# The Spectral Gap of a Gaussian Quantum Markovian Generator

F. Fagnola(1), D. Poletti(2), E. Sasso(2), V. Umanità(2)

###### Abstract

Gaussian quantum Markov semigroups are the natural non-commutative extension
of classical Ornstein-Uhlenbeck semigroups. They arise in open quantum systems
of bosons where canonical non-commuting random variables of positions and
momenta come into play. If there exits a faithful invariant density we
explicitly compute the optimal exponential convergence rate, namely the
spectral gap of the generator, in non-commutative $L^{2}$ spaces determined by
the invariant density showing that the exact value is the lowest eigenvalue of
a certain matrix determined by the diffusion and drift matrices. The spectral
gap turns out to depend on the non-commutative $L^{2}$ space considered,
whether the one determined by the so-called GNS or KMS multiplication by the
square root of the invariant density. In the first case, it is strictly
positive if and only if there is the maximum number of linearly independent
noises. While, we exhibit explicit examples in which it is strictly positive
only with KMS multiplication. We do not assume any symmetry or quantum
detailed balance condition with respect to the invariant density.

Keywords: Open quantum systems, Gaussian Markov Semigroup, Spectral Gap,
quantum Ornstein-Uhlenbeck process.

##  1 Introduction

Classical Ornstein-Uhlenbeck semigroups $(T_{t})_{t\geq 0}$ on bounded
measurable functions on $\mathbb{R}^{d}$ have the following explicit formula

$(T_{t}f)(x)=\frac{1}{\sqrt{(2\pi)^{d}\operatorname{det}(\Sigma_{t})}}\int_{\mathbb{R}^{d}}f\left(\mathrm{e}^{tZ^{*}}x-y\right)\exp\left(-\langle
y,\Sigma_{t}^{-1}y\rangle/2\right)\mathrm{d}y$

where $Z$ is the drift matrix and $\Sigma_{t}$ the covariance matrix at time
$t$ given by
$\Sigma_{t}=\int_{0}^{t}\mathrm{e}^{sZ^{*}}C\mathrm{e}^{sZ}\mathrm{d}s$ in
terms of the symmetric matrix $C$ of diffusion coefficients. Here, for
simplicity, we assume that $\Sigma_{t}$ is invertible and that the mean of the
Markov process is zero. If we consider the function
$x\mapsto\exp(\mathrm{i}\langle z,x\rangle)$ the above formula becomes

$T_{t}\exp(\mathrm{i}\langle
z,\cdot\rangle)=\exp\left(-\frac{1}{2}\int_{0}^{t}\left\langle\mathrm{e}^{sZ}z,C\mathrm{e}^{sZ}z\right\rangle\mathrm{d}s\right)\exp(\mathrm{i}\langle\mathrm{e}^{tZ}z,\cdot\rangle)\,.$

In applications to quantum theory one has to deal with random variables that
do not commute, typically positions and momenta, vector spaces are complex and
symplectic structures come into play, but a similar formula holds (see [15,
33] and the references therein). More precisely, for a system of $d$ bosons,
the natural variables are $d$ position $q_{1},\dots,q_{d}$ and $d$ momentum
$p_{1}=-\mathrm{i}\mathrm{d}/\mathrm{d}q_{1},\dots,p_{d}=-\mathrm{i}\mathrm{d}/\mathrm{d}q_{d}$
operators. Exponential functions are replaced by unitary Weyl operators
$W(x+\mathrm{i}y)=\exp\left(-\mathrm{i}\sqrt{2}\sum_{k=1}^{d}(x_{k}p_{k}-y_{k}q_{k})\right)$
($x,y\in\mathbb{R}^{d}$) and one finds the explicit formula (9), Section 2
(with the further inclusion of a non-zero mean vector
$\zeta\in\mathbb{C}^{d}$). There, $z=x+\mathrm{i}y$, $Z,C$ are _real linear_
operators on $\mathbb{C}^{d}$ that can be viewed as complex linear operators
on $\mathbb{C}^{2d}$ as explained in Section 2, Remark 5. In this way, one can
define a Gaussian quantum Markov semigroup (QMS), namely a weakly∗-continuous
semigroups of completely positive, identity preserving maps on the von Neumann
algebra of all bounded operators a Fock space (see Section 2 for precise
definitions).

QMSs are essential tools in the mathematical modeling of open quantum systems
(see [8, 27]). Starting from the seminal papers by Gorini, Kossakowski,
Lindblad and Sudharshan (GKLS) in the seventies, they have been used in the
physical literature ([15, 33]) and are now established as non-commutative
extensions of classical Markov semigroups (see [11, 14, 22, 34, 35]. Gaussian
QMSs, that are considered in this article, are a notable class of semigroups
for many reasons: they include many semigroups that are interesting for
physical applications (see [8, 21, 32] and the references therein), they are
the natural non-commutative extension of classical Ornstein-Uhlenbeck
semigroups (see Section 2 and [1, 2, 10, 11, 12, 13, 14, 32]), they arise from
quantum master equations and Fokker-Planck equations [4, 5] and they provide a
well-behaved and workable class of semigroups acting on the algebra
$\mathcal{B}(\mathsf{h})$ of all bounded operators on a Hilbert space
$\mathsf{h}$. Furthermore, they have countless applications to open quantum
systems of bosons, but rather little is known about their properties,
especially if one thinks of the wealth of results established in the classical
case (see [25, 26] and the references therein).

Several properties of gaussian QMS have been recently established: the
characterization as the unique class of quantum Markov semigroups preserving
quantum gaussian states [30], the structure of the subalgebra where they act
as ∗-homomorphisms [1], the characterization of irreducibility (under a
regularity assumption [17]) and, in the symmetric case, gradient flow
properties and entropy inequalities [11], and the expression of the generator
as a sum of squares of derivations [34].

The main result of this paper is the explicit calculation of the spectral gap
of a gaussian QMS as the lowest eigenvalue of a certain matrix determined by
the diffusion and drift parameters (Theorem 18). As for the classical
Ornstein-Uhlenbeck process, assuming that the matrix $Z$ is stable, namely all
eigenvalues have strictly negative real part, every initial state converges to
the invariant one [18]. It is therefore natural to study the speed of
convergence towards that invariant state in view of applications to problems
such as return to equilibrium and limit theorems and dynamical phase
transitions (see [23]).

A natural approach, motivated by the analogue in classical Markov semigroups
theory, is to embed the set of bounded operators in the Hilbert space of
Hilbert-Schmidt operators, $\mathfrak{I}_{2}(\mathsf{h})$, by multiplying them
by the square root of the invariant density. In this way, one obtains a
contraction semigroup (see [9]) and one can analyse its generator by spectral
analytic tools. However, due to the non-commutative nature of operators,
multiplication by the invariant density can be done in multiple ways,
obtaining different embeddings. In this paper, first of all, we work with GNS
embedding

$i_{2}:\mathcal{B}(\mathsf{h})\hookrightarrow\mathfrak{I}_{2}(\mathsf{h}),\quad\
i_{2}(x)=x\rho^{1/2}$

We also consider the KMS embedding (see [9])

$i_{1/2,2}:\mathcal{B}(\mathsf{h})\hookrightarrow\mathfrak{I}_{2}(\mathsf{h}),\quad\
i_{1/2,2}(x)=\rho^{1/4}x\rho^{1/4}$

and show that, in this case we get a different result (Theorem 26) for the
explicit expression of the spectral gap. In particular, the spectral gap could
be strictly positive for the KMS embedding and zero for the GNS embedding (see
the example in subsection 7.1). This is a typical feature of non-
commutativity. Our results are first presented for the $i_{2}$ embedding
because computations are somewhat more direct from the explicit formula of the
characteristic function of a quantum gaussian state.

Once chosen the GNS embedding, since $i_{2}$ has dense range, the next step is
to extend $\mathcal{T}$ to a strongly continuous contraction semigroup
$(T_{t})_{t\geq 0}$ on $\mathfrak{I}_{2}(\mathsf{h})$ by setting

$T_{t}(x\rho^{1/2})=\mathcal{T}_{t}(x)\rho^{1/2}\qquad\forall\,x\in\mathcal{B}(\mathsf{h}),\
t\geq 0.$

Clearly $\rho^{1/2}$ is an invariant vector for $(T_{t})_{t\geq 0}$. Denoting
by $\langle x,y\rangle_{2}=\hbox{\rm tr}(x^{*}y)$ the Hilbert-Schmidt scalar
product and by $\|\cdot\|_{2}=\langle x,x\rangle_{2}^{1/2}$ the norm, we
define the _spectral gap_ of this semigroup in the following way

######  Definition 1.

If there exists $g>0$ such that

$\left\|T_{t}(x)-\langle\rho^{1/2},x\rangle_{2}\,\rho^{1/2}\right\|_{2}\leq\mathrm{e}^{-gt}\left\|x-\langle\rho^{1/2},x\rangle_{2}\,\rho^{1/2}\right\|_{2}\quad\forall\,x\in\mathfrak{I}_{2}(\mathsf{h}),\,\forall\,t>0$
(1)

we say the semigroup $T$ has a spectral gap. In that case, we call the
_spectral gap_ of the semigroup the biggest constant $g>0$ that satisfies (1).

The spectral gap for the KMS embedding is defined in a similar way.

Note that, as usual in the theory of Markov processes and with a little abuse
of terminology, the name spectral gap for the above constant $g$, should be
more precisely referred to as the spectrum of the symmetric part of the
generator of $T$. In this case, indeed, it is the distance from $0$ of the
remaining part of the spectrum.

The spectral gap $g$ is explicitly computed for both the above embeddings
(Theorem 18, and Theorem 26). In the first case, it turns out to be strictly
positive under the assumption that there is noise in all the non-commutative
coordinates. In the second case, this condition in no more necessary. In
particular, we give explicit examples in which it strictly positive only for
the KMS embedding in subsection 7.1.

It is important to note that, due to the nature of gaussian QMSs, the
computation of the spectral gap, as other problems (see [1, 2, 3]), can be
reduced to linear algebraic one. However, it is worth noticing that the final
result is not a straightforward generalization of the classical one because
one can find classical Ornstein-Uhlenbeck subprocesses with strictly positive
spectral gap of a quantum Ornstein-Uhlenbeck process with $g=0$ (see the
example at the end of Section 6).

The paper is organized as follows. Section 2 is devoted to the formal
introduction of all the necessary mathematical preliminaries, including the
definition of gaussian QMSs. In Section 3 we discuss the relationship between
gaussian QMSs and classical Ornstein-Uhlenbeck semigroups, showing how the
latter emerge from suitable restrictions of the former. Section 4 contains the
main result on the spectral gap for the GNS embedding. Section 5 contains some
further insight on the sufficient conditions used for the spectral gap,
showing that they are not very restrictive when looking for a positive
spectral gap. In Section 6 we present a class of examples with $d=1$ depending
on some parameters. The study of the spectral gap for the KMS embedding is
undertaken in Section 7. More technical material and some computations are
collected in Appendixes A, B and C.

##  2 Gaussian Quantum Markov Semigroups

In this section we introduce gaussian QMSs starting from their generators and
fix some notation.

Let $\mathsf{h}$ be the Fock space $\mathsf{h}=\Gamma(\mathbb{C}^{d})$ which
is isometrically isomorphic to
$\Gamma(\mathbb{C})\otimes\cdots\otimes\Gamma(\mathbb{C})$ with canonical
orthonormal basis $(e(n_{1},\ldots,n_{d}))_{n_{1},\ldots,n_{d}\geq 0}$ (with
$e(n_{1},\ldots,n_{d})$ the symmetrized version of
$e_{n_{1}}\otimes\ldots\otimes e_{n_{d}}$). Let $a_{j},a_{j}^{\dagger}$ be the
creation and annihilation operators of the Fock representation of the
$d$-dimensional Canonical Commutation Relations (CCR)

$\displaystyle a_{j}\,e(n_{1},\ldots,n_{d})$ $\displaystyle=$
$\displaystyle\sqrt{n_{j}}\ e(n_{1},\ldots,n_{j-1},n_{j}-1,\ldots,n_{d}),$
$\displaystyle a_{j}^{\dagger}\,e(n_{1},\ldots,n_{d})$ $\displaystyle=$
$\displaystyle\sqrt{n_{j}+1}\ e(n_{1},\ldots,n_{j-1},n_{j}+1,\ldots,n_{d}),$

The CCRs are written as $[a_{j},a_{k}^{\dagger}]=\delta_{jk}\mathbbm{1}$,
where $[\cdot,\cdot]$ denotes the commutator, or, more precisely,
$[a_{j},a_{k}^{\dagger}]\subseteq\delta_{jk}\mathbbm{1}$ because the domain of
the operator in the left-hand side is smaller.

Linear combinations of both creation and annihilation operators are denoted as
follows:

$a(v)=\sum\limits_{j=1}^{d}\overline{v}_{j}a_{j},\quad
a^{\dagger}(u)=\sum\limits_{j=1}^{d}u_{j}a_{j}^{\dagger}$

for all
$u^{T}=[u_{1},\ldots,u_{d}],v^{T}=[v_{1},\ldots,v_{d}]\in\mathbb{C}^{d}$.

The above operators are obviously defined on the linear manifold $D$ spanned
by the elements $(e(n_{1},\ldots,n_{d}))_{n_{1},\ldots,n_{d}\geq 0}$ of the
canonical orthonormal basis of $\mathsf{h}$ that turns out to be an essential
domain for all the operators considered so far. This also happens for field
operators

$q(u)=\left(a(u)+a^{\dagger}(u)\right)/\sqrt{2}\qquad u\in\mathbb{C}^{d}$ (2)

that are symmetric and essentially self-adjoint on the domain $D$ by Nelson’s
theorem on analytic vectors ([31] Th. X.39 p. 202). If the vector $u$ has real
(resp. purely imaginary) components one finds position (resp. momentum)
operators and the commutation relation
$[q(u),q(v)]\subseteq\mathrm{i}\Im\langle u,v\rangle\hbox{\rm 1\kern-2.8ptl}$
(where $\Im$ and $\Re$ denote the imaginary and real part of complex number
and vectors). Momentum operators, i.e. fields $q(\mathrm{i}r)$ with
$r\in\mathbb{R}^{d}$ are also denoted by $p(r)=\sum_{1\leq j\leq d}r_{j}p_{j}$
where $p_{j}=\mathrm{i}(a^{\dagger}_{j}-a_{j})/\sqrt{2}$. In a similar way we
write $q(r)=\sum_{1\leq j\leq d}r_{j}q_{j}$ with
$q_{j}=q(e_{j})=(a^{\dagger}_{j}+a_{j})/\sqrt{2}$.

Another set of operators that will play an important role in this paper are
the Weyl operators $W(z),z\in\mathbb{C}^{d}$, which generate the entirety of
$\mathcal{B}(\mathsf{h})$. They satisfy the CCR in the exponential form,
namely, for every $z,z^{\prime}\in\mathbb{C}^{d}$,

$W(z)W(z^{\prime})=\hbox{\rm e}^{-\mathrm{i}\Im\left\langle
z,z^{\prime}\right\rangle}W(z+z^{\prime}).$ (3)

It is well-known that $W(z)$ is the exponential of the anti self-adjoint
operator $-\mathrm{i}\sqrt{2}\,q(\mathrm{i}z)$

$W(z)=\mathrm{e}^{-\mathrm{i}\sqrt{2}\,q(\mathrm{i}z)}=\mathrm{e}^{za^{\dagger}-\overline{z}a}.$
(4)

A QMS $\mathcal{T}=(\mathcal{T}_{t})_{t\geq 0}$ is a weakly∗-continuous
semigroup of completely positive, identity preserving, weakly∗-continuous maps
on $\mathcal{B}(\mathsf{h})$. The predual semigroup
$\mathcal{T}_{*}=(\mathcal{T}_{*t})_{t\geq 0}$ on the predual space of trace
class operators on $\mathsf{h}$ is a strongly continuous contraction
semigroup.

Gaussian QMSs can be defined either through their explicit action on Weyl
operators or through their generator, as we will do. Let $L_{\ell},H$ be the
operators on $\mathsf{h}$ defined on the domain $D$ by

$\displaystyle H$
$\displaystyle=\sum_{j,k=1}^{d}\left(\Omega_{jk}a_{j}^{\dagger}a_{k}+\frac{\kappa_{jk}}{2}a_{j}^{\dagger}a_{k}^{\dagger}+\frac{\overline{\kappa_{jk}}}{2}a_{j}a_{k}\right)+\sum_{j=1}^{d}\left(\frac{\zeta_{j}}{2}a_{j}^{\dagger}+\frac{\overline{\zeta_{j}}}{2}a_{j}\right),$
(5) $\displaystyle L_{\ell}$
$\displaystyle=\sum_{k=1}^{d}\left(\overline{v_{\ell k}}a_{k}+u_{\ell
k}a_{k}^{\dagger}\right)$ (6)

where $1\leq m\leq 2d$, $\Omega:=(\Omega_{jk})_{1\leq j,k\leq d}=\Omega^{*}$
and $\kappa:=(\kappa_{jk})_{1\leq j,k\leq d}=\kappa^{T}\in M_{d}(\mathbb{C})$,
are $d\times d$ complex matrices with $\Omega$ Hermitian and $\kappa$
symmetric, $V=(v_{\ell k})_{1\leq\ell\leq m,1\leq k\leq d},U=(u_{\ell
k})_{1\leq\ell\leq m,1\leq k\leq d}\in M_{m\times d}(\mathbb{C})$ are $m\times
d$ matrices and $\zeta=(\zeta_{j})_{1\leq j\leq d}\in\mathbb{C}^{d}$. We
assume also $\ker(V^{*})\cap\ker(U^{T})=\\{0\\}$ so that the operators
$L_{\ell}$, called Kraus’ operators, are linearly independent (see [1],
Proposition 2.2).The above operators are closable and we will denote their
closure by the same symbol.

For all $x\in\mathcal{B}(\mathsf{h})$ consider the quadratic form with domain
$D\times D$

$\begin{split}\texttt{{\pounds}}(x)[\xi^{\prime},\xi]&=\mathrm{i}\left\langle
H\xi^{\prime},x\xi\right\rangle-\mathrm{i}\left\langle\xi^{\prime},xH\xi\right\rangle\\\
&-\frac{1}{2}\sum_{\ell=1}^{m}\left(\left\langle\xi^{\prime},xL_{\ell}^{*}L_{\ell}\xi\right\rangle-2\left\langle
L_{\ell}\xi^{\prime},xL_{\ell}\xi\right\rangle+\left\langle
L_{\ell}^{*}L_{\ell}\xi^{\prime},x\xi\right\rangle\right)\end{split}$ (7)

This is a natural way to make sense of a Gorini, Kossakowski, Lindblad-
Sudarshan (GKLS) representation of the generator (see [30] Theorems 5.1, 5.2
and also [15, 33]) in a generalized form since operators $L_{\ell},H$ are
unbounded. Indeed, in other models, when these operators are bounded, one
writes the generator as

$\mathcal{L}(x)=\mathrm{i}\left[H,x\right]-\frac{1}{2}\sum_{\ell=1}^{m}\left(L_{\ell}^{*}L_{\ell}\,x-2L_{\ell}^{*}xL_{\ell}+x\,L_{\ell}^{*}L_{\ell}\right),\qquad
x\in{\mathcal{B}(\mathsf{h})}.$ (8)

Gaussian QMSs are then defined by the following Theorem, whose proof can be
found in [2], Appendix A.

