac{1}{2}L_{k}^{*}\star L_{k}\star
W_{\rho}-\frac{1}{2}W_{\rho}\star L_{k}^{*}\star L_{k}\right)$ (A.2)

are both covariant under linear symplectic transformations. This means (A.1)
is unchanged under $H\to H^{\mathcal{Z}}$, $G\to G^{\mathcal{Z}}$, $D\to
D^{\mathcal{Z}}$, and $f\to f^{\mathcal{Z}}$ because
$\partial_{a}f^{\mathcal{Z}}=\mathcal{Z}^{b}_{\phantom{a}a}\partial_{a}f$.
Likewise (A.2) is unchanged under under $H\to H^{\mathcal{Z}}$, $L_{k}\to
L_{k}^{\mathcal{Z}}$, and $W_{\rho}\to W_{\rho}^{\mathcal{Z}}$ because
$A^{\mathcal{Z}}\star B^{\mathcal{Z}}=(A\star B)^{\mathcal{Z}}$, a basic
property of the Moyal product.

If the data $(H,\\{L_{k}\\}_{k=1}^{K})$ are admissible under Assumption 3.1,
then the transformed data
$(H^{\mathcal{Z}},\\{L_{k}^{\mathcal{Z}}\\}_{k=1}^{K})$ are also admissible.
Now suppose an the initial state can expressed as a mixture of NTS states
($\sigma\geq\frac{g}{2}\sigma_{*}$) _and also_ as a mixture of
$\mathcal{Z}$-transformed NTS states
($\sigma\geq\frac{g_{\mathcal{Z}}}{2}\mathcal{Z}^{-1}\sigma_{*}\mathcal{Z}^{-\top}$),
where $g$ and $g_{\mathcal{Z}}$ are the relative diffusion strengths (3.10)
computed with the original and transformed data. Then we can apply Theorem 3.1
to the original and transformed data and _both_ sets of bounds will hold.
These will generally be distinct bounds because the anharmonicity measures
$B^{\mathrm{anh}}_{\mathrm{c}}$, $B^{\mathrm{anh}}_{\mathrm{q}}$, and
$B^{\mathrm{anh}}_{\mathrm{q}^{\prime}}$ are not invariant under linear
symplectic transformations of the data.585858Note that $H^{\mathcal{Z}}$ is
quadratic if and only if $H$ is quadratic, and $L_{k}^{\mathcal{Z}}$ is linear
if and only if $L_{k}$ is linear, so whether the anharmonicity measures vanish
will generally be invariant under linear symplectic transformations.

###  A.2 Units

Given a phase-space vector with mixed units like
$\tilde{\alpha}=(x_{0}\,\mathrm{m},p_{0}\,\mathrm{kg}\\!\cdot\\!\mathrm{m}/\mathrm{s})$,
where $x_{0}$ and $p_{0}$ are dimensionless numbers, we can transform this to
a vector with uniform units using a symplectic matrix like
$\mathcal{Z}^{-1}=\operatorname{diag}(\eta,\eta^{-1})$ with
$\eta=\sqrt{(\mathrm{kg}\\!\cdot\\!\mathrm{m}/\mathrm{s})/(\mathrm{m})}=\sqrt{\mathrm{kg}/\mathrm{s}}$.
Specifically,
$\mathcal{Z}^{-1}\tilde{\alpha}=(x_{0},p_{0})[\mathrm{m}\sqrt{\mathrm{kg}/\mathrm{s}}]$.
Given such a choice of $\mathcal{Z}$ and a real, physical Hamiltonian $H(x,p)$
taking as input dimensionful positions $x$ and momenta $p$, we can then apply
Theorem 3.1 to the transformed functions
$H^{\mathcal{Z}}(\alpha):=H(\mathcal{Z}\alpha)$ and
$L_{k}^{\mathcal{Z}}(\alpha):=L_{k}(\mathcal{Z}\alpha)$, which will accept
vectors with uniform units. Now, there is generally no symplectic matrix that
can make an arbitrary unitful vector unitless. Still, a choice like
$\mathcal{Z}$ above is sufficient to ensure that all of the manipulations in
this paper (such as taking the Euclidean595959An alternative way to think
about this is that the choice $\mathcal{Z}$ defines an inner product:
$\langle\tilde{\alpha},\tilde{\beta}\rangle_{\mathcal{Z}}:={\tilde{\alpha}}^{\top}(\mathcal{Z}\mathcal{Z}^{\top})^{-1}\tilde{\beta}=\alpha^{\top}\cdot\beta$
for uniform-unit vectors $\alpha=\mathcal{Z}^{-1}\tilde{\alpha}$,
$\beta=\mathcal{Z}^{-1}\tilde{\beta}$. norm of $\mathcal{Z}\alpha$ for mixed-
unit vector $\alpha$) are physically meaningful once such a choice of units
has been made.

