A Meta Logarithmic-Sobolev Inequality for Phase-Covariant Gaussian Channels

Abstract

We introduce a meta logarithmic-Sobolev (log-Sobolev) inequality for the Lindbladian of all single-mode phase-covariant Gaussian channels of bosonic quantum systems and prove that this inequality is saturated by thermal states. We show that our inequality provides a general framework to derive information theoretic results regarding phase-covariant Gaussian channels. Specifically, by using the optimality of thermal states, we explicitly compute the optimal constant $\alpha _p$, for $1\le p\le 2$, of the p-log-Sobolev inequality associated with the quantum Ornstein–Uhlenbeck semigroup. Prior to our work, the optimal constant was only determined for $p=1$. Our meta log-Sobolev inequality also enables us to provide an alternative proof for the constrained minimum output entropy conjecture in the single-mode case. Specifically, we show that for any single-mode phase-covariant Gaussian channel $\Phi$, the minimum of the von Neumann entropy $S\big (\Phi (\rho )\big )$ over all single-mode states $\rho$ with a given lower bound on $S(\rho )$ is achieved at a thermal state.

1 Introduction

Quantum Gaussian channels are prototype noise models for the transmission of quantum information in current quantum communication technologies and information processing. In particular, they model transmission channels for sending quantum data encoded in bosonic systems through optical fibers and free space. Despite several developments in the past decades, there are still wide open conjectures regarding information theoretic properties of these channels. These conjectures are important not only due to the significance of quantum Gaussian channels in quantum information science, but also as quantum generalizations of some influential results in functional analysis.

In functional analysis and information theory, the optimality of Gaussian functions and Gaussian distributions has been established for various optimization problems involving Gaussian kernels or Gaussian channels. To name a few results, it is shown in [7] that only Gaussian functions saturate Gross’s celebrated logarithmic-Sobolev (log-Sobolev) inequality [25]. Also, it is shown in [36] that “Gaussian kernels have only Gaussian maximizers.” In information theory, it is shown that the optimal input distribution for Gaussian broadcast channels is Gaussian [20]. More generally, various inequalities including Brascamp–Lieb inequalities, the Loomis–Whitney inequality, the Prékopa–Leindler inequality, sharp form of Young’s inequality for convolution of functions as well as the entropy power inequality are tight for Gaussian functions and distributions; see [1] and references therein. Thus, it is tempting to verify the validity of these results in the non-commutative case for quantum Gaussian channels.

Single-mode quantum Gaussian channels have been classified in [30]; see also [31]. Among these classes are the three important physical classes of attenuator, amplifier and additive-noise channels, which have two main features. First, these channels are phase-covariant meaning, that they are covariant with respect to the group of phase operators, and second, they form quantum Markov semigroups [28]. These two features make these channels suitable candidates for generalizing some of the aforementioned properties of Gaussian kernels in functional analysis to the quantum case. In particular, as we show in this paper, these channels satisfy a new inequality that is saturated by Gaussian states, which we refer to as meta log-Sobolev inequality and provides a general framework for deriving quantum information-theoretic results regarding bosonic systems.

One application of our meta log-Sobolev inequality is to compute the optimal constants of the p-log-Sobolev inequalities for a semigroup of attenuator channels, called the quantum (bosonic) Ornstein–Uhlenbeck semigroup. This semigroup is of particular interest as it resembles its classical counterpart when acting on certain operators; see [9, Equation (7.5)]. The classical Ornstein–Uhlenbeck semigroup is an important example of a Markov semigroup, which can be understood as the convolution of an input distribution by a Gaussian one. Its associated log-Sobolev inequality takes the form

$$\begin{aligned} \frac{1}{2}{{\textrm{Ent}}}_2(f)\le \Vert f'\Vert _{2}^2, \qquad \forall f>0, \end{aligned}$$

(1)

where the entropy ${{\textrm{Ent}}}_2(f) = {\mathbb {E}}[f^2\log f^2] - {\mathbb {E}}[f^2]\log {\mathbb {E}}[f^2]$ and the 2-norm $\Vert f\Vert _2={\mathbb {E}}[f^2]^{1/2}$ are defined with respect to the standard normal distribution. This inequality was first established by Gross in his seminal work [25].

Given the log-Sobolev inequality (1) for the classical Ornstein–Uhlenbeck semigroup and the development of the theory of hypercontractivity for quantum semigroups [39], it is natural to ask if the quantum Ornstein–Uhlenbeck semigroup is hypercontractive and satisfies a log-Sobolev inequality.¹ The 2-log-Sobolev inequality for the quantum Ornstein–Uhlenbeck semigroup takes the form

$$\begin{aligned} \alpha _2\,{{\textrm{Ent}}}_{2, \sigma _\beta } (X)\le {{\textrm{tr}}}\Big (\!\sqrt{\sigma _\beta } [{\textbf{a}}, X]^\dagger \sqrt{\sigma _\beta } [{\textbf{a}}, X]\Big ),\quad \forall X>0,\; {{\textrm{tr}}}\Big (\sigma ^{\frac{1}{4}}_\beta X \sigma _\beta ^{\frac{1}{4}}\Big )^2 <+\infty . \end{aligned}$$

(2)

Here, $\sigma _\beta =(1-e^{-\beta }) e^{-\beta a^\dagger a}$ is a thermal (and then Gaussian) state with parameter $\beta >0$, $[{\textbf{a}}, X]={\textbf{a}}X-X{\textbf{a}}$ is the commutator of the annihilation operator with X, and in the special case that $\rho = \Big (\sigma ^{\frac{1}{4}}_\beta X \sigma _\beta ^{\frac{1}{4}}\Big )^2$ is a quantum state

$$\begin{aligned} {{\textrm{Ent}}}_{2, \sigma _\beta } (X)=D(\rho \Vert \sigma _\beta )= {{\textrm{tr}}}(\rho \log \rho )- {{\textrm{tr}}}(\rho \log \sigma _\beta ). \end{aligned}$$

Moreover, $\alpha _2$ is called the 2-log-Sobolev constant whose optimal value, depending on $\beta$, is one of our main problems of study in this work.

Using the spectral properties of the Lindbladian of the quantum Ornstein–Uhlenbeck semigroup established in [9], a bound on the 2-log-Sobolev constant $\alpha _2$ is derived in [6]. This bound confirms the hypercontractivity of the quantum Ornstein–Uhlenbeck semigroup, yet it seems far from being optimal. Moreover, the bound of [6] works only in the single-mode case and is not clear to satisfy the so called tensorization property.

Although 2-log-Sobolev inequalities (assuming some regularity conditions [35, 39]) imply hypercontractivity inequalities, for a more refined characterization of the latter inequalities we need to establish generalizations of the above 2-log-Sobolev inequality for all values of $1\le p\le 2$. These generalizations, called p-log-Sobolev inequalities, are all equivalent and ignorant of the parameter p for the classical Ornstein–Uhlenbeck semigroup, yet they differ in the quantum case.²

The 1-log-Sobolev (also called the modified log-Sobolev) inequality for the quantum Ornstein–Uhlenbeck semigroup is first studied in [34] where it is examined for Gaussian states and it is conjectured that Gaussian states are optimal for the 1-log-Sobolev inequality. This conjecture is first proven in [8]. An alternative proof of this inequality and an extension are also derived in [12].

Our meta log-Sobolev inequality is also related to computing the classical capacity of attenuators, amplifiers and additive-noise channels, i.e., finding the maximum rate of reliable transmission of classical data through these channels. Since these channels are phase-covariant, their classical capacity is related to their minimum output entropy, i.e., the minimum entropy of their output states; see, e.g., [31]. It was a long-standing open problem that the minimum output entropy of these channels is additive and is achieved over Gaussian states [21, 23, 33]. This conjecture was answered in the affirmative in [22]. Extensions of this result have also been proven for which we refer to [18, 32] and references therein.

Generalizing the aforementioned minimum output entropy conjecture, the constrained minimum output entropy (CMOE) conjecture states that the minimum output entropy of the above phase-covariant Gaussian channels over input states with a given entropy is achieved by Gaussian states [27]. More precisely, the CMOE conjecture states that if $\Phi$ is a single-mode phase-covariant Gaussian channel, which by the above discussion is either attenuator, amplifier or additive-noise, then for any m-mode state $\rho$ we have

$$\begin{aligned} \frac{1}{m} S\big ( \Phi ^{\otimes m}(\rho )\big )\ge S(\Phi (\tau )), \end{aligned}$$

(3)

where $\tau$ is a single-mode thermal state satisfying $S(\tau ) = \frac{1}{\,}m S(\rho )$. This conjecture is in particular related to the problem of computing the capacity of quantum Gaussian broadcast channels [27]. Using tools developed in [14] the CMOE conjecture is proven in [15, 16] in the special case of $m=1$. It is also proven for arbitrary m in the parameter regime that the channel $\Phi$ becomes entanglement breaking [11].

Some other conjectures regarding the optimality of Gaussian states have also been proposed including an entropy photon-number inequality [13, 26] and a sharp Young’s inequality for the beam-splitter [18]. For more details on quantum Gaussian optimizer conjectures in quantum information theory we refer to [18].

1.1 Our Contributions

In this paper, we introduce a meta log-Sobolev inequality for phase-covariant Gaussian quantum channels. In the Schrödinger picture, the Lindbladian of the semigroups of attenuator, amplifier and additive-noise channels mentioned above, takes the form

$$\begin{aligned} {\mathcal {L}}(\rho ) = \nu _0\Big (\frac{1}{2} \{{\textbf{a}}{\textbf{a}}^\dagger , \rho \} - {\textbf{a}}^\dagger \rho {\textbf{a}}\Big )+ \nu _1\Big ( \frac{1}{2} \{{\textbf{a}}^\dagger {\textbf{a}}, \rho \} - {\textbf{a}}\rho {\textbf{a}}^\dagger \Big ), \end{aligned}$$

where $\nu _0, \nu _1\ge 0$ are non-negative constants and $\{X, Y\} = XY+YX$ denotes anti-commutator. Then, given any $\omega \ge 0$, we define the function

$$\begin{aligned} \Upsilon (\rho )= \frac{p}{p-1}{{\textrm{tr}}}\Big ( {\mathcal {L}}\big (\rho ^{1/p}\big ) \rho ^{1-1/p}\Big ) + \omega \,{{\textrm{tr}}}\big (\rho {\textbf{a}}^\dagger {\textbf{a}}\big ) + S(\rho ), \end{aligned}$$

over single-mode states $\rho$. Later, we will see that for the choice of

$$\begin{aligned} X=\sigma _\beta ^{-\frac{1}{2p}} \rho ^{\frac{1}{p}} \sigma _\beta ^{-\frac{1}{2p}}, \end{aligned}$$

and $p=2$, the first term in $\Upsilon (\rho )$ resembles the Dirichlet form appearing on the right hand side of (2). In this case, ${{\textrm{Ent}}}_{p, \sigma _\beta }(X)=D(\rho \Vert \sigma _\beta )$ can be written in terms of the other two terms in $\Upsilon (\rho )$. Thus, computing the infimum of $\Upsilon (\rho )$ is useful in obtaining the optimal log-Sobolev constant $\alpha _p$.

Our main result, which we call a meta log-Sobolev inequality, states that optimal states in the minimization of $\Upsilon (\rho )$ are thermal, i.e.,

$$\begin{aligned} \Upsilon (\rho ) \ge \inf _{\tau : {\textrm{thermal}}} \Upsilon \big (\tau \big ), \end{aligned}$$

(4)

for any state $\rho$ of a single-mode bosonic system.

By using (4), we turn the problem of computing the optimal p-log-Sobolev constant $\alpha _p$ into an optimization problem over thermal states that are characterized by a single real parameter. We explicitly compute the optimal constant for the quantum Ornstein–Uhlenbeck semigroup for any $1\le p\le 2$:

$$\begin{aligned} \alpha _p = \frac{p{\hat{p}}}{4\beta } e^{\beta /2}\big (1-e^{-\beta /p}\big )\big (1-e^{-\beta /{\hat{p}}}\big ). \end{aligned}$$

(5)

This result for $p=1$ recovers the modified log-Sobolev inequality of [8], and for $p=2$ is an improvement over the bound of [6].

We also study the quantum Ornstein–Uhlenbeck semigroup in the multimode case. It is well known that log-Sobolev constants for classical Markov semigroups have the tensorization property (see, e.g., [38]), meaning that considering the m-fold tensor product of the semigroup, the log-Sobolev constants of the resulting semigroup are the same as those of the original one. This property is not known to hold for arbitrary quantum Markov semigroups. Thus, we also consider the m-mode version of (2):

$$\begin{aligned} {{\widehat{\alpha }}}_{2}{{\textrm{Ent}}}_{2, \sigma _{\beta }^{\otimes m}} (X)\le \sum _{j=1}^m {{\textrm{tr}}}\Big (\!\sqrt{\sigma _\beta ^{\otimes m}} [{\textbf{a}}_j, X]^\dagger \sqrt{\sigma _\beta ^{\otimes m}} [{\textbf{a}}_j, X]\Big ), \end{aligned}$$

where $X>0$ runs over m-mode operators, and ${{\widehat{\alpha }}}_{2}$ is the 2-log-Sobolev constant for the m-fold tensor product semigroup. We show that for any $m>1$,

$$\begin{aligned} {{\widehat{\alpha }}}_{2} \ge \bigg (\frac{2+ \log (2m+1)}{\sinh (\beta /2)} + \frac{1}{\alpha _{2}}\bigg )^{-1}\!, \end{aligned}$$

(6)

where $\alpha _2$ is given by (5).

To prove our result (6), we follow the approach of [6]. We use properties of the spectrum of the generator ${\mathcal {L}}$ of the quantum Ornstein–Uhlenbeck semigroup to reduce the problem for arbitrary m-mode operators X to operators that are diagonal in the number basis. To this end, we prove an entropic inequality which might be of independent interest. Next, restricting to diagonal operators, we obtain an essentially classical Markov semigroup that is known to satisfy the tensorization property. Thus, the problem for m-mode states is reduced to that of single-mode states for which we have already established the optimal log-Sobolev constant in (5). We note that, although, our first step of the proof is inspired by [6], our penalty term that compensates for the reduction to diagonal states is smaller than that of [6]. Moreover, as mentioned above we compute the exact value of the 2-log-Sobolev constant for diagonal states while [6] derives only a lower bound. Thus, our estimate on the 2-log-Sobolev inequality is not just an m-mode generalization of [6], but a twofold improvement thereof.

Our meta log-Sobolev inequality (4) provides a general framework that can be applied in other settings as well. In this paper, using (4) we also give an alternative proof of the CMOE conjecture (3) in the case of $m=1$.

Outline of the paper The rest of this paper is organized as follows. In Sect. 2, we review the class of single-mode phase-covariant Gaussian channels. We argue that these channels form semigroups and compute their Lindbladians. Section 3 is devoted to our meta log-Sobolev inequality (4) and its proof. The log-Sobolev constants for the quantum Ornstein–Uhlenbeck semigroup are computed in Sect. 4. Also, our proof of the CMOE conjecture is given in Sect. 5. Final remarks are discussed in Sect. 6, and some detailed computations are left for the appendices.

2 Phase-Covariance Gaussian Channels

A single-mode bosonic continuous-variable system is described by the annihilation operator ${\textbf{a}}$ and it hermitian conjugate, the creation operator ${\textbf{a}}^\dagger$. These operators satisfy the bosonic commutation relation

$$\begin{aligned} {[}{\textbf{a}}, {\textbf{a}}^\dagger ]={\textbf{a}}{\textbf{a}}^\dagger - {\textbf{a}}^\dagger {\textbf{a}}=1. \end{aligned}$$

Fock states $\{|n\rangle :\, n\ge 0\}$, also called number states, form an orthonormal basis for the corresponding Hilbert space and are eigenvectors of ${\textbf{a}}^\dagger {\textbf{a}}$ called the number operator: ${\textbf{a}}^\dagger {\textbf{a}}|n\rangle =n|n\rangle$. We indeed have ${\textbf{a}}|n\rangle = \sqrt{n}|n-1\rangle$ and ${\textbf{a}}^\dagger |n\rangle = \sqrt{n+1}|n+1\rangle$. Single-mode quantum states $\rho$ are density operators ($\rho \succeq 0$ and ${{\textrm{tr}}}(\rho )=1$) acting on this Hilbert space.