######  Theorem 2.

There exists a unique QMS, $\mathcal{T}=(\mathcal{T}_{t})_{t\geq 0}$ such
that, for all $x\in\mathcal{B}(\mathsf{h})$ and $\xi,\xi^{\prime}\in D$, the
function $t\mapsto\left\langle\xi^{\prime},\mathcal{T}_{t}(x)\xi\right\rangle$
is differentiable and

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\left\langle\xi^{\prime},\mathcal{T}_{t}(x)\xi\right\rangle=\texttt{{\pounds}}(\mathcal{T}_{t}(x))[\xi^{\prime},\xi]\qquad\forall\,t\geq
0.$

One can derive as in [2], Theorem 2.4, the following formula for the action of
a gaussian semigroup on Weyl operators.

######  Theorem 3.

Let $(\mathcal{T}_{t})_{t\geq 0}$ be the quantum Markov semigroup with
generalized GKLS generator associated with $H,L_{\ell}$ as above. For all Weyl
operator $W(z)$ we have

$\mathcal{T}_{t}(W(z))=\exp\left(-\frac{1}{2}\int_{0}^{t}\Re\left\langle\mathrm{e}^{sZ}z,C\mathrm{e}^{sZ}z\right\rangle\mathrm{d}s+\mathrm{i}\int_{0}^{t}\Re\left\langle\zeta,\mathrm{e}^{sZ}z\right\rangle\mathrm{d}s\right)W\left(\mathrm{e}^{tZ}z\right)$
(9)

where the _real linear_ operators $Z,C$ on $\mathbb{C}^{d}$ are

$\displaystyle Zz$
$\displaystyle=\left[\left(U^{T}\,\overline{U}-V^{T}\,\overline{V}\right)/2+\mathrm{i}\Omega\right]z+\left[\left(U^{T}V-V^{T}U\right)/2+\mathrm{i}\kappa\right]\overline{z},$
(10) $\displaystyle Cz$
$\displaystyle=\left(U^{T}\,\overline{U}+V^{T}\,\overline{V}\right)z+\left(U^{T}V+V^{T}U\right)\overline{z}.$
(11)

Note that the operators $Z,C$ that appear in the previous Theorem are only
real linear since they inherit the real vector space structure of
$\mathbb{C}^{d}$ as argument of the Weyl operators.

For $d=1$ special cases of gaussian QMSs have been considered as quantum
analogues of classical Ornstein-Uhlenbeck semigroups (see [10, 14]). The
multidimensional case has also been extensively studied [11, 34, 35] under
symmetry or detailed balance conditions.

The natural duality between $\mathcal{B}(\mathsf{h})$ and trace class
operators on $\mathsf{h}$ given by $(\rho,x)\mapsto\hbox{\rm tr}(\rho\,x)$,
together with the weak∗ continuity of the QMS $(\mathcal{T}_{t})_{t\geq 0}$
allows one to define the pre-adjoint QMS $(\mathcal{T}_{*t})_{t\geq 0}$ on
trace class operators on $\mathsf{h}$. Operators $\mathcal{T}_{*t}$ act on
positive operators with unit trace, called density matrices and representing
quantum states, that play the role of classical probability densities on
$\mathbb{R}^{2d}$. A state is called _faithful_ if $\rho\,\xi=0$ implies
$\xi=0$; this is the analogue of a classical density with full support.

One of the defining properties of gaussian QMSs is that operators
$\mathcal{T}_{*t}$ preserve the set of quantum gaussian states. A quantum
state is gaussian if its quantum characteristic function $\widehat{\rho}\,$
has a similar expression to the one of classic gaussian multivariate random
variables

$\widehat{\rho}(z):=\hbox{\rm tr}(\rho
W(z))=\exp\left(-\mathrm{i}\Re\left\langle\mu,z\right\rangle-\frac{1}{2}\Re\left\langle
z,Sz\right\rangle\right)\qquad\forall\,z\in\mathbb{C}^{d}$ (12)

for some $\mu\in\mathsf{h}$ and real linear, bounded, positive and invertible
operator $S$ on $\mathsf{h}$ (we write
$S\in\mathcal{B}_{\mathbb{R}}(\mathsf{h})$). We denote this state as
$\rho=\rho_{{(\mu,S)}}$. One can prove as in [30] Theorem 5.1 that

######  Proposition 4.

If $\mathcal{T}$ is a gaussian QMS, then
$\mathcal{T}_{*t}(\rho_{(\mu,S)})=\rho_{(\mu_{t},S_{t})}$ with

$\mu_{t}=\hbox{\rm e}^{tZ^{\sharp}}\mu-\int_{0}^{t}\hbox{\rm
e}^{sZ^{\sharp}}\zeta\hbox{\rm d}s,\quad S_{t}=\hbox{\rm
e}^{tZ^{\sharp}}S\hbox{\rm e}^{tZ}+\int_{0}^{t}\hbox{\rm
e}^{sZ^{\sharp}}C\hbox{\rm e}^{sZ}\hbox{\rm d}s$ (13)

where $Z^{\sharp}$ denotes the adjoint of $Z$ with respect to the scalar
product $\Re\langle\cdot,\cdot\rangle$.

The adjoint $Z^{\sharp}$ of $Z$ with respect to the scalar product
$\Re\langle\cdot,\cdot\rangle$ is explicitly given by
$Z^{\sharp}z=Z_{1}^{*}z+Z_{2}^{T}\overline{z}$ where $Z_{1}$ and $Z_{2}$ are
the operators in (10) acting on $z$ and $\overline{z}$. Here, however, further
clarifications and remarks on properties of real linear operators that are
useful to better explain the computations performed in the following sections
are in order.

######  Remark 5.

In general a real linear operator $A$ on $\mathbb{C}^{d}$ can be written as

$Az=A_{1}z+A_{2}\overline{z},\quad\forall z\in\mathbb{C}^{d},$

where now $A_{1},A_{2}$ are complex linear operators. In order to better
understand the action of such operators we can exploit the isomorphism of
$(\mathbb{C}^{d},\Re\left\langle\cdot,\cdot\right\rangle)$ as a real vector
space with $(\mathbb{R}^{2d},\left\langle\cdot,\cdot\right\rangle)$, through
the decomposition of a complex vector $z=x+\mathrm{i}y$ in its real and
imaginary part. On this new vector space $A$ takes the form

$A_{2d}=\begin{bmatrix}\Re A_{1}+\Re A_{2}&\Im A_{2}-\Im A_{1}\\\ \Im
A_{1}+\Im A_{2}&\Re A_{1}-\Re A_{2}\end{bmatrix}$

and $A_{2d}$ is a _real linear_ operator on $\mathbb{R}^{2d}$. This
isomorphism will also be useful when considering the complexification of the
operator $A$. Indeed it is much simpler to work with the complexification
$\mathbf{A}$ of $A_{2d}$

$\mathbf{A}=\begin{bmatrix}\Re A_{1}+\Re A_{2}&\Im A_{2}-\Im A_{1}\\\ \Im
A_{1}+\Im A_{2}&\Re A_{1}-\Re A_{2}\end{bmatrix},$

now acting on $(\mathbb{C}^{2d},\left\langle\cdot,\cdot\right\rangle)$, than
it is to consider the complexification of $A$. For this reason whenever the
complexification of a real linear operator is needed we will think of it as
acting on $(\mathbb{C}^{2d},\left\langle\cdot,\cdot\right\rangle)$. We also
have

$\Re\left\langle z,Aw\right\rangle=\left\langle\begin{bmatrix}\Re z\\\ \Im
z\end{bmatrix},A_{2d}\begin{bmatrix}\Re w\\\ \Im
w\end{bmatrix}\right\rangle=\left\langle\begin{bmatrix}\Re z\\\ \Im
z\end{bmatrix},\mathbf{A}\begin{bmatrix}\Re w\\\ \Im
w\end{bmatrix}\right\rangle\quad\forall\,z,w\in\mathbb{C}^{d},$ (14)

where the scalar products are the canonical ones on
$\mathbb{C}^{d},\mathbb{R}^{2d},\mathbb{C}^{2d}$ respectively. The adjoint
operations are denoted $A^{\sharp},A_{2d}^{T},\mathbf{A}^{*}$, respectively.

A notable example of real linear operator is $Jz=-\mathrm{i}z$ to which is
associated

$\mathbf{J}=\begin{bmatrix}0&\hbox{\rm 1\kern-2.8ptl}\,\\\ -\hbox{\rm
1\kern-2.8ptl}&0\end{bmatrix}\in M_{2d}(\mathbb{C}).$

It is easy to prove that the following equality holds

$\Re\left\langle z,Aw\right\rangle+\mathrm{i}\Im\left\langle
z,w\right\rangle=\left\langle\begin{bmatrix}\Re z\\\ \Im
z\end{bmatrix},(\mathbf{A}+\mathrm{i}\mathbf{J})\begin{bmatrix}\Re w\\\ \Im
w\end{bmatrix}\right\rangle\qquad\forall\,z,w\in\mathbb{C}^{d},$ (15)

and

$\left\langle\begin{bmatrix}z_{1}\\\
z_{2}\end{bmatrix},\mathbf{A}\begin{bmatrix}w_{1}\\\
w_{2}\end{bmatrix}\right\rangle=\Re\left\langle\zeta_{1},A\xi_{1}\right\rangle+\Re\left\langle\zeta_{2},A\xi_{2}\right\rangle+\mathrm{i}\Re\left\langle\zeta_{1},A\xi_{2}\right\rangle-\mathrm{i}\Re\left\langle\zeta_{2},A\xi_{1}\right\rangle$
(16)

for all $z_{1},z_{2},w_{1},w_{2}\in\mathbb{C}^{d}$, where

$\begin{array}[]{ll}\zeta_{1}=\Re z_{1}+\mathrm{i}\Re z_{2}&\ \zeta_{2}=\Im
z_{1}+\mathrm{i}\Im z_{2}\\\ \xi_{1}=\Re w_{1}+\mathrm{i}\Re w_{2}&\
\xi_{2}=\Im w_{1}+\mathrm{i}\Im w_{2}\end{array}.$

We can then conclude that two _real_ linear operators $A$ and $B$ on
$\mathbb{C}^{d}$ coincide if and only if their complexification $\mathbf{A}$
and $\mathbf{B}$ coincide (as complex linear operators on $\mathbb{C}^{2d}$).

The following Proposition now collects useful formulae on the operators $Z,C$
and introduces the matrix

$\mathbf{C_{Z}}:=\mathbf{C}-\mathrm{i}(\mathbf{Z}^{*}\mathbf{J}+\mathbf{J}\mathbf{Z})$

which will have a central role in the paper.

######  Proposition 6.

Let $Z,C$ be given by (10) and (11).

  1. 1.

We can write

$C=\sqrt{C}^{T}\sqrt{C}\geq 0,$

where $\sqrt{C}z=\overline{U}z+V\overline{z}$, for $z\in\mathbb{C}^{d}$.

  2. 2.

It holds

$\displaystyle\mathbf{Z}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\begin{bmatrix}\Re\left(\left(U-\overline{V}\right)^{*}\left(U+\overline{V}\right)\right)&\Im\left(\left(U-\overline{V}\right)^{*}\left(U-\overline{V}\right)\right)\\\
-\Im\left(\left(U+\overline{V}\right)^{*}\left(U+\overline{V}\right)\right)&\Re\left(\left(U+\overline{V}\right)^{*}\left(U-\overline{V}\right)\right)\end{bmatrix}$
(17) $\displaystyle+$
$\displaystyle\begin{bmatrix}-\Im\left(\Omega+\kappa\right)&\Re\left(\kappa-\Omega\right)\\\
\Re\left(\Omega+\kappa\right)&\Im\left(\kappa-\Omega\right)\end{bmatrix}$
$\displaystyle\mathbf{C_{Z}}$ $\displaystyle=$
$\displaystyle\begin{bmatrix}\left(U+\overline{V}\right)^{*}\left(U+\overline{V}\right)&-\mathrm{i}\left(U+\overline{V}\right)^{*}\left(U-\overline{V}\right)\\\
\mathrm{i}\left(U-\overline{V}\right)^{*}\left(U+\overline{V}\right)&\left(U-\overline{V}\right)^{*}\left(U-\overline{V}\right)\end{bmatrix}\geq
0.$ (18)

  3. 3.

The semigroup has exactly $2d$ (linearly independent) Kraus’ operators if and
only if $\mathbf{C_{Z}}>0$.

###### Proof.

Equalities in statements $1$ and $2$ follow from direct computation starting
from the expression (10), (11) (see Appendix A). The inequality instead
follows from noticing that, setting

$M=\begin{bmatrix}U+\overline{V}&-\mathrm{i}\left(U-\overline{V}\right)\end{bmatrix}\in
M_{m,2d}(\mathbb{C})$

one has

$\mathbf{C_{Z}}=M^{*}M\geq 0.$

Eventually the inequality holds strict if and only if $\ker M=\\{0\\}$ or
equivalently $\operatorname{dim}\operatorname{Ran}M=2d$. However we have

$\operatorname{dim}\operatorname{Ran}M=m-\operatorname{dim}(\operatorname{Ran}M)^{\perp}=m-\operatorname{dim}\ker
M^{*}=m,$

since $\ker M^{*}=\ker(U+\overline{V})^{*}\cap\ker(U-\overline{V})^{*}$ and by
taking linear combinations of the two operators we get to $\ker M^{*}=\ker
U^{T}\cap\ker V^{*}=\\{0\\}$ by the conditions we imposed on the matrices
$U,V$. ∎

Note that we introduced the notation $\mathbf{C_{Z}}$ using bold-face
characters even though the operator is not the complexification of any real
linear operator. This is to emphasize the fact that the operator acts on
$\mathbb{C}^{2d}$.

We end this section by recalling the following result on invariant densities,
namely, those positive operators $\rho$ with unit trace such that
$\mathcal{T}_{*t}(\rho)=\rho$ for all $t\geq 0$. This parallels known results
([13] Section 11.2.3) for classical Ornstein-Uhlenbeck processes. The real
linear operator $\mathbf{Z}$ is called _stable_ if all its eigenvalues have
strictly negative real part. It is clear from formula (14) that $Z$ is stable
if and only if $\mathbf{Z}$ is stable.

######  Theorem 7.

Let $\mathcal{T}$ be a gaussian QMS and suppose $Z$ is stable. Then there is a
unique gaussian state $\rho=\rho_{(\mu,S)}$ which is invariant for
$\mathcal{T}$ whose parameters are given by:

$\mu=(Z^{\sharp})^{-1}\zeta,\quad S=\int_{0}^{\infty}\hbox{\rm
e}^{sZ^{\sharp}}C\hbox{\rm e}^{sZ}\hbox{\rm d}s.$ (19)

If also $\mathbf{C_{Z}}>0$ then $\rho$ is faithful and it is the unique
invariant state for $\mathcal{T}$.

###### Proof.

Let us start by showing the existence of an invariant state under the
stability condition. Let $\rho=\rho_{(\mu,S)}$ a gaussian state. For it to be
an invariant state, from Proposition 4, we must have

$\mu=\hbox{\rm e}^{tZ^{\sharp}}\mu-\int_{0}^{t}\hbox{\rm
e}^{sZ^{\sharp}}\zeta\hbox{\rm d}s,\quad S_{t}=\hbox{\rm
e}^{tZ^{\sharp}}S\hbox{\rm e}^{tZ}+\int_{0}^{t}\hbox{\rm
e}^{sZ^{\sharp}}C\hbox{\rm e}^{sZ}\hbox{\rm d}s,\quad\forall t\geq 0.$

The previous equations can be equivalently rewritten as

$\int_{0}^{t}\hbox{\rm
e}^{sZ^{\sharp}}\left(Z^{\sharp}\mu-\zeta\right)\hbox{\rm
d}s=0,\quad\int_{0}^{t}\hbox{\rm
e}^{sZ^{\sharp}}\left(Z^{\sharp}S+SZ+C\right)\hbox{\rm e}^{sZ}\hbox{\rm
d}s=0,\quad\forall t\geq 0.$

Since the equations hold for every $t\geq 0$ their derivative must vanish at
every $t\geq 0$. Moreover both $\hbox{\rm e}^{tZ^{\sharp}}$ and $\hbox{\rm
e}^{tZ}$ are invertible so the equations are eventually equivalent to

$Z^{\sharp}\mu=\zeta,\quad Z^{\sharp}S+SZ=-C.$ (20)

Now it is easy to show that the pair $\mu,S$ given by (19) solves the previous
equations. Indeed the integral in the definition for $S$ converges and

$\displaystyle Z^{\sharp}S+SZ=\int_{0}^{\infty}\hbox{\rm
e}^{sZ^{\sharp}}\left(Z^{\sharp}C+CZ\right)\hbox{\rm e}^{sZ}\hbox{\rm
d}s=\left[\hbox{\rm e}^{sZ^{\sharp}}C\hbox{\rm
e}^{sZ}\right]_{0}^{\infty}=-C.$

where the last equality is due to the stability of $Z$.

Uniqueness of the invariant state among gaussian states is due to the fact
that $Z$ is stable and rewriting equation (20) as a linear system for the
entries of $S$ (see [6] Theorem 12.4 and the discussion preceding it) one
obtains a system matrix which is invertible.