The symplectic matrices are closed under multiplication (being a group) and so
the choice of units does not exhaust the freedom to choose $\mathcal{Z}$. For
instance, the skewing matrix $\mathcal{Z}=\left(\begin{smallmatrix}1&1\\\
0&1\end{smallmatrix}\right)$ is symplectic and mixes position and momentum in
a way that does not correspond to a choice of units.606060In order that an
arbitrary symplectic matrix $\mathcal{Z}^{-1}$ correctly makes all units
uniform, the entries $\mathcal{Z}_{\mathrm{x}_{i},\mathrm{x}_{j}}$ and
$\mathcal{Z}_{\mathrm{p}_{i},\mathrm{x}_{j}}$ must have units of
[length/momentum]1/2 and the entries
$\mathcal{Z}_{\mathrm{x}_{i},\mathrm{p}_{j}}$ and
$\mathcal{Z}_{\mathrm{p}_{i},\mathrm{p}_{j}}$ must have units of
[momentum/length]1/2. This ensures that all elements of the vector
$\mathcal{Z}\alpha$ have units of [length $\cdot$ momentum]1/2, where $\alpha$
is a physical phase-space vector with units of [length] in the first $d$
elements and [momentum] in the other $d$ elements. Thus, the fact that
multiple bounds can be derived using Theorem 3.1 by applying difference
choices of $\mathcal{Z}$ is not removed by a choice of units. Although the
trajectory $\tilde{\rho}(t)$ constructed in different cases will generally be
different, note that the existence of such a trajectory is a units-independent
statement; the bounds (3.20), (3.21) are on unitless norms, which in turn
constrains the maximum (unitless) difference in outcome probability for any
measurement.

Ideally the choice of $\mathcal{Z}$ could be optimized for the best bound
(since coherent states $\hat{\tau}_{\alpha}$ would correspond to covariance
matrices satisfying $\sigma=\frac{\hbar}{2}\mathcal{Z}^{\top}\mathcal{Z}$).
Alternatively, to name a concrete choice, one might choose e.g.

$\begin{split}\mathcal{Z}{}_{0}=\left(\begin{matrix}r\mathds{1}_{d}&0\\\
0&r^{-1}\mathds{1}_{d}\end{matrix}\right)\end{split}$ (A.3)

where $r=\sqrt{r_{x}/r_{p}}$ for characteristic scales $r_{x},r_{p}$ defined
by $r_{x}^{-2}=\sup_{\alpha}\sup_{w\in\mathbb{R}^{d},\left\lVert
w\right\rVert=1}(w_{a}\partial_{x_{a}})^{2}H(\alpha)$ and
$r_{p}^{-2}=\sup_{\alpha}\sup_{w\in\mathbb{R}^{d},\left\lVert
w\right\rVert=1}(w_{a}\partial_{p_{a}})^{2}H(\alpha)$, which quantify the
maximum second derivatives of $H$ with respect to position and momentum,
respectively.

###  A.3 Corollary

To illustrate the above informal discussion, we will apply Theorem 3.1 to a
special case with linear Lindblad operators (diffusion matrix homogeneous over
phase space) and no friction. We will ask: given a physical system with
(unitful) Hamiltonian $H$, how much environmental noise must we add to ensure
that the quantum and classical dynamics cannot be distinguished up to some
tolerable error? We will first identify the relevant characteristic timescales
and action scales of the Hamiltonian. They will be constructed from the
Hamiltonian’s derivatives, and in particular we recall from Eq. (3.16) the
seminorm