All states that we consider in this paper are assumed to be physical and hence have finite mean photon number. More precisely, for a pure state $\rho =|\psi \rangle \langle \psi |$ we assume that $|\psi \rangle$ belongs to the domain of ${\textbf{a}}$ and $\langle \psi |{\textbf{a}}^\dagger {\textbf{a}}|\psi \rangle = \Vert {\textbf{a}}|\psi \rangle \Vert ^2 < +\infty$, and more generally for a mixed state with eigen-decomposition $\rho = \sum _{j=0}^\infty \lambda _j |\psi _j\rangle \langle \psi _j|$ we assume that $|\psi _j\rangle$, for any j with $\lambda _j>0$, belongs to the domain of ${\textbf{a}}$ and $\sum _j \lambda _j \Vert {\textbf{a}}|\psi _j\rangle \Vert ^2 < +\infty$. The latter expression is indeed equal to ${{\textrm{tr}}}({\textbf{a}}\rho {\textbf{a}}^\dagger )$, yet we often write it as ${{\textrm{tr}}}(\rho {\textbf{a}}^\dagger {\textbf{a}})$, the conventional notation for the mean photon number, considering the following formal definition for the expectation value of an operator in quantum mechanics.³ For any positive semidefinite operator X, we formally define ${{\textrm{tr}}}(\rho X) = \sum _j \lambda _j \Vert X^{1/2} |\psi _j\rangle \Vert ^2$ if $|\psi _j\rangle$ belongs to the domain of $X^{1/2}$ for all j with $\lambda _j>0$, and let ${{\textrm{tr}}}(\rho X) = +\infty$ otherwise.⁴ Also, for a self-adjoint operator X with decomposition $X=X_+-X_-$ where both $X_+, X_-$ are positive semidefinite, we define ${{\textrm{tr}}}(\rho X) = {{\textrm{tr}}}(\rho X_+) - {{\textrm{tr}}}(\rho X_-)$.

Quantum states can be represented in terms of displacement (Weyl) operators $D_\xi =\exp (\xi {\textbf{a}}^\dagger -{{\bar{\xi }}}{\textbf{a}})$ with $\xi \in {\mathbb {C}}$ as

$$\begin{aligned} \rho =\frac{1}{\pi } \int _{{\mathbb {C}}} \chi _\rho (\xi ) D_{-\xi }\, {\textrm{d}}^{2}\xi , \end{aligned}$$

(7)

where $\chi _\rho (\xi )={{\textrm{tr}}}(\rho D_\xi )$ is called the characteristic function [5]. The characteristic function is the Fourier transform of the Wigner function [29].

A quantum state is called Gaussian if its characteristic and equivalently Wigner functions are Gaussian. Therefore, Gaussian states can be simply described in terms of the first-order and second-order moments of their Wigner function. Thermal states are an important class of Gaussian states. A thermal state with parameter $\beta >0$ that is proportional to the inverse of temperature, is diagonal in the Fock basis and is given by

$$\begin{aligned} \sigma _\beta =\frac{1}{{{\textrm{tr}}}(e^{-\beta {\textbf{a}}^\dagger {\textbf{a}}})}e^{-\beta {\textbf{a}}^\dagger {\textbf{a}}}=(1-e^{-\beta })\sum _{n=0}^{\infty } e^{-n\beta }|n\rangle \langle n|. \end{aligned}$$

(8)

The characteristic function of this thermal state equals

$$\begin{aligned} \chi _{\sigma _\beta }(\xi ) = e^{-\frac{1}{2} \coth (\beta /2)|\xi |^2}. \end{aligned}$$

(9)

At zero temperature ($\beta =\infty$), we obtain the vacuum state $\sigma _\infty =|0\rangle \langle 0|$.

The evolution of quantum systems, in general, is described by quantum channels $\Phi$ that are linear completely positive and trace-preserving superoperators. Quantum channels that transform Gaussian states into Gaussian states are known as Gaussian channels [3, 31, 33, 42]. Gaussian channels that are unitary are called Gaussian unitaries.

A single-mode Gaussian channels is called phase-covariant⁵ [22, 24] if it satisfies

$$\begin{aligned} \Phi \big (U_\theta \rho \, U_{\theta }^\dagger \big )=U_\theta \Phi (\rho ) U_{\theta }^\dagger , \end{aligned}$$

(10)

for any state $\rho$ and $\theta \in [0,2\pi )$, where $U_\theta =e^{i\theta \, {\textbf{a}}^\dagger {\textbf{a}}}$ is the phase-rotation unitary. By the classification of single-mode Gaussian channels [30, 31], a single-mode phase-covariant Gaussian channel can be described in terms of its action on the characteristic function. For any such channel there are parameters $\gamma , \lambda \ge 0$ such that

$$\begin{aligned} \chi _{\Phi (\rho )}(\xi )=e^{-\frac{1}{2} \gamma |\xi |^2}\chi _\rho \big (\!\sqrt{\lambda }\xi \big ), \end{aligned}$$

(11)

where the complete positivity condition implies $\gamma \ge |1-\lambda |$ [33]. The channel $\Phi$ is called quantum limited if $\gamma =|1-\lambda |$. By using the characteristic function of thermal states (9) in (11), one can verify that phase-covariant Gaussian channels transform thermal states to thermal states.

Single-mode phase-covariant Gaussian channels consist of three classes of attenuator channels corresponding to $0\le \lambda <1$, additive-noise channels corresponding to $\lambda =1$, and amplifier channels corresponding $1<\lambda$. These are important physical channels in describing dynamics of continuous-variable quantum systems.

A crucial property of single-mode phase-covariant Gaussian channels is that they admit semigroup structures [23, 28]. Specifically, any such channel can be written as $\Phi _{t_0}=e^{-t_0{\mathcal {L}}}$ for some $t_0\ge 0$ where $\big \{\Phi _{t} = e^{-t{\mathcal {L}}}:\, t\ge 0\big \}$ is a semigroup of single-mode phase-covariant Gaussian channels. We show in Appendix A that the corresponding Lindbladian ${\mathcal {L}}$ of such a semigroup takes the form

$$\begin{aligned} {\mathcal {L}}(\rho ) = \nu _0\Big (\frac{1}{2} \{{\textbf{a}}{\textbf{a}}^\dagger , \rho \} - {\textbf{a}}^\dagger \rho \,{\textbf{a}}\Big )+ \nu _1\Big ( \frac{1}{2} \{{\textbf{a}}^\dagger {\textbf{a}}, \rho \} - {\textbf{a}}\, \rho \,{\textbf{a}}^\dagger \Big ), \end{aligned}$$

(12)

where $\{X, Y\} = XY+YX$ denotes the anti-commutator, and $\nu _0, \nu _1\ge 0$ are parameters determining the semigroup. We argue in Appendix A that three ranges for the parameters $\nu _0, \nu _1$ give the three classes of single-mode phase-covariant channels: $\nu _1>\nu _0$ corresponds to attenuator channels, $\nu _0=\nu _1$ corresponds to additive-noise channels, and $\nu _1<\nu _0$ corresponds to amplifier channels.

Attenuator channels, in general, can be physically modeled by applying a beam splitter unitary with the transmissivity of $0\le \lambda \le 1$ on the system and an auxiliary system in the thermal state $\sigma _\beta$, and then tracing out the second subsystem:

$$\begin{aligned} \Phi ^{\text {att}}_\lambda (\rho )={{\textrm{tr}}}_2\Big ( U_{\text {BS},\lambda } (\rho \otimes \sigma _\beta ) U_{\text {BS},\lambda }^\dagger \Big ). \end{aligned}$$

(13)

The characteristic function of the output state is given by

$$\begin{aligned} \chi _{\Phi ^{\text {att}}_\lambda (\rho )} (\xi )=e^{-\frac{1}{2} \coth (\beta /2)(1-\lambda )|\xi |^2}\chi _\rho (\sqrt{\lambda }\xi ). \end{aligned}$$

Choosing the transmissivity parameter $\lambda _t=e^{-2ct}$ as a function of time, where $c>0$ is some constant, we obtain a semigroup of attenuation channels. The generator of this semigroup given in (12) has parameters $\nu _0 = c \left( \coth (\beta /2)-1\right)$ and $\nu _1 =c \left( \coth (\beta /2)+1\right)$. Note that for the special choice of $c=\sinh (\beta /2)$ we have $\nu _0=e^{-\beta /2}$ and $\nu _1=e^{\beta /2}$. This semigroup of attenuator channels is sometimes called the quantum Ornstein–Uhlenbeck semigroup.

Amplifier channels can be described by replacing the beam splitter unitary in the above model by a two-mode squeezing unitary to get

$$\begin{aligned} \Phi ^{\text {amp}}_\lambda (\rho )={{\textrm{tr}}}_2\Big ( U_{\text {2S},\lambda } (\rho \otimes \sigma _\beta ) U_{\text {2S},\lambda }^\dagger \Big ), \end{aligned}$$

(14)

where $\lambda \ge 1$ is the squeezing parameter. In this case, the relation between input and output characteristic functions becomes

$$\begin{aligned} \chi _{\Phi ^{\text {att}}_\lambda (\rho )} (\xi )=e^{-\frac{1}{2} \coth (\beta /2)(\lambda -1)|\xi |^2}\chi _\rho (\sqrt{\lambda }\xi ). \end{aligned}$$

Choosing $\lambda _t=e^{2ct}$ with $c>0$ to be a function of time, we obtain a semigroup with the generator corresponding to parameters $\nu _0=c\left( \coth (\beta /2)+1\right)$ and $\nu _1=c\left( \coth (\beta /2)-1\right)$ in (12).

Additive-noise channels can be modeled by applying a displacement operator whose parameter is chosen at random according to a Gaussian probability distribution:

$$\begin{aligned} \Phi ^{\text {add}}_\gamma (\rho )=\frac{2}{\pi \gamma }\int e^{-\frac{2}{\gamma }|\xi |^2} D_\xi \rho D^{\dagger }_\xi \, {\textrm{d}}^2\xi . \end{aligned}$$

This channel in terms of characteristic functions can be written as

$$\begin{aligned} \chi _{\Phi ^{\text {add}}_\gamma (\rho )} (\xi )=e^{-\frac{1}{2}\gamma |\xi |^2}\chi _\rho (\xi ). \end{aligned}$$

Again, by setting $\gamma _t=2ct$ with $c>0$ we obtain a semigroup whose generator given by (12) has parameters $\nu _0=\nu _1=c$. This semigroup is sometimes called the quantum heat semigroup.

We emphasize that by the above discussion any single-mode phase-covariant channel (11) can be viewed as a member of one of the above three semigroups. The point is that, the above choices of parameters $\lambda _t, \gamma _t$ cover all the valid ranges of $\lambda , \gamma$ in (11).

In this paper, we also consider m-mode bosonic systems, described by m pairs of annihilation and creation operators $\{{\textbf{a}}_1, {\textbf{a}}_1^\dagger , \dots , {\textbf{a}}_m, {\textbf{a}}_m^\dagger \}$ satisfying $[{\textbf{a}}_{i}, {\textbf{a}}_j]=0$ and $[{\textbf{a}}_{i}, {\textbf{a}}_j^\dagger ]=\delta _{i,j}$. Vectors of the associated tensor product Hilbert space can be expressed in terms of m-mode number states $\{|n_1,\dots ,n_m\rangle :\, n_1,\dots , n_m\ge 0\}$. Also, the characteristic function for an m-mode state $\rho$ is given by $\chi _\rho (\xi )={{\textrm{tr}}}\big (\rho D_{\xi }\big )$, where $\xi =(\xi _1,\dots ,\xi _m)\in {\mathbb {C}}^m$ and $D_\xi = D_{\xi _1}\otimes \dots \otimes D_{\xi _m}$. As in the single-mode case, physical multimode states $\rho$ considered in this paper have finite mean photon number: ${{\textrm{tr}}}(\rho H_m)<+\infty$, where $H_m=\sum _{j=1}^m {\textbf{a}}_j^\dagger {\textbf{a}}_j$ is the m-mode number operator. The definitions of Gaussian channels and unitaries can be extended to the multimode case. Any Gaussian unitary can be decomposed into displacement operators, multimode passive transformations, which preserve the mean photon number, and single-mode squeezing transformations; see [41] for more details.

3 Meta Log-Sobolev Inequality

As discussed in the previous section, all the Lindbladians associated with single-mode phase-covariant channels have the form ${\mathcal {L}}=\nu _0{\mathcal {L}}_{0} + \nu _1 {\mathcal {L}}_1$ for some $\nu _0, \nu _1\ge 0$ where

$$\begin{aligned} {\mathcal {L}}_0(X) = \frac{1}{2} \{{\textbf{a}}{\textbf{a}}^\dagger , X\} - {\textbf{a}}^\dagger X{\textbf{a}}, \qquad {\mathcal {L}}_1(X) = \frac{1}{2} \{{\textbf{a}}^\dagger {\textbf{a}}, X\} - {\textbf{a}}X{\textbf{a}}^\dagger . \end{aligned}$$

(15)

Let $\langle \cdot , \cdot \rangle$ denote the Hilbert–Schmidt inner product:

$$\begin{aligned} \langle X, Y\rangle = {{\textrm{tr}}}(X^\dagger Y). \end{aligned}$$

Also, let ${\mathcal {L}}^*$ be the adjoint of the generator ${\mathcal {L}}$ with respect to this inner product: $\langle X, {\mathcal {L}}(Y)\rangle = \langle {\mathcal {L}}^*(X), Y\rangle$. Indeed, ${\mathcal {L}}^*$ is the generator in the Heisenberg picture given by

$$\begin{aligned} {\mathcal {L}}^*(X) = \nu _0\Big ( \frac{1}{2}\{{\textbf{a}}{\textbf{a}}^\dagger , X\}-{\textbf{a}}X{\textbf{a}}^\dagger \Big )+\nu _1\Big ( \frac{1}{2}\{{\textbf{a}}^\dagger {\textbf{a}}, X\} -{\textbf{a}}^\dagger X{\textbf{a}}\Big ). \end{aligned}$$

Let $p\ge 1$ and ${\hat{p}}$ be the Hölder conjugate of p given by⁶

$$\begin{aligned} \frac{1}{p} + \frac{1}{{\hat{p}}}=1. \end{aligned}$$

Then, for any single-mode quantum state $\rho$ define

$$\begin{aligned} \Upsilon (\rho )&:={\hat{p}}\big \langle {\mathcal {L}}\big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle + \omega {{\textrm{tr}}}(\rho {\textbf{a}}^\dagger {\textbf{a}}) + S(\rho )\nonumber \\&={\hat{p}}\Big (\!\nu _0\big \langle {\mathcal {L}}_{0}\big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle +\nu _1\big \langle {\mathcal {L}}_{1}\big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle \!\Big ) + \omega {{\textrm{tr}}}(\rho {\textbf{a}}^\dagger {\textbf{a}}) +S(\rho ), \end{aligned}$$

(16)

where $\omega \ge 0$ is a fixed parameter and $S(\rho )= -{{\textrm{tr}}}(\rho \log \rho )$ is the von Neumann entropy of the state. The factor ${\hat{p}}$ in the first term of $\Upsilon (\rho )$ is for the sake of normalization in the limiting case of $p\rightarrow 1^+$. Indeed, since ${\mathcal {L}}^*(I)=0$, we have

$$\begin{aligned} \lim _{p\rightarrow 1^+} {\hat{p}}\big \langle {\mathcal {L}}\big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle =\lim _{p\rightarrow 1^+} {\hat{p}}\,\big \langle \rho ^{1/p}, {\mathcal {L}}^*\big (\rho ^{1/{\hat{p}}}\big )\big \rangle = \langle \rho , {\mathcal {L}}^*(\log \rho )\rangle = \langle {\mathcal {L}}(\rho ), \log \rho \rangle . \end{aligned}$$

Thus, by convention for $p=1$ we let ${\hat{p}}\, \big \langle {\mathcal {L}}\big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle = \langle {\mathcal {L}}(\rho ), \log \rho \rangle$.