Suppose now that $\mathbf{C_{Z}}>0$. It is easy to see that $\mathbf{C_{Z}}$,
explicitly given in Proposition 6, is unitarily equivalent to the Kossakowski
matrix of the semigroup (see [3]). Therefore, by [3] Theorem 4, the semigroup
is irreducible.
Recall now that a gaussian state is faithful if and only if
$\mathbf{S}-\mathrm{i}\mathbf{J}>0$, or equivalently, by conjugation,
$\mathbf{S}+\mathrm{i}\mathbf{J}>0$ (see [29, Theorem 4]) and note that,
thanks to the stability of $Z$ we can write

$-J=\int_{0}^{\infty}\hbox{\rm e}^{sZ^{\sharp}}(Z^{\sharp}J+JZ)\hbox{\rm
e}^{sZ}\hbox{\rm d}s.$

Using (19) we have therefore

$\displaystyle\mathbf{S}+\mathrm{i}\mathbf{J}$
$\displaystyle=\int_{0}^{\infty}\hbox{\rm
e}^{s\mathbf{Z}^{*}}\mathbf{C}\hbox{\rm e}^{s\mathbf{Z}}\hbox{\rm
d}s-\mathrm{i}\int_{0}^{\infty}\hbox{\rm
e}^{s\mathbf{Z}^{*}}(\mathbf{Z}^{*}\mathbf{J}+\mathbf{JZ})\hbox{\rm
e}^{s\mathbf{Z}}\hbox{\rm d}s$ $\displaystyle=\int_{0}^{\infty}\hbox{\rm
e}^{s\mathbf{Z}^{*}}\mathbf{C_{Z}}\hbox{\rm e}^{s\mathbf{Z}}\hbox{\rm d}s.$

Since, from Proposition 6, $\mathbf{C_{Z}}\geq 0$, we have that $\left\langle
w,\left(\mathbf{S}+\mathrm{i}\mathbf{J}\right)w\right\rangle=0$ for some
$w\in\mathbb{C}^{2d}$ if and only if

$\left\langle\hbox{\rm e}^{s\mathbf{Z}}w,\mathbf{C_{Z}}\hbox{\rm
e}^{s\mathbf{Z}}w\right\rangle=0,\quad\forall s\geq 0.$

which in turn is equivalent to $\hbox{\rm e}^{tZ}w\in\ker\mathbf{C_{Z}}$ for
every $t\geq 0$. However $\mathbf{C_{Z}}>0$ means this cannot happen. We can
now end the proof by observing that irreducibility and the existence of an
invariant faithful state imply the invariant state is unique (see [22]). ∎

##  3 Relationship with Classical Ornstein-Uhlenbeck semigroups

Special cases of gaussian QMSs have been considered as quantum analogues of
classical Ornstein-Uhlenbeck semigroups (see [10, 14]) for $d=1$. The
multidimensional case, under the further assumption of a detailed balance
condition, has been extensively studied [11, 34]. The key observation is that
it is possible to restrict a QMS on $\mathcal{B}(\Gamma(\mathbb{C}^{d}))$
described in Theorem 3 to an abelian subalgebra generated by commuting (i.e.
classical) observables. Here we show that, with this procedure, one can
recover some classical Ornstein-Uhlenbeck semigroups starting from gaussian
QMSs. A preliminary clarification is now in order. Using Schrödinger
representation (see Example 5.2.16 in [7]) we can represent
$\mathcal{B}(\mathsf{h})$ onto $L^{2}(\mathbb{R}^{d})$ via the identifications

$\displaystyle W(cf_{j})\psi(x_{1},\dots,x_{d})=\hbox{\rm
e}^{\mathrm{i}cf_{j}}\psi(x_{1},\dots,x_{d}),$ $\displaystyle
W(\mathrm{i}cf_{j})\psi(x_{1},\dots,x_{d})=\psi(x_{1},\dots,x_{j}-c,\dots
x_{d}),$

with $c\in\mathbb{R}$ and $(f_{j})$ the canonical orthonormal basis of
$\mathbb{R}^{d}$. Under this representation we have then

$\displaystyle q_{j}\psi(x_{1},\dots,x_{d})=x_{j}\psi(x_{1},\dots,x_{d}),$
$\displaystyle
p_{j}\psi(x_{1},\dots,x_{d})=-\mathrm{i}\partial_{j}\psi(x_{1},\dots,x_{d}).$

When position operators, in the different coordinates, are the commuting
observables of interest, the algebra they generate is particularly simple to
describe. Indeed a (bounded) function of field operators is simply the
multiplication operator for that function, i.e.

$f(q_{1},\dots,q_{d})\psi(x_{1},\dots,x_{d})=f(x_{1},\dots,x_{d})\psi(x_{1},\dots,x_{d}).$

We restrict ourselves to the case where $H=0$ and $U,V\in M_{m\times
d}(\mathbb{R})$, i.e. all the coefficients of the Kraus operators are real.
This is not at all a necessary restraint but it is sufficient to show the link
between classical Ornstein-Uhlenbeck and gaussian quantum Markov semigroups.
Indeed we can rewrite the Kraus’ operators (6) as

$L_{\ell}=\sum_{j=1}^{d}\left(r_{\ell j}q_{j}+\mathrm{i}s_{\ell
j}p_{j}\right),\quad r_{\ell j}=\frac{\overline{v_{\ell j}}+u_{\ell
j}}{\sqrt{2}},s_{\ell j}=\frac{\overline{v_{\ell j}}-u_{\ell j}}{\sqrt{2}}$

with $X=(s_{jk})_{jk},R=(r_{jk})_{jk}\in M_{md}(\mathbb{R})$, and obtain the
following.

######  Proposition 8.

Let $H=0$ and $U,V$ be real matrices, then the commutative algebra of bounded
functions of $q_{1},\dots,q_{d}$ is invariant for the semigroup and the
generator acts on them as

$\texttt{{\pounds}}f(q_{1},\dots,q_{d})=\frac{1}{2}\sum_{j,k=1}^{d}(X^{*}X)_{jk}(\partial_{j}\partial_{k}f)(q_{1},\dots,q_{d})-\sum_{j,k=1}^{d}(R^{*}X)_{jk}q_{j}(\partial_{k}f)(q_{1},\dots,q_{d}),$

where $X^{*}X$ and $R^{*}X$ correspond, respectively, to the bottom-right
$d\times d$ blocks of $\mathbf{C_{Z}}$ and $\mathbf{Z}$, (17) (18) .

###### Proof.

All the calculations in this proof are assumed to be performed using the
quadratic form on $D\times D$ but, in order to avoid cluttering notation, we
leave it implicit. On every bounded function of $q_{1},\dots,q_{d}$, again at
least on the dense domain $D$, we can compute

$[q_{j},f(q_{1},\dots,q_{d})]=0,\quad[p_{j},f(q_{1},\dots,q_{d})]=-\mathrm{i}(\partial_{j}f)(q_{1},\dots,q_{d}).$

Using these computations we can evaluate the generator on such functions
starting from (8)

$\displaystyle\texttt{{\pounds}}\left(f(q_{1},\dots,q_{d})\right)$
$\displaystyle=-\frac{1}{2}\sum_{\ell=1}^{m}\left(L_{\ell}^{*}[L_{\ell},f(q_{1},\dots,q_{d})]-[L_{\ell}^{*},f(q_{1},\dots,q_{d})]L_{\ell}\right)$
$\displaystyle=-\frac{1}{2}\sum_{\ell=1}^{m}\sum_{j,k=1}^{d}\left(r_{\ell
j}q_{j}-\mathrm{i}s_{\ell j}p_{j}\right)\left[r_{\ell
k}q_{k}+\mathrm{i}s_{\ell k}p_{k},f(q_{1},\dots,q_{d})\right]$
$\displaystyle+\frac{1}{2}\sum_{\ell=1}^{m}\sum_{j,k=1}^{d}\left[r_{\ell
j}q_{j}-\mathrm{i}s_{\ell j}p_{j},f(q_{1},\dots,q_{d})\right]\left(r_{\ell
k}q_{k}+\mathrm{i}s_{\ell k}p_{k}\right)$
$\displaystyle=-\sum_{j,k=1}^{d}(R^{*}X)_{jk}q_{j}(\partial_{k}f)(q_{1},\dots,q_{d})$
$\displaystyle+\frac{1}{2}\sum_{j,k=1}^{d}(X^{*}X)_{jk}(\partial_{j}\partial_{k}f)(q_{1},\dots,q_{d}).$

∎

The previous result shows indeed that starting from a particular subset of
gaussian QMSs one can recover a Ornstein-Uhlenbeck semigroup. What one can
also show, in the special case of a non-degenerate diffusion matrix $X^{*}X$,
that a converse result also holds.

######  Proposition 9.

Let

$\mathscr{L}=\frac{1}{2}\sum_{j,k=1}^{d}Q_{jk}\partial_{j}\partial_{k}+\sum_{j,k=1}^{d}A_{jk}\partial_{k},$

be a Ornstein-Uhlenbeck generator with $Q>0$. Then the semigroup generated by
$\mathscr{L}$ can be recovered as a restriction of a suitable gaussian QMS.

###### Proof.

From Proposition 8 we just need to show that there is a suitable choice of
$R,X$ (corresponding uniquely to a choice of $U,V$) such that

$Q=X^{*}X,\quad A=-R^{*}X.$

However, since $Q>0$, it is immediate to obtain $X=\sqrt{Q}$. Moreover, from
invertibility of $Q$ we infer invertibility of $X$, therefore we can set
$R=-{X^{*}}^{-1}A^{*}$ and conclude the proof. ∎

The previous Proposition shows that we can actually recover all Ornstein-
Uhlenbeck semigroups with non-degenerate diffusion matrix starting from a
gaussian QMS. The case of a degenerate diffusion matrix is more complicated
and is beyond the scope of this work.

##  4 The spectral gap for the GNS embedding

In this section we explicitly compute the spectral gap for the $i_{2}$
embedding. Even when not explicitly stated we will assume $Z$ stable and
$\mathbf{C_{Z}}>0$, or equivalently that the semigroup has $2d$ Kraus’
operators, thanks to Proposition 6. We will prove in section 5 that the these
two requirements are not really restrictive since they are necessary
conditions for the semigroup to have a spectral gap $g>0$.

######  Remark 10.

In the following we will always assume $\zeta=0$ to simplify calculations even
though all the results remain true without this assumption.

We will show the existence of a spectral gap by first finding a candidate
$g>0$ and then proving it actually satisfies (1). We will work on a dense
subspace of $\mathfrak{I}_{2}(\mathsf{h})$, the space of Hilbert-Schmidt
operators on $\mathsf{h}$, which is the range of the immersion $i_{2}$ of the
space of linear combination of Weyl operators

$\mathcal{W}=\left\\{\sum_{j=1}^{l}\eta_{j}W(z_{j})\mid l\geq
1,z_{1},\dots,z_{l}\in\mathbb{C}^{d},\eta_{1},\dots,\eta_{l}\in\mathbb{C}\right\\}.$

In the following, for any $x\in\mathcal{B}(\mathsf{h})$ we will denote by
$\widetilde{x}$ the projection of $x\rho^{1/2}$ on the orthogonal of
$\rho^{1/2}$ in $\mathfrak{I}_{2}(\mathsf{h})$ given by

$\widetilde{x}:=x\rho^{1/2}-\left\langle\rho^{1/2},x\rho^{1/2}\right\rangle_{2}\rho^{1/2}.$
(21)

Note that for $x\in\mathcal{W}$ we have

$\left\langle\rho^{1/2},x\rho^{1/2}\right\rangle_{2}=\sum_{j=1}^{l}\eta_{j}\,\hbox{\rm
tr}\left(\rho
W(z_{j})\right)=\sum_{j=1}^{l}\eta_{j}\,\mathrm{e}^{-\frac{1}{2}\Re\langle
z_{j},Sz_{j}\rangle},$

so that the orthogonalization $\widetilde{x}$ yields

$\widetilde{x}=\sum_{j=1}^{l}\eta_{j}\left(W(z_{j})-\mathrm{e}^{-\frac{1}{2}\Re\langle
z_{j},Sz_{j}\rangle}\hbox{\rm 1\kern-2.8ptl}\right)\rho^{1/2}.$ (22)

We start with the following

######  Lemma 11.

Let $x\in\mathcal{W}$ and set $\xi_{j}=\mathrm{e}^{-\frac{1}{2}\langle
z_{j},Sz_{j}\rangle}\eta_{j}$ for all $j=1,\ldots,n$. Then

$\left\|T_{t}(\tilde{x})\right\|_{2}^{2}=\sum_{j,k=1}^{l}\overline{\xi}_{j}\xi_{k}\left(\mathrm{exp}\left\langle\hbox{\rm
e}^{t\mathbf{Z}}\begin{bmatrix}\Re z_{j}\\\
\Im{z_{j}}\end{bmatrix},\left(\mathbf{S}+\mathrm{i}\mathbf{J}\right)\hbox{\rm
e}^{t\mathbf{Z}}\begin{bmatrix}\Re z_{k}\\\
\Im{z_{k}}\end{bmatrix}\right\rangle-1\right)$ (23)

for all $t\geq 0$.

###### Proof.

By formula (22) we have

$\displaystyle\left\|T_{t}(\widetilde{x})\right\|_{2}^{2}$
$\displaystyle=\left\|\sum_{j=1}^{l}\eta_{j}\left(\mathcal{T}_{t}(W(z_{j}))-\mathrm{e}^{-\frac{1}{2}\Re\langle
z_{j},Sz_{j}\rangle}\hbox{\rm 1\kern-2.8ptl}\right)\rho^{1/2}\right\|_{2}^{2}$
$\displaystyle=\sum_{j,k=1}^{l}\overline{\eta}_{j}\eta_{k}\hbox{\rm
tr}\left(\rho\left(\mathcal{T}_{t}(W(-z_{j}))-\mathrm{e}^{-\frac{1}{2}\Re\langle
z_{j},Sz_{j}\rangle}\hbox{\rm
1\kern-2.8ptl}\right)\left(\mathcal{T}_{t}(W(z_{k}))-\mathrm{e}^{-\frac{1}{2}\Re\langle
z_{k},Sz_{k}\rangle}\hbox{\rm 1\kern-2.8ptl}\right)\right)$

which is equal to

$\displaystyle\ \sum_{j,k=1}^{l}\overline{\eta}_{j}\eta_{k}\,\hbox{\rm
tr}\left(\rho\left(\mathcal{T}_{t}(W(-z_{j}))\mathcal{T}_{t}(W(z_{k}))-\mathrm{e}^{-\frac{1}{2}\Re\langle
z_{j},Sz_{j}\rangle-\frac{1}{2}\Re\langle z_{k},Sz_{k}\rangle}\hbox{\rm
1\kern-2.8ptl}\right)\right)$
$\displaystyle=\sum_{j,k=1}^{l}\overline{\eta}_{j}\eta_{k}\Big{[}\hbox{\rm
tr}\left(\rho\,\mathrm{e}^{-\frac{1}{2}\Re\int_{0}^{t}\left(\left\langle\mathrm{e}^{sZ}z_{j},C\mathrm{e}^{sZ}z_{j}\right\rangle+\left\langle\mathrm{e}^{sZ}z_{k},C\mathrm{e}^{sZ}z_{k}\right\rangle\right)\mathrm{d}s}\,W(-\mathrm{e}^{tZ}z_{j})W(\mathrm{e}^{tZ}z_{k})\right)$
$\displaystyle\qquad\qquad-\,\mathrm{e}^{-\frac{1}{2}\Re\langle
z_{j},Sz_{j}\rangle-\frac{1}{2}\Re\langle z_{k},Sz_{k}\rangle}\Big{]}.$

Keeping into account the CCR relation for Weyl operators (3) we get

$\displaystyle\left\|T_{t}(\widetilde{x})\right\|_{2}^{2}$ $\displaystyle=$
$\displaystyle\sum_{j,k}\overline{\eta}_{j}\eta_{k}\Big{[}\mathrm{e}^{-\frac{1}{2}\Re\int_{0}^{t}\left(\left\langle\mathrm{e}^{sZ}z_{j},C\mathrm{e}^{sZ}z_{j}\right\rangle+\left\langle\mathrm{e}^{sZ}z_{k},C\mathrm{e}^{sZ}z_{k}\right\rangle\right)\mathrm{d}s}\,$
(24)
$\displaystyle\qquad\cdot\,\mathrm{e}^{\mathrm{i}\Im\langle\mathrm{e}^{tZ}z_{j},\mathrm{e}^{tZ}z_{k}\rangle-\frac{1}{2}\Re\left\langle\mathrm{e}^{tZ}(z_{k}-z_{j}),S\mathrm{e}^{tZ}(z_{k}-z_{j})\right\rangle}$
$\displaystyle\qquad-\mathrm{e}^{-\frac{1}{2}\Re\langle
z_{j},Sz_{j}\rangle-\frac{1}{2}\Re\langle z_{k},Sz_{k}\rangle}\Big{]}.$

Now note that using the expression (19) for $S$ we have

$\Re\left\langle\mathrm{e}^{tZ}(z_{k}-z_{j}),S\mathrm{e}^{tZ}(z_{k}-z_{j})\right\rangle=\int_{t}^{\infty}\Re\left\langle\mathrm{e}^{sZ}(z_{k}-z_{j}),C\mathrm{e}^{sZ}(z_{k}-z_{j})\right\rangle\mathrm{d}s$

for all $j,k$, which allows us to rewrite

$\displaystyle-\frac{1}{2}\Re\int_{0}^{t}\left(\left\langle\mathrm{e}^{sZ}z_{j},C\mathrm{e}^{sZ}z_{j}\right\rangle+\left\langle\mathrm{e}^{sZ}z_{k},C\mathrm{e}^{sZ}z_{k}\right\rangle\right)\mathrm{d}s+\mathrm{i}\Im\langle\mathrm{e}^{tZ}z_{j},\mathrm{e}^{tZ}z_{k}\rangle$
$\displaystyle-\frac{1}{2}\Re\left\langle\mathrm{e}^{tZ}(z_{k}-z_{j}),S\mathrm{e}^{tZ}(z_{k}-z_{j})\right\rangle$
$\displaystyle=-\frac{1}{2}\Re\langle
z_{j},Sz_{j}\rangle-\frac{1}{2}\Re\langle
z_{k},Sz_{k}\rangle+\Re\langle\mathrm{e}^{tZ}z_{j},S\mathrm{e}^{tZ}z_{k}\rangle+\mathrm{i}\Im\langle\mathrm{e}^{tZ}z_{j},\mathrm{e}^{tZ}z_{k}\rangle.$

Summing up, we can write $\left\|T_{t}(\widetilde{x})\right\|_{2}^{2}$ as

$\displaystyle\sum_{j,k}\overline{\eta}_{j}\eta_{k}\Big{[}$
$\displaystyle\mathrm{e}^{-\frac{1}{2}\Re\langle
z_{j},Sz_{j}\rangle-\frac{1}{2}\Re\langle
z_{k},Sz_{k}\rangle+\Re\langle\mathrm{e}^{tZ}z_{j},S\mathrm{e}^{tZ}z_{k}\rangle+\mathrm{i}\Im\langle\mathrm{e}^{tZ}z_{j},\mathrm{e}^{tZ}z_{k}\rangle}$
$\displaystyle-\mathrm{e}^{-\frac{1}{2}\Re\langle
z_{j},Sz_{j}\rangle-\frac{1}{2}\Re\langle z_{k},Sz_{k}\rangle}\Big{]}.$

Letting $\xi_{j}=\mathrm{e}^{-\frac{1}{2}\Re\langle
z_{j},Sz_{j}\rangle}\eta_{j}$ for all $j=1,\ldots,n$, we obtain

$\displaystyle\left\|T_{t}(\widetilde{x})\right\|_{2}^{2}$
$\displaystyle=\sum_{j,k=1}^{l}\overline{\xi}_{j}\xi_{k}\left(\mathrm{e}^{\Re\langle\mathrm{e}^{tZ}z_{j},S\mathrm{e}^{tZ}z_{k}\rangle+\mathrm{i}\Im\langle\mathrm{e}^{tZ}z_{j},\mathrm{e}^{tZ}z_{k}\rangle}-1\right)$
$\displaystyle\left\|\widetilde{x}\right\|_{2}^{2}$
$\displaystyle=\sum_{j,k=1}^{l}\overline{\xi}_{j}\xi_{k}\left(\mathrm{e}^{\Re\langle
z_{j},Sz_{k}\rangle+\mathrm{i}\Im\langle z_{j},z_{k}\rangle}-1\right).$

Finally, recalling equation (15), we can write

$\Re\langle\mathrm{e}^{tZ}z_{j},S\mathrm{e}^{tZ}z_{k}\rangle+\mathrm{i}\Im\langle\mathrm{e}^{tZ}z_{j},\mathrm{e}^{tZ}z_{k}\rangle=\left\langle\hbox{\rm
e}^{t\mathbf{Z}}\begin{bmatrix}\Re z_{j}\\\
\Im{z_{j}}\end{bmatrix},\left(\mathbf{S}+\mathrm{i}\mathbf{J}\right)\hbox{\rm
e}^{t\mathbf{Z}}\begin{bmatrix}\Re z_{k}\\\
\Im{z_{k}}\end{bmatrix}\right\rangle.$ (25)

This completes the proof. ∎

The previous lemma allows us to rewrite equation (1) for $\tilde{x}$ with
$x\in\mathcal{W}$. In this linear space of operators the condition for the
spectral gap reduces to positive definiteness of a kernel as highlighted in
the following proposition.