$\displaystyle\mathcal{Q}^{q,r}_{s}[E]$
$\displaystyle:=\sum_{j=q}^{r}s^{(j-q)/2}\sup_{\alpha}\left\|\nabla^{j}E(\beta)\right\|_{\mathrm{gop}}=\sum_{j=q}^{r}s^{(j-q)/2}\left|E\right|_{C^{j}}$
(A.4)

which bounds the $q$-th through $r$-th derivatives, where
$\left|H\right|_{C^{j}}=\sup_{\alpha}\left\|\nabla^{j}H(\alpha)\right\|_{\mathrm{gop}}$
is the $C^{k}$ seminorm and where

$\displaystyle\left\|A\right\|_{\mathrm{gop}}:=\sup_{\|\beta_{i}\|=1}\left|\left(\bigotimes_{i=1}^{j}\beta_{j}\right)\cdot
A\right|=\sup_{\|\beta_{i}\|=1}\left|\beta^{a_{1}}_{1}\cdots\beta^{a_{j}}_{j}A_{a_{1}\cdots
a_{j}}\right|$ (A.6)

is the generalized operator norm for a $j$-th order tensor $A$, as discussed
in Section 4.3. For our main result, we used $s=\hbar$ to get the tightest
bound, but here it will be instructive to use a (macroscopic) action scale
$s\geq\hbar$.

616161In a future version I’d probably want to reduce the amount of notation
we have here, if possible. Lots of slightly different ways of writing the same
thing.

######  Definition A.1 (Characteristic classical scales).

For any classical Hamiltonian $H$ with bounded partial derivatives of degree
$k=2,\ldots,2d+4$ and any symplectic matrix $\mathcal{Z}$, we define the
harmonic time $\tau_{\mathrm{H}}$ as the inverse of the maximum operator norm
taken by the Hessian of $H^{\mathcal{Z}}$ over phase space,626262For linear
Hamiltonians, $\tau_{\mathrm{H}}=\infty$ by convention. This indicates that
there is no natural time scale because the (real or imaginary) frequency of
the local harmonic dynamics is zero everywhere in phase space.

$\displaystyle\frac{1}{\tau_{\mathrm{H}}}:=|H^{\mathcal{Z}}|_{C^{2}}=\mathcal{Q}^{2,2}[H^{\mathcal{Z}}]=\sup_{\alpha}\left\|(\nabla^{2}H^{\mathcal{Z}})(\alpha)\right\|_{\mathrm{gop}}$
(A.7)

The anharmonic action $s_{\mathrm{A}}$ of $H$ is given by the ratio of the
largest second and third directional derivative:

$\displaystyle
s_{\mathrm{A}}:=\left(\tau_{\mathrm{H}}|H^{\mathcal{Z}}|_{C^{3}}\right)^{-2}=\left(\frac{|H^{\mathcal{Z}}|_{C^{2}}}{|H^{\mathcal{Z}}|_{C^{3}}}\right)^{2}=\left(\frac{\mathcal{Q}^{2,2}[H^{\mathcal{Z}}]}{\mathcal{Q}^{3,3}[H^{\mathcal{Z}}]}\right)^{2},$
(A.8)

We use the anharmonic action $s_{\mathrm{A}}$ to define the modified
anharmonic action636363For quadratic Hamiltonians,
$s_{\mathrm{A}}=\tilde{s}_{\mathrm{A}}=\infty$ by convention. This is the case
of harmonic dynamics where quantum and classical evolution is identical on
phase space. of $H$

$\displaystyle\tilde{s}_{\mathrm{A}}:=\left(\tau_{\mathrm{H}}\mathcal{Q}^{3,2d+4}_{s_{\mathrm{A}}}[H^{\mathcal{Z}}]\right)^{-2}=\left(\frac{\mathcal{Q}^{2,2}[H^{\mathcal{Z}}]}{\mathcal{Q}^{3,2d+4}_{s_{\mathrm{A}}}[H^{\mathcal{Z}}]}\right)^{2}.$
(A.9)

Given $\tau_{\mathrm{H}}$, $\mathcal{Z}$, and $\tilde{s}_{\mathrm{A}}$, we
furthermore have a natural choice of characteristic diffusion matrix given by
$D_{\mathrm{C}}=(\tilde{s}_{\mathrm{A}}/\tau_{\mathrm{H}})\mathcal{Z}\mathcal{Z}^{\top}$.