A straightforward computation verifies that

$$\begin{aligned} \Upsilon (\rho )&={\hat{p}}\bigg (\!\nu _0 \Big [{{\textrm{tr}}}\big (\rho {\textbf{a}}{\textbf{a}}^\dagger \big ) - {{\textrm{tr}}}\big ( \rho ^{1/p} {\textbf{a}}\rho ^{1/{\hat{p}}} {\textbf{a}}^\dagger \big ) \Big ] +\nu _1\Big [ {{\textrm{tr}}}\big (\rho {\textbf{a}}^\dagger {\textbf{a}}\big )- {{\textrm{tr}}}\big ( \rho ^{1/p} {\textbf{a}}^\dagger \rho ^{1/{\hat{p}}} {\textbf{a}}\big )\!\Big ]\!\bigg ) \nonumber \\&\quad + \omega {{\textrm{tr}}}\big (\rho {\textbf{a}}^\dagger {\textbf{a}}\big ) + S(\rho ). \end{aligned}$$

(17)

We may think of this equation as the starting definition of $\Upsilon (\rho )$. In Appendix B, we show that if $\rho$ has a finite mean photon number, then all the terms in the above equation are finite and $\Upsilon (\rho )$ given by (17) is well defined.

One of the main technical contributions of our work is that the infimum of $\Upsilon (\rho )$ over states $\rho$ is achieved at single-mode thermal states. To this end, it would be beneficial to compute $\Upsilon (\rho )$ for thermal states. Using (17), for a single-mode thermal state,⁷

$$\begin{aligned} \tau =\tau _x=(1-x)\sum _n x^n|n\rangle \langle n|, \end{aligned}$$

where we put $0<x=e^{-\beta }<1$ in (8), we have

$$\begin{aligned} \Upsilon (\tau )&= {\hat{p}}(1-x) \sum _n \Big ( \nu _0 (1-x^{1/{\hat{p}}}) (n+1) x^n + \nu _1 (1-x^{-1/{\hat{p}}}) n x^n \Big )\nonumber \\&\quad + \omega (1-x) \sum _n nx^n - (1-x)\sum _n x^n\log \big ((1-x)x^n\big )\nonumber \\&= {\hat{p}}(1-x)\Big (\nu _0 (1-x^{1/{\hat{p}}}) \frac{1}{(1-x)^2} + \nu _1 (1-x^{-1/{\hat{p}}}) \frac{x}{(1-x)^2} \Big )\nonumber \\&\quad + \omega (1-x) \frac{x}{(1-x)^2} - (1-x) \frac{x}{(1-x)^2} \log x- \log (1-x) \nonumber \\&= \frac{{\hat{p}}}{1-x} \Big (\nu _0 (1-x^{1/{\hat{p}}})+ \nu _1(x- x^{1/p}) \Big )\nonumber \\&\quad + \omega \frac{x}{1-x} - \frac{x}{1-x} \log x- \log (1-x). \end{aligned}$$

(18)

Optimizing over the choice of the thermal state $\tau$, we define

$$\begin{aligned} \eta _{{{{\textrm{th}}}}}&:= \inf _{\tau : {\textrm{thermal}}} \Upsilon (\tau )\nonumber \\&=\inf _{0<x<1}\, \frac{{\hat{p}}}{1-x} \Big (\nu _0 (1-x^{1/{\hat{p}}}) + \nu _1(x- x^{1/p}) \Big ) \nonumber \\&\quad + \omega \frac{x}{1-x} - \frac{x}{1-x} \log x- \log (1-x). \end{aligned}$$

(19)

Theorem 1

(Meta log-Sobolev inequality). For any $\nu _0, \nu _1, \omega \ge 0$ and $p\ge 1$ define $\Upsilon (\rho )$ by (16). Let $\eta _{{{{\textrm{th}}}}}$ be the infimum of $\Upsilon (\tau )$ over thermal states as in (19). Then, for any single-mode quantum state $\rho$ with finite mean photon number we have

$$\begin{aligned} \Upsilon (\rho )\ge \eta _{{{\textrm{th}}}}. \end{aligned}$$

(20)

The proof of this theorem is broken into two steps:

• (i) The first step is to reduce the problem for arbitrary states $\rho$ to states that are diagonal in the Fock basis. To this end, we use ideas developed in [14]. We show that, fixing the eigenvalues of $\rho$ and rotating its eigen-basis, we can obtain a diagonal state ${{\widehat{\rho }}}$ that satisfies $\Upsilon (\rho )\ge \Upsilon ({{\widehat{\rho }}})$. We then conclude that diagonal states are sufficient when minimizing $\Upsilon (\cdot )$.

• (ii) In the second step, we show that the optimal diagonal states are thermal. The proof idea in this step is extracted from tensorization type arguments. Tensorization was first used by Gross [25] for proving his celebrated log-Sobolev inequality. The idea in [25] is that by the central limit theorem, an expectation value with respect to a Gaussian distribution can be understood as the limit $n\rightarrow +\infty$ of the expectation value of some lifted function on the product space $\{0,1\}^n$ with respect to a product probability measure. Here, being interested in thermal states, our distributions of reference are geometric distributions, and geometric distributions can be understood in terms of the first success in a sequence of independent Bernoulli trials. Thus, it is natural to employ a tensorization type argument in order to reduce the optimization over general single-mode diagonal states to thermal ones. In our proof, we do not directly refer to Bernoulli trials, yet our intuition on why and how it works is really rooted in a tensorization argument as described here.

Proof

(i) In the first step of the proof, we show that for any state $\rho$, there is a state ${{\widehat{\rho }}}$ that is diagonal in the Fock basis and satisfies $\Upsilon (\rho )\ge \Upsilon ({{\widehat{\rho }}})$.

Let

$$\begin{aligned} \rho = \sum _{n=0}^\infty \lambda _n |\psi _n\rangle \langle \psi _n|, \end{aligned}$$

be the eigen-decomposition of $\rho$ where $\lambda _0 \ge \lambda _1\ge \cdots \ge 0$ are the eigenvalues of $\rho$ and $\{|\psi _n\rangle :\, n\ge 0\}$ is its eigen-basis. In this case, there are $r_k=\lambda _k-\lambda _{k+1}\ge 0$ such that

$$\begin{aligned} \rho = \sum _{k=0}^{\infty } r_k P_k, \end{aligned}$$

where $P_k = \sum _{n=0}^{k} |\psi _n\rangle \langle \psi _n|$ is the projection operator on the first $k+1$ eigenvectors of $\rho$. Now we define

$$\begin{aligned} {{\widehat{\rho }}} = \sum _{k=0}^\infty r_k \Pi _k, \end{aligned}$$

where $\Pi _k = \sum _{n=0}^{k} |n\rangle \langle n|$ is the projection operator on the first $k+1$ vectors in the Fock basis. We note that by definition, ${{\widehat{\rho }}} = \sum _{k=0}^n \lambda _k |n\rangle \langle n|$ is diagonal in the Fock basis and shares the same eigenvalues with $\rho$. Therefore, they have the same entropy

$$\begin{aligned} S({{\widehat{\rho }}})= S(\rho ). \end{aligned}$$

(21)

Next, we note that the infimum of ${{\textrm{tr}}}(Q_k {\textbf{a}}^\dagger {\textbf{a}})$ over projectors $Q_k$ with rank $k+1$, equals the sum of the first $k+1$ smallest eigenvalues of ${\textbf{a}}^\dagger {\textbf{a}}$, and is achieved by $Q_k = \Pi _k$. This implies that the mean photon number of $\rho$ is lower bounded by that of ${{\widehat{\rho }}}$:

$$\begin{aligned} {{\textrm{tr}}}(\rho {\textbf{a}}^\dagger {\textbf{a}})&= \sum _k r_k {{\textrm{tr}}}(P_k {\textbf{a}}^\dagger {\textbf{a}}) \ge \sum _k r_k {{\textrm{tr}}}(\Pi _k {\textbf{a}}^\dagger {\textbf{a}}) = {{\textrm{tr}}}({{\widehat{\rho }}} {\textbf{a}}^\dagger {\textbf{a}}). \end{aligned}$$

(22)

We also have

$$\begin{aligned} \big \langle {\mathcal {L}}_{0}\big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle= & {{\textrm{tr}}}(\rho {\textbf{a}}{\textbf{a}}^\dagger ) - {{\textrm{tr}}}\big (\rho ^{1/p} {\textbf{a}}\rho ^{1/{\hat{p}}} {\textbf{a}}^\dagger \big ) \\= & {{\textrm{tr}}}\big (\rho ^{1/p}\rho ^{1/{\hat{p}}} {\textbf{a}}{\textbf{a}}^\dagger \big ) - {{\textrm{tr}}}\big (\rho ^{1/p} {\textbf{a}}\rho ^{1/{\hat{p}}} {\textbf{a}}^\dagger \big ). \end{aligned}$$

We note that $\rho ^{1/p} = \sum _n \lambda _n^{1/p} |\psi _n\rangle \langle \psi _n|$ with $\lambda _0^{1/p}\ge \lambda _1^{1/p}\ge \cdots \ge 0$. Then, there are $r_{p,k} \ge 0$ such that $\rho ^{1/p} = \sum _k r_{p, k} P_k$. Similarly, there are $r_{{\hat{p}}, k}\ge 0$ such that $\rho ^{1/{\hat{p}}} = \sum _k r_{{\hat{p}}, k} P_k$. We also have ${{\widehat{\rho }}}^{1/p} = \sum _k r_{p, k} \Pi _k$ and ${{\widehat{\rho }}}^{1/{\hat{p}}} = \sum _k r_{{\hat{p}}, k} \Pi _k$. Therefore, we have

$$\begin{aligned} \big \langle {\mathcal {L}}_0 \big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle&= \sum _{k, \ell } r_{p,k}\,r_{{{\hat{p}}}, \ell } \Big (\! {{\textrm{tr}}}\big (P_kP_\ell {\textbf{a}}{\textbf{a}}^\dagger \big ) - {{\textrm{tr}}}\big ( P_k {\textbf{a}}P_\ell {\textbf{a}}^\dagger \big )\!\Big )\\&= \sum _{k\ge \ell } r_{p,k}\,r_{{{\hat{p}}}, \ell } \Big (\! {{\textrm{tr}}}\big (P_kP_\ell {\textbf{a}}{\textbf{a}}^\dagger \big ) - {{\textrm{tr}}}\big ( P_k {\textbf{a}}P_\ell {\textbf{a}}^\dagger \big )\!\Big ) \\&\quad + \sum _{k< \ell } r_{p,k}\,r_{{{\hat{p}}}, \ell } \Big (\! {{\textrm{tr}}}\big (P_kP_\ell {\textbf{a}}{\textbf{a}}^\dagger \big ) - {{\textrm{tr}}}\big ( P_k {\textbf{a}}P_\ell {\textbf{a}}^\dagger \big )\!\Big )\\&= \sum _{k\ge \ell } r_{p,k}\,r_{{{\hat{p}}}, \ell } \Big (\! {{\textrm{tr}}}\big (P_\ell {\textbf{a}}{\textbf{a}}^\dagger \big ) - {{\textrm{tr}}}\big ( P_k {\textbf{a}}P_\ell {\textbf{a}}^\dagger \big )\!\Big ) \\&\quad + \sum _{k< \ell } r_{p,k}\,r_{{{\hat{p}}}, \ell } \Big (\! {{\textrm{tr}}}\big (P_k {\textbf{a}}{\textbf{a}}^\dagger \big ) - {{\textrm{tr}}}\big ( P_k {\textbf{a}}P_\ell {\textbf{a}}^\dagger \big )\!\Big )\\&\ge \sum _{k\ge \ell } r_{p,k}\,r_{{{\hat{p}}}, \ell } \Big (\! {{\textrm{tr}}}\big (P_\ell {\textbf{a}}{\textbf{a}}^\dagger \big ) - {{\textrm{tr}}}\big ( {\textbf{a}}P_\ell {\textbf{a}}^\dagger \big )\!\Big ) \\&\quad + \sum _{k< \ell } r_{p,k}\,r_{{{\hat{p}}}, \ell } \Big (\! {{\textrm{tr}}}\big (P_k {\textbf{a}}{\textbf{a}}^\dagger \big ) - {{\textrm{tr}}}\big ( P_k {\textbf{a}}{\textbf{a}}^\dagger \big )\!\Big ), \end{aligned}$$

where in the last inequality we use $P_k, P_\ell \preceq I$. Thus, using the commutation relation $[{\textbf{a}}, {\textbf{a}}^\dagger ]= 1$, we get

$$\begin{aligned} \big \langle {\mathcal {L}}_0 \big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle \ge \sum _{k\ge \ell } r_{p,k}\,r_{{{\hat{p}}}, \ell }\, {{\textrm{tr}}}(P_k ) = \sum _{k\ge \ell } r_{p,k}\,r_{{{\hat{p}}}, \ell } (k+1). \end{aligned}$$

Repeating the same computation for ${{\widehat{\rho }}}$, we observe that the support of ${\textbf{a}}\Pi _\ell {\textbf{a}}^\dagger$ is included in the support of $\Pi _k$ if $k\ge \ell$. This implies ${{\textrm{tr}}}\big ( \Pi _k {\textbf{a}}\Pi _\ell {\textbf{a}}^\dagger \big ) = {{\textrm{tr}}}\big ( {\textbf{a}}\Pi _\ell {\textbf{a}}^\dagger \big )$. Similarly, if $k<\ell$, then the support of ${\textbf{a}}^\dagger \Pi _k{\textbf{a}}$ is inside the support of $\Pi _\ell$, which implies ${{\textrm{tr}}}\big ( \Pi _k {\textbf{a}}\Pi _\ell {\textbf{a}}^\dagger \big )={{\textrm{tr}}}\big ( {\textbf{a}}^\dagger \Pi _k {\textbf{a}}\Pi _\ell \big ) = {{\textrm{tr}}}\big ( \Pi _k {\textbf{a}}{\textbf{a}}^\dagger \big )$. Therefore, we have

$$\begin{aligned} \big \langle {\mathcal {L}}_0 \big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle \ge \sum _{k\ge \ell } r_{p,k}\,r_{{{\hat{p}}}, \ell } (k+1) = \big \langle {\mathcal {L}}_0 \big ({{\widehat{\rho }}}^{1/p}\big ), {{\widehat{\rho }}}^{1/{\hat{p}}}\big \rangle . \end{aligned}$$

(23)

By using the same argument, we also have

$$\begin{aligned} \big \langle {\mathcal {L}}_1 \big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle \ge \big \langle {\mathcal {L}}_1 \big ({{\widehat{\rho }}}^{1/p}\big ), {{\widehat{\rho }}}^{1/{\hat{p}}}\big \rangle . \end{aligned}$$

(24)

Putting (21)–(24) together and using the fact that $\nu _0, \nu _1, \omega \ge 0$ we conclude that $\Upsilon (\rho )\ge \Upsilon ({{\widehat{\rho }}})$ for $p>1$. The same inequality for $p=1$ is obtained by taking the limit $p\rightarrow 1^+$. Therefore, to prove $\Upsilon (\rho )\ge \eta _{{{{\textrm{th}}}}}$ in Theorem 1, it suffices to restrict to diagonal states in the Fock basis.

