 the lowering operator~$A$ and the raising operator~$A^\dagger$ are (formally) adjoint to each other and its components fulfill the canonical communication relations,
\[
[A_k,A_\ell^\dagger] = [\partial_k,x_\ell] = \delta_{k,\ell}\quad\text{for all}\ k,\ell=1,\ldots,d.
\]
Moreover, $A$ annihilates the Gaussian in the sense that, $A\varphi_0 = 0$.
Following the Bargman formalism \cite{Barg}, we use the raising operator $A^\dagger$ to translate the
Gaussian in phase space. We denote
\[
z\cdot w = z_1w_1+ \cdots + z_dw_d
\]
for any complex vectors  $z,w\in\C^d$, but also for the vector of raising operators such that
$z\cdot A^\dagger = z_1A^\dagger_1 + \cdots + z_dA^\dagger_d$.

\begin{lemma}[Gaussian wave-packet]\label{upro} Let $\delta>0$. For $z=q+ip\in \mathbb{C}^d$ with $q,p\in\R^d$ we define
\[
u(x)=\exp\! \left(-|z|^2/2+z\cdot A^\dagger\right)\varphi_0(x).
\]
Then,
\[
u(x)=(2\pi\delta^2)^{-d/4}\exp\!{\Big(-\frac{1}{4\delta^2}|x-2\delta q|^2+\frac{i}{\delta} p \cdot (x-\delta q)\Big)}.
\]
In particular,
\begin{align}
&\|u\|=1,\quad \lag u\,|\,\widehat q u\rag=2\delta q,\quad \lag u\,|\,\widehat p u\rag=\hbar/\delta\, p,\\*[1ex]
&\lag x_mu\,|\,x_nu\rag=\delta^2(\delta_{mn}+4q_m q_n),\quad
\|xu\|^2=\delta^2(d+4|q|^2),\label{upro1}\\*[1ex]
& Au=zu,\,\,\,A^\dagger u=(x/\delta-z)u.\label{upro2}
\end{align}
\end{lemma}

\begin{proof} We view the function $u$ as the solution at time $t=1$ of the Cauchy problem
\[
\partial_t\Psi = \Big(-\frac12|z|^2 + z\cdot A^\dagger\Big)\Psi,\qquad \Psi(0) = \varphi_0.
\]
We have
\[
\Psi(t,x) = \exp\left(-\frac{t}{2}\,|z|^2-\frac{t^2}{4}\, z\cdot z+\frac{t}{2\delta} \,z\cdot x\right) \varphi_0(x-t \delta z),
\]
as can be verified by direct calculation, that is,
\[
\partial_t\Psi
= \Big( -\frac12|z|^2-\frac{t}{2} z\cdot z + \frac{1}{2\delta}z\cdot x -\delta z\cdot\nabla_x + \frac{t}{2}z\cdot z\Big)\Psi= \Big( -\frac12|z|^2 + z\cdot A^\dagger\Big)\Psi.\]
Therefore,
\[
u(x) = \exp\left(-\frac{1}{2}\,|z|^2-\frac{1}{4}\, z\cdot z+\frac{1}{2\delta} \,z\cdot x\right) \varphi_0(x-\delta z).
\]
Since the quadratic form in this exponential function can be rewritten as
\[
-\frac12|z|^2-\frac14 z\cdot z + \frac{1}{2\delta}z\cdot x
- \frac{1}{4\delta^2}(x-\delta z)\cdot(x-\delta z)= -\frac{1}{4\delta^2}|x-2\delta q|^2 + \frac{i}{\delta}p\cdot (x-\delta q),
\]
thn $u$ indeed has the claimed form. For the position expectation we get
\begin{align*}
\lag u\,|\widehat{q}u\rag&=(2\pi\delta^2)^{-\frac{d}{2}}\int_{\mathbb{R}^d}x\exp\left(
-\frac{1}{2\delta^2}|x-2\delta q|^2\right)\D x\\
&=2\delta q \int_{\mathbb{R}^d} |\varphi_0(x)|^2 \,\D x =2\delta q.\end{align*}
Since the gradient satisfies
\[
\nabla u(x)=u(x)\left(-\frac{x-2\delta q}{2\delta^2}+\frac{ip}{\delta}\right)=u(x)\left(-\frac{1}{2\delta^2}x+\frac{1}{\delta}z\right),
\]
we calculate the position expectation as
\begin{align*}
\lag u\,|\,\widehat p u\rag &= -i\hbar\lag u\,|\,\nabla u\rag
= -i\hbar\left(-\frac{1}{2\delta^2}\lag u\,|\,xu\rag + \frac{1}{\delta}z\lag u\,|\,u\rag\right)\\
&=\frac{-i\hbar}{\delta}\left(-q + z\right) = \frac{\hbar}{\delta}\,p.
\end{align*}
For the second moments we have
\begin{align*}
\lag x_mu\,|\,x_nu\rag&=(2\pi\delta^2)^{-\frac{d}{2}}\int_{\mathbb{R}^d}x_m x_n\exp\left(-\frac{1}{2\delta^2}|x-2\delta q|^2\right)\D x\\
&=\int_{\mathbb{R}^d}(x_m+2\delta q_m)(x_n+2\delta q_n)\,|\varphi_0(x)|^2\,\D x\\
&=4\delta^2q_m q_n+\int_{\mathbb{R}^d} x_mx_n|\varphi_0(x)|^2\, \D x\\*[1ex]
&=4\delta^2q_m q_n+\delta^2\delta_{mn}.
\end{align*}
In particular,
\[
\|xu\|^2=\sum\limits_{n=1}^d\lag x_n u\,|\, x_nu\rag=
\delta^2\sum\limits_{n=1}^d(1+4q_n^2)=\delta^2(d+4|q|^2).
\]
which completes the proof for \eqref{upro1}. Using the formula for $\nabla u$
once more, we obtain
\[
Au=\frac{1}{2\delta}xu+\delta\nabla u=zu,\quad
A^\dagger u=\frac{1}{2\delta}xu-\delta\nabla u=\big(\frac{x}{\delta}-z\big)u,
\]
which concludes \eqref{upro2}.
\end{proof}

Now, for a fixed width parameter $\delta>0$, we consider the variational \emph{frozen} Gaussian approximation on the manifold
\begin{align*}
\mathcal{M}&=\left\{u(\theta,q,p)\mid  \theta\in\mathbb{R},\ p, q\in\mathbb{R}^d\right\}\subset L^2(\R^d),
\end{align*}
where $u(\theta,q,p)$ denotes the wave-packet
\be\label{eq:frozen}
u(\theta, q, p)(x) =
(2\pi\delta^2)^{-\frac{d}{4}}\exp\!{\left(i\theta-\frac{1}{4\delta^2}|x-2\delta q|^2+\frac{i}{\delta} p \cdot (x-\delta q)\right)}.
\ee
Note that $\|u(\theta,q,p)\| = 1$ for all choices of the parameters $\theta\in\R$, $q,p\in\R^d$. The tangent spaces for this manifold can easily be calculated as follows:
\begin{lemma}[Tangent space]\label{lem:tang}
For $\forall u\in \calM$, the tangent