Intuitively, the harmonic time $\tau_{\mathrm{H}}$ is the _shortest_ timescale
associated with the local harmonic approximation at any point in phase space.
When the harmonic time is long, the classical dynamics are slow compared to
the quantum scale set by $\hbar$, and we expect they well approximate the
quantum dynamics they correspond to.

The anharmonic action $s_{\mathrm{A}}$ and modified anharmonic action
$\tilde{s}_{\mathrm{A}}$ are _not_ measures of the accessible phase space.
Rather, they measure the phase-space scale on which the anharmonicity of the
potential is important over the harmonic time $\tau_{\mathrm{H}}$. For cubic
potentials, $\tilde{s}_{\mathrm{A}}=s_{\mathrm{A}}$ because the higher-order
seminorms in (A.9) vanish. Introducing the higher order terms increases the
denominator in $\tilde{s}_{\mathrm{A}}$, so $\tilde{s}_{\mathrm{A}}\leq
s_{\mathrm{A}}$ always holds.

We now prove a corollary of Theorem 3.1 making use of the scales defined in
Definition A.1. The seminorm $\mathcal{Q}^{3,2d+4}_{\hbar}[H^{\mathcal{Z}}]$
naturally arises in Theorem 3.1, but through $\tilde{s}_{\mathrm{A}}$ we will
upper bound it with the factor of
$\mathcal{Q}^{3,2d+4}_{s_{\mathrm{A}}}[H^{\mathcal{Z}}]\geq\mathcal{Q}^{3,2d+4}_{\hbar}[H^{\mathcal{Z}}$
(assuming $s_{\mathrm{A}}\geq\hbar$). This loosens the bound, essentially
throwing out the detailed information about how the anharmonic factors depend
on higher powers of the action scale, but has the benefit of isolating the
leading $\hbar^{4/3}$ dependence, with everything else expressed in terms of
($\hbar$-independent) macroscopic properties of the classical Hamiltonian.

######  Corollary A.1 (Minimum diffusion for correspondence).

For $d$ degrees of freedom, let $\hat{H}$ be a quantum Hamiltonian function
with bounded partial derivatives of degree $j=2,\ldots,2d+4$ and corresponding
classical Hamiltonian $H=\mathcal{W}_{\hbar}[\hat{H}]$. Let $\mathcal{Z}$ a
symplectic matrix and let
$\tau_{\mathrm{H}}=1/\mathcal{Q}^{2,2}[H^{\mathcal{Z}}]$,
$s_{\mathrm{A}}=(\mathcal{Q}^{2,2}[H^{\mathcal{Z}}]/\mathcal{Q}^{3,3}[H^{\mathcal{Z}}])^{2}$,
$\tilde{s}_{\mathrm{A}}=(\mathcal{Q}^{2,2}[H^{\mathcal{Z}}]/\mathcal{Q}^{3,2d+4}_{s_{\mathrm{A}}}[H^{\mathcal{Z}}])^{2}$,
$D_{\mathrm{C}}=(\tilde{s}_{\mathrm{A}}/\tau_{\mathrm{H}})\mathcal{Z}\mathcal{Z}^{\top}$
be the harmonic time, anharmonic action, modified anharmonic action, and
characteristic diffusion matrix of $H$ from Definition A.1. Assume
$s_{\mathrm{A}}\geq\hbar$. Assume an initial state quantum state
$\rho(t\mathord{=}0)$ given as a mixture of Gaussian states with covariance
matrix $\sigma=(\hbar/2)\mathcal{Z}\mathcal{Z}^{\top}$. Finally, let $\rho(t)$
and $f(t)$ be the corresponding quantum and classical trajectory for the
frictionless dynamics specified by $H$ and a homogenous (i.e, constant over
phase space) diffusion matrix $D$. Then for any tolerable error growth rate
$\epsilon$ satisfying

$\epsilon\geq\frac{C_{d}}{\tau_{\mathrm{H}}}\sqrt{\frac{\hbar}{\tilde{s}_{\mathrm{A}}}},$
(A.10)

the diffusion strength constraint

$\displaystyle
D\geq\left(\frac{C_{d}}{\tau_{\mathrm{H}}\epsilon}\right)^{\frac{2}{3}}\left(\frac{\hbar}{\tilde{s}_{\mathrm{A}}}\right)^{\frac{4}{3}}D_{\mathrm{C}}$
(A.11)

guarantees there exists a quantum trajectory $\tilde{\rho}(t)$ with strictly
positive Wigner function $\mathcal{W}_{\hbar}[\tilde{\rho}(t)]$ such that

  1. a.