(ii) We now, in the second step, show that the optimal diagonal states that minimize $\Upsilon (\rho )$ are thermal. Let

$$\begin{aligned} \rho = \sum _n \lambda _n |n\rangle \langle n|, \end{aligned}$$

be the eigen-decomposition of $\rho$. A straightforward computation yields

$$\begin{aligned} \Upsilon (\rho )&= {\hat{p}}\sum _n \Big ( \nu _0 (n+1)\Big (\lambda _n -\lambda _n^{1/p}\lambda _{n+1}^{1/{{\hat{p}}}}\Big ) + \nu _1 n \Big (\lambda _n - \lambda _n^{1/p}\lambda _{n-1}^{1/{{\hat{p}}}} \Big )\!\Big )\nonumber \\&\quad + \omega \sum _n n\,\lambda _n -\sum _n \lambda _n\log \lambda _n. \end{aligned}$$

(25)

By the definition of $\eta _{{{\textrm{th}}}}$ in (19), for any $0\le x\le 1$ we have⁸

$$\begin{aligned} \eta _{{{{\textrm{th}}}}} (1-x)\le {\hat{p}}\Big (\nu _0 (1-x^{1/{\hat{p}}})+ \nu _1 (x-x^{1/p}) \Big ) + \omega x - x \log x- (1-x)\log (1-x). \end{aligned}$$

For any $\ell \ge 0$ let

$$\begin{aligned} s_\ell = \sum _{n=\ell }^\infty \lambda _n, \qquad x_\ell = \frac{s_{\ell +1}}{s_\ell }. \end{aligned}$$

We note that $s_\ell = \lambda _\ell + s_{\ell +1}$ and $0\le x_\ell \le 1$. Therefore,

$$\begin{aligned} \eta _{{{{\textrm{th}}}}} (1-x_\ell )&\le {\hat{p}}\Big (\nu _0 \Big (1-x_\ell ^{1/{\hat{p}}}\Big )+ \nu _1 \Big (x-x_\ell ^{1/p}\Big ) \! \Big ) + \omega x_\ell \\&\quad - x_\ell \log x_\ell - (1-x_\ell )\log (1-x_\ell ). \end{aligned}$$

Multiplying both sides by $s_\ell$ and summing over $\ell$, we obtain

$$\begin{aligned} \eta _{{{{\textrm{th}}}}} \sum _{\ell =0}^\infty s_\ell (1-x_\ell )&\le {\hat{p}}\sum _{\ell =0}^\infty s_\ell \Big (\nu _0 \Big (1-x_\ell ^{1/ {\hat{p}}}\Big )+ \nu _1\Big (x_\ell -x_\ell ^{1/p}\Big ) \! \Big ) \nonumber \\&\quad + \omega \sum _{\ell =0}^\infty s_\ell x_\ell - \sum _{\ell =0}^\infty s_\ell \big (x_\ell \log x_\ell + (1-x_\ell )\log (1-x_\ell )\big ). \end{aligned}$$

(26)

We compute and estimate each term in the above equation as follows. First, we have

$$\begin{aligned} \sum _{\ell =0}^\infty s_\ell (1-x_\ell ) = \sum _{\ell =0}^\infty \lambda _\ell = 1. \end{aligned}$$

Second, by applying Hölder’s inequality, we obtain

$$\begin{aligned} \sum _{\ell =0}^\infty s_\ell (1-x_\ell ^{1/{{\hat{p}}}})&= \sum _{\ell =0}^\infty s_\ell -\sum _{\ell =0}^\infty s_\ell ^{1/p} s_{\ell +1}^{1/{{\hat{p}}}} \nonumber \\&= \sum _{\ell =0}^\infty s_\ell -\sum _{\ell =0}^\infty \Big (\sum _{n=\ell }^\infty \lambda _n\Big )^{\!1/p} \Big (\sum _{n=\ell }^\infty \lambda _{n+1}\Big )^{\!1/{\hat{p}}} \nonumber \\&\le \sum _{\ell =0}^\infty \sum _{n=\ell }^\infty \lambda _n - \sum _{\ell =0}^\infty \sum _{n=\ell }^\infty \lambda _n^{1/p} \lambda _{n+1}^{1/{\hat{p}}} \nonumber \\&= \sum _{n=0}^\infty (n+1)\Big (\lambda _n - \lambda _n^{1/p} \lambda _{n+1}^{1/{\hat{p}}}\Big ). \end{aligned}$$

(27)

Similarly, we obtain

$$\begin{aligned} \sum _{\ell =0}^\infty s_\ell (x_\ell -x_\ell ^{1/p}) \le \sum _{n=0}^\infty n \big (\lambda _n- \lambda _n^{1/p} \lambda _{n-1}^{1/{\hat{p}}}\big ). \end{aligned}$$

We also have

$$\begin{aligned} \sum _{\ell =0}^\infty s_\ell x_\ell = \sum _{\ell =0}^\infty s_{\ell +1} =\sum _{\ell =0}^\infty \sum _{n=\ell +1}^\infty \lambda _n = \sum _{n=0}^\infty n\lambda _n. \end{aligned}$$

Finally, the last term of (26) becomes

$$\begin{aligned}&\sum _{\ell =0}^\infty s_\ell \big (x_\ell \log x_\ell + (1-x_\ell )\log (1-x_\ell )\big )\\&\quad = \sum _{\ell =0}^\infty s_\ell \Big (\frac{s_{\ell +1}}{s_\ell } \log \frac{s_{\ell +1}}{s_\ell }+ \frac{\lambda _{\ell }}{s_\ell } \log \frac{\lambda _{\ell }}{s_\ell }\Big )\\&\quad = \sum _{\ell =0}^\infty \Big (s_{\ell +1} \log s_{\ell +1}- s_{\ell +1} \log s_\ell + \lambda _\ell \log \lambda _\ell -\lambda _{\ell }\log s_\ell \Big )\\&\quad = \sum _{\ell =0}^\infty \Big (s_{\ell +1} \log s_{\ell +1}- s_{\ell } \log s_\ell + \lambda _\ell \log \lambda _\ell \Big )\\&\quad = \sum _{\ell =0}^\infty \lambda _\ell \log \lambda _\ell + \lim _{k\rightarrow +\infty } \sum _{\ell =0}^{k-1} \Big (s_{\ell +1} \log s_{\ell +1}- s_{\ell }\log s_\ell \Big )\\&\quad = \sum _{\ell =0}^\infty \lambda _\ell \log \lambda _\ell - s_0 \log s_0 + \lim _{k\rightarrow +\infty } s_{k} \log s_{k}\\&\quad = \sum _{\ell =0}^\infty \lambda _\ell \log \lambda _\ell , \end{aligned}$$

where in the last line we use $s_0=\sum _n \lambda _n=1$ and $\lim _{k\rightarrow +\infty } s_k=0$. Using these equations in (26) and compared to (25) we arrive at $\Upsilon (\rho )\ge \eta _{{{\textrm{th}}}}$.

The proof for $p=1$ is similar; we only need to replace the Hölder inequality in (27) with

$$\begin{aligned} -s_\ell \log x_{\ell }\le \sum _{n=\ell }^\infty \lambda _n\big (\log \lambda _{n}-\log \lambda _{n+1}\big ), \end{aligned}$$

that is derived from Hölder’s inequality by taking an appropriate limit.⁹$\square$

We remark that in part (ii) of the above proof, when the diagonal state $\rho$ is thermal, the parameter $x_\ell$ is independent of $\ell$. Moreover, in this case, the Hölder inequality applied in (27) is tight.

Note that our meta log-Sobolev inequality can be viewed as a generalization of a log-Sobolev inequality introduced in [17, Theorem 3.1] which holds only for the generator of the quantum-limited attenuator channel.

3.1 Meta Log-Sobolev Inequality for Multimode States

For some applications, it is useful to consider the function $\Upsilon (\cdot )$ in the multimode case and prove a generalization of Theorem 1 for multimode quantum states. In this subsection, we establish such a generalization but for the special cases of multimode states that can be prepared by applying a Gaussian unitary on multimode states that are diagonal in the Fock basis, which include all multimode Gaussian states, and multimode states that can be prepared by applying a passive Gaussian unitary on any product state.

For any m-mode quantum state $\rho$ with finite mean photon number define

$$\begin{aligned} \Upsilon _{m}(\rho )&:=\frac{{\hat{p}}}{m} \sum _{j=1}^m \big \langle {\mathcal {L}}_j\big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle + \frac{\omega }{m}\sum _{j=1}^m{{\textrm{tr}}}(\rho {\textbf{a}}_j^\dagger {\textbf{a}}_j) +\frac{1}{m} S(\rho ), \end{aligned}$$

(28)

where as before $\nu _{0}, \nu _{1},\omega \ge 0$ and ${\mathcal {L}}_j = \nu _{0}{\mathcal {L}}_{0, j}+\nu _{1}{\mathcal {L}}_{1, j}$ is the Lindbladian acting on the j-th mode. Note that we still use $\Upsilon (\rho )=\Upsilon _{1}(\rho )$ for the single-mode case. To establish our results on the m-mode generalizations of Theorem 1, we first state a lemma.

Lemma 2

For any m-mode state $\rho$ the followings hold:

• (i) Let $\rho ' = D_{ \xi } \rho D_\xi ^\dagger$ where $D_\xi = D_{\xi _1}\otimes \dots \otimes D_{\xi _m}$ is an m-mode displacement operator. Suppose that ${{\textrm{tr}}}(\rho {\textbf{a}}_j)=0$ for all j. Then, $\Upsilon _m(\rho ')\ge \Upsilon _m(\rho )$.

• (ii) Let $\rho ' = U\rho U^\dagger$ where U is a passive transformation, i.e., U is a Gaussian unitary that commutes with $H_m=\sum _j {\textbf{a}}_j^\dagger {\textbf{a}}_j$. Then, $\Upsilon _m(\rho ')= \Upsilon _m(\rho )$.

• (iii) Let $\rho ' = S_r\rho S_r^\dagger$ where $S_r = e^{\frac{1}{2}\sum _j r_j({\textbf{a}}_j^2 - ({\textbf{a}}_j^\dagger )^2)}$ is a squeezing transformation. Suppose that ${{\textrm{tr}}}\big (\rho {\textbf{a}}_j^2\big )={{\textrm{tr}}}\big ( \rho ^{1/p} {\textbf{a}}_j \rho ^{1/{\hat{p}}} {\textbf{a}}_j\big )=0$ for all j. Then, $\Upsilon _m(\rho ')\ge \Upsilon _m(\rho )$.

Proof

(i) We note that $D_\xi ^\dagger {\textbf{a}}_j D_\xi = {\textbf{a}}_j + \xi _j$. Therefore, $D_\xi ^\dagger {\textbf{a}}_j^\dagger {\textbf{a}}_j D_\xi = {\textbf{a}}_j^\dagger {\textbf{a}}_j + \xi _j {\textbf{a}}_j^\dagger + {{\bar{\xi }}}_j {\textbf{a}}_j + |\xi _j|^2$. Moreover,

$$\begin{aligned} {{\textrm{tr}}}(\rho ' {\textbf{a}}_j^\dagger {\textbf{a}}_j) = {{\textrm{tr}}}(\rho D_\xi ^\dagger {\textbf{a}}_j^\dagger {\textbf{a}}_k D_\xi ) = {{\textrm{tr}}}(\rho {\textbf{a}}_j^\dagger {\textbf{a}}_j) + |\xi _j|^2, \end{aligned}$$

where for the second equality we use the assumption ${{\textrm{tr}}}(\rho {\textbf{a}}_j)=0$. We similarly have

$$\begin{aligned} {{\textrm{tr}}}\big ({\rho '}^{1/p} {\textbf{a}}_j {\rho '}^{1/{\hat{p}}}{\textbf{a}}_j^\dagger \big ) = {{\textrm{tr}}}\big ({\rho }^{1/p} {\textbf{a}}_j {\rho }^{1/{\hat{p}}}{\textbf{a}}_j^\dagger \big ) + |\xi _j|^2. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \big \langle {\mathcal {L}}_{0,j}(\rho '^{1/p}), \rho '^{1/{\hat{p}}}\big \rangle ={{\textrm{tr}}}(\rho ' {\textbf{a}}_j{\textbf{a}}_j^\dagger )-{{\textrm{tr}}}\big ({\rho '}^{1/p} {\textbf{a}}_j {\rho '}^{1/{\hat{p}}}{\textbf{a}}_j^\dagger \big ) =\big \langle {\mathcal {L}}_{0,j}(\rho ^{1/p}), \rho ^{1/{\hat{p}}}\big \rangle . \end{aligned}$$

By using similar computations, we can also show that $\big \langle {\mathcal {L}}_{1,j}(\rho '^{1/p}), \rho '^{1/{\hat{p}}}\big \rangle = \big \langle {\mathcal {L}}_{1,j}(\rho ^{1/p}), \rho ^{1/{\hat{p}}}\big \rangle$. On the other hand, $S(\rho ') = S(D_\xi \rho D_\xi ^\dagger ) = S(\rho )$. Putting these together, we find that

$$\begin{aligned} \Upsilon _m(\rho ') = \Upsilon _m(\rho )+ \frac{\omega }{m} \sum _j |\xi _j|^2\ge \Upsilon _m(\rho ). \end{aligned}$$

(ii) For any passive transformation U, there is an $m\times m$ unitary matrix $(u_{jk})_{j,k}$ such that $U^\dagger {\textbf{a}}_j U = \sum _{k} u_{jk} {\textbf{a}}_{k}$. Employing this relation yields

$$\begin{aligned} \sum _{j=1}^m{{\textrm{tr}}}(\rho ' {\textbf{a}}_j^\dagger {\textbf{a}}_j)= & \sum _{j=1}^m{{\textrm{tr}}}\big (\rho \, U^\dagger {\textbf{a}}_j^\dagger U U^\dagger {\textbf{a}}_j U\big )=\sum _{j,k,k'=1}^m u_{jk}{\bar{u}}_{jk'}\, {{\textrm{tr}}}\big (\rho {\textbf{a}}_k^\dagger {\textbf{a}}_{k'}\big )\\= & \sum _{k=1}^m{{\textrm{tr}}}(\rho {\textbf{a}}_k^\dagger {\textbf{a}}_k), \end{aligned}$$

where we used $\sum _{j}u_{jk}{\bar{u}}_{jk'}=\delta _{k,k'}$. Applying similar computations, one can also verify that $\big \langle {\mathcal {L}}_{b,j}(\rho '^{1/p}), \rho '^{1/{\hat{p}}}\big \rangle = \big \langle {\mathcal {L}}_{b,j}(\rho ^{1/p}), \rho ^{1/{\hat{p}}}\big \rangle$ for $b\in \{0,1\}$. Therefore, using these relations together with $S(\rho ') = S(\rho )$, we obtain $\Upsilon _m(\rho ')=\Upsilon _m(\rho )$.