######  Proposition 12.

For all $t\geq 0$, defining

$s_{t}(z,w):=\left\langle\hbox{\rm e}^{t\mathbf{Z}}\begin{bmatrix}\Re z\\\
\Im{z}\end{bmatrix},\left(\mathbf{S}+\mathrm{i}\mathbf{J}\right)\hbox{\rm
e}^{t\mathbf{Z}}\begin{bmatrix}\Re w\\\
\Im{w}\end{bmatrix}\right\rangle\qquad\forall\,t\geq 0,\ z,w\in\mathbb{C}^{d}$

for all $x\in\mathcal{W}$, setting $\xi_{j}=\hbox{\rm
e}^{-\frac{1}{2}\left\langle z_{j},Sz_{j}\right\rangle}\eta_{j}$ as in Lemma
11, we have

$\left\|T_{t}(\tilde{x})\right\|_{2}^{2}=\sum_{n\geq
1}\sum_{j,k=1}^{l}\frac{\overline{\xi_{j}}\xi_{k}}{n!}\left(s_{t}(z_{j},z_{k})\right)^{n}.$
(26)

In particular, if the kernel

$K_{n,t}(z,w)=\hbox{\rm
e}^{-2gt}\left(s_{0}(z,w)\right)^{n}-\left(s_{t}(z,w)\right)^{n},\quad
z,w\in\mathbb{C}^{d},$

is positive definite for every $n\geq 1$, the spectral gap inequality (1)
holds for any $\tilde{x}$ with $x\in\mathcal{W}$.

###### Proof.

The power series expansion of the exponential immediately yields (26) from
(23). This formula for $t>0$ and for $t=0$ allows us to rewrite (1) as

$\displaystyle 0\leq\hbox{\rm
e}^{-2gt}\left\|\tilde{x}\right\|_{2}^{2}-\left\|T_{t}(\tilde{x})\right\|_{2}^{2}$
$\displaystyle=\sum_{n\geq
1}\frac{1}{n!}\sum_{j,k=1}^{l}\overline{\xi_{j}}\xi_{k}\left(\hbox{\rm
e}^{-2gt}\left(s_{0}(z_{j},z_{k})\right)^{n}-\left(s_{t}(z_{j},z_{k})\right)^{n}\right),$

which proves the claim. ∎

The characterization of Proposition 12 can be exploited to find a suitable
candidate for the spectral gap. We begin by the following Lemma.

######  Lemma 13.

Let $Y\in M_{n}(\mathbb{C})$, $n\geq 1$, and let $\omega_{0}$ be the maximum
eigenvalue of $Y+Y^{*}$. Then

$\left\|\hbox{\rm e}^{tY}v\right\|^{2}\leq\hbox{\rm
e}^{t\omega_{0}}\left\|v\right\|^{2}\quad\forall\,t\geq 0,\
v\in\mathbb{C}^{n}.$ (27)

Moreover $\omega_{0}$ is the optimal choice in the inequality, i.e.

$\omega_{0}=\min\\{\omega\in\mathbb{R}\,:\,\left\|\hbox{\rm
e}^{tY}v\right\|^{2}\leq\hbox{\rm
e}^{t\omega}\left\|v\right\|^{2}\forall\,t\geq 0,\ v\in\mathbb{C}^{n}\\}.$

###### Proof.

Differentiating the function $f_{v}(t)=\left\|\hbox{\rm e}^{tZ}v\right\|^{2}$
we find

$\frac{\hbox{\rm d}}{\hbox{\rm d}t}f_{v}(t)=\left\langle\hbox{\rm
e}^{tY}z,(Y+Y^{*})\hbox{\rm e}^{tY}z\right\rangle\leq\omega_{0}f_{v}(t)$ (28)

and Gronwall’s Lemma implies (27). On the other hand, if we choose
$\omega<\omega_{0}$ and we consider the eigenvector $v$ of $Y+Y^{*}$
associated with the eigenvalue $\omega_{0}$, then

$\left.\frac{\hbox{\rm d}}{\hbox{\rm d}t}\hbox{\rm
e}^{-t\omega}f_{v}(t)\right|_{t=0}=(\omega_{0}-\omega)\left\|v\right\|^{2}>0.$

In particular we have $\hbox{\rm e}^{-t\omega}f_{v}(t)>\left\|v\right\|^{2}$
in a right neighbourhood of $t=0$ which means $\omega_{0}$ is the optimal
bound in inequality (27). ∎

Let $\tilde{\mathbf{S}}=\mathbf{S}+\mathrm{i}\mathbf{J}$. Proposition 12 with
$n=1$ and $x=W(z)$ yields

$K_{1,t}(z,z)=\hbox{\rm e}^{-2gt}s_{0}(z,z)-s_{t}(z,z)=\hbox{\rm
e}^{-2gt}\left\|\tilde{\mathbf{S}}^{1/2}\begin{bmatrix}\Re z\\\ \Im
z\end{bmatrix}\right\|^{2}-\left\|\hbox{\rm
e}^{t\tilde{\mathbf{S}}^{1/2}\mathbf{Z}\tilde{\mathbf{S}}^{-1/2}}\tilde{\mathbf{S}}^{1/2}\begin{bmatrix}\Re
z\\\ \Im z\end{bmatrix}\right\|^{2}$ (29)

for all $t\geq 0$, and we get a natural candidate for $2g$ from Lemma 13
applied to $Y=\tilde{\mathbf{S}}^{1/2}\mathbf{Z}\tilde{\mathbf{S}}^{-1/2}$ as
a $2d\times 2d$ matrix.

######  Proposition 14.

Suppose that $Z$ is stable and $\mathbf{C_{Z}}>0$. Let $\omega_{0}$ be the
greatest eigenvalue of the $2d\times 2d$ matrix
$\tilde{\mathbf{S}}^{\frac{1}{2}}\mathbf{Z}\tilde{\mathbf{S}}^{-\frac{1}{2}}+\tilde{\mathbf{S}}^{-\frac{1}{2}}\mathbf{Z}^{*}\tilde{\mathbf{S}}^{\frac{1}{2}}$,
then $-\omega_{0}$ is the smallest eigenvalue of
$\mathbf{C_{\mathbf{Z}}\tilde{S}}^{-1}$ and $\omega_{0}<0$.

###### Proof.

The condition on the covariance operator $\mathbf{S}$ of an invariant state
$\mathbf{Z}^{*}\mathbf{S}+\mathbf{SZ}=-\mathbf{C}$ immediately yields

$\mathbf{Z}^{*}\tilde{\mathbf{S}}+\tilde{\mathbf{S}}\mathbf{Z}=-\mathbf{C}+\mathrm{i}(\mathbf{Z}^{*}\mathbf{J}+\mathbf{J}\mathbf{Z})=-\mathbf{C}_{\mathbf{Z}}.$

Left and right multiplying by $\tilde{\mathbf{S}}^{-\frac{1}{2}}$ we get

$\tilde{\mathbf{S}}^{\frac{1}{2}}\mathbf{Z}\tilde{\mathbf{S}}^{-\frac{1}{2}}+\tilde{\mathbf{S}}^{-\frac{1}{2}}\mathbf{Z}^{*}\tilde{\mathbf{S}}^{\frac{1}{2}}=-\tilde{\mathbf{S}}^{-\frac{1}{2}}\mathbf{C}_{\mathbf{Z}}\tilde{\mathbf{S}}^{-\frac{1}{2}}.$
(30)

Since $\mathbf{C}_{\mathbf{Z}}>0$ by our assumptions, we get $\omega_{0}<0$.
On the other hand,
$\tilde{\mathbf{S}}^{-\frac{1}{2}}\mathbf{C}_{\mathbf{Z}}\tilde{\mathbf{S}}^{-\frac{1}{2}}$
is similar to $\mathbf{C}_{\mathbf{Z}}\tilde{\mathbf{S}}^{-1}$ and then
$-\omega_{0}$ is the smallest eigenvalue of
$\mathbf{C}_{\mathbf{Z}}\tilde{\mathbf{S}}^{-1}$. ∎

Our goal now is to prove that $g=-\omega_{0}/2$ is actually the spectral gap.
We begin by showing that equation (1) holds for $\tilde{x}$ with
$x\in\mathcal{W}$, i.e. that $K_{n,t}$ is a definite positive kernel. As a
preliminary result we prove the following

######  Lemma 15.

Let $n\geq 1$ and $t\geq 0$. The kernel

$K^{\prime}_{n,t}(z,w)=\hbox{\rm e}^{t\omega_{0}/n}s_{0}(z,w)-s_{t}(z,w),\quad
z,w\in\mathbb{C}^{d}$ (31)

is positive definite.

###### Proof.

Note that we just need to prove the proposition for $n=1$ because we have

$\sum_{j,k}\overline{c_{j}}c_{k}K^{\prime}_{n,t}(z_{j},z_{k})=\sum_{j,k}\overline{c_{j}}c_{k}\left(K^{\prime}_{1,t}(z_{j},z_{k})+\left(\hbox{\rm
e}^{-2tg/n}-\hbox{\rm e}^{-2tg}\right)s_{0}(z_{j},z_{k})\right),$

($c_{j}\in\mathbb{C},z_{j}\in\mathbb{C}^{d}$) which is positive if
$K^{\prime}_{1,t}$ is positive definite, since $s_{0}$ is positive definite
and $\hbox{\rm e}^{-2tg/n}-\hbox{\rm e}^{-2tg}>0$. Therefore, fix $n=1$ and
let us prove positive definiteness of (31). By sesquilinearity of $s_{t}$ we
have

$\sum_{j,k}\overline{c_{j}}c_{k}K^{\prime}_{1,t}(z_{j},z_{k})=\hbox{\rm
e}^{-2tg}s_{0}\left(\xi,\xi\right)-s_{t}(\xi,\xi),$

where $\xi=\sum_{j}c_{j}z_{j}$. Since this quantity is positive by our choice
of $2g$ and Lemma 13, $K^{\prime}_{1,t}$ is positive definite and the proof is
complete. ∎

We are now in a position to prove the key inequality

######  Proposition 16.

For all $x\in\mathcal{W}$ and $t\geq 0$ we have

$\left\|T_{t}(\tilde{x})\right\|_{2}^{2}\leq\hbox{\rm
e}^{-2gt}\left\|\tilde{x}\right\|_{2}^{2}\,.$

###### Proof.

Thanks to Proposition 12 it suffices to show that the kernel

$K_{n,t}(z,w)=\hbox{\rm
e}^{t\omega_{0}}\left(s_{0}(z,w)\right)^{n}-\left(s_{t}(z,w)\right)^{n},\quad
z,w\in\mathbb{C}^{d},$

is positive definite. Recalling the kernel defined in (31), we have

$\displaystyle K_{n,t}(z,w)=K^{\prime}_{n,t}(z,w)\sum_{m=0}^{n-1}\hbox{\rm
e}^{\frac{n-1-m}{n}\,t\omega_{0}}\left(s_{0}(z,w)\right)^{n-1-m}\left(s_{t}(z,w)\right)^{m}.$

Since both $s_{t}$ and $s_{0}$ are positive definite kernels, their powers and
their linear combinations with positive coefficients are still positive
definite by [27] Corollary 15.2, and so is $K_{n,t}$. ∎

Finally we prove that the bound for the spectral gap given by Proposition 16
is actually optimal, even if we just consider Hilbert-Schmidt operators
$\tilde{x}$ with $x\in\mathcal{W}$.

######  Proposition 17.

Let $\omega<\omega_{0}$. There exist $\delta>0$ and ${x}\in\mathcal{W}$ such
that

$\left\|T_{t}(\tilde{x})\right\|_{2}^{2}>\hbox{\rm
e}^{t\omega}\left\|\tilde{x}\right\|_{2}^{2}\ \quad\forall\,t\in(0,\delta).$

###### Proof.

Let $z\in\mathbb{C}^{2d}$ such that $\tilde{\mathbf{S}}^{\frac{1}{2}}z$ is an
eigenvector of
$\tilde{\mathbf{S}}^{\frac{1}{2}}\mathbf{Z}\tilde{\mathbf{S}}^{-\frac{1}{2}}+\tilde{\mathbf{S}}^{-\frac{1}{2}}\mathbf{Z}^{*}\tilde{\mathbf{S}}^{\frac{1}{2}}$
associated with the eigenvalue $\omega_{0}$. Write $z=[z_{p},z_{q}]^{T}$ with
$z_{p},z_{q}\in\mathbb{C}^{d}$ and consider $z_{1},z_{2}\in\mathbb{C}^{d}$
defined by $z_{1}=\Re z_{p}+\mathrm{i}\Re z_{q}$, $z_{2}=\Im
z_{p}+\mathrm{i}\Im z_{q}$, so that $[\Re z_{1},\Im z_{1}]^{T}=\Re z$ and
$[\Re z_{2},\Im z_{2}]^{T}=\Im z$. For all $r\in\mathbb{R}$, considering
$x_{r}=W(rz_{1})+\mathrm{i}W(rz_{2})=\sum_{j=1}^{2}c_{j}W(rz_{j})$, with
$c_{1}=1,c_{2}=\mathrm{i}$, we have

$\widetilde{x_{r}}=\sum_{j=1}^{2}c_{j}\left(W(rz_{j})-\hbox{\rm
e}^{-\frac{r^{2}}{2}s_{0}(z_{j},z_{j})}\hbox{\rm
1\kern-2.8ptl}\right)\rho^{1/2}$

by equation (22). Setting again $\xi_{j,r}=\hbox{\rm
e}^{-\frac{r^{2}}{2}s_{0}(z_{j},z_{j})}c_{j}$, using equation (23) and
recalling the definition of $s_{t}(z,w)$ in Proposition 12 we then have

$\left\|T_{t}(\tilde{x})\right\|_{2}^{2}=\sum_{j,k=1}^{l}\overline{\xi}_{j,r}\,\xi_{k,r}\left(\mathrm{e}^{r^{2}s_{t}(z_{j},z_{k})}-1\right)$

Taking the derivative at $t=0$ we find

$\left.\frac{\hbox{\rm d}}{\hbox{\rm
d}t}s_{t}(z_{j},z_{k})\right|_{t=0}=\left\langle\tilde{\mathbf{S}}^{\frac{1}{2}}\begin{bmatrix}\Re
z_{j}\\\ \Im
z_{j}\end{bmatrix},\left(\tilde{\mathbf{S}}^{1/2}\mathbf{Z}\tilde{\mathbf{S}}^{-1/2}+\tilde{\mathbf{S}}^{-1/2}\mathbf{Z}^{*}\tilde{\mathbf{S}}^{1/2}\right)\tilde{\mathbf{S}}^{\frac{1}{2}}\begin{bmatrix}\Re
z_{k}\\\ \Im z_{k}\end{bmatrix}\right\rangle$

Consider now

$\displaystyle f(r)$ $\displaystyle:=\left.\frac{\hbox{\rm d}}{\hbox{\rm
d}t}\left\|T_{t}(\tilde{x_{r}})\right\|_{2}^{2}-\hbox{\rm
e}^{t\omega}\left\|\tilde{x_{r}}\right\|_{2}^{2}\right|_{t=0}$
$\displaystyle=\sum_{j,k=1}^{2}\overline{\xi}_{j,r}\,\xi_{k,r}\left(\left(r^{2}\left.\frac{\hbox{\rm
d}}{\hbox{\rm d}t}s_{t}(z_{j},z_{k})\right|_{t=0}-\omega\right)\hbox{\rm
e}^{r^{2}s_{0}(z_{j},z_{k})}+\omega\right)$

It holds $f(0)=f^{\prime}(0)=0$ while

$f^{\prime\prime}(0)=2\sum_{j,k=1}^{2}\overline{c_{j}}c_{k}\left(\left.\frac{\hbox{\rm
d}}{\hbox{\rm d}t}s_{t}(z_{j},z_{k})\right|_{t=0}-\omega
s_{0}(z_{j},z_{k})\right)=2(\omega_{0}-\omega)s_{0}(z,z)$

where the last equality follows from sesquilinearity of the derivative of
$s_{t}$ and $s_{0}$ as well as the fact that
$\tilde{\mathbf{S}}^{\frac{1}{2}}z$ is an eigenvector. In particular,
recalling that $\omega<\omega_{0}<0$, we find $f^{\prime\prime}(0)>0$ because
the invariant state is faithful and $z\neq 0$. This implies that there exists
$\delta_{0}>0$ such that $f(r)>0$ for every $r\in(0,\delta_{0})$. This implies
in turn that for every $r\in(0,\delta_{0})$ there exists a $\delta>0$ such
that

$\left\|T_{t}(\tilde{x_{r}})\right\|_{2}^{2}>\hbox{\rm
e}^{t\omega}\left\|\tilde{x_{r}}\right\|_{2}^{2},\quad\forall t\in(0,\delta).$

∎

Proposition 16 proves the inequality (1) for the spectral gap for elements
$\tilde{x}$ with $x\in\mathcal{W}$. The following Theorem shows that it holds
for all the remaining elements.

######  Theorem 18.

Let $\mathcal{T}$ be a gaussian QMS with $Z$ stable, $\mathbf{C_{Z}}>0$ and
$\zeta=0$. Then $\mathcal{T}$ has a spectral gap $g=-\omega_{0}/2>0$.

###### Proof.

By Proposition 16 we have
$\left\|T_{t}(\tilde{x})\right\|_{2}^{2}\leq\hbox{\rm
e}^{-2gt}\left\|\tilde{x}\right\|_{2}^{2}$ for all $t\geq 0$ and all
$\tilde{x}\in\mathfrak{I}_{2}(\mathsf{h})$ written in the form (22). To show
that it also holds for any element of $\mathfrak{I}_{2}(\mathsf{h})$
orthogonal to $\rho^{1/2}$ we show that the set of those $\tilde{x}$ is dense
in the orthogonal subspace of $\rho^{1/2}$. To this end, consider
$y\in\mathfrak{I}_{2}(\mathsf{h})$ orthogonal to $\rho^{1/2}$ such that
$\operatorname{tr}\left(y^{*}\tilde{x}\right)=0$ for all
$\tilde{x}=x\rho^{1/2}-\left\langle\rho^{1/2},x\rho^{1/2}\right\rangle_{2}\rho^{1/2}$
with $x\in\mathcal{W}$. In particular, taking $x=W(z)$ we have

$0=\operatorname{tr}\left(y^{*}\tilde{x}\right)=\operatorname{tr}\left(y^{*}\left(W(z)-\operatorname{tr}\left(\rho\,W(z)\right)\hbox{\rm
1\kern-2.8ptl}\right)\rho^{1/2}\right)=\operatorname{tr}\left(\rho^{1/2}y^{*}W(z)\right)$

for all $z\in\mathbb{C}^{d}$. Since $\rho^{1/2}y^{*}$ is a trace class
operator, the weak∗ density of Weyl operators in $\mathcal{B}(\mathsf{h})$
implies $\rho^{1/2}y^{*}=0$, namely $y=0$ because $\rho$ is faithful.