$\tilde{\rho}(t)$ cannot be distinguished from $\rho(t)$ with probability
greater than $\epsilon t$ by _any_ quantum measurement; and

  2. b.

$\mathcal{W}_{\hbar}[\tilde{\rho}(t)]$ cannot be distinguished from $f(t)$
with probability greater than $\epsilon t$ by _any_ classical variable.

Above, $C_{d}$ is the same universal dimensionless constant depending only on
$d$ from Theorem 3.1.

###### Proof.

We can consult the definitions of the anharmonic factors in Eqs. (3.15),
(3.18), and (3.19) to see that in the special case of linear Lindblad
operators ($L_{k}=\ell_{k,a}\alpha^{a}$) they reduce to just

$\displaystyle
B^{\mathrm{anh}}_{\mathrm{c}}[H^{\mathcal{Z}},\ell_{k,a}\alpha^{a}]$
$\displaystyle=\mathcal{Q}^{3,3}[H^{\mathcal{Z}}]$ (A.12) $\displaystyle
B^{\mathrm{anh}}_{\mathrm{q}}[H^{\mathcal{Z}},\ell_{k,a}\alpha^{a},\hbar]$
$\displaystyle=\mathcal{Q}^{3,2d+4}_{\hbar}[H^{\mathcal{Z}}]$ (A.13)
$\displaystyle
B^{\mathrm{anh}}_{\mathrm{q}^{\prime}}[\ell_{k,a}\alpha^{a},\hbar,\nu]$
$\displaystyle=0$ (A.14)

The dynamics are taken to be frictionless, so we will apply Theorem 3.1 with
Eq. (3.11) from Definition 3.2 for the relative diffusion strength. The key
quantity is

$\displaystyle\inf_{\alpha}\frac{\lambda_{\mathrm{min}}[\Delta^{\mathcal{Z}}(\alpha)]}{\lambda_{\mathrm{max}}[(\nabla^{2}H)^{\mathcal{Z}}(\alpha)]}$
$\displaystyle\geq\tau_{\mathrm{H}}\hbar^{-1}\lambda_{\mathrm{min}}[D^{\mathcal{Z}}(\alpha)]$
(A.15)
$\displaystyle\geq\tau_{\mathrm{H}}\hbar^{-1}\left(\frac{C_{d}}{\tau_{\mathrm{H}}\epsilon}\right)^{2/3}\left(\frac{\hbar}{\tilde{s}_{\mathrm{A}}}\right)^{4/3}\lambda_{\mathrm{min}}[D_{\mathrm{C}}^{\mathcal{Z}}]$
(A.16)
$\displaystyle\geq\left(\frac{C_{d}^{2}\hbar}{\epsilon^{2}\tau_{\mathrm{H}}^{2}\tilde{s}_{\mathrm{A}}}\right)^{1/3}$
(A.17) $\displaystyle\geq 1$ (A.18)

where we have sequentially applied the definitions
$\tau_{\mathrm{H}}=\mathcal{Q}^{2,2}[H^{\mathcal{Z}}]$ and $D=\hbar\Delta$,
the diffusion constraint (A.11), the fact that
$\lambda_{\mathrm{min}}[D_{\mathrm{C}}^{\mathcal{Z}}]=\lambda_{\mathrm{min}}[(\tilde{s}_{\mathrm{A}}/\tau_{\mathrm{H}})\mathds{1}]=\tilde{s}_{\mathrm{A}}/\tau_{\mathrm{H}}$,
and the error rate constraint (A.10). This means that the relative diffusion
strength $g=1$ so that, by Eqs. (3.20) and (3.21) of Theorem 3.1 we have that
$\left\lVert\tilde{\rho}(t)-\rho(t)\right\rVert_{\mathrm{Tr}}$ and
$\left\lVert\mathcal{W}_{\hbar}[\tilde{\rho}(t)]-f(t)\right\rVert_{L^{1}}$ are
both upper bounded by