(iii) The proof of this part is more involved. First, we note that $S_r^\dagger {\textbf{a}}_j S_r = \cosh (r_j){\textbf{a}}_j - \sinh (r_j){\textbf{a}}_j^\dagger$, which using $\cosh ^2(r_j) - \sinh ^2(r_j)=1$ implies

$$\begin{aligned} S_r^\dagger {\textbf{a}}_j^\dagger {\textbf{a}}_j S_r = {\textbf{a}}_j^\dagger {\textbf{a}}_j + \sinh ^2(r_j)({\textbf{a}}_j^\dagger {\textbf{a}}_j + {\textbf{a}}_j{\textbf{a}}_j^\dagger ) - \cosh (r_j)\sinh (r_j) \big (({\textbf{a}}_j^\dagger )^2 + {\textbf{a}}_j^2\big ). \end{aligned}$$

(29)

Using these relations and applying the assumption on $\rho$, we find that

$$\begin{aligned} \big \langle {\mathcal {L}}_{0,j}\big (\rho '^{1/p}\big ), \rho '^{1/{\hat{p}}}\big \rangle&= \big \langle {\mathcal {L}}_{0,j}\big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle \\&\quad + \sinh ^2(r_j)\Big ( {{\textrm{tr}}}(\rho {\textbf{a}}_j^\dagger {\textbf{a}}_j) + {{\textrm{tr}}}(\rho {\textbf{a}}_j{\textbf{a}}_j^\dagger ) - {{\textrm{tr}}}(\rho ^{1/p} {\textbf{a}}_j \rho ^{1/{\hat{p}}} {\textbf{a}}_j^\dagger ) \\&\quad - {{\textrm{tr}}}(\rho ^{1/p}{\textbf{a}}_j^\dagger \rho ^{1/{\hat{p}}} {\textbf{a}}_j) \Big ). \end{aligned}$$

Let $\rho =\sum _k \lambda _k|\psi _k\rangle \langle \psi _k|$ be the eigen-decomposition of $\rho$. Then, by Young’s inequality we have

$$\begin{aligned}&{{\textrm{tr}}}(\rho ^{1/p} {\textbf{a}}_j \rho ^{1/{\hat{p}}} {\textbf{a}}_j^\dagger ) + {{\textrm{tr}}}(\rho ^{1/p} {\textbf{a}}_j^\dagger \rho ^{1/{\hat{p}}} {\textbf{a}}_j) \\&\quad = \sum _{k, \ell } \Big ( \lambda _k^{1/p}\lambda _\ell ^{1/{\hat{p}}} + \lambda _k^{1/{\hat{p}}}\lambda _\ell ^{1/p}\Big )\big | \langle \psi _k| {\textbf{a}}_j |\psi _\ell \rangle \big |^2 \\&\quad \ge \sum _{k, \ell } \Big ( \frac{1}{p}\lambda _k + \frac{1}{{\hat{p}}} \lambda _\ell + \frac{1}{{\hat{p}}}\lambda _k+ \frac{1}{p}\lambda _\ell \Big ) \big | \langle \psi _k| {\textbf{a}}_j |\psi _\ell \rangle \big |^2\\&\quad = \sum _{k, \ell } \big ( \lambda _k + \lambda _\ell \big )\big | \langle \psi _k| {\textbf{a}}_j |\psi _\ell \rangle \big |^2\\&\quad = \sum _{k} \lambda _k \big \Vert {\textbf{a}}_j^\dagger |\psi _k\rangle \big \Vert ^2 + \sum _{ \ell } \lambda _\ell \big \Vert {\textbf{a}}_j |\psi _\ell \rangle \big \Vert ^2\\&\quad = {{\textrm{tr}}}\big (\rho {\textbf{a}}_j{\textbf{a}}_j^\dagger \big ) + {{\textrm{tr}}}\big (\rho {\textbf{a}}_j^\dagger {\textbf{a}}_j\big ). \end{aligned}$$

Therefore, we can see that

$$\begin{aligned} \big \langle {\mathcal {L}}_{0,j}\big (\rho '^{1/p}\big ), \rho '^{1/{\hat{p}}}\big \rangle \ge \big \langle {\mathcal {L}}_{0,j}\big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle , \end{aligned}$$

and similarly $\big \langle {\mathcal {L}}_{1,j}\big (\rho '^{1/p}\big ), \rho '^{1/{\hat{p}}}\big \rangle \ge \big \langle {\mathcal {L}}_{1,j}\big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}}\big \rangle$. Moreover, once again using (29) we have

$$\begin{aligned} {{\textrm{tr}}}(\rho ' H_m) ={{\textrm{tr}}}(\rho H_m) + \sum _j \sinh ^2(r_j) \big ( {{\textrm{tr}}}(\rho {\textbf{a}}_j^\dagger {\textbf{a}}_j) + {{\textrm{tr}}}(\rho {\textbf{a}}_j{\textbf{a}}_j^\dagger ) \big ) \ge {{\textrm{tr}}}(\rho H_m). \end{aligned}$$

We also have $S(\rho ')=S(\rho )$. Putting these together we arrive at $\Upsilon _m(\rho ')\ge \Upsilon _m(\rho )$. $\square$

Theorem 3

For any $\nu _0, \nu _1, \omega \ge 0$ let $\eta _{{{\textrm{th}}}}$ be given by (19). Then, for any multimode state $\rho '=U_{\textrm{G}}\, \rho U_{\textrm{G}}^\dagger$ where $U_{\textrm{G}}$ is an m-mode Gaussian unitary, we have

$$\begin{aligned} \Upsilon _m(\rho ')\ge \eta _{{{\textrm{th}}}}, \end{aligned}$$

(30)

assuming that one of the following conditions is satisfied:

• (a) $\rho =\rho _{1}\otimes \cdots \otimes \rho _{m}$ is a product state and $U_G$ is passive.

• (b) $\rho =\rho _{{\textrm{diag}}}$ is diagonal in the Fock basis.

Proof

We first note that by the definition of $\Upsilon _m(\cdot )$ and using Theorem 1 we have

$$\begin{aligned} \Upsilon (\rho _1\otimes \cdots \otimes \rho _m) = \frac{1}{m} \sum _j \Upsilon (\rho _j)\ge \eta _{{{{\textrm{th}}}}}. \end{aligned}$$

Then, (a) follows from part (ii) of Lemma 2. To prove (b), we use the Bloch-Messiah decomposition to express the Gaussian unitary $U_G$ as $U_{\textrm{G}}=D_\xi U S_r V$, where $D_\xi$ is a multimode displacement operator, U, V are passive Gaussian transformations, and $S_r$ is a product of single-mode squeezing transformations [41]. Therefore, the multimode state can be written as

$$\begin{aligned} \rho ' = D_\xi U S_r V \rho _{{\textrm{diag}}} V^\dagger S_r^\dagger U^\dagger D_\xi ^\dagger , \end{aligned}$$

where $\rho _{{\textrm{diag}}}$ is diagonal in the Fock basis. We use this expression to prove (30) in the following steps:

• By Theorem 1, the meta log-Sobolev inequality holds for all single-mode diagonal states. Then, applying a standard tensorization argument in the classical case, i.e., using the chain rule and the concavity of the entropy function (see, e.g., [38, Proposition 3.7]), we find that $\Upsilon _m(\rho _{\textrm{diag}})\ge \eta _{{{{\textrm{th}}}}}$. Then, by part (ii) of Lemma 2, we have $\Upsilon _m(V \rho _{{\textrm{diag}}} V^\dagger )=\Upsilon _m(\rho _{{\textrm{diag}}})\ge \eta _{{{\textrm{th}}}}$.

• Next we use part (iii) of Lemma 2 to establish (30) for $S_r V \rho _{{\textrm{diag}}} V^\dagger S_r^\dagger$. To this end, we need to verify the required assumptions ${{\textrm{tr}}}\big (V \rho _{{\textrm{diag}}} V^\dagger {\textbf{a}}_j^2\big )={{\textrm{tr}}}\big ( (V \rho _{{\textrm{diag}}} V^\dagger )^{1/p} {\textbf{a}}_j (V \rho _{{\textrm{diag}}} V^\dagger )^{1/{\hat{p}}} {\textbf{a}}_j\big )=0$. We note that ${{\textrm{tr}}}\big (V \rho _{{\textrm{diag}}} V^\dagger {\textbf{a}}_j^2\big ) = {{\textrm{tr}}}\big (\rho _{{\textrm{diag}}} (V^\dagger {\textbf{a}}_j V)^2\big )$ and $V^\dagger {\textbf{a}}_j V= \sum _{k} v_{jk} {\textbf{a}}_{k}$ is a linear combination of ${\textbf{a}}_{k}$’s. Moreover, ${{\textrm{tr}}}(\rho _{{\textrm{diag}}} {\textbf{a}}_k{\textbf{a}}_{k'})=0$ for any $k, k'$ simply because $\rho _{{\textrm{diag}}}$ is diagonal in the Fock basis. Therefore, we have ${{\textrm{tr}}}\big (V \rho _{{\textrm{diag}}} V^\dagger {\textbf{a}}_j^2\big )=0$. By the same argument, we also have ${{\textrm{tr}}}\big ( (V \rho _{{\textrm{diag}}} V^\dagger )^{1/p} {\textbf{a}}_j (V \rho _{{\textrm{diag}}} V^\dagger )^{1/{\hat{p}}} {\textbf{a}}_j\big )=0$. Hence, $\Upsilon _m(S_r V \rho _{{\textrm{diag}}} V^\dagger S_r^\dagger )=\Upsilon _m(V \rho _{{\textrm{diag}}} V^\dagger )=\Upsilon _m(\rho _{{\textrm{diag}}})\ge \eta _{{{\textrm{th}}}}$.

• Once again, using part (ii) of Lemma 2 we find that (30) holds for $U S_r V \rho _{{\textrm{diag}}} V^\dagger S_r^\dagger U^\dagger$ since we have already verified it for $S_r V \rho _{{\textrm{diag}}} V^\dagger S_r^\dagger$ and U is a passive transformation.

• Finally, we note that $V^\dagger S_r^\dagger U^\dagger {\textbf{a}}_j U S_r V$ is a linear combination of ${\textbf{a}}_k$’s and ${\textbf{a}}_k^\dagger$’s. Moreover, ${{\textrm{tr}}}(\rho _{{\textrm{diag}}} {\textbf{a}}_k) ={{\textrm{tr}}}(\rho _{{\textrm{diag}}} {\textbf{a}}_k^\dagger )=0$. Therefore, ${{\textrm{tr}}}(U S_r V \rho _{{\textrm{diag}}} V^\dagger S_r^\dagger U^\dagger {\textbf{a}}_j)=0$ for all j. Thus, by part (i) of Lemma 2, inequality (30) holds for $\rho = D_\xi U S_r V \rho _{{\textrm{diag}}} V^\dagger S_r^\dagger U^\dagger D_\xi ^\dagger$. $\square$

Notice that Theorem 3 implies $\Upsilon _m(\rho )\ge \eta _{{{{\textrm{th}}}}}$ for all multimode Gaussian states. The point is that by Williamson’s theorem any Gaussian state can be transformed into a tensor product of thermal states using a Gaussian unitary, and thermal states are diagonal in the Fock basis [41].

4 Log-Sobolev Inequalities for the Quantum Ornstein–Uhlenbeck Semigroup

As discussed in Sect. 2, the semigroup of attenuator channels is sometimes called the quantum (bosonic) Ornstein–Uhlenbeck semigroup.¹⁰ In this section, we explicitly compute the optimal p-log-Sobolev constant for this semigroup for any $1\le p\le 2$, and derive a quantum variant of the celebrated log-Sobolev inequality of Gross [25]. Due to the equivalence of log-Sobolev inequalities and hypercontractivity inequalities for quantum Markov semigroups [2, 35, 39], our results provide the optimal hypercontractivity inequalities for the quantum Ornstein–Uhlenbeck semigroup.

We need to develop some notations to present our results. Let $\Phi _t=e^{-t{\mathcal {L}}}$ be the quantum attenuator channel with

$$\begin{aligned} {\mathcal {L}}(\rho ) = e^{-\beta /2}\Big ( \frac{1}{2} \big \{{\textbf{a}}{\textbf{a}}^\dagger , \rho \big \} - {\textbf{a}}^\dagger \rho {\textbf{a}}\Big ) + e^{\beta /2}\Big ( \frac{1}{2} \big \{{\textbf{a}}^\dagger {\textbf{a}}, \rho \big \} - {\textbf{a}}\rho {\textbf{a}}^\dagger \Big ), \end{aligned}$$

(31)

where $\beta >0$ is some parameter. The adjoint of the channel with respect to the Hilbert–Schmidt inner product is denoted by $\Phi _t^*$ and describes the evolution in the Heisenberg picture. Then, $\Phi _t^*=e^{-t{\mathcal {L}}^*}$ where¹¹

$$\begin{aligned} {\mathcal {L}}^*(X) = e^{-\beta /2}\Big ( \frac{1}{2}\big \{{\textbf{a}}{\textbf{a}}^\dagger , X\big \}-{\textbf{a}}X{\textbf{a}}^\dagger \Big )+e^{\beta /2}\Big ( \frac{1}{2}\big \{{\textbf{a}}^\dagger {\textbf{a}}, X\big \} -{\textbf{a}}^\dagger X{\textbf{a}}\Big ). \end{aligned}$$

The semigroup $\{\Phi _t:\, t\ge 0\}$ has a fixed point which we denote by $\sigma =\sigma _\beta$:

$$\begin{aligned} \sigma =\sigma _\beta = (1-e^{-\beta })\sum _{n=0}^\infty e^{-\beta n}|n\rangle \langle n|. \end{aligned}$$

Then, it is natural to define a weighted inner product with respect to this state:

$$\begin{aligned} \langle X, Y\rangle _{\sigma } = {{\textrm{tr}}}\big ( \sigma ^{1/2} X^\dagger \sigma ^{1/2} Y\big )=\big \langle \Gamma _\sigma (X), Y\big \rangle , \end{aligned}$$

(32)

where

$$\begin{aligned} \Gamma _\sigma (X) = \sigma ^{1/2} X\sigma ^{1/2}. \end{aligned}$$

We emphasize that the weighted inner product $\langle \cdot , \cdot \rangle _\sigma$ should not be confused with the Hilbert–Schmidt inner product $\langle \cdot , \cdot \rangle$ that has no subscript. This inner product induces the 2-norm

$$\begin{aligned} \Vert X\Vert _{2, \sigma }^2 ={{\textrm{tr}}}\big ( \big |\Gamma _\sigma ^{1/2}(X)\big |^2 \big ) ={{\textrm{tr}}}\big ( |\sigma ^{1/4}X\sigma ^{1/4}|^2 \big ), \end{aligned}$$

where $|Y| = \sqrt{Y^\dagger Y}$. It can be verified that ${\mathcal {L}}=\Gamma _\sigma \circ {\mathcal {L}}^*\circ \Gamma _\sigma ^{-1}$ and that ${\mathcal {L}}^*$ is self-adjoint with respect to this weighted inner product. Thus, we may consider the Dirichlet form associated with ${\mathcal {L}}^*$ given by

$$\begin{aligned} {\mathcal {E}}_p(\rho ) = \frac{p{\hat{p}}}{4}\Big \langle \Gamma _\sigma ^{-\frac{1}{{\hat{p}}}}\big (\rho ^{\frac{1}{{\hat{p}}}}\big ), {\mathcal {L}}^*\circ \Gamma _\sigma ^{-\frac{1}{p}}\big (\rho ^{\frac{1}{p}}\big ) \Big \rangle _{\sigma }. \end{aligned}$$

Here, $p\ge 1$ and the case of $p=1$ is understood in the limit as

$$\begin{aligned} {\mathcal {E}}_1(\rho ) = \frac{1}{4}\big \langle \log \rho -\log \sigma , {\mathcal {L}}^*\circ \Gamma _\sigma ^{- 1}(\rho ) \big \rangle _{\sigma }. \end{aligned}$$

Now, a p-log-Sobolev inequality with parameters $c\ge 0$ takes the form:

$$\begin{aligned} c D(\rho \Vert \sigma ) \le {\mathcal {E}}_p(\rho ), \quad \forall \rho , \end{aligned}$$

where $D(\rho \Vert \sigma )={{\textrm{tr}}}(\rho \log \rho ) - {{\textrm{tr}}}(\rho \log \sigma )$ is Umegaki’s relative entropy. The optimal constant c for which the above inequality holds is usually denoted by $\alpha _p$, i.e.,

$$\begin{aligned} \alpha _p = \inf _{\rho } \frac{{\mathcal {E}}_p(\rho )}{D(\rho \Vert \sigma )}, \end{aligned}$$

where the infimum is taken over all states with finite mean photon number. We note that, as it will become clear in the proof of the following theorem, similar to $\Upsilon (\rho )$, ${\mathcal {E}}_p(\rho )$ is well defined for any state $\rho$ with finite mean photon number.