Proposition 17 now concludes the proof, showing that $g$ is the optimal
choice. ∎

##  5 Conditions on the invariant state

In this section we want to discuss the assumptions we made on $Z$ and
$\mathbf{C_{Z}}$ to show that they are not restrictive when looking for a
positive spectral gap. We have indeed the following result.

######  Theorem 19.

Let $\mathcal{T}$ be a gaussian QMS and suppose it has a spectral gap $g$.
Then $Z$ is stable and $\mathbf{C_{Z}}>0$.

We will split the proof in two steps. We start by showing the stability of
$Z$, which is due to the requirement of the existence of an invariant state
and the asymptotic stability-like requirement in (1).

######  Lemma 20.

If $Z$ is not stable then there exists $z\in\mathbb{C}^{d}$ that satisfies one
of the following:

  1. 1.

$\hbox{\rm e}^{tZ}z\in\ker C$ for every $t\geq 0$;

  2. 2.

$\lim_{t\to\infty}\int_{0}^{t}\Re\left\langle\hbox{\rm e}^{sZ}z,C\hbox{\rm
e}^{sZ}z\right\rangle\hbox{\rm d}s=+\infty.$

###### Proof.

Suppose $\lambda$ is an eigenvalue for $\mathbf{Z}$ with $\Re\lambda\geq 0$
and let $w\in\mathbb{C}^{2d}$ be its associated eigenvector. This implies that
$\overline{w}$ is an eigenvector for $\overline{\lambda}$. Consider
$x=w+\overline{w}$ and $y=\mathrm{i}(w-\overline{w})$, both belonging to
$\mathbb{R}^{2d}$ and satisfying

$\mathbf{Z}x=\Re\lambda x+\Im\lambda y,\quad\mathbf{Z}y=-\Im\lambda
x+\Re\lambda y.$

If $\Im\lambda=0$ we can set $z\in\mathbb{C}^{d}$ as the element corresponding
to $w$ and have $\hbox{\rm e}^{tZ}z=\hbox{\rm e}^{t\Re\lambda}z$, in
particular

$\int_{0}^{t}\Re\left\langle\hbox{\rm e}^{sZ}z,C\hbox{\rm
e}^{sZ}z\right\rangle\hbox{\rm d}s=\Re\left\langle
z,Cz\right\rangle\int_{0}^{t}\hbox{\rm e}^{2s\Re\lambda}\hbox{\rm d}s.$

This implies either $z\in\ker C$ or it converges to infinity, proving the
result in the case $\Im\lambda=0$. Suppose then $\Im\lambda\neq 0$. On the
space generated by $x,y$ we have

$Z=\begin{bmatrix}\Re\lambda&\Im\lambda\\\
-\Im\lambda&\Re\lambda\end{bmatrix},\quad\hbox{\rm e}^{tZ}=\hbox{\rm
e}^{t\Re\lambda}\begin{bmatrix}\cos(t\Im\lambda)&\sin(t\Im\lambda)\\\
-\sin(t\Im\lambda)&\cos(t\Im\lambda)\end{bmatrix}.$

For any $z\in\mathbb{C}^{d}$ corresponding to a real linear combination of
$x,y$ we have

$\displaystyle\int_{0}^{t}$ $\displaystyle\Re\left\langle\hbox{\rm
e}^{sZ}z,C\hbox{\rm e}^{sZ}z\right\rangle\hbox{\rm d}s$
$\displaystyle=\int_{0}^{t}\left\langle\begin{bmatrix}\cos(s\Im\lambda)&\sin(s\Im\lambda)\\\
-\sin(s\Im\lambda)&\cos(s\Im\lambda)\end{bmatrix}\begin{bmatrix}\Re z\\\ \Im
z\end{bmatrix},\mathbf{C}\begin{bmatrix}\cos(s\Im\lambda)&\sin(s\Im\lambda)\\\
-\sin(s\Im\lambda)&\cos(s\Im\lambda)\end{bmatrix}\begin{bmatrix}\Re z\\\ \Im
z\end{bmatrix}\right\rangle\hbox{\rm e}^{2s\Re\lambda}\hbox{\rm d}s$
$\displaystyle\geq\int_{0}^{t}\left\langle\begin{bmatrix}\cos(s\Im\lambda)&\sin(s\Im\lambda)\\\
-\sin(s\Im\lambda)&\cos(s\Im\lambda)\end{bmatrix}\begin{bmatrix}\Re z\\\ \Im
z\end{bmatrix},\mathbf{C}\begin{bmatrix}\cos(s\Im\lambda)&\sin(s\Im\lambda)\\\
-\sin(s\Im\lambda)&\cos(s\Im\lambda)\end{bmatrix}\begin{bmatrix}\Re z\\\ \Im
z\end{bmatrix}\right\rangle\hbox{\rm d}s$

The scalar product inside the integral is non-negative and periodic as a
function of $s$, therefore either it is identically zero or the integral
diverges. Eventually, the scalar product being identically zero is equivalent
to $\hbox{\rm e}^{tZ}z\in\ker C$ for every $t\geq 0$, completing the proof. ∎

######  Proposition 21.

Let $\mathcal{T}$ be a gaussian QMS with an invariant state $\rho$ and a
positive spectral gap $g$. Then $Z$ is stable.

###### Proof.

Suppose $\mathbf{Z}$ has an eigenvalue $\lambda$ with $\Re\lambda\geq 0$. From
Lemma 20 we can find $z\in\mathbb{C}^{d}$ such that either
$\lim_{t\to\infty}\int_{0}^{t}\Re\left\langle\hbox{\rm e}^{sZ}z,C\hbox{\rm
e}^{sZ}z\right\rangle\hbox{\rm d}s=+\infty$ or $\hbox{\rm e}^{tZ}z\in\ker C$
for every $t\geq 0$. In the former case, the constant $c_{t}(z)$ multiplying
the Weyl operator in (9) satisfies

$c_{t}(z)=\exp\left(-\frac{1}{2}\int_{0}^{t}\Re\left\langle\hbox{\rm
e}^{sZ}z,C\hbox{\rm e}^{sZ}z\right\rangle\hbox{\rm d}s\right)\longrightarrow
0.$ (32)

We can explicitly compute

$|\hat{\rho}(z)|=|\hbox{\rm tr}(\rho W(z)|=|\hbox{\rm
tr}(\rho\mathcal{T}_{t}(W(z))|=|c_{t}(z)\hat{\rho}(\hbox{\rm
e}^{tZ}z)|=c_{t}(z)|\hat{\rho}(\hbox{\rm e}^{\lambda t}z)|\leq c_{t}(z),$ (33)

and,, for every $r>0$, equation (33) reads $|\hat{\rho}(rz)|=0.$ This is a
contradiction since the characteristic function is continuous and
$\hat{\rho}(0)=1$. On the other hand, if $z\in\ker C$, from explicit
computations

$\mathbbm{1}=\mathcal{T}_{t}(W(z))\mathcal{T}_{t}(W(z))^{*}.$

Therefore

$\left\|T_{t}(W(z)\rho^{1/2})\right\|^{2}_{2}=\hbox{\rm
tr}\left(\rho\mathcal{T}_{t}(W(z))^{*}\mathcal{T}_{t}(W(z))\right)=1$

and

$\left\|T_{t}(W(z)\rho^{1/2})-\left\langle\rho^{1/2},W(z)\rho^{1/2}\right\rangle_{2}\rho^{1/2}\right\|^{2}_{2}\geq
1-\left|\hat{\rho}(z)\right|^{2}.$

For the semigroup to have a spectral gap and inequality (1) to hold we must
have $\left|\hat{\rho}(z)\right|=1$ which is a contradiction since then

$\left\|W(z)\rho^{1/2}-\left\langle\rho^{1/2},W(z)\rho^{1/2}\right\rangle_{2}\rho^{1/2}\right\|_{2}^{2}=1-\left|\hat{\rho}(z)\right|^{2}=0.$

∎

To prove the necessity of $\mathbf{C_{Z}}>0$, by contradiction, one can show
that if $\ker\mathbf{C_{Z}}\neq\\{0\\}$ then (1) does not hold for some
$x\in\mathfrak{I}_{2}(\mathsf{h})$ and $t>0$ and so there is no strictly
positive spectral gap. The intuition behind the proof is that if
$(z_{p},z_{q})\in\ker\mathbf{C_{Z}}$ with $z_{p},z_{q}\in\mathbb{C}^{d}$ then
(1) fails when considering
$x=\sum_{j=1}^{d}(z_{p,j}p_{j}-z_{q,j}q_{j})\rho^{1/2}$. This choice for $x$
however is not rigorous since
$\sum_{j=1}^{d}(z_{p,j}p_{j}-z_{q,j}q_{j})\not\in\mathcal{B}(\mathsf{h})$ and
we have to work out computations through a limiting procedure. However, we can
prove the following proposition whose technical details are deferred to
Appendix B.

######  Proposition 22.

Let $\mathcal{T}$ be a gaussian QMS with a faithful invariant state $\rho$. If
$\ker\mathbf{C_{Z}}\neq\\{0\\}$. Then the semigroup has no spectral gap.

###### Proof of Theorem 19.

From Proposition 21 we have stability of $Z$, while Proposition 22 gives
$\mathbf{C_{Z}}>0$. ∎

##  6 A one-dimensional case

The detailed analysis of a one-dimensional case with the explicit computation
of $g$ in terms of a few parameters clarifies why the condition
$\mathbf{C_{Z}}>0$ is necessary for $g>0$. Moreover, it shows that one can
find a quantum Ornstein-Uhlenbeck process with $g=0$ with a classical
subprocess, obtained by restriction to an abelian subalgebra, with strictly
positive spectral gap.

Fix $d=1$. In order to further simplify the notation we consider

$L_{1}=\mu a,\qquad L_{2}=\lambda a^{\dagger},\qquad
H=\Omega\,a^{\dagger}a+\kappa(a^{\dagger\,2}+a^{2})/2$

with $0<\lambda<\mu$, $\Omega,\kappa\in\mathbb{R}$. Define
$\gamma=(\mu^{2}-\lambda^{2})/2$. In this framework a faithful invariant state
exists if and only if $\gamma^{2}+\Omega^{2}-\kappa^{2}>0$ (see [1] Theorem 9
and Proposition 5). Moreover

$\mathbf{C}=\left[\begin{array}[]{cc}\mu^{2}+\lambda^{2}&0\\\
0&\mu^{2}+\lambda^{2}\end{array}\right],\quad\mathbf{Z}=\left[\begin{array}[]{cc}-\gamma&\kappa-\Omega\\\
\kappa+\Omega&-\gamma\end{array}\right],\quad\mathbf{C}_{\mathbf{Z}}=\left[\begin{array}[]{cc}\mu^{2}+\lambda^{2}&2\mathrm{i}\gamma\\\
-2\mathrm{i}\gamma&\mu^{2}+\lambda^{2}\end{array}\right]$

Solving $\mathbf{Z}^{*}\mathbf{S}+\mathbf{S}\mathbf{Z}+\mathbf{C}=0$ we find
the covariance matrix of the unique invariant state

$\mathbf{S}=\frac{\mu^{2}+\lambda^{2}}{2\gamma(\gamma^{2}+\Omega^{2}-\kappa^{2})}\left[\begin{array}[]{cc}\gamma^{2}+\Omega(\Omega+\kappa)&\kappa\gamma\\\
\kappa\gamma&\gamma^{2}+\Omega(\Omega-\kappa)\end{array}\right]$ (34)

Since
$\tilde{\mathbf{S}}^{\frac{1}{2}}\mathbf{Z}\tilde{\mathbf{S}}^{-\frac{1}{2}}+\tilde{\mathbf{S}}^{-\frac{1}{2}}\mathbf{Z}^{*}\tilde{\mathbf{S}}^{\frac{1}{2}}=-\tilde{\mathbf{S}}^{-\frac{1}{2}}\mathbf{C}_{\mathbf{Z}}\tilde{\mathbf{S}}^{-\frac{1}{2}}$,
we have
$\mathbf{C}_{\mathbf{Z}}\tilde{\mathbf{S}}^{-1}=-\mathbf{Z}^{*}-\tilde{\mathbf{S}}\mathbf{Z}\tilde{\mathbf{S}}^{-1}$,
therefore

$\operatorname{tr}\left(\mathbf{C}_{\mathbf{Z}}\tilde{\mathbf{S}}^{-1}\right)=4\gamma.$

In addition

$\operatorname{det}\left(\mathbf{C}_{\mathbf{Z}}\tilde{\mathbf{S}}^{-1}\right)=\frac{\operatorname{det}\left(\mathbf{C}_{\mathbf{Z}}\right)}{\operatorname{det}\left(\tilde{\mathbf{S}}\right)}=\frac{4\mu^{2}\lambda^{2}}{\operatorname{det}(\mathbf{S})-1},$

yielding eventually

$\operatorname{det}\left(\mathbf{C}_{\mathbf{Z}}\tilde{\mathbf{S}}^{-1}\right)=\frac{4\mu^{2}\lambda^{2}}{\frac{(\gamma^{2}+\Omega^{2})(\mu^{2}+\lambda^{2})^{2}}{4\gamma^{2}(\gamma^{2}+\Omega^{2}-\kappa^{2})}-1}=\frac{4\mu^{2}\lambda^{2}\gamma^{2}(\gamma^{2}+\Omega^{2}-\kappa^{2})}{\mu^{2}\lambda^{2}(\gamma^{2}+\Omega^{2})+\gamma^{2}\kappa^{2}}$

Eigenvalues of $\mathbf{C_{Z}}\tilde{\mathbf{S}}^{-1}$ are the roots of
$r^{2}-\operatorname{tr}\left(\mathbf{C}_{\mathbf{Z}}\tilde{\mathbf{S}}^{-1}\right)r+{\operatorname{det}\left(\mathbf{C}_{\mathbf{Z}}\right)}/{\operatorname{det}\left(\tilde{\mathbf{S}}\right)}=0$
and the spectral gap is one-half of

$\gamma-\sqrt{\gamma^{2}-\frac{\mu^{2}\lambda^{2}\gamma^{2}(\gamma^{2}+\Omega^{2}-\kappa^{2})}{\mu^{2}\lambda^{2}(\gamma^{2}+\Omega^{2})+\gamma^{2}\kappa^{2}}}=\gamma\left(1-\frac{|\kappa|(\mu^{2}+\lambda^{2})}{2\sqrt{(\mu^{2}\lambda^{2}(\gamma^{2}+\Omega^{2})+\gamma^{2}\kappa^{2})}}\right)$

Note that, for $\lambda=0$ and $\kappa\not=0$, even the invariant state is
faithful by

$\operatorname{det}\left(\tilde{\mathbf{S}}\right)=\operatorname{det}(\mathbf{S})-1=\frac{\kappa^{2}}{4\gamma^{2}(\mu^{4}+\Omega^{2})}>0\,,$

but the spectral gap is zero because $\operatorname{det}(\mathbf{C_{Z}})=0$.
If, in addition $\kappa=2\Omega$ so that $H=\Omega(q^{2}+\hbox{\rm
1\kern-2.8ptl})/2$

$\mathcal{L}(f(q))=\frac{\mu^{2}+\lambda^{2}}{4}f^{\prime\prime}(q)-\frac{\mu^{2}-\lambda^{2}}{2}qf^{\prime}(q)$

which is the generator of a classical Ornstein-Uhlenbeck process which is also
symmetric. The density of the invariant measure (up to the normalization
constant) is $\mathrm{e}^{-(\mu^{2}-\lambda^{2})q^{2}/(\mu^{2}+\lambda^{2})}$
and the spectral gap is $(\mu^{2}-\lambda^{2})/2>0$.

##  7 The spectral gap for the KMS embedding

As we already recalled in the introduction, the invariant density can induce
different semigroups $T$ on the space of Hilbert-Schmidt operators due to non-
commutativity. The $i_{1/2,2}$ embedding has the advantage of producing a
semigroup $T$ of self-adjoint operators in more cases (see [20]), but at the
cost of longer calculations because of the computations for quantities of the
kind of $\hbox{\rm tr}(\rho^{1/2}W(w)\rho^{1/2}W(z))$. These can be simplified
by symplectic diagonalization of $S$ but their explicit derivation is deferred
to Appendix C.

In this section we present the computation of the spectral gap when we
consider the KMS embedding, highlighting the differences with the GNS case and
the changes to be made for the calculations to go through. The spectral gap is
again defined through condition (1), however the semigroup $T_{t}$ is no more
defined by $T_{t}(x\rho^{1/2})=\mathcal{T}_{t}(x)\rho^{1/2}$ but instead by an
adjusted version for the KMS-embedding

$T_{t}(\rho^{\frac{1}{4}}x\rho^{\frac{1}{4}})=\rho^{\frac{1}{4}}\mathcal{T}_{t}(x)\rho^{\frac{1}{4}},\quad\forall
x\in\mathcal{B}(\mathbf{h}).$

We will show that many results obtained in Section 4 follow similarly with
this new embedding, apart for some adjustments and many more calculations. The
overall result can be briefly summarized by saying that the spectral gap is no
longer $-1/2$ times the greatest eigenvalue of
$\mathbf{\tilde{S}}^{\frac{1}{2}}\mathbf{Z}\mathbf{\tilde{S}}^{-\frac{1}{2}}+\mathbf{\tilde{S}}^{-\frac{1}{2}}\mathbf{Z}^{*}\mathbf{\tilde{S}}^{\frac{1}{2}}$
but $-1/2$ times the greatest eigenvalue of an alike matrix obtained by
replacing $\mathbf{\tilde{S}}$ by another operator $\breve{S}$, still linked
with $S$. Remarkably, there will be no need to consider complexifications of
matrices since, in this case, real linear operators will be sufficient to
describe the behaviour of the semigroup.

We start by considering $x\in\mathcal{W}$, this time we denote with
$\breve{x}$ the projection of $\rho^{\frac{1}{4}}x\rho^{\frac{1}{4}}$ onto the
orthogonal of $\rho^{\frac{1}{2}}$, namely

$\breve{x}=\rho^{\frac{1}{4}}\left[\sum_{j=1}^{l}\eta_{j}\left(W(z_{j})-\hbox{\rm
e}^{-\frac{1}{2}\Re\left\langle
z_{j},Sz_{j}\right\rangle}\mathbbm{1}\right)\right]\rho^{\frac{1}{4}}.$ (35)

An analogous result to Lemma 11 and Proposition 12 holds, with some slight
adjustments. First of all we recall that every covariance matrix of a gaussian
state can be symplectically diagonalized [29]. Explicitly, there exists a
symplectic transformation $M$ on $\mathbb{C}^{d}$, i.e. a real linear operator
satisfying $\Im\left\langle Mz,Mw\right\rangle=\Im\left\langle
z,w\right\rangle$ for every $z,w\in\mathbb{C}^{d}$, such that
$S=M^{T}D_{\sigma}M$ with $D_{\sigma}$ the diagonal matrix with entries
$\sigma_{1},\dots,\sigma_{d}\in(1,+\infty)$. We can now state the following
result.