$\displaystyle
C_{d}\,t\,\hbar^{\frac{1}{2}}\mathcal{Q}^{3,2+4d}_{\hbar}[H^{\mathcal{Z}}]\leq\epsilon\,t$
(A.19)

because
$\tilde{s}_{\mathrm{A}}=(\tau_{\mathrm{H}}\mathcal{Q}^{3,2+4d}_{s_{\mathrm{A}}}[H^{\mathcal{Z}}])^{-2}$
and $\mathcal{Q}^{q,r}_{\tilde{s}_{\mathrm{A}}}$ is an increasing function of
an action $\tilde{s}_{\mathrm{A}}\geq\hbar$. ∎

This corollary is more general than the result in our companion paper [45]
because here we do not restrict to Hamiltonians of the form $H=p^{2}/2m+V(x)$,
but it is weaker in that special case because it involves an unknown constant
$C_{d}$.

##  Appendix B Gaussian derivatives and integrals

###  B.1 Gaussian derivatives

The Gaussian probability distribution with mean $\alpha$ and covariance matrix
$\sigma$ is

$\displaystyle\tau_{\alpha,\sigma}(\alpha+\beta)=\frac{e^{-\beta^{\top}\sigma^{-1}\beta/2}}{(2\pi)^{d}\sqrt{\det\sigma}}=\frac{1}{(2\pi)^{d}\sqrt{\det\sigma}}\exp\left(-\frac{1}{2}\beta^{a}\sigma^{-1}_{ab}\beta^{b}\right)$
(B.1)

Let us consider this a real-valued function of any vector $\beta$ and any
invertible matrix $\sigma$, including non-symmetric ones, so that
$\sigma_{ab}$ and $\sigma_{ba}$ are independent variables for the purposes of
partial derivatives. However, at the end we will evaluate these derivatives on
the subspace where $\sigma$ is symmetric. Recalling our notation
$\partial_{c}=\partial/\partial\beta^{c}$ so
$\partial_{c}\beta^{a}=\delta_{c}^{\phantom{a}a}$, we have

$\displaystyle\partial_{d}\left(\beta^{a}\sigma^{-1}_{ab}\beta^{b}\right)\,$
$\displaystyle=\sigma^{-1}_{db}\beta^{b}+\beta^{a}\sigma^{-1}_{ad}$ (B.2)
$\displaystyle\partial_{c}\partial_{d}\left(\beta^{a}\sigma^{-1}_{ab}\beta^{b}\right)\,$
$\displaystyle=\sigma^{-1}_{dc}+\sigma^{-1}_{cd}.$ (B.3)

We also deploy the standard [104] matrix derivative identities646464Some
sources express will express this for a not-necessarily invertible matrix $Z$
using the matrix adjugate $\mathrm{adj}(Z)$. A property of the adjugate is
that $Z\,\mathrm{adj}(Z)=(\det Z)\mathds{1}$ so that, when $Z$ is invertible,
$\mathrm{adj}(Z)=(\det Z)Z^{-1}.$

$\displaystyle\frac{\partial\det Z}{\partial y}$ $\displaystyle=(\det
Z)\operatorname{Tr}\left[Z^{-1}\frac{\partial Z}{\partial y}\right],$ (B.4)
$\displaystyle\frac{\partial Z^{-1}}{\partial y}$
$\displaystyle=-Z^{-1}\frac{\partial Z}{\partial y}Z^{-1}$ (B.5)

for an invertible matrix $Z$, so in particular

$\displaystyle\frac{\partial\det Z}{\partial Z^{ab}}$ $\displaystyle=(\det
Z)Z^{-1}_{ba},$ (B.6) $\displaystyle\frac{\partial Z^{-1}_{cd}}{\partial
Z^{ab}}$ $\displaystyle=-Z^{-1}_{ca}Z^{-1}_{bd}.$ (B.7)