Theorem 4

For any $\beta >0$, the p-log-Sobolev constant of the quantum Ornstein–Uhlenbeck semigroup is given by

$$\begin{aligned} \alpha _p = \frac{p{\hat{p}}}{4\beta } e^{\beta /2}\big (1-e^{-\beta /p}\big )\big (1-e^{-\beta /{\hat{p}}}\big ), \end{aligned}$$

(33)

if $p>1$ and $\alpha _1 = \lim _{p\rightarrow 1^+}\alpha _p = \frac{1}{2} \sinh (\beta /2)$.

Proof

We note that $\sigma ^s{\textbf{a}}= e^{s\beta } {\textbf{a}}\sigma ^s$ for every s, and $\sigma$ commutes with ${\textbf{a}}^\dagger {\textbf{a}}$. Then, for a single-mode state $\rho$ we have

$$\begin{aligned} {\mathcal {E}}_p(\rho )&= \frac{p{\hat{p}}}{4}\Big \langle \Gamma _\sigma ^{-\frac{1}{{\hat{p}}}}\big (\rho ^{\frac{1}{{\hat{p}}}}\big ) , {\mathcal {L}}^*\circ \Gamma _\sigma ^{-\frac{1}{p}} \big (\rho ^{\frac{1}{p}}\big ) \! \Big \rangle _{\sigma } \nonumber \\&= \frac{p{\hat{p}}}{4} {{\textrm{tr}}}\Big ( \rho ^{\frac{1}{{\hat{p}}}} \Gamma _\sigma ^{\frac{1}{p}} \circ {\mathcal {L}}^*\circ \Gamma _\sigma ^{-\frac{1}{p}} (\rho ^{\frac{1}{p}}) \Big ) \nonumber \\&= \frac{p{\hat{p}}}{4} \bigg (\!e^{-\beta /2}\Big [ {{\textrm{tr}}}\big (\rho \,{\textbf{a}}\, {\textbf{a}}^\dagger \big ) - {{\textrm{tr}}}\big ( \rho ^{\frac{1}{{\hat{p}}}} \sigma ^{\frac{1}{2p}} {\textbf{a}}\, \sigma ^{-\frac{1}{2p}} \rho ^{\frac{1}{p}} \sigma ^{-\frac{1}{2p}} {\textbf{a}}^\dagger \sigma ^{\frac{1}{2p}} \big ) \Big ] \nonumber \\&\quad + e^{\beta /2}\Big [ {{\textrm{tr}}}\big (\rho \,{\textbf{a}}^\dagger {\textbf{a}}\big ) - {{\textrm{tr}}}\big ( \rho ^{\frac{1}{{\hat{p}}}} \sigma ^{\frac{1}{2p}} {\textbf{a}}^\dagger \sigma ^{-\frac{1}{2p}} \rho ^{\frac{1}{p}} \sigma ^{-\frac{1}{2p}} {\textbf{a}}\, \sigma ^{\frac{1}{2p}} \big ) \Big ]\!\bigg ) \nonumber \\&= \frac{p{\hat{p}}}{4} \bigg (\!e^{-\beta /2}\Big [ {{\textrm{tr}}}\big (\rho \, {\textbf{a}}\,{\textbf{a}}^\dagger \big ) - e^{\frac{\beta }{p}} {{\textrm{tr}}}\big ( \rho ^{\frac{1}{{\hat{p}}}} {\textbf{a}}\, \rho ^{\frac{1}{p}} {\textbf{a}}^\dagger \big ) \Big ] \nonumber \\&\quad + e^{\beta /2}\Big [ {{\textrm{tr}}}\big (\rho \,{\textbf{a}}^\dagger {\textbf{a}}\big ) - e^{-\frac{\beta }{p}}{{\textrm{tr}}}\big ( \rho ^{\frac{1}{{\hat{p}}}} {\textbf{a}}^\dagger \rho ^{\frac{1}{p}} {\textbf{a}}\big ) \Big ] \!\bigg ) \nonumber \\&= {\hat{p}}\Big ( \nu '_0\big \langle {\mathcal {L}}_0(\rho ^{1/p}), \rho ^{1/{\hat{p}}} \big \rangle + \nu '_1\big \langle {\mathcal {L}}_1\big (\rho ^{1/p}\big ), \rho ^{1/{\hat{p}}} \big \rangle \Big ) \nonumber \\&\quad + \omega '{{\textrm{tr}}}\big (\rho \,{\textbf{a}}^\dagger {\textbf{a}}\big ) + \frac{p{\hat{p}}}{4}\Big (e^{-\beta /2} - e^{-\big (\frac{1}{2} -\frac{1}{p}\big )\beta }\Big ), \end{aligned}$$

(34)

where ${\mathcal {L}}_0, {\mathcal {L}}_1$ are given in (15) and

$$\begin{aligned} \nu '_0&= \frac{p}{4} e^{\big (\frac{1}{2} -\frac{1}{p}\big )\beta }, \qquad \nu '_1=\frac{p}{4}e^{-\big (\frac{1}{2} -\frac{1}{p}\big )\beta },\\ \omega '&=\frac{p{\hat{p}}}{4}\Big (e^{\beta /2} + e^{-\beta /2} - e^{\big (\frac{1}{2} -\frac{1}{p}\big )\beta }-e^{-\big (\frac{1}{2} -\frac{1}{p}\big )\beta } \Big )\\&=\frac{p{\hat{p}}}{4}e^{\beta /2} \big (1-e^{-\beta /p}\big )\big (1-e^{-\beta /{\hat{p}}}\big ). \end{aligned}$$

On the other hand, since $\sigma = (1-e^{-\beta }) e^{-\beta {\textbf{a}}^\dagger {\textbf{a}}}$, we have

$$\begin{aligned} D(\rho \Vert \sigma ) ={{\textrm{tr}}}(\rho \log \rho ) - {{\textrm{tr}}}(\rho \log \sigma ) = -S(\rho ) +\beta {{\textrm{tr}}}(\rho \,{\textbf{a}}^\dagger {\textbf{a}}) - \log (1-e^{-\beta }). \end{aligned}$$

Putting these together, we find that for $\alpha _p$ given by (33) we have

$$\begin{aligned} \frac{1}{\alpha _p}{\mathcal {E}}_p(\rho )- D(\rho \Vert \sigma ) -\log (1-e^{-\beta })-\frac{p{\hat{p}}}{4}\Big (e^{-\beta /2} - e^{-\big (\frac{1}{2} -\frac{1}{p}\big )\beta }\Big )=\Upsilon (\rho ), \end{aligned}$$

where $\Upsilon (\rho )$ is given in (16) with $\nu _0 = \frac{\nu '_0}{\alpha _p}$, $\nu _1=\frac{\nu '_1}{\alpha _p}$ and $\omega =\frac{\omega '}{\alpha _p}-\beta =0$. Therefore, to prove the theorem we can use Theorem 1 to conclude that it suffices to show that for any thermal state $\tau$ we have

$$\begin{aligned} \alpha _p D(\tau \Vert \sigma ) \le {\mathcal {E}}_p(\tau ), \end{aligned}$$

(35)

where $\alpha _p$ is given by (33), and it is the best possible such constant.

By the above computations, we have

$$\begin{aligned} {\mathcal {E}}_p(\tau )&= \frac{p{\hat{p}}}{4} \bigg (\!e^{-\beta /2}\Big [ {{\textrm{tr}}}(\tau {\textbf{a}}\,{\textbf{a}}^\dagger ) - e^{\frac{\beta }{p}} {{\textrm{tr}}}\big ( \tau ^{\frac{1}{{\hat{p}}}} {\textbf{a}}\, \rho ^{\frac{1}{p}} {\textbf{a}}^\dagger \big ) \Big ] \\&\quad + e^{\beta /2}\Big [ {{\textrm{tr}}}(\tau {\textbf{a}}^\dagger {\textbf{a}}) - e^{-\frac{\beta }{p}}{{\textrm{tr}}}\big ( \tau ^{\frac{1}{{\hat{p}}}} {\textbf{a}}^\dagger \tau ^{\frac{1}{p}} {\textbf{a}}\big ) \Big ]\!\bigg )\\&= \frac{p{\hat{p}}}{4} \bigg (\! \big (e^{\beta /2}+e^{-\beta /2}\big ) {{\textrm{tr}}}(\tau {\textbf{a}}^\dagger {\textbf{a}}) +e^{-\beta /2} - e^{\beta \big (\frac{1}{p}-\frac{1}{2}\big )} {{\textrm{tr}}}\big ( \tau ^{\frac{1}{{\hat{p}}}} {\textbf{a}}\tau ^{\frac{1}{p}} {\textbf{a}}^\dagger \big ) \\&\quad - e^{\beta \big (\frac{1}{{\hat{p}}} - \frac{1}{2}\big )}{{\textrm{tr}}}\big ( \tau ^{\frac{1}{{\hat{p}}}} {\textbf{a}}^\dagger \tau ^{\frac{1}{p}} {\textbf{a}}\big ) \! \bigg ). \end{aligned}$$

Here, for convenience we use a new parametrization for thermal states:

$$\begin{aligned} \tau = (1-y^2)\sum _{n} y^{2n}|n\rangle \langle n|. \end{aligned}$$

Using this, we have

$$\begin{aligned}&{{\textrm{tr}}}(\tau {\textbf{a}}^\dagger {\textbf{a}}) = \frac{y^2}{1-y^2}, \qquad {{\textrm{tr}}}(\tau \log \tau ) = \frac{y^2}{1-y^2}\log y^2+ \log (1-y^2),\\&{{\textrm{tr}}}\big ( \tau ^{\frac{1}{p}} {\textbf{a}}^\dagger \tau ^{\frac{1}{{\hat{p}}}} {\textbf{a}}\big )= \frac{y^{\frac{2}{p}}}{1-y^2} , \qquad {{\textrm{tr}}}\big ( \tau ^{\frac{1}{{\hat{p}}}} {\textbf{a}}^\dagger \tau ^{\frac{1}{p}} {\textbf{a}}\big ) = \frac{y^{\frac{2}{{\hat{p}}}}}{1-y^2}. \end{aligned}$$

Therefore, we get

$$\begin{aligned} {\mathcal {E}}_p(\tau )&= \frac{p{\hat{p}}}{4(1-y^2)} \bigg ( \! \Big (e^{\beta /2}+e^{-\beta /2}\Big ) y^2 +e^{-\beta /2}(1-y^2) - e^{\beta \big (\frac{1}{p}-\frac{1}{2}\big )} y^{\frac{2}{p}} - e^{\beta \big (\frac{1}{{\hat{p}}} - \frac{1}{2}\big )}y^{\frac{2}{{\hat{p}}}}\bigg )\\&= \frac{p{\hat{p}}}{4(1-y^2)} \bigg (\! e^{\beta /2} y^2+e^{-\beta /2} - e^{\beta \big (\frac{1}{p}-\frac{1}{2}\big )} y^{\frac{2}{p}} - e^{\beta \big (\frac{1}{{\hat{p}}} - \frac{1}{2}\big )}y^{\frac{2}{{\hat{p}}}}\bigg )\\&= \frac{p{\hat{p}}}{4(1-y^2)} e^{\beta /2} \Big ( y^{\frac{2}{p}}-e^{-\frac{\beta }{p}} \Big )\Big ( y^{\frac{2}{{\hat{p}}}}-e^{-\frac{\beta }{{\hat{p}}} }\Big ). \end{aligned}$$

We also have

$$\begin{aligned} D(\tau \Vert \sigma )&= {{\textrm{tr}}}(\tau \log \tau ) + \beta {{\textrm{tr}}}(\tau {\textbf{a}}^\dagger {\textbf{a}}) - \log (1-e^{-\beta })\\&= \frac{1}{1-y^2}\Big ( y^2\log y^2 + (1-y^2)\log (1-y^2) +\beta y^2 -(1-y^2)\log (1-e^{-\beta }) \Big )\\&= \frac{1}{1-y^2} d(y^2\Vert e^{-\beta }), \end{aligned}$$

where $d(y^2\Vert x^2) = y^2\log y^2 + (1-y^2)\log (1-y^2) -y^2\log x^2-(1-y^2)\log (1-x^2)$ is the binary relative entropy function. Hence, (35) is equivalent to $\phi (x, y)\ge 0$ for all $0<x, y<1$ where $x^2=e^{-\beta }$ and

$$\begin{aligned} \phi (x, y)&= (1-y^2)\frac{{\mathcal {E}}_p(\tau )}{\alpha _p} - (1-y^2)D(\tau \Vert \sigma ) \\&= -\frac{\big ( x^{\frac{2}{p}}-y^{\frac{2}{p}} \big )\big ( x^{\frac{2}{{\hat{p}}}}-y^{\frac{2}{{\hat{p}}} }\big )\log x^2}{\big (1-x^{\frac{2}{p}}\big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )} - d(y^2\Vert x^2). \end{aligned}$$

Now the idea is to fix y and think of $\phi (x, y)$ as a function of x. It is shown in Appendix C that

$$\begin{aligned} \frac{{\textrm{d}}}{{\textrm{d}}x}\phi (x, y)&= \frac{2x\Big (\! \big (1- x^{\frac{2}{{\hat{p}}}} \big )\big (1-y^{\frac{2}{{\hat{p}}}}\big ) \big (x^{\frac{2}{p}} - y^{\frac{2}{p}} \big ) + \big (1- x^{\frac{2}{p}}\big )\big (1-y^{\frac{2}{p}}\big ) \big (x^{\frac{2}{{\hat{p}}}} - y^{\frac{2}{{\hat{p}}}}\big ) \! \Big )}{(1-x^2) \big (1-x^{\frac{2}{p}}\big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )} \nonumber \\&\quad - \frac{2x^{\frac{2}{p}}\big (1-y^{\frac{2}{p}}\big )\big (x^{\frac{2}{{\hat{p}}}} - y^{\frac{2}{{\hat{p}}}}\big )}{x\big (1-x^{\frac{2}{p}}\big )\big (1 -x^{\frac{2}{{\hat{p}}}}\big )} \bigg ( \frac{\log x^2}{p(1-x^{\frac{2}{p}})}+1 \bigg ) \nonumber \\&\quad - \frac{2x^{\frac{2}{{\hat{p}}}}\big (1-y^{\frac{2}{{\hat{p}}}}\big ) \big (x^{\frac{2}{p}} - y^{\frac{2}{p}}\big )}{ x\big (1-x^{\frac{2}{p}}\big ) \big (1-x^{\frac{2}{{\hat{p}}}}\big )} \bigg ( \frac{\log x^2}{{\hat{p}}(1-x^{\frac{2}{{\hat{p}}}})}+1 \bigg ) . \end{aligned}$$

(36)

We argue that this derivative vanishes on the interval (0, 1) only if $x=y$. To this end, we use $\log t< t-1$ for $1\ne t\in \big \{x^{\frac{2}{p}}, x^{\frac{2}{{\hat{p}}}}\big \}$ to conclude that for $0<x<1$,

$$\begin{aligned} -\bigg (\frac{\log x^2}{p(1-x^{\frac{2}{p}})}+1\bigg ), -\bigg (\frac{\log x^2}{{\hat{p}}(1-x^{\frac{2}{{\hat{p}}}})}+1\bigg )>0. \end{aligned}$$

Therefore, $\frac{{\textrm{d}}}{{\textrm{d}}x}\phi (x, y)$ is nonzero for $0<x\ne y<1$ and its sign depends on whether $x>y$ or $x<y$. In fact, if $0<x_1<y<x_2<1$, then

$$\begin{aligned} \frac{{\textrm{d}}}{{\textrm{d}}x}\phi (x, y) \Big |_{x=x_1}< \frac{{\textrm{d}}}{{\textrm{d}}x}\phi (x, y) \Big |_{x=y}=0 < \frac{{\textrm{d}}}{{\textrm{d}}x}\phi (x, y) \Big |_{x=x_2}. \end{aligned}$$

Thus, the minimum of $\phi (x, y)$, as a function of x, is achieved at $x=y$ and we have $\phi (y, y)=0$. This means that $\phi (x, y)\ge 0$ for all $0<x, y<1$, and (35) holds for any thermal state $\tau$. Also, the limiting case of $y\rightarrow 1^-$ confirms that the constant $\alpha _p$ in (35) is optimal. $\square$

We remark that some authors define the quantum Ornstein–Uhlenbeck semigroup by considering arbitrary parameters $\nu _0, \nu _1$ satisfying $\nu _1>\nu _0>0$, while here we study only the case of $\nu _0=\nu _1^{-1}=e^{-\beta /2}$. Nevertheless, by a rescaling argument all the log-Sobolev constants of these semigroups can also be computed using Theorem 4.