######  Lemma 23.

Let $x\in\mathcal{W}$ and set $\xi_{j}=\mathrm{e}^{-\frac{1}{2}\langle
z_{j},Sz_{j}\rangle}\eta_{j}$ for all $j=1,\ldots,n$. Then

$\displaystyle\left\|T_{t}(\breve{x})\right\|_{2}^{2}$
$\displaystyle=\sum_{j,k=1}^{l}\overline{\xi}_{j}\xi_{k}\left(\mathrm{exp}\left(\Re\left\langle\hbox{\rm
e}^{tZ}z_{j},\breve{S}\hbox{\rm e}^{tZ}z_{k}\right\rangle\right)-1\right)$
(36)
$\displaystyle=\sum_{j,k=1}^{l}\overline{\xi_{j}}\xi_{k}\left(\mathrm{e}^{\breve{s}_{t}(z_{j},z_{k})}-1\right)=\sum_{n\geq
1}\sum_{j,k=1}^{l}\frac{\overline{\xi_{j}}\xi_{k}}{n!}\left(\breve{s}_{t}(z_{j},z_{k})\right)^{n}$
(37)

for all $t\geq 0$, where $\breve{S}=M^{T}D_{\nu}M$, $M$ is the above
symplectic transformation, $\nu_{j}=\operatorname{csch}\coth^{-1}(\sigma_{j})$
and

$\breve{s}_{t}(z,w):=\Re\left\langle\hbox{\rm e}^{tZ}z,\breve{S}\hbox{\rm
e}^{tZ}w\right\rangle\qquad\forall\,t\geq 0,\ z,w\in\mathbb{C}^{d}.$

In particular we have

$\breve{s}_{t}(z,z)=\left\|\hbox{\rm
e}^{t\breve{S}^{1/2}Z\breve{S}^{-1/2}}\breve{S}^{1/2}z\right\|^{2}$ (38)

for all $t\geq 0$ and $z\in\mathbb{C}^{d}$.

###### Proof.

We defer the proof to Appendix C. ∎

The previous Lemma joins in a single statement results similar to those of
Lemma 11 and Proposition 12, simply replacing all instances of
$\mathbf{\tilde{S}}$ with $\breve{S}$.The matrices $\breve{S}$ and $S$ are
diagonalized by the same symplectic transformation $M$ and we know the
explicit relation between their symplectic eigenvalues $\sigma_{j},\nu_{j}$.
Furthermore, from the condition $\sigma_{j}>1$, we infer $\nu_{j}>0$ showing
that $\breve{S}$ is still a positive and invertible matrix, which were the
relevant properties that also $\mathbf{\tilde{S}}$ enjoyed. Letting
$\breve{\omega}_{0}$ the greatest eigenvalue of
$\breve{S}^{\frac{1}{2}}Z\breve{S}^{-\frac{1}{2}}+\breve{S}^{-\frac{1}{2}}Z^{\sharp}\breve{S}^{\frac{1}{2}}$
and $\breve{g}=-\breve{\omega}_{0}/2$, we can now proceed to compute the
spectral gap, in total analogy to what we did for the GNS embedding.

With these new definition a new version for Proposition 16 holds, yielding a
lower bound for the spectral gap with the KMS embedding, namely $2\breve{g}$.

######  Proposition 24.

Let $x\in\mathcal{W}$. It holds

$\left\|T_{t}(\breve{x})\right\|_{2}^{2}\leq\hbox{\rm
e}^{-2\breve{g}t}\left\|\breve{x}\right\|_{2}^{2},\quad\forall t\geq 0.$

The proof follows precisely the lines of the one in Section 4 except that
Lemma 15 and all the computations have to be performed with the
$\breve{\cdot}$ counterparts of the quantities involved.

In the same way we can prove optimality of the spectral gap by following the
proof of Proposition 17 and using the $\breve{\cdot}$ counterparts of the
quantities. We get the following result.

######  Proposition 25.

Let $g>\breve{g}$. There exists $\delta>0$ and $x\in\mathcal{W}$ such that

$\left\|T_{t}(\breve{x})\right\|_{2}^{2}>\hbox{\rm
e}^{-2tg}\left\|\breve{x}\right\|_{2}^{2},\quad\forall t\in(0,\delta).$

Eventually, following again Theorem 18, we arrive to the following result.

######  Theorem 26.

Let $\mathcal{T}$ be a gaussian QMS with a unique invariant faithful gaussian
state and $\zeta=0$. If $\ker Z^{\sharp}\breve{S}+\breve{S}Z\neq\\{0\\}$ then
$\mathcal{T}$ has the spectral gap $\breve{g}$.

As foreshadowed in the introduction to this section, the only difference
between Theorems 18 and 26 is the need to replace $\mathbf{\tilde{S}}$ with
$\breve{S}$. Even the condition on $\ker Z^{\sharp}\breve{S}+\breve{S}Z$ is
the translation of the condition $\mathbf{C_{Z}}>0$ for the GNS embedding
since $-\mathbf{C_{Z}}=\mathbf{Z^{*}\tilde{S}}+\mathbf{\tilde{S}Z}$. However
the explicit conditions we had in Theorem 18 are now replaced by the implicit
requirement of a unique invariant faithful gaussian state. This is due to the
fact that the sufficient conditions we had in the GNS case, namely
$\mathbf{C_{Z}}>0$ and $Z$ stable, are no more necessary for the existence of
a spectral gap (see the last comment of subsection 7.1 below).

Another difference of note, is the lack of a counterpart of Proposition 14 in
the KMS embedding. Indeed we lack a simple formula to connect the matrix
$\breve{S}$ to the operators $Z$ and $C$, that describe the QMS. For this very
reason it is also difficult to compare the spectral gaps that we obtain with
respect to the two different embeddings. We believe that $\breve{g}$ is bigger
than $g$, however we do not have a definite proof. The next example support
this claim as the prototypical example of a gaussian QMS.

###  7.1 one dimensional case: KMS embedding

In Section 6, we computed the spectral gap for the GNS embedding for a typical
model with $d=1$. Here, we consider the same gaussian QMS and we analyse the
situation for the KMS embedding. Starting from the expression for $\mathbf{S}$
in (34) we may symplectically diagonalize it writing
$\mathbf{S}=\mathbf{M}^{*}\mathbf{D_{\sigma}}\mathbf{M}$ with

$\mathbf{M}=\begin{bmatrix}\sqrt{\gamma^{2}+\Omega^{2}}&0\\\
\frac{\kappa\gamma}{\left(\gamma^{2}+\Omega^{2}-\kappa^{2}\right)\sqrt{\gamma^{2}+\Omega^{2}}}&\frac{1}{\sqrt{\gamma^{2}+\Omega^{2}}}\end{bmatrix},\quad\mathbf{D_{\sigma}}=\begin{bmatrix}\sigma&0\\\
0&\sigma\end{bmatrix},\quad\sigma=\frac{\mu^{2}+\lambda^{2}}{2\gamma}\sqrt{\frac{\gamma^{2}+\Omega^{2}}{\gamma^{2}+\Omega^{2}-\kappa^{2}}}.$

Since $d=1$, both the matrix $\mathbf{D_{\sigma}}$ in the previous equation
and $\mathbf{D_{\nu}}$ coming from Lemma 23 are actually multiple of the
identity therefore we can write

$\mathbf{\breve{S}}=\frac{\operatorname{csch}\coth^{-1}(\sigma)}{\sigma}\mathbf{S}=\frac{\operatorname{csch}\coth^{-1}(\sigma)}{\sigma}\mathbf{M}^{*}\left(\sigma\mathbbm{1}\right)\mathbf{M}.$
(39)

In particular, contrary to what happens in the GNS embedding with
$\mathbf{\tilde{S}}$, the operator $\mathbf{\breve{S}}$ does not involve any
operations that require the complexification for its definition. Therefore
equation (39) can simply be formulated using real linear operators, i.e.

$\breve{S}=\frac{\operatorname{csch}\coth^{-1}(\sigma)}{\sigma}S.$

This allows us to also avoid consider complexification of the operators $Z,C$
so that the computations for the spectral gap can proceed, using just real
linear operators, as follows

$\displaystyle\breve{S}^{\frac{1}{2}}Z\breve{S}^{-\frac{1}{2}}+\breve{S}^{-\frac{1}{2}}Z^{\sharp}\breve{S}^{\frac{1}{2}}$
$\displaystyle=S^{\frac{1}{2}}ZS^{-\frac{1}{2}}+S^{-\frac{1}{2}}Z^{\sharp}S^{\frac{1}{2}}$
$\displaystyle=-S^{-\frac{1}{2}}CS^{-\frac{1}{2}},$

which is similar to $-CS^{-1}$. Notice that in the calculations we used once
again the algebraic characterization of the covariance matrix $S$, namely
$SZ+Z^{\sharp}S=-C$. The eigenvalues of $CS^{-1}$ satisfy now the equation
$r^{2}-\hbox{\rm tr}(CS^{-1})r+\frac{\det C}{\det S}=0$. Borrowing some of the
calculations from Section 6 we have

$\hbox{\rm
tr}(CS^{-1})=4\gamma,\quad\det(CS^{-1})=\frac{4\gamma^{2}(\gamma^{2}+\Omega^{2}-\kappa^{2})}{\Omega^{2}+\gamma^{2}}$

In particular $2\breve{g}$ is the smallest eigenvalue of $CS^{-1}$ and so

$\breve{g}=\gamma\left(1-\frac{\left|\kappa\right|}{\sqrt{\Omega^{2}+\gamma^{2}}}\right).$

Comparing this quantity with $g$ obtained in Section 6 we have $\breve{g}>g$
if and only if

$\frac{\left|\kappa\right|}{\sqrt{\Omega^{2}+\gamma^{2}}}<\frac{\left|\kappa\right|(\mu^{2}+\lambda^{2})}{2\sqrt{\mu^{2}\lambda^{2}(\Omega^{2}+\gamma^{2})+\gamma^{2}\kappa^{2}}}$

which simplifies down to (recalling that $\gamma=(\mu^{2}-\lambda^{2})/2$)

$\gamma^{2}(\gamma^{2}+\Omega^{2}-\kappa^{2})>0.$

The inequality is now always satisfied since $\gamma>0$ and
$\Omega^{2}+\gamma^{2}-\kappa^{2}>0$ are necessary and sufficient conditions
for the semigroup to have a gaussian invariant state (see [1]).

In particular we can set $\lambda=0$, resulting in the non-invertibility of
$\mathbf{C_{Z}}$, and still have $\breve{g}>0$. This shows that the conditions
for the existence of a unique faithful invariant gaussian state of Theorem 18,
namely $Z$ stable and $\mathbf{C_{Z}}>0$, are only sufficient for the
existence of a spectral gap in the KMS embedding, unlike in the GNS one.

Acknowledgements The authors are members of GNAMPA-INdAM. FF acknowledges the
support of the MUR grant “Dipartimento di Eccellenza 2023–2027” of
Dipartimento di Matematica, Politecnico di Milano and “Centro Nazionale di
ricerca in HPC, Big Data and Quantum Computing”. DP, ES and VU have been
supported by the MUR grant “Dipartimento di Eccellenza 2023–2027” of
Dipartimento di Matematica, Università di Genova.

##  Appendix A Calculations on $\mathbf{C_{Z}}$

######  Lemma 27.

Matrices $\mathbf{Z}$ and $\mathbf{C_{Z}}$ are given by (17) and (18).

###### Proof.

Using the formula of the complexification of a real linear operator we have

$\displaystyle\mathbf{Z}$
$\displaystyle=\frac{1}{2}\begin{bmatrix}\Re(U^{T}\overline{U}-V^{T}\overline{V}+U^{T}V-V^{T}U)&\Im(-U^{T}\overline{U}+V^{T}\overline{V}+U^{T}V-V^{T}U)\\\
\Im(U^{T}\overline{U}-V^{T}\overline{V}+U^{T}V-V^{T}U)&\Re(U^{T}\overline{U}-V^{T}\overline{V}-U^{T}V+V^{T}U)\end{bmatrix}$
$\displaystyle+\begin{bmatrix}-\Im\left(\Omega+\kappa\right)&\Re\left(\kappa-\Omega\right)\\\
\Re\left(\Omega+\kappa\right)&\Im\left(\kappa-\Omega\right)\end{bmatrix}$
$\displaystyle=\frac{1}{2}\begin{bmatrix}\Re\left(\left(U-\overline{V}\right)^{*}\left(U+\overline{V}\right)\right)&\Im\left(\left(U-\overline{V}\right)^{*}\left(U-\overline{V}\right)\right)\\\
-\Im\left(\left(U+\overline{V}\right)^{*}\left(U+\overline{V}\right)\right)&\Re\left(\left(U+\overline{V}\right)^{*}\left(U-\overline{V}\right)\right)\end{bmatrix}+\begin{bmatrix}-\Im\left(\Omega+\kappa\right)&\Re\left(\kappa-\Omega\right)\\\
\Re\left(\Omega+\kappa\right)&\Im\left(\kappa-\Omega\right)\end{bmatrix}.$

Similarly

$\displaystyle\mathbf{C}$
$\displaystyle=\begin{bmatrix}\Re(\overline{U^{*}U+V^{*}V}+U^{T}V+V^{T}U)&-\Im(\overline{U^{*}U+V^{*}V}-U^{T}V-V^{T}U)\\\
\Im(\overline{U^{*}U+V^{*}V}+U^{T}V+V^{T}U)&\Re(\overline{U^{*}U+V^{*}V}-U^{T}V-V^{T}U)\end{bmatrix}$
$\displaystyle=\begin{bmatrix}\Re\left(\left(U+\overline{V}\right)^{*}\left(U+\overline{V}\right)\right)&\Im\left(\left(U+\overline{V}\right)^{*}\left(U-\overline{V}\right)\right)\\\
-\Im\left(\left(U-\overline{V}\right)^{*}\left(U+\overline{V}\right)\right)&\Re\left(\left(U-\overline{V}\right)^{*}\left(U-\overline{V}\right)\right)\end{bmatrix}.$

Moreover

$\mathbf{Z}^{*}\mathbf{J}+\mathbf{JZ}=\mathbf{JZ}-(\mathbf{JZ})^{*}$

and

$\displaystyle\mathbf{JZ}$
$\displaystyle=\frac{1}{2}\begin{bmatrix}-\Im\left(\left(U+\overline{V}\right)^{*}\left(U+\overline{V}\right)\right)&\Re\left(\left(U+\overline{V}\right)^{*}\left(U-\overline{V}\right)\right)\\\
-\Re\left(\left(U-\overline{V}\right)^{*}\left(U+\overline{V}\right)\right)&-\Im\left(\left(U-\overline{V}\right)^{*}\left(U-\overline{V}\right)\right)\end{bmatrix}$
$\displaystyle+\begin{bmatrix}\Re(\Omega+\kappa)&\Im(\kappa-\Omega)\\\
\Im(\kappa+\Omega)&\Re(\Omega-\kappa)\end{bmatrix}$

Now note that the first matrix in the expression for $\mathbf{JZ}$ is anti-
selfadjoint while the second one is selfadjoint. In particular then

$\mathbf{Z}^{*}\mathbf{J}+\mathbf{JZ}=\begin{bmatrix}-\Im\left(\left(U+\overline{V}\right)^{*}\left(U+\overline{V}\right)\right)&\Re\left(\left(U+\overline{V}\right)^{*}\left(U-\overline{V}\right)\right)\\\
-\Re\left(\left(U-\overline{V}\right)^{*}\left(U+\overline{V}\right)\right)&-\Im\left(\left(U-\overline{V}\right)^{*}\left(U-\overline{V}\right)\right)\end{bmatrix}.$

Using the definition of
$\mathbf{C_{Z}}=\mathbf{C}-\mathrm{i}\left(\mathbf{Z}^{*}\mathbf{J}+\mathbf{JZ}\right)$
we get the desired result. ∎

##  Appendix B Proof of Proposition 22

This appendix is devoted to the proof of Proposition 22 Let
$(z_{p},z_{q})\in\ker\mathbf{C_{Z}}$ with $z_{p},z_{q}\in\mathbb{C}^{d}$ and
consider $x=\sum_{j=1}^{d}(z_{p,j}p_{j}-z_{q,j}q_{j})\rho^{1/2}$. Note that

$\displaystyle\sum_{j=1}^{d}\left(z_{p,j}p_{j}-z_{q,j}q_{j}\right)$
$\displaystyle=\sum_{j=1}^{d}\left(\Re z_{p,j}p_{j}-\Re
z_{q,j}q_{j}\right)+\mathrm{i}\sum_{j=1}^{d}\left(\Im z_{p,j}p_{j}-\Im
z_{q,j}q_{j}\right)$
$\displaystyle=\sqrt{2}\left(p(z_{1})+\mathrm{i}p(z_{2})\right)$

where $\sqrt{2}z_{1}=\Re z_{p}+\mathrm{i}\Re z_{q}$, $\sqrt{2}z_{2}=\Im
z_{p}+\mathrm{i}\Im z_{q}$. Which motivates the introduction of the following
notation

$\tilde{W}(z_{p},z_{q})=W(z_{1})+\mathrm{i}W(z_{2})$

and the ratios of increments

$R_{z}(r)=\frac{W(rz)-\mathbbm{1}}{-\mathrm{i}r},\quad\tilde{R}_{z_{p},z_{q}}(r)=\frac{\tilde{W}(rz_{p},rz_{q})-(1+\mathrm{i})\mathbbm{1}}{-\mathrm{i}r}=R_{z_{1}}(r)+\mathrm{i}R_{z_{2}}(r).$
(40)

In particular it holds

$\sum_{j=1}^{d}\left(z_{p,j}p_{j}-z_{q,j}q_{j}\right)=\lim_{r\to
0^{+}}\frac{W(rz_{1})+\mathrm{i}W(rz_{2})-(1+\mathrm{i})\mathbbm{1}}{-\mathrm{i}r}=\lim_{r\to
0^{+}}\tilde{R}_{z_{p},z_{q}}(r)$

where the limits holds in the strong operator topology and on a suitable dense
domain. We can now prove that the existence of a spectral gap implies
$\mathbf{C_{Z}}>0$, starting from a lemma.