Combining these we get

$\displaystyle\partial_{a}\partial_{b}\tau_{\alpha,\sigma}(\alpha+\beta)=(\sigma^{-1}_{ac}\beta^{c}\sigma^{-1}_{bd}\beta^{d}-\sigma^{-1}_{ab})\tau_{\alpha,\sigma}(\alpha+\beta)=2\frac{\partial}{\partial\sigma^{ab}}\tau_{\alpha,\sigma}(\alpha+\beta),$
(B.8)

when evaluated for symmetric $\sigma$. (As expected, this is singular when
$\sigma$ is non-invertible.) Weyl quantizing both sides with
$\operatorname{Op}_{\hbar}=\mathcal{W}_{\hbar}^{-1}$ gives the corresponding
quantum expression
$\partial_{a}\partial_{b}\hat{\tau}_{\alpha,\sigma}=2\frac{\partial}{\partial\sigma_{ab}}\hat{\tau}_{\alpha,\sigma}$.

###  B.2 Gaussian integrals

Here we recall the evaluation of some Gaussian integrals, as can be done with
Wick’s theorem. We define the shorthand:

$\displaystyle\begin{split}\langle(\beta^{\top}A\beta)\rangle_{\sigma}:=&\int\mathrm{d}\beta\tau_{0,\sigma}(\beta)(\beta^{\top}A\beta)\\\
=&\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)(\beta^{\top}A\beta)\\\
=&A_{ab}\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)\beta^{a}\beta^{b}\\\
=&A_{ab}\sigma^{ab}\\\ =&\operatorname{Tr}[\sigma A].\end{split}$ (B.9)

for any positive semi-definite matrix $A$. ($\sigma$ is also positive, of
course.) Likewise, for $B$, $C$, and $D$ also positive semi-definite, we have

$\displaystyle\begin{split}\langle(\beta^{\top}A\beta)(\beta^{\top}B\beta)\rangle_{\sigma}:=&\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)(\beta^{\top}A\beta)(\beta^{\top}B\beta)\\\
=&A_{ab}B_{cd}\left[\sigma^{ab}\sigma^{cd}+2\sigma^{ad}\sigma^{bc}\right]\\\
=&\operatorname{Tr}[\sigma A]\operatorname{Tr}[\sigma
B]+2\operatorname{Tr}[\sigma A\sigma B]\end{split}$ (B.10)

and

$\displaystyle\begin{split}\langle(\beta^{\top}A\beta)(\beta^{\top}B\beta)(\beta^{\top}C\beta)\rangle_{\sigma}:=&\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)(\beta^{\top}A\beta)(\beta^{\top}B\beta)(\beta^{\top}C\beta)\\\
=&A_{ab}B_{cd}C_{ef}\Big{[}\sigma^{ab}\sigma^{cd}\sigma^{ef}+2\left(\sigma^{ab}\sigma^{cf}\sigma^{de}+\sigma^{af}\sigma^{cd}\sigma^{be}+\sigma^{ad}\sigma^{bc}\sigma^{ef}\right)\\\
&\qquad\qquad\qquad+4\left(\sigma^{ad}\sigma^{be}\sigma^{cf}+\sigma^{af}\sigma^{bc}\sigma^{de}\right)\Big{]}\\\
=&\operatorname{Tr}[\sigma A]\operatorname{Tr}[\sigma
B]\operatorname{Tr}[\sigma C]+2\operatorname{Tr}[\sigma
A]\operatorname{Tr}[\sigma B\sigma C]+2\operatorname{Tr}[\sigma
B]\operatorname{Tr}[\sigma C\sigma A]\\\ &\qquad+2\operatorname{Tr}[\sigma
C]\operatorname{Tr}[\sigma B\sigma A]+8\operatorname{Tr}[\sigma A\sigma
B\sigma C].\end{split}$ (B.11)