Extending the definition of the weighted 2-norm, for any $p\ge 1$ we may define

$$\begin{aligned} \Vert X\Vert _{p, \sigma } = {{\textrm{tr}}}\Big ( \Big | \Gamma _{ \sigma }^{\frac{1}{p}}(X) \Big |^{p} \Big )^{\frac{1}{p}}. \end{aligned}$$

(37)

We let $L_p(\sigma )$ to be the closure of the space of bounded operators under this norm.

The following corollary is a consequence of the above theorem, and [39, Theorem 3.8] (see also [2, Theorem 11]).

Corollary 5

Let $\Phi _{t} = e^{-t{\mathcal {L}}}$ where ${\mathcal {L}}$ is given by (3138). Also, let $\Phi ^*_{t} = e^{-t{\mathcal {L}}^*}$. Then, for any $1< q\le p< +\infty$ and operator $X\in L_q(\sigma )$ we have

$$\begin{aligned} \big \Vert \Phi ^*_{t}(X)\big \Vert _{p, \sigma }\le \big \Vert X\big \Vert _{q,\sigma }, \qquad \quad \forall t\ge \frac{1}{4\alpha _{2}} \log \frac{p-1}{q-1}, \end{aligned}$$

where $\alpha _2= \frac{4}{\beta }\sinh ^2(\beta /4)$.

Proof

It is well known that it suffices restrict to operators that are positive semidefinite (see, e.g., [2]). For $p\ge q$ define $t(p):= \frac{1}{4\alpha _2} \log \frac{p-1}{q-1}$ and note that $t(q)=0$. For a positive semidefinite operator X let $X_p:= \Phi _{t(p)}^*(X)$ and define $f(p):=\Vert X_p\Vert _{p, \sigma } - \Vert X\Vert _{q, \sigma }$. Our goal is to show that $f(p)\le 0$ for any $p\ge q$. To this end, since $f(q)=0$, it suffices to verify that $f'(p)\le 0$. Computing the derivative (see the details in the proof of [2, Theorem 11]) we find that

$$\begin{aligned} f'(p)=\frac{1}{p^2}\Vert X_p\Vert _{p, \sigma }^p \Big ( D(\rho _p\Vert \sigma ) - \frac{1}{\alpha _2}{\mathcal {E}}_p(\rho _p)\Big ), \end{aligned}$$

where

$$\begin{aligned} \rho _p:= \frac{1}{\Vert X_p\Vert _{p, \sigma }^p} \Gamma _\sigma ^{1/p}(X_p)^p. \end{aligned}$$

It is shown in Appendix D that $\alpha _p\ge \alpha _2$ for all p. Then, the negativity of $f'(p)$ follows from the log-Sobolev inequality of Theorem 4.

For the above argument, we need to make sure that $\Phi _{t}^*(X)$ is a well-defined operator. To this end, we notice that the quantum Ornstein–Uhlenbeck semigroup satisfies the Feller property with respect to the algebra of compact operators [9, Theorem 5.1]. In particular, $\Phi _t^*(X)$ is a well-defined compact operator if X is compact. Thus, as also done in the proof of [39, Theorem 3.8], in the above argument we may restrict to compact operators X, and after proving the inequality for such an operator, generalize to arbitrary $X\in L_q(\sigma )$ by a continuity argument. We also note that since $\sigma$ is Gaussian, when X is compact and then bounded, the corresponding density operator $\rho _p$ in the above argument has a finite mean photon number. Thus, the log-Sobolev inequality of Theorem 4 can indeed be employed. $\square$

4.1 Log-Sobolev Inequality for Multimode States

It is well known that log-Sobolev constants for classical Markov semigroups satisfy the tensorization property, meaning that for generators ${\mathcal {K}}_1, {\mathcal {K}}_2$ of two classical Markov semigroups we have $\alpha _p({\mathcal {K}}_1\otimes I + I\otimes {\mathcal {K}}_2)= \min \{\alpha _p({\mathcal {K}}_1), \alpha _p({\mathcal {K}}_2)\}$, where ${\mathcal {K}}_1\otimes {\mathcal {I}} + {\mathcal {I}}\otimes {\mathcal {K}}_2$ is the generator of the tensor product semigroup $\{e^{-t{\mathcal {K}}_1}\otimes e^{-t{\mathcal {K}}_2}:\, t\ge 0\}$. This tensorization property is known only for some special quantum Markov semigroups; see [2] and references therein for more details. Thus, a natural question is whether the tensorization property holds for the quantum Ornstein–Uhlenbeck semigroup. To explore this problem we first develop some notations.

For parameters $\beta _1, \dots , \beta _m>0$ let

$$\begin{aligned} {{\widehat{{\mathcal {L}}}}} = {\mathcal {L}}_1+\cdots + {\mathcal {L}}_m, \end{aligned}$$

be the generator of an m-mode quantum Markov semigroup where

$$\begin{aligned} {\mathcal {L}}_j(\rho ) = e^{-\beta _j/2}\Big ( \frac{1}{2} \{{\textbf{a}}_j{\textbf{a}}_j^\dagger , \rho \} - {\textbf{a}}^\dagger _j \rho {\textbf{a}}_j\Big ) + e^{\beta _j/2}\Big ( \frac{1}{2} \{{\textbf{a}}_j^\dagger {\textbf{a}}_j, X\} - {\textbf{a}}_j \rho {\textbf{a}}_j^\dagger \Big ). \end{aligned}$$

(38)

Then, the quantum channels

$$\begin{aligned} {{\widehat{\Phi }}}_t = e^{-t{{\widehat{{\mathcal {L}}}}}} = e^{-t{\mathcal {L}}_1}\otimes \cdots \otimes e^{-t{\mathcal {L}}_m}, \end{aligned}$$

form an m-mode semigroup. The fixed point of this semigroup is

$$\begin{aligned} {{\widehat{\sigma }}} = \sigma _{\beta _1}\otimes \cdots \otimes \sigma _{\beta _m}, \end{aligned}$$

and its corresponding Dirichlet form is equal to

$$\begin{aligned} {{\mathcal {E}}}_p(\rho ) = \frac{p{\hat{p}}}{4} \Big \langle \Gamma _{{{\widehat{\sigma }}}}^{-\frac{1}{{\hat{p}}}}\big (\rho ^{\frac{1}{{\hat{p}}}}\big ), {{\widehat{{\mathcal {L}}}}}^*\circ \Gamma _{{{\widehat{\sigma }}}}^{-\frac{1}{p}}\big (\rho ^{\frac{1}{p}}\big ) \Big \rangle _{{{\widehat{\sigma }}}}=\frac{p{\hat{p}}}{4} \sum _{j=1}^m \Big \langle \Gamma _{{{\widehat{\sigma }}}}^{-\frac{1}{{\hat{p}}}}\big (\rho ^{\frac{1}{{\hat{p}}}}\big ), {\mathcal {L}}_j^*\circ \Gamma _{{{\widehat{\sigma }}}}^{-\frac{1}{p}}\big (\rho ^{\frac{1}{p}}\big ) \Big \rangle _{{{\widehat{\sigma }}}}. \end{aligned}$$

Now, as in the single mode case we are interested in the p-log-Sobolev constant

$$\begin{aligned} {{\widehat{\alpha }}}_p = \inf _{\rho } \frac{{{\mathcal {E}}}_p(\rho )}{D(\rho \Vert {{\widehat{\sigma }}})}. \end{aligned}$$

Assuming that the optimal states in the above optimization problem are Gaussian, the following corollary is an evidence for the tensorization property.

Corollary 6

Suppose that $\beta _1=\cdots =\beta _m= \beta$. Then, for any m-mode Gaussian state $\rho$ we have

$$\begin{aligned} \alpha _{p} D(\rho \Vert {{\widehat{\sigma }}})\le \widehat{{\mathcal {E}}}_p(\rho ), \end{aligned}$$

where $\alpha _{p} = \frac{p{\hat{p}}}{4\beta } e^{\beta /2}\big (1-e^{-\beta /p}\big )\big (1-e^{-\beta /{\hat{p}}}\big )$.

Proof

This is a direct consequence of Theorem 3 as well as Theorem 4 and its proof. $\square$

Our next goal is to prove a lower bound on ${{\widehat{\alpha }}}_2$ which probably is not optimal, yet may be useful for some applications. To this end, it is useful to consider a correspondence between density operators $\rho$ and positive semidefinite operators X as follows:

$$\begin{aligned} X=\Gamma _{{{\widehat{\sigma }}}}^{-1/2}(\rho ^{1/2}). \end{aligned}$$

Then, we note that $\Vert X\Vert _{2, {{\widehat{\sigma }}}}^2 = {{\textrm{tr}}}(\rho )=1$. Indeed, $\Gamma _{{{\widehat{\sigma }}}}^{-1/2}$ is an isometric embedding of the space of Hilbert–Schmidt operators into $L_2({{\widehat{\sigma }}})$, the Hilbert space of operators equipped with norm $\Vert \cdot \Vert _{2, {{\widehat{\sigma }}}}$. With this correspondence, we have

$$\begin{aligned} {\mathcal {E}}_2(\rho ) = \big \langle X, {{\widehat{{\mathcal {L}}}}}^*(X)\big \rangle _{{{\widehat{\sigma }}}}. \end{aligned}$$

From this equation, we realize that the spectrum of ${{\widehat{{\mathcal {L}}}}}^*$ is a relevant object in the study of the Dirichlet form. Here, we consider ${{\widehat{{\mathcal {L}}}}}^*$ as an operator acting on the Hilbert space $L_2({{\widehat{\sigma }}})$. We note that, since ${{\widehat{\sigma }}}$ is a Gaussian state, any operator X that is a polynomial of operators ${\textbf{a}}_j, {\textbf{a}}_j^\dagger$ belongs to $L_2({{\widehat{\sigma }}})$. Letting ${\mathcal {P}}$ be the space of these polynomial operators, we find that indeed ${\mathcal {P}} \subset L_2({{\widehat{\sigma }}})$ is a dense subset. Moreover, by definition, ${{\widehat{{\mathcal {L}}}}}^*$ leaves this subspace invariant: ${{\widehat{{\mathcal {L}}}}}^*({\mathcal {P}})\subseteq {\mathcal {P}}$. Thus, the domain of ${{\widehat{{\mathcal {L}}}}}^*$ contains the dense subset ${\mathcal {P}}$ [9]. In the following proposition, we summarize Theorem 7.2 of [9] regarding the spectrum of ${{\widehat{{\mathcal {L}}}}}^*$ acting on $L_{2}({{\widehat{\sigma }}})$, and for the sake of completeness present its main proof idea in Appendix E.¹²

Proposition 7

[9]. Consider ${\widehat{{\mathcal {L}}}}^*$ as an operator acting on the Hilbert space $L_2({{\widehat{\sigma }}})$ equipped with norm $\Vert \cdot \Vert _{2, {{\widehat{\sigma }}}}$. For $z_1, \dots , z_m\in {\mathbb {C}}$ with $|z_j|=1$ define the quadrature operator ${\textbf{q}}_{j,z_j} = \frac{1}{\sqrt{2}} (z_j{\textbf{a}}_j^\dagger + {{\bar{z}}}_j{\textbf{a}}_j)$. Also, for any j, let $\{h_{j,k_j}(t):\, k_j\ge 0\}$ be the set of Hermite polynomials specified by

$$\begin{aligned} e^{st-\frac{\coth (\beta _j/2)}{4}s^2} = \sum _{k_j=0}^\infty \frac{s^{k_j}}{k_j!}h_{j,k_j}(t). \end{aligned}$$

(39)

Then, for any tuple $k=(k_1, \dots , k_m)$ of non-negative integers, the operator $V_k:=\bigotimes _{j=1}^m h_{j,k_j}({\textbf{q}}_{j,z_j})$ is an eigenvector of ${{\widehat{{\mathcal {L}}}}}^*$ with eigenvalue $\sum _{j=1}^m\sinh (\beta _j/2)k_j$. Moreover, these operators (and their appropriate linear combinations) are all the eigenvectors of ${{\widehat{{\mathcal {L}}}}}^*$.

For any $\ell =(\ell _1, \dots , \ell _m) \in {\mathbb {Z}}^m$, let ${\mathcal {F}}_\ell \subset L_2({{\widehat{\sigma }}})$ be the space of operators X such that $\Vert X\Vert _{2, {{\widehat{\sigma }}}}<+\infty$ and for any $n=(n_1, \dots , n_m)$ their matrix entries in the Fock basis satisfy

$$\begin{aligned} \langle n_1, \dots , n_m| X|n_1+\ell '_1, \dots , n_m+\ell '_m\rangle =0, \qquad \forall \ell '\ne \ell . \end{aligned}$$

Then, any operator X with $\Vert X\Vert _{2, {{\widehat{\sigma }}}}<+\infty$ can be decomposed as

$$\begin{aligned} X= \sum _{\ell \in {\mathbb {Z}}^m} X_\ell , \qquad \quad X_\ell \in {\mathcal {F}}_\ell . \end{aligned}$$

(40)

Following [6] we call (40) the diagonal decomposition of X. A crucial property of the diagonal decomposition is that ${\mathcal {F}}_\ell$’s are orthogonal subspaces, and $\langle X_\ell , X_{\ell '}\rangle _{{{\widehat{\sigma }}}}=0$ if $\ell \ne \ell '$. This is a consequence of the fact that ${{\widehat{\sigma }}}$ is diagonal in the Fock basis.