######  Lemma 28.

Let $\mathcal{T}$ be a gaussian QMS with an invariant state $\rho$. For every
$z,w\in\mathbb{C}^{d}$ we have

$\left.\frac{\hbox{\rm d}}{\hbox{\rm d}t}\hbox{\rm
tr}\left(\rho\mathcal{T}_{t}(R_{z}(r))^{*}\mathcal{T}_{t}(R_{w}(r))\right)\right|_{t=0}=-\left\langle\begin{pmatrix}\Re
z\\\ \Im z\end{pmatrix},\mathbf{C_{Z}}\begin{pmatrix}\Re w\\\ \Im
w\end{pmatrix}\right\rangle\hbox{\rm tr}\left(\rho W(-rz)W(rw)\right)$

###### Proof.

Note at first that, from (9),

$\displaystyle\mathcal{T}_{t}(W(z))\mathcal{T}_{t}(W(w))$
$\displaystyle=\alpha_{t}(z)\alpha_{t}(w)W(\hbox{\rm e}^{tZ}z)W(\hbox{\rm
e}^{tZ}w)$ $\displaystyle=\alpha_{t}(z)\alpha_{t}(w)\hbox{\rm
e}^{-\mathrm{i}\Im\left\langle\hbox{\rm e}^{tZ}z,\hbox{\rm
e}^{tZ}w\right\rangle}W(\hbox{\rm e}^{tZ}(z+w))$
$\displaystyle=\exp\left\\{\int_{0}^{t}\Re\left\langle\hbox{\rm
e}^{sZ}z,C\hbox{\rm e}^{sZ}w\right\rangle\hbox{\rm
d}s-\mathrm{i}\Im\left\langle\hbox{\rm e}^{tZ}z,\hbox{\rm
e}^{tZ}w\right\rangle\right\\}\mathcal{T}_{t}(W(z+w)),$

where we set

$\alpha_{t}(z)=\exp\left(-\frac{1}{2}\int_{0}^{t}\Re\left\langle\mathrm{e}^{sZ}z,C\mathrm{e}^{sZ}z\right\rangle\mathrm{d}s\right).$

Tracing both sides of the previous equation against $\rho$ and taking the
derivative at $t=0$ one gets

$\displaystyle\left.\frac{\hbox{\rm d}}{\hbox{\rm d}t}\hbox{\rm
tr}\left(\rho\mathcal{T}_{t}(W(z))\mathcal{T}_{t}(W(w))\right)\right|_{t=0}$
$\displaystyle=\left(\Re\left\langle
z,Cw\right\rangle-\mathrm{i}\Im\left\langle
Zz,w\right\rangle-\mathrm{i}\Im\left\langle z,Zw\right\rangle\right)$
$\displaystyle\cdot\hbox{\rm e}^{-\mathrm{i}\Im\left\langle
z,w\right\rangle}\hbox{\rm tr}\left(\rho W(z+w)\right)$
$\displaystyle=\left\langle\begin{pmatrix}\Re z\\\ \Im
z\end{pmatrix},\mathbf{C_{Z}}\begin{pmatrix}\Re w\\\ \Im
w\end{pmatrix}\right\rangle\hbox{\rm tr}\left(\rho W(z)W(w)\right).$

where the first equality is due to explicit computations and the invariant
property for $\rho$, while the second one is due to (3) and (15). Eventually

$\displaystyle\hbox{\rm
tr}\left(\rho\mathcal{T}_{t}(R_{z}(r))^{*}\mathcal{T}_{t}(R_{w}(r))\right)$
$\displaystyle=\frac{1}{r^{2}}\hbox{\rm
tr}\left(\rho\mathcal{T}_{t}(W(-rz)-\mathbbm{1})\mathcal{T}_{t}(W(rw)-\mathbbm{1})\right)$
$\displaystyle=\frac{1}{r^{2}}\hbox{\rm
tr}\left(\rho\mathcal{T}_{t}(W(-rz))\mathcal{T}_{t}(W(rw))\right)$
$\displaystyle+\frac{1}{r^{2}}\left(-\hbox{\rm tr}(\rho W(-rz))-\hbox{\rm
tr}(\rho W(rw))+\hbox{\rm tr}(\rho)\right)$

Therefore, when taking the derivative at $t=0$ only the first summand does not
vanish and the result is proven. ∎

######  Lemma 29.

Let $\mathcal{T}$ be a gaussian QMS with a faithful gaussian invariant state
$\rho$. Then for every $z,w\in\mathbb{C}^{d}$ it holds

$\lim_{r\to
0^{+}}\left\|\tilde{R}_{z_{p},z_{q}}(r)\rho^{1/2}\right\|_{2}^{2}>0$

###### Proof.

Note at first that, for $z,w\in\mathbb{C}^{d}$

$\displaystyle\hbox{\rm tr}(\rho R_{z}(r)^{*}R_{w}(r))$
$\displaystyle=\frac{1}{r^{2}}\hbox{\rm
tr}\left(\rho(W(-rz)-\mathbbm{1})(W(rw)-\mathbbm{1})\right)$
$\displaystyle=\frac{\hbox{\rm e}^{\mathrm{i}r^{2}\Im\left\langle
z,w\right\rangle}\hat{\rho}(-r(z-w))-1}{r^{2}}-\frac{\hat{\rho}(-rz)-1}{r^{2}}-\frac{\hat{\rho}(rw)-1}{r^{2}}$

and the limit as $r\to 0^{+}$ exists and is finite since the characteristic
function of $\rho$ is differentiable. In particular it holds

$\displaystyle\lim_{r\to 0^{+}}\hbox{\rm tr}(\rho R_{z}(r)^{*}R_{w}(r))$
$\displaystyle=\left(\mathrm{i}\Im\left\langle
z,w\right\rangle-\frac{1}{2}\Re\left\langle(z-w),S(z-w)\right\rangle\right)$
$\displaystyle+\frac{1}{2}\Re\left\langle
z,Sz\right\rangle+\frac{1}{2}\Re\left\langle w,Sw\right\rangle$
$\displaystyle=\Re\left\langle z,Sw\right\rangle+\mathrm{i}\Im\left\langle
z,w\right\rangle=\left\langle\begin{bmatrix}\Re z\\\ \Im
z\end{bmatrix},(\mathbf{S}+\mathrm{i}\mathbf{J})\begin{bmatrix}\Re w\\\ \Im
w\end{bmatrix}\right\rangle$

thanks to (15). To conclude that the desired limit exists and is finite it
sufficient to note that, using definition (40),

$\displaystyle\left\|\tilde{R}_{z_{p},z_{q}}(r)\rho^{1/2}\right\|_{2}^{2}$
$\displaystyle=\hbox{\rm
tr}\left(\rho\left(\tilde{R}_{z_{p},z_{q}}(r)\right)^{*}\tilde{R}_{z_{p},z_{q}}(r)\right)$
$\displaystyle=\hbox{\rm tr}\left(\rho
R_{z_{1}}(r)^{*}R_{z_{1}}(r)\right)+\hbox{\rm tr}\left(\rho
R_{z_{2}}(r)^{*}R_{z_{2}}(r)\right)$ $\displaystyle+\mathrm{i}\,\hbox{\rm
tr}\left(\rho R_{z_{1}}(r)^{*}R_{z_{2}}(r)\right)-\mathrm{i}\,\hbox{\rm
tr}\left(\rho R_{z_{2}}(r)^{*}R_{z_{1}}(r)\right)$

and taking the limit as $r\to 0^{+}$

$\displaystyle\lim_{r\to
0^{+}}\left\|\tilde{R}_{z_{p},z_{q}}(r)\rho^{1/2}\right\|_{2}^{2}$
$\displaystyle=\left\langle\begin{bmatrix}\Re z_{1}+\mathrm{i}\Re z_{2}\\\ \Im
z_{1}+\mathrm{i}\Im
z_{2}\end{bmatrix},(\mathbf{S}+\mathrm{i}\mathbf{J})\begin{bmatrix}\Re
z_{1}+\mathrm{i}\Re z_{2}\\\ \Im z_{1}+\mathrm{i}\Im
z_{2}\end{bmatrix}\right\rangle$
$\displaystyle=\frac{1}{2}\left\langle\begin{bmatrix}z_{p}\\\
z_{q}\end{bmatrix},(\mathbf{S}+\mathrm{i}\mathbf{J})\begin{bmatrix}z_{p}\\\
z_{q}\end{bmatrix}\right\rangle>0$

Where positivity of the last quantity follows from the fact that
$\mathbf{S}+\mathrm{i}\mathbf{J}=\overline{\mathbf{S}-\mathrm{i}\mathbf{J}}>0$.
∎

We are now ready to prove Proposition 22.

###### Proof of Proposition 22.

Let $[z_{p},z_{q}]^{T}$ be an eigenvector for $\mathbf{C_{Z}}$ associated with
the eigenvalue $0$. For every $g,r>0$ consider the function

$f_{g,r}(t)=\left\|\mathcal{T}_{t}\left(\tilde{R}_{z_{p},z_{q}}(r)\right)\rho^{1/2}\right\|_{2}^{2}-\hbox{\rm
e}^{-2gt}\left\|\tilde{R}_{z_{p},z_{q}}(r)\rho^{1/2}\right\|_{2}^{2}$

Clearly $f_{g,r}(0)=0$ and we will show that for every $g>0$ there exists
$r_{g}>0$ such that $f^{\prime}_{g,r_{g}}(0)>0$. In this way, at least for
$t>0$ small enough, $f_{g,r_{g}}(t)>0$ contradicting the existence of a
spectral gap. The derivative at $t=0$ of $f_{g,r}(t)$ is

$f^{\prime}_{g,r}(0)=\left.\frac{\hbox{\rm d}}{\hbox{\rm
d}t}\left\|\mathcal{T}_{t}\left(\tilde{R}_{z_{p},z_{q}}(r)\right)\rho^{1/2}\right\|_{2}^{2}\right|_{t=0}+2g\left\|\tilde{R}_{z_{p},z_{q}}(r)\rho^{1/2}\right\|_{2}^{2}.$

The first summand can be further expanded using

$\displaystyle\left\|\mathcal{T}_{t}\left(\tilde{R}_{z_{p},z_{q}}(r)\right)\rho^{1/2}\right\|_{2}^{2}$
$\displaystyle=\hbox{\rm
tr}\left(\rho\mathcal{T}_{t}(\tilde{R}_{z_{p},z_{q}}(r))^{*}\mathcal{T}_{t}(\tilde{R}_{z_{p},z_{q}}(r))\right)$
$\displaystyle=\hbox{\rm
tr}\left[\rho\mathcal{T}_{t}(R_{z_{1}}(r))^{*}\mathcal{T}_{t}(R_{z_{1}}(r)))\right]+\hbox{\rm
tr}\left[\rho\mathcal{T}_{t}(R_{z_{2}}(r))^{*}\mathcal{T}_{t}(R_{z_{2}}(r))\right]$
$\displaystyle+\mathrm{i}\hbox{\rm
tr}\left[\rho\mathcal{T}_{t}(R_{z_{1}}(r))^{*}\mathcal{T}_{t}(R_{z_{2}}(r))\right]-\mathrm{i}\hbox{\rm
tr}\left[\rho\mathcal{T}_{t}(R_{z_{2}}(r))^{*}\mathcal{T}_{t}(R_{z_{1}}(r))\right]$

and then taking the derivative at $t=0$, using Lemma 28

$\displaystyle\left.\frac{\hbox{\rm d}}{\hbox{\rm
d}t}\left\|\mathcal{T}_{t}\left(\tilde{R}_{z_{p},z_{q}}(r)\right)\rho^{1/2}\right\|_{2}^{2}\right|_{t=0}$
$\displaystyle=-\left\langle\begin{pmatrix}\Re z_{1}\\\ \Im
z_{1}\end{pmatrix},\mathbf{C_{Z}}\begin{pmatrix}\Re z_{1}\\\ \Im
z_{1}\end{pmatrix}\right\rangle\hbox{\rm tr}\left(\rho
W(-rz_{1})W(rz_{1})\right)$ $\displaystyle-\left\langle\begin{pmatrix}\Re
z_{2}\\\ \Im z_{2}\end{pmatrix},\mathbf{C_{Z}}\begin{pmatrix}\Re z_{2}\\\ \Im
z_{2}\end{pmatrix}\right\rangle\hbox{\rm tr}\left(\rho
W(-rz_{2})W(rz_{2})\right)$
$\displaystyle+\mathrm{i}\left\langle\begin{pmatrix}\Re z_{2}\\\ \Im
z_{2}\end{pmatrix},\mathbf{C_{Z}}\begin{pmatrix}\Re z_{1}\\\ \Im
z_{1}\end{pmatrix}\right\rangle\hbox{\rm tr}\left(\rho
W(-rz_{2})W(rz_{1})\right)$
$\displaystyle-\mathrm{i}\left\langle\begin{pmatrix}\Re z_{1}\\\ \Im
z_{1}\end{pmatrix},\mathbf{C_{Z}}\begin{pmatrix}\Re z_{2}\\\ \Im
z_{2}\end{pmatrix}\right\rangle\hbox{\rm tr}\left(\rho
W(-rz_{1})W(rz_{2})\right).$

Letting $r\to 0^{+}$ we get

$\lim_{r\to 0^{+}}\left.\frac{\hbox{\rm d}}{\hbox{\rm
d}t}\left\|\mathcal{T}_{t}\left(\tilde{R}_{z_{p},z_{q}}(r)\right)\rho^{1/2}\right\|_{2}^{2}\right|_{t=0}=\left\langle\begin{pmatrix}z_{p}\\\
z_{q}\end{pmatrix},\mathbf{C_{Z}}\begin{pmatrix}z_{p}\\\
z_{q}\end{pmatrix}\right\rangle=0.$

Eventually, using Lemma 29, we get

$\displaystyle\lim_{r\to 0^{+}}f^{\prime}_{g,r}(0)$ $\displaystyle=\lim_{r\to
0^{+}}\left\|\tilde{R}_{z_{p},z_{q}}(r)\rho^{1/2}\right\|_{2}^{2}>0.$

∎

##  Appendix C Proof of Lemma 23

This appendix contains the proof of Lemma 23. The first part revolves around
the computation of the quantity

$\hbox{\rm tr}(\rho^{\frac{1}{2}}W(z)\rho^{\frac{1}{2}}W(w)),$

where $z,w\in\mathbb{C}^{d}$ and $\rho=\rho_{(0,S)}$ is a faithful gaussian
state. The starting point is to simplify the expression for $\rho$ as one does
for gaussian vectors in classical probability. Let us start by introducing
some notation. For $c\in\mathbb{R}^{d}$, we denote with $D_{c}$ the real
linear operator defined by

$D_{c}z=\operatorname{diag}(c_{1},\dots,c_{d})z,\qquad\mathbf{D_{c}}\begin{bmatrix}x\\\
y\end{bmatrix}=\operatorname{diag}(c_{1},\dots,c_{d},c_{1},\dots,c_{d})\begin{bmatrix}x\\\
y\end{bmatrix}$

From [29] every faithful gaussian state $\rho=\rho_{(0,S)}$ has the form

$\rho=\Gamma(M)^{-1}\prod_{j}(1-\hbox{\rm e}^{-s_{j}})\hbox{\rm
e}^{-\sum_{j}s_{j}a_{j}^{\dagger}a_{j}}\Gamma(M),$

where $M$ is a symplectic, or Bogoliubov, transformation (i.e. $M^{T}JM=J$)
such that ${M}^{T}D_{\sigma}{M}=S$, having set $\sigma\in\mathbb{R}^{d}$ and
$\sigma_{j}=\coth(s_{j}/2)$. Eventually $\Gamma(M)$ is a unitary operator
satisfying

$\Gamma(M)W(z)\Gamma(M)^{-1}=W(Mz).$

The previous discussion allows us to have an expression for
$\rho^{\frac{1}{2}}$ that one can work with, explicitly

$\rho^{-\frac{1}{2}}=\Gamma(M)^{-1}\prod_{j}(1-\hbox{\rm
e}^{-s_{j}})^{\frac{1}{2}}\hbox{\rm
e}^{-\sum_{j}s_{j}/2a_{j}^{\dagger}a_{j}}\Gamma(M).$ (41)

Before moving to the actual calculations let us recall first _exponential
vectors_ in $\mathsf{h}$. For any $f\in\mathbb{C}^{d}$ we have

$e(f)=\hbox{\rm e}^{\frac{\left|z\right|^{2}}{2}}W(f)e(0,\dots,0)=\sum_{n\geq
0}\sum_{\alpha\in\mathbb{N}^{d},\left|\alpha\right|=n}\frac{f_{1}^{\alpha_{1}}\cdot\dots\cdot
f_{d}^{\alpha_{d}}}{\alpha_{1}!\dots\alpha_{d}!}e(\alpha_{1},\dots,\alpha_{d}).$

In particular for any $z\in\mathbb{C}^{d}$ it holds

$W(z)e(f)=\exp\left\\{-\frac{\left|z\right|^{2}}{2}-\left\langle
z,f\right\rangle\right\\}e(z+f).$

Exponential vectors are a total set in $\mathsf{h}$ and provide an alternative
way to computing traces, as highlighted in the following Lemma, whose proof
can be found in [28].

######  Lemma 30.

For $z=x+\mathrm{i}y$

$\mathbbm{1}=\frac{1}{\pi^{d}}\int_{\mathbb{R}^{2d}}\hbox{\rm
e}^{-\left|z\right|^{2}}\left|e(z)\right\rangle\\!\left\langle
e(z)\right|\hbox{\rm d}x\hbox{\rm d}y,$

also

$\hbox{\rm tr}(\rho)=\frac{1}{\pi^{d}}\int_{\mathbb{R}^{2d}}\hbox{\rm
e}^{-\left|z\right|^{2}}\left\langle e(z),\rho e(z)\right\rangle\hbox{\rm
d}x\hbox{\rm d}y$

We can now start the computations with the following two Lemmas.