and

$\displaystyle\begin{split}\langle(\beta^{\top}A\beta)&(\beta^{\top}B\beta)(\beta^{\top}C\beta)(\beta^{\top}D\beta)\rangle_{\sigma}\\\
:=&\int\mathrm{d}\beta\tau_{\alpha,\sigma}(\alpha+\beta)(\beta^{\top}A\beta)(\beta^{\top}B\beta)(\beta^{\top}C\beta)(\beta^{\top}D\beta)\\\
=&A_{ab}B_{cd}C_{ef}D_{gh}\Big{[}\sigma^{ab}\sigma^{cd}\sigma^{ef}\sigma^{gh}\\\
&+2\big{(}\sigma^{ab}\sigma^{cd}\sigma^{eg}\sigma^{fh}+\sigma^{ab}\sigma^{ce}\sigma^{df}\sigma^{gh}+\sigma^{ac}\sigma^{bd}\sigma^{ef}\sigma^{gh}+\\\
&\qquad\sigma^{ab}\sigma^{cg}\sigma^{ef}\sigma^{dh}+\sigma^{ae}\sigma^{cd}\sigma^{bf}\sigma^{gh}+\sigma^{ag}\sigma^{cd}\sigma^{ef}\sigma^{bh}\big{)}\\\
&+4\big{(}\sigma^{ac}\sigma^{bd}\sigma^{eg}\sigma^{fh}+\sigma^{ae}\sigma^{cg}\sigma^{bf}\sigma^{dh}+\sigma^{ag}\sigma^{ce}\sigma^{df}\sigma^{bh}\big{)}\\\
&+8\big{(}\sigma^{ac}\sigma^{de}\sigma^{fb}\sigma^{gh}+\sigma^{ac}\sigma^{dg}\sigma^{ef}\sigma^{gb}+\sigma^{ae}\sigma^{cd}\sigma^{fg}\sigma^{hb}+\sigma^{ab}\sigma^{ce}\sigma^{fg}\sigma^{hd}\big{)}\\\
&+16\big{(}\sigma^{ac}\sigma^{de}\sigma^{fg}\sigma^{hb}+\sigma^{ae}\sigma^{fc}\sigma^{dg}\sigma^{hb}+\sigma^{ac}\sigma^{dg}\sigma^{he}\sigma^{fb}\big{)}\Big{]}\\\
=&\operatorname{Tr}[\sigma A]\operatorname{Tr}[\sigma
B]\operatorname{Tr}[\sigma C]\operatorname{Tr}[\sigma D]+\\\
&+2\big{(}\operatorname{Tr}[\sigma A]\operatorname{Tr}[\sigma
B]\operatorname{Tr}[\sigma C\sigma D]+\operatorname{Tr}[\sigma
A]\operatorname{Tr}[\sigma D]\operatorname{Tr}[\sigma B\sigma
C]+\operatorname{Tr}[\sigma C]\operatorname{Tr}[\sigma
D]\operatorname{Tr}[\sigma A\sigma B]+\\\ &\qquad\operatorname{Tr}[\sigma
A]\operatorname{Tr}[\sigma C]\operatorname{Tr}[\sigma B\sigma
D]+\operatorname{Tr}[\sigma B]\operatorname{Tr}[\sigma
D]\operatorname{Tr}[\sigma A\sigma C]+\operatorname{Tr}[\sigma
B]\operatorname{Tr}[\sigma C]\operatorname{Tr}[\sigma A\sigma D]\big{)}\\\
&+4\big{(}\operatorname{Tr}[\sigma A\sigma B]\operatorname{Tr}[\sigma C\sigma
D]+\operatorname{Tr}[\sigma A\sigma C]\operatorname{Tr}[\sigma B\sigma
D]+\operatorname{Tr}[\sigma A\sigma D]\operatorname{Tr}[\sigma B\sigma
C]\big{)}\\\ &+8\big{(}\operatorname{Tr}[\sigma A\sigma B\sigma
C]\operatorname{Tr}[\sigma D]+\operatorname{Tr}[\sigma A\sigma B\sigma
D]\operatorname{Tr}[\sigma C]+\operatorname{Tr}[\sigma A\sigma C\sigma
D]\operatorname{Tr}[\sigma B]+\operatorname{Tr}[\sigma B\sigma C\sigma
D]\operatorname{Tr}[\sigma A]\big{)}\\\ &+16\big{(}\operatorname{Tr}[\sigma
A\sigma B\sigma C\sigma D]+\operatorname{Tr}[\sigma A\sigma C\sigma B\sigma
D]+\operatorname{Tr}[\sigma A\sigma B\sigma D\sigma
C]\big{)}\Big{]}\end{split}$ (B.12)

## References

  * [1] Yurii Vladimirovich Egorov.  The canonical transformations of pseudodifferential operators.  Uspekhi Matematicheskikh Nauk, 24(5):235–236, 1969.