######  Lemma 31.

For any $f,g\in\mathbb{C}^{d}$ and $c_{1},\dots,c_{d}\in(0,\infty)$ it holds

$\left\langle e(f),\hbox{\rm
e}^{-\sum_{j}c_{j}a_{j}^{\dagger}a_{j}}e(g)\right\rangle=\exp\left\\{\left\langle
f,\hbox{\rm e}^{-D_{c}}g\right\rangle\right\\}.$

###### Proof.

Note at first that, for some fixed index $j$,

$\displaystyle\hbox{\rm e}^{-c_{j}a_{j}^{\dagger}a_{j}}e(f)$
$\displaystyle=\sum_{m\geq 0}\sum_{n\geq
0}\sum_{|\alpha|=n}\frac{f_{1}^{\alpha_{1}}\cdot\dots\cdot
f_{d}^{\alpha_{d}}}{\alpha_{1}!\dots\alpha_{d}!}\frac{1}{m!}(-c_{j}a_{j}^{\dagger}a_{j})^{m}e(\alpha_{1},\dots,\alpha_{d})$
$\displaystyle=\sum_{m\geq 0}\sum_{n\geq
0}\sum_{|\alpha|=n}\frac{f_{1}^{\alpha_{1}}\cdot\dots\cdot
f_{d}^{\alpha_{d}}}{\alpha_{1}!\dots\alpha_{d}!}\frac{1}{m!}(-c_{j}\alpha_{j})^{m}e(\alpha_{1},\dots,\alpha_{d})$
$\displaystyle=\sum_{n\geq
0}\sum_{|\alpha|=n}\frac{f_{1}^{\alpha_{1}}\cdot\dots\cdot
f_{d}^{\alpha_{d}}}{\alpha_{1}!\dots\alpha_{d}!}\hbox{\rm
e}^{-c_{j}\alpha_{j}}e(\alpha_{1},\dots,\alpha_{d})$
$\displaystyle=\sum_{n\geq
0}\sum_{|\alpha|=n}\frac{f_{1}^{\alpha_{1}}\cdot\dots\cdot(\hbox{\rm
e}^{-c_{j}}f_{j})^{\alpha_{j}}\cdot\dots\cdot
f_{d}^{\alpha_{d}}}{\alpha_{1}!\dots\alpha_{d}!}e(\alpha_{1},\dots,\alpha_{d})$
$\displaystyle=e(\operatorname{diag}(1,\dots,1,\hbox{\rm
e}^{-c_{j}},1,\dots,1)f).$

In this way

$\displaystyle\left\langle e(f),\hbox{\rm
e}^{\sum_{j}-c_{j}a_{j}^{\dagger}a_{j}}e(g)\right\rangle$
$\displaystyle=\left\langle e(f),\prod_{j}\left(\hbox{\rm
e}^{-c_{j}a_{j}^{\dagger}a_{j}}\right)e(g)\right\rangle$
$\displaystyle=\left\langle e(f),e(\hbox{\rm
e}^{-D_{c}}g)\right\rangle=\exp\left\\{\left\langle f,\hbox{\rm
e}^{-D_{c}}g\right\rangle\right\\}.$

∎

######  Lemma 32.

Let $f,g,z,w\in\mathbb{C}^{d}$ and $c_{1},\dots,c_{d}\in(0,+\infty)$ it holds

$\displaystyle\left\langle e(f),W(z)\hbox{\rm
e}^{-\sum_{j}c_{j}a_{j}^{\dagger}a_{j}}W(w)e(g)\right\rangle$
$\displaystyle=\exp\left\\{-\frac{\left|z\right|^{2}+\left|w\right|^{2}}{2}-\left\langle
z,\hbox{\rm e}^{-D_{c}}w\right\rangle+\left\langle f,\hbox{\rm
e}^{-D_{c}}g\right\rangle\right\\}$ $\displaystyle\exp\left\\{\left\langle
f,z+\hbox{\rm e}^{-D_{c}}w\right\rangle-\left\langle w+\hbox{\rm
e}^{-D_{c}}z,g\right\rangle\right\\}$

###### Proof.

We have

$\displaystyle\left\langle e(f),W(z)\hbox{\rm
e}^{-\sum_{j}c_{j}a_{j}^{\dagger}a_{j}}W(w)e(g)\right\rangle$
$\displaystyle=\left\langle W(-z)e(f),\hbox{\rm
e}^{-\sum_{j}c_{j}a_{j}^{\dagger}a_{j}}W(w)e(g)\right\rangle$
$\displaystyle=\hbox{\rm e}^{-\frac{\left|z\right|^{2}}{2}+\left\langle
f,z\right\rangle}\hbox{\rm e}^{-\frac{\left|w\right|^{2}}{2}-\left\langle
w,g\right\rangle}$ $\displaystyle\cdot\left\langle e(f-z),\hbox{\rm
e}^{-\sum_{j}c_{j}a_{j}^{\dagger}a_{j}}e(g+w)\right\rangle.$

Now using Lemma 31 and rearranging the terms we get the desired result. ∎

We are now ready to prove the formula

######  Proposition 33.

For $z,w\in\mathbb{C}^{d}$ it holds

$\displaystyle\hbox{\rm tr}(\rho^{1/2}W(z)\rho^{1/2}W(w))$
$\displaystyle=\exp\left\\{-\frac{1}{2}\left(\Re\left\langle
z,Sz\right\rangle+\Re\left\langle w,Sw\right\rangle+2\Re\left\langle
z,\breve{S}w\right\rangle\right)\right\\}$

with $\breve{S}=M^{T}D_{\xi}M$, $\xi_{j}=\operatorname{csch}(s_{j}/2)$ and $M$
coming from (41).

###### Proof.

We start by noticing that using expression (41), the commutation rule for Weyl
operators, the cyclic property of the trace and Lemma 30 we can write

tr
$\displaystyle(\rho^{\frac{1}{2}}W(z)\rho^{\frac{1}{2}}W(w))=\prod_{j}(1-\hbox{\rm
e}^{-s_{j}})\hbox{\rm tr}\left(\hbox{\rm
e}^{-\frac{1}{2}\sum_{j}s_{j}a_{j}^{\dagger}a_{j}}W(Mz)\hbox{\rm
e}^{-\frac{1}{2}\sum_{j}s_{j}a_{j}^{\dagger}a_{j}}W(Mw)\right)$
$\displaystyle=\frac{\prod_{j}(1-\hbox{\rm
e}^{-s_{j}})}{\pi^{2d}}\int_{\mathbb{R}^{4d}}\hbox{\rm
e}^{-(\left|f\right|^{2}+\left|g\right|^{2})}\left\langle
e(f),W\left(\frac{Mw}{2}\right)\hbox{\rm
e}^{-\frac{1}{2}\sum_{j}s_{j}a_{j}^{\dagger}a_{j}}W\left(\frac{Mz}{2}\right)e(g)\right\rangle$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\cdot\left\langle
e(g),W\left(\frac{Mz}{2}\right)\hbox{\rm
e}^{-\frac{1}{2}\sum_{j}s_{j}a_{j}^{\dagger}a_{j}}W\left(\frac{Mw}{2}\right)e(f)\right\rangle\hbox{\rm
d}f\hbox{\rm d}g$

Eventually, using Lemma 32, we obtain that

$\hbox{\rm
tr}(\rho^{\frac{1}{2}}W(z)\rho^{\frac{1}{2}}W(w))=\frac{\prod_{j}(1-\hbox{\rm
e}^{-s_{j}})}{\pi^{2d}}\int_{\mathbb{R}^{4d}}e^{\alpha_{(Mz,Mw)}(f,g)}\hbox{\rm
d}f\hbox{\rm d}g,$

where

$\displaystyle\alpha_{(z,w)}(f,g)$
$\displaystyle=\exp\left\\{-\frac{\left|z\right|^{2}+\left|w\right|^{2}}{4}-\frac{1}{2}\Re\left\langle
z,\hbox{\rm e}^{-\frac{1}{2}D_{s}}w\right\rangle\right\\}$
$\displaystyle\exp\left\\{-\left|f\right|^{2}-\left|g\right|^{2}+2\Re\left\langle
f,\hbox{\rm e}^{-\frac{1}{2}D_{s}}g\right\rangle\right\\}$
$\displaystyle\cdot\exp\left\\{\mathrm{i}\Im\left\langle f,w+\hbox{\rm
e}^{-\frac{1}{2}D_{s}}z\right\rangle+\mathrm{i}\Im\left\langle g,z+\hbox{\rm
e}^{-\frac{1}{2}D_{s}}w\right\rangle\right\\}$

We can rewrite the previous expression in matrix form in the following way

$\displaystyle\alpha_{(z,w)}(f,g)$
$\displaystyle=\exp\left\\{-\frac{1}{4}\left\langle\begin{pmatrix}\Re z\\\ \Im
z\\\ \Re w\\\ \Im w\end{pmatrix},\begin{pmatrix}\mathbbm{1}&\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\\\ \hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}&\mathbbm{1}\end{pmatrix}\begin{pmatrix}\Re
z\\\ \Im z\\\ \Re w\\\ \Im w\end{pmatrix}\right\rangle\right\\}$
$\displaystyle\exp\left\\{-\left\langle\begin{pmatrix}\Re f\\\ \Im f\\\ \Re
g\\\ \Im g\end{pmatrix},\begin{pmatrix}\mathbbm{1}&-\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\\\ -\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}&\mathbbm{1}\end{pmatrix}\begin{pmatrix}\Re
f\\\ \Im f\\\ \Re g\\\ \Im g\end{pmatrix}\right\rangle\right\\}$
$\displaystyle\cdot\exp\left\\{\mathrm{i}\left\langle\begin{pmatrix}\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\mathbf{J}&\mathbf{J}\\\ \mathbf{J}&\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\mathbf{J}\end{pmatrix}\begin{pmatrix}\Re z\\\
\Im z\\\ \Re w\\\ \Im w\end{pmatrix},\begin{pmatrix}\Re f\\\ \Im f\\\ \Re g\\\
\Im g\end{pmatrix}\right\rangle\right\\}$

Recall now for $x,b\in\mathbb{R}^{n}$ and $A\in M_{n}(\mathbb{R})$ we have the
formula for the gaussian integral

$\int_{\mathbb{R}^{n}}\hbox{\rm e}^{-\frac{1}{2}\left\langle
x,Ax\right\rangle+\mathrm{i}\left\langle b,x\right\rangle}\hbox{\rm
d}x=\sqrt{\frac{(2\pi)^{n}}{\det A}}\exp\left\\{-\frac{1}{2}\left\langle
b,A^{-1}b\right\rangle\right\\}.$

We want to use it with

$A=2\begin{pmatrix}\mathbbm{1}&-\hbox{\rm e}^{-\frac{1}{2}\mathbf{D_{s}}}\\\
-\hbox{\rm e}^{-\frac{1}{2}\mathbf{D_{s}}}&\mathbbm{1}\end{pmatrix},\quad
b=\begin{pmatrix}\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\mathbf{J}&\mathbf{J}\\\ \mathbf{J}&\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\mathbf{J}\end{pmatrix}\begin{pmatrix}\Re z\\\
\Im z\\\ \Re w\\\ \Im w\end{pmatrix}.$

Note that, by rearranging rows and columns of $A$ we get

$\frac{1}{2}A=\begin{pmatrix}A_{1}&&&\\\ &A_{2}&&\\\ &&\ddots&\\\
&&&A_{d}\end{pmatrix},\quad A_{j}=\begin{pmatrix}1&0&-\hbox{\rm
e}^{-\frac{s_{j}}{2}}&0\\\ 0&1&0&-\hbox{\rm e}^{-\frac{s_{j}}{2}}\\\
-\hbox{\rm e}^{-\frac{s_{j}}{2}}&0&1&0\\\ 0&-\hbox{\rm
e}^{-\frac{s_{j}}{2}}&0&1\end{pmatrix},$

and for each block we can easily compute $\det A_{j}=(1-\hbox{\rm
e}^{-s_{j}})^{2}$ while

$A_{j}^{-1}=\frac{1}{1-\hbox{\rm e}^{-s_{j}}}\begin{pmatrix}1&0&\hbox{\rm
e}^{-\frac{s_{j}}{2}}&0\\\ 0&1&0&\hbox{\rm e}^{-\frac{s_{j}}{2}}\\\ \hbox{\rm
e}^{-\frac{s_{j}}{2}}&0&1&0\\\ 0&\hbox{\rm
e}^{-\frac{s_{j}}{2}}&0&1\end{pmatrix}.$

In particular $\det A=2^{4d}\prod_{j}(1-\hbox{\rm e}^{-s_{j}})^{2}$ and
rearranging rows and columns to the initial configuration we obtain

$A^{-1}=\frac{1}{2}\begin{pmatrix}\mathbbm{1}&\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\\\ \hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}&\mathbbm{1}\end{pmatrix}\begin{pmatrix}\mathbbm{1}-\hbox{\rm
e}^{-\mathbf{D_{s}}}&0\\\ 0&\mathbbm{1}-\hbox{\rm
e}^{-\mathbf{D_{s}}}\end{pmatrix}^{-1}.$

Therefore we have proved

$\hbox{\rm tr}(\rho^{\frac{1}{2}}W(z)\rho^{\frac{1}{2}}W(w))=\hbox{\rm
e}^{-\frac{1}{2}\beta(Mz,Mw)}$

with

$\displaystyle\beta(z,w)$ $\displaystyle=\left\langle\begin{pmatrix}\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\mathbf{J}&\mathbf{J}\\\ \mathbf{J}&\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\mathbf{J}\end{pmatrix}\begin{pmatrix}\Re z\\\
\Im z\\\ \Re w\\\ \Im w\end{pmatrix},A^{-1}\begin{pmatrix}\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\mathbf{J}&\mathbf{J}\\\ \mathbf{J}&\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\mathbf{J}\end{pmatrix}\begin{pmatrix}\Re z\\\
\Im z\\\ \Re w\\\ \Im w\end{pmatrix}\right\rangle$
$\displaystyle+\frac{1}{2}\left\langle\begin{pmatrix}\Re z\\\ \Im z\\\ \Re
w\\\ \Im w\end{pmatrix},\begin{pmatrix}\mathbbm{1}&\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\\\ \hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}&\mathbbm{1}\end{pmatrix}\begin{pmatrix}\Re
z\\\ \Im z\\\ \Re w\\\ \Im w\end{pmatrix}\right\rangle$
$\displaystyle=\left\langle\begin{pmatrix}\Re z\\\ \Im z\\\ \Re w\\\ \Im
w\end{pmatrix},B\begin{pmatrix}\Re z\\\ \Im z\\\ \Re w\\\ \Im
w\end{pmatrix}\right\rangle+\frac{1}{2}\left\langle\begin{pmatrix}\Re z\\\ \Im
z\\\ \Re w\\\ \Im w\end{pmatrix},\begin{pmatrix}\mathbbm{1}&\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\\\ \hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}&\mathbbm{1}\end{pmatrix}\begin{pmatrix}\Re
z\\\ \Im z\\\ \Re w\\\ \Im w\end{pmatrix}\right\rangle$

where we set

$B=\begin{pmatrix}\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\mathbf{J}&\mathbf{J}\\\ \mathbf{J}&\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\mathbf{J}\end{pmatrix}^{*}A^{-1}\begin{pmatrix}\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\mathbf{J}&\mathbf{J}\\\ \mathbf{J}&\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}\mathbf{J}\end{pmatrix}.$

Now note that every matrix that appears as a block in the previous expression
commutes with the others (e.g. $\mathbf{J}$ with $\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}$ or $\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}$ with $(\mathbbm{1}-\hbox{\rm
e}^{-\mathbf{D_{s}}})^{-1}$). This observation, alongside the explicit
expression we obtained for $A^{-1}$ leads to

$B=\frac{1}{2}\begin{pmatrix}\mathbbm{1}+3\hbox{\rm
e}^{-\mathbf{D_{s}}}&\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}(3\mathbbm{1}+\hbox{\rm
e}^{-\mathbf{D_{s}}})\\\ \hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}(3\mathbbm{1}+\hbox{\rm
e}^{-\mathbf{D_{s}}})&\mathbbm{1}+3\hbox{\rm
e}^{-\mathbf{D_{s}}}\end{pmatrix}\begin{pmatrix}\mathbbm{1}-\hbox{\rm
e}^{-\mathbf{D_{s}}}&0\\\ 0&\mathbbm{1}-\hbox{\rm
e}^{-\mathbf{D_{s}}}\end{pmatrix}^{-1}.$

Putting it all back together we obtain

$\displaystyle\beta(z,w)$ $\displaystyle=\left\langle\begin{pmatrix}\Re z\\\
\Im z\\\ \Re w\\\ \Im w\end{pmatrix},\begin{pmatrix}\mathbbm{1}+\hbox{\rm
e}^{-\mathbf{D_{s}}}&2\hbox{\rm e}^{-\frac{1}{2}\mathbf{D_{s}}}\\\ 2\hbox{\rm
e}^{-\frac{1}{2}\mathbf{D_{s}}}&\mathbbm{1}+\hbox{\rm
e}^{-\mathbf{D_{s}}}\end{pmatrix}\begin{pmatrix}\mathbbm{1}-\hbox{\rm
e}^{-\mathbf{D_{s}}}&0\\\ 0&\mathbbm{1}-\hbox{\rm
e}^{-\mathbf{D_{s}}}\end{pmatrix}^{-1}\begin{pmatrix}\Re z\\\ \Im z\\\ \Re
w\\\ \Im w\end{pmatrix}.\right\rangle$

Performing the final calculation we obtain

$\beta(z,w)=\left\langle\begin{pmatrix}\Re z\\\ \Im z\\\ \Re w\\\ \Im
w\end{pmatrix},\begin{pmatrix}\mathbf{D_{\sigma}}&\mathbf{D_{\xi}}\\\
\mathbf{D_{\xi}}&\mathbf{D_{\sigma}}\end{pmatrix}\begin{pmatrix}\Re z\\\ \Im
z\\\ \Re w\\\ \Im w\end{pmatrix}\right\rangle=\Re\left\langle
z,D_{\sigma}z\right\rangle+\Re\left\langle
w,D_{\sigma}w\right\rangle+2\Re\left\langle z,D_{\xi}w\right\rangle$

where $\xi_{j}=\operatorname{csch}(s_{j}/2)$, ending the proof. ∎

We are now ready to prove Lemma 23 following the same lines of Lemma 11 but
using the new formula for the trace.

######  Proof of Lemma 23.

Using (35) we have

$\displaystyle\left\|T_{t}(\breve{x})\right\|^{2}$
$\displaystyle=\sum_{j,k}\overline{\eta_{j}}\eta_{k}\left[\hbox{\rm
tr}(\rho^{\frac{1}{2}}\mathcal{T}_{t}(W(-z_{j}))\rho^{\frac{1}{2}}\mathcal{T}_{t}(W(z_{k})))-\hbox{\rm
e}^{-\frac{1}{2}\Re\left\langle
z_{j},Sz_{j}\right\rangle-\frac{1}{2}\Re\left\langle
z_{k},Sz_{k}\right\rangle}\right]$
$\displaystyle=\sum_{j,k}\overline{\eta_{j}}\eta_{k}\left[c_{t}(z_{j})c_{t}(z_{k})\hbox{\rm
tr}(\rho^{\frac{1}{2}}W(-\hbox{\rm e}^{tZ}z_{j})\rho^{\frac{1}{2}}W(\hbox{\rm
e}^{tZ}z_{k}))-\hbox{\rm e}^{-\frac{1}{2}\Re\left\langle
z_{j},Sz_{j}\right\rangle-\frac{1}{2}\Re\left\langle
z_{k},Sz_{k}\right\rangle}\right]$

where $c_{t}(z_{j}),c_{t}(z_{k})$ are defined as in (32). Recalling that

$\Re\left\langle\hbox{\rm e}^{tZ}z,S\hbox{\rm
e}^{tZ}z\right\rangle=\int_{t}^{\infty}\Re\left\langle\hbox{\rm
e}^{sZ}z,C\hbox{\rm e}^{sZ}z\right\rangle\hbox{\rm d}s$

we obtain

$c_{t}(z_{j})c_{t}(z_{k})\hbox{\rm tr}(\rho^{\frac{1}{2}}W(-\hbox{\rm
e}^{tZ}z_{j})\rho^{\frac{1}{2}}W(\hbox{\rm e}^{tZ}z_{k}))=\hbox{\rm
e}^{-\frac{1}{2}\Re\left\langle
z_{j},Sz_{j}\right\rangle-\frac{1}{2}\Re\left\langle
z_{k},Sz_{k}\right\rangle}\hbox{\rm e}^{\Re\left\langle\hbox{\rm
e}^{tZ}z_{j},\breve{S}\hbox{\rm e}^{tZ}z_{k}\right\rangle}.$

Therefore we get

$\left\|T_{t}(\breve{x})\right\|^{2}=\sum_{j,k}\overline{\xi_{j}}\xi_{k}\left[\hbox{\rm
e}^{\Re\left\langle\hbox{\rm e}^{tZ}z_{j},\breve{S}\hbox{\rm
e}^{tZ}z_{k}\right\rangle}-1\right].$

The remaining part of the Lemma follows immediately. ∎

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Authors’ addresses:

  * (1)

Mathematics Department, Politecnico di Milano, Piazza Leonardo da Vinci 32, I
- 20133 Milano, Italy<EMAIL_ADDRESS>
  * (2)

Mathematics Department, University of Genova, Via Dodecaneso, I - 16146
Genova, Italy