Corollary 8

[9]. For any $\ell \in {\mathbb {Z}}^m$ the subspace ${\mathcal {F}}_\ell$ is invariant under ${{\widehat{{\mathcal {L}}}}}^*$, and the eigenvalues of ${{\widehat{{\mathcal {L}}}}}^*$ restricted to ${\mathcal {F}}_\ell$ are contained in $\big \{\sum _j \sinh (\beta _j/2)k_j:~ k_j\ge |\ell _j| \big \}$. In particular, for any $X_\ell \in {\mathcal {F}}_\ell$ we have

$$\begin{aligned} \bigg (\sum _{j=1}^m \sinh (\beta _j/2)|\ell _j| \bigg ) \Vert X_\ell \Vert _{2, {{\widehat{\sigma }}}}\le \big \langle X_\ell , {{\widehat{{\mathcal {L}}}}}^*(X_\ell )\big \rangle _{2, {{\widehat{\sigma }}}}. \end{aligned}$$

Proof

The fact that ${\mathcal {L}}^*({\mathcal {F}}_\ell )\subseteq {\mathcal {F}}_\ell$ is easily verified using the definition of ${{\widehat{{\mathcal {L}}}}}^*$. Therefore, considering the diagonal decomposition $V_k= \sum _\ell V_{k, \ell }$ of the eigenvector $V_k$ defined in Proposition 7, we find that if $V_{k, \ell }\ne 0$, then it is also an eigenvector of ${{\widehat{{\mathcal {L}}}}}^*$ with the same eigenvalue as that of $V_k$. On the other hand, by definition, $V_k$ is a polynomial of degree $k_j$ in terms of ${\textbf{a}}_j$ and ${\textbf{a}}_j^\dagger$. This means that we have $V_{k, \ell }=0$ if $|\ell _j|>k_j$ for some j. We also note that since by Proposition 7, the closure of the span of $V_k$’s is the whole space $L_2({{\widehat{\sigma }}})$, the closure of the span of operators $V_{k, \ell }$ equals ${\mathcal {F}}_\ell$. Indeed, ${\mathcal {F}}_\ell = \overline{\text {span}\{V_{k, \ell }:\, k_j\ge |\ell _j|, \forall j\}}$. Putting these together the desired result is implied. $\square$

To state our next lemma, it is convenient to extend the definition of the entropy function for operators that are not necessarily normalized. For any positive operator $X=\Gamma _{{{\widehat{\sigma }}}}^{-1/2}(\rho ^{1/2})$ where $\rho$ is a density operator, we define

$$\begin{aligned} {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}(X) = D(\rho \Vert {{\widehat{\sigma }}}). \end{aligned}$$

Extending this definition to non-normalized operators satisfying $\Vert X\Vert _{2, {{\widehat{\sigma }}}}<+\infty$, we let

$$\begin{aligned} {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}(X) = \Vert X\Vert _{2, {{\widehat{\sigma }}}}^2 \,{{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}\Big (\frac{X}{\Vert X\Vert _{2, {{\widehat{\sigma }}}}}\Big ). \end{aligned}$$

The significance of this entropy function is in its connection to the derivative of $p\mapsto \Vert X\Vert _{p, {{\widehat{\sigma }}}}$ at $p=2$ for which we refer to [2, 35, 39].

Lemma 9

Let X be a positive operator satisfying ${{\textrm{tr}}}\big (\rho H_m \big )<+\infty$ where $\rho = \Gamma _{{{\widehat{\sigma }}}}^{1/2}(X)^2$ and $H_m=\sum _{j=1}^m {\textbf{a}}_j^\dagger {\textbf{a}}_j$ is the m-mode number operator. Then, for any vector $w=(w_\ell )_\ell$ of positive real numbers we have

$$\begin{aligned} {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}(X) \le \sum _{\ell } \big ( \log \Vert w\Vert _2^2 - \log w_{\ell }^2\big ) \Vert X_\ell \Vert _{2, {{\widehat{\sigma }}}}^2 + \sum _\ell {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}\big (I_{2,2}(X_\ell )\big ) , \end{aligned}$$

(41)

where $X_\ell$’s are given by the diagonal decomposition $X= \sum _\ell X_\ell$ with $X_\ell \in {\mathcal {F}}_\ell$. Also, $I_{2,2}(X_\ell ) = \Gamma _{{{\widehat{\sigma }}}}^{-1/2}\big (\big |\Gamma _{{{\widehat{\sigma }}}}^{1/2}(X_\ell ) \big |\big )$ and $\Vert w\Vert _2^2 = \sum _\ell w_\ell ^2$.

Proof

Fix a finite subset $S\subset {\mathbb {Z}}^m$ satisfying $-\ell \in S$ for all $\ell \in S$. For some technical reason, in the following we restrict to the subspace of operators X satisfying $X_\ell =0$ if $\ell \notin S$. Later, we will relax this assumption.

For any $z\in {\mathbb {C}}$ define the map $T_z$ by

$$\begin{aligned} T_z(X) = \sum _{\ell } w_\ell ^zX_\ell , \end{aligned}$$

where as before $X=\sum _\ell X_\ell$ is the diagonal decomposition of X. Note that by the above assumption this sum is indeed a finite sum and over $\ell \in S$. By the orthogonality of the subspaces ${\mathcal {F}}_\ell$ for any $t\in {\mathbb {R}}$ we have

$$\begin{aligned} \big \Vert T_{it}(X)\big \Vert _{2, {{\widehat{\sigma }}}}^2 = \Big \Vert \sum _\ell w_{\ell }^{it} X_\ell \Big \Vert _{2, {{\widehat{\sigma }}}}^2= \sum _\ell |w_{\ell }^{it}|^2\cdot \Vert X_\ell \Vert _{2, {{\widehat{\sigma }}}}^2= \sum _\ell \Vert X_\ell \Vert _{2, \sigma }^2. \end{aligned}$$

(42)

Next, recall the definition of norm $\Vert \cdot \Vert _{p, {{\widehat{\sigma }}}}$ given by (37). Using the triangle and Cauchy–Schwarz inequalities, for any $t\in {\mathbb {R}}$ we have

$$\begin{aligned} \big \Vert T_{1+it}(X)\big \Vert _{\infty , {{\widehat{\sigma }}}}^2&=\Big \Vert \sum _\ell w_{\ell }^{1+it} X_\ell \Big \Vert _{\infty , {{\widehat{\sigma }}}}^2 \nonumber \\&\le \Big ( \sum _\ell |w_{\ell }^{1+it}|\cdot \Vert X_k\Vert _{\infty , {{\widehat{\sigma }}}}\Big )^2\nonumber \\&\le \Big (\sum _\ell w_{\ell }^2\Big )\Big ( \sum _{\ell } \Vert X_\ell \Vert _{\infty , {{\widehat{\sigma }}}}^2 \Big ). \end{aligned}$$

(43)

For any $1\le p\le +\infty$ define

$$\begin{aligned} \Vert X\Vert _{2, p, {{\widehat{\sigma }}}}:= \Big ( \sum _\ell \Vert X_\ell \Vert ^2_{p,{{\widehat{\sigma }}}} \Big )^{1/2}. \end{aligned}$$

It is well known and can be easily verified that $\Vert \cdot \Vert _{2, p, {{\widehat{\sigma }}}}$ satisfies the triangle inequality and is really a norm. Moreover, these norms form an interpolation family [40]. Now, with this notation, (42) and (43) imply that

$$\begin{aligned} \Vert T_{it}\Vert _{(2, 2, {{\widehat{\sigma }}})\rightarrow (2, {{\widehat{\sigma }}})}=1, \qquad \Vert T_{1+it}\Vert _{(2, \infty , {{\widehat{\sigma }}})\rightarrow (\infty , {{\widehat{\sigma }}})}\le \Vert w\Vert _2, \qquad \quad \forall t\in {\mathbb {R}}. \end{aligned}$$

Therefore, by the interpolation inequality [37], we have

$$\begin{aligned} \Vert T_{1-2/p}\Vert _{(2, p, \sigma )\rightarrow (p, \sigma )}\le \Vert w\Vert _2^{1-2/p}, \qquad \quad \forall ~2\le p\le +\infty . \end{aligned}$$

This means that

$$\begin{aligned} \Big \Vert \sum _\ell w_{\ell }^{1-2/p} X_\ell \Big \Vert _p \le \Vert w\Vert _2^{1-2/p}\cdot \Big (\sum _\ell \Vert X_\ell \Vert _p^2\Big )^{1/2}. \end{aligned}$$

We note that this inequality turns into an equality for $p=2$. Therefore, we must have

$$\begin{aligned} \frac{{\textrm{d}}}{{\textrm{d}}p}\bigg (\Big \Vert \sum _\ell w_{\ell }^{1-2/p} X_\ell \Big \Vert _p \bigg )\bigg |_{p=2} \le \frac{{\textrm{d}}}{{\textrm{d}}p}\bigg (\Vert w\Vert _2^{1-2/p}\cdot \Big (\sum _\ell \Vert X_\ell \Vert _p^2\Big )^{1/2} \bigg )\bigg |_{p=2}. \end{aligned}$$

Computing the derivative of both sides by [2, Proposition 3] and using the fact that $X_{\ell }^\dagger = X_{-\ell }$, the desired inequality is obtained.

Now we relax the assumption that $X_\ell$ is nonzero only for finitely many $\ell$. Assume that $X= \Vert X\Vert _{2, {{\widehat{\sigma }}}} \Gamma _{{{\widehat{\sigma }}}}^{-1/2}\big (\rho ^{1/2}\big )$ where $\rho$ is a quantum state with finite mean photon number, in which case

$$\begin{aligned} {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}(X) = \Vert X\Vert ^2_{2, {{\widehat{\sigma }}}} D( \rho \Vert {{\widehat{\sigma }}}) = -\Vert X\Vert ^2_{2, {{\widehat{\sigma }}}}\big (S(\rho ) + {{\textrm{tr}}}(\rho \log {{\widehat{\sigma }}})\big ). \end{aligned}$$

Let $\Pi _k$ be the projection on the span of basis vectors $|n_1, \dots , n_m\rangle$ satisfying $n_1+\cdots + n_m\le k$. Let $X^{(k)} = \Pi _k X \Pi _k$ and consider the quantum state $\rho ^{(k)}=\Vert X^{(k)}\Vert ^{-2}_{2, {{\widehat{\sigma }}}} \Gamma _{{{\widehat{\sigma }}}}^{-1/2}\big (X^{(k)}\big )^2$. Observe that $X_\ell ^{(k)}$ is nonzero only for finitely many values of $\ell$. Then, by the above argument (41) holds for $X^{(k)}$. Taking the limit of $k\rightarrow +\infty$, clearly $\big \Vert X_\ell ^{(k)}\big \Vert _{2, {{\widehat{\sigma }}}}$ tends to $\Vert X_\ell \Vert _{2, {{\widehat{\sigma }}}}$, which also implies that the 2-norm of $X^{(k)}$ tends to that of X. For the entropy terms, it can be verified that the mean photon number of $\rho ^{(k)}$ tends to that of $\rho$. This implies that ${{\textrm{tr}}}\big (\rho ^{(k)} \log {{\widehat{\sigma }}}\big )$ tends to ${{\textrm{tr}}}(\rho \log {{\widehat{\sigma }}})$. Next, using the assumption ${{\textrm{tr}}}(\rho H_m)<+\infty$, we can apply the continuity bound for the entropy function [43, Lemma 18] to conclude that $S\big (\rho ^{(k)}\big )$ tends to $S(\rho )$. Therefore, ${{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}\big (X^{(k)}\big )\rightarrow {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}(X)$ and similarly ${{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}\big (I_{2,2}\big (X^{(k)}_\ell \big ) \rightarrow {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}\big (I_{2,2}(X_\ell )$ as $k\rightarrow +\infty$. Putting these together the desired inequality for X is implied. $\square$

We remark that Lemma 9 holds beyond the diagonal decomposition considered above and works for any decomposition of the space of operators into orthogonal subspaces. Thus, this lemma is of independent interest and may find other applications.

Theorem 10

For any $\beta _1, \dots , \beta _m>0$ we have

$$\begin{aligned} {{\widehat{\alpha }}}_{2} \ge \bigg (\frac{2+ \log (2m+1)}{\sinh (\beta _{\min }/2)} + \frac{1}{\alpha _{2, \min }}\bigg )^{-1}, \end{aligned}$$

where $\beta _{\min } = \min _j \beta _j$ and $\alpha _{2, \min } = \min _j \frac{4\sinh ^2(\beta _j/4)}{\beta _j} = \frac{4\sinh ^2(\beta _{\min }/4)}{\beta _{\min }}$.

Proof

We need to show that

$$\begin{aligned} {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}(X) \le \bigg (\frac{2+ \log (2m+1)}{\sinh (\beta _{\min }/2)} + \frac{1}{\alpha _{2, \min }}\bigg ) \big \langle X, {{\widehat{{\mathcal {L}}}}}^*(X)\big \rangle _{{{\widehat{\sigma }}}}, \end{aligned}$$

(44)

for any X satisfying ${{\textrm{tr}}}\big (\Gamma _{{{\widehat{\sigma }}}}^{1/2}(X)^2 H_m\big )<+\infty$. We use Lemma 9 for the choice of $w_\ell =e^{-\frac{c}{2} |\ell |}$ where $c>0$ and $|\ell | = \sum _{j=1}^m |\ell _j|$. We have $\log \Vert w\Vert _2^2 - \log w_\ell ^2 = c|\ell | +m\log \frac{e^c+1}{e^c-1}$ and

$$\begin{aligned} {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}(X) \le m\log \frac{e^c+1}{e^c-1}\Vert X\Vert _{2, {{\widehat{\sigma }}}}^2 + c\sum _{\ell } |\ell |\cdot \Vert X_\ell \Vert _{2, {{\widehat{\sigma }}}}^2 + \sum _\ell {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}(I_{2,2}(X_\ell )). \end{aligned}$$

(45)

Now, since $X_\ell \in {\mathcal {F}}_\ell$, by Corollary 8 we have

$$\begin{aligned} \sum _\ell |\ell | \cdot \Vert X_\ell \Vert _{2, {{\widehat{\sigma }}}}^2\le \frac{1}{\sinh (\beta _{\min }/2)} \sum _\ell \big \langle X_\ell , {{\widehat{{\mathcal {L}}}}}^*(X_\ell ) \big \rangle _{{{\widehat{\sigma }}}} = \frac{1}{\sinh (\beta _{\min }/2)} \big \langle X, {{\widehat{{\mathcal {L}}}}}^*(X)\big \rangle _{{{\widehat{\sigma }}}}. \end{aligned}$$

Next, using the fact that $X_\ell \in {\mathcal {F}}_\ell$, it is easily verified that the operator $I_{2,2}(X_\ell )$ is diagonal. On the other hand, the restriction of ${{\widehat{{\mathcal {L}}}}}^*$ to diagonal operators is essentially a classical Markov semigroup for which the tensorization property holds. Thus, the 2-log-Sobolev constant of ${{\widehat{{\mathcal {L}}}}}^*$ restricted to diagonal operators equals the minimum of the 2-log-Sobolev constants of ${\mathcal {L}}^*_j$’s restricted to (single-mode) diagonal operators. The latter quantity is computed in Theorem 4. Putting these together, we conclude that

$$\begin{aligned} \alpha _{2, \min } {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}(I_{2, 2}(X_\ell )) \le \big \langle I_{2, 2}(X_\ell ), {{\widehat{{\mathcal {L}}}}}^*\big (I_{2, 2}(X_\ell )\big )\big \rangle _{{{\widehat{\sigma }}}}. \end{aligned}$$

Using the above two inequalities in (45), we find that

$$\begin{aligned} {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}(X)&\le m\log \frac{e^c+1}{e^c-1} \Vert X\Vert _{2, {{\widehat{\sigma }}}}^2 + \frac{c}{\sinh (\beta _{\min }/2)}\big \langle X, {{\widehat{{\mathcal {L}}}}}^*(X)\big \rangle _{{{\widehat{\sigma }}}} \nonumber \\&\quad + \frac{1}{\alpha _{2, \min }}\sum _\ell \big \langle I_2(X_\ell ), {{\widehat{{\mathcal {L}}}}}^*\big (I_2(X_\ell )\big )\big \rangle _{{{\widehat{\sigma }}}}. \end{aligned}$$

It is shown in [2, Lemma 23] that for any operator Y,

$$\begin{aligned} & \big \langle I_{2,2}(Y),{{\widehat{{\mathcal {L}}}}}^*(I_{2,2}(Y))\big \rangle _{{{\widehat{\sigma }}}} + \big \langle I_{2,2}(Y^\dagger ), {{\widehat{{\mathcal {L}}}}}^*(I_{2,2}(Y^\dagger ))\big \rangle _{{{\widehat{\sigma }}}}\\ & \quad \le \big \langle Y, {{\widehat{{\mathcal {L}}}}}^*(Y)\big \rangle _{{{\widehat{\sigma }}}}+\big \langle Y^\dagger , {{\widehat{{\mathcal {L}}}}}^*(Y^\dagger )\big \rangle _{{{\widehat{\sigma }}}}. \end{aligned}$$

Therefore, using $X_\ell ^\dagger = X_{-\ell }$, the above inequalities imply

$$\begin{aligned} {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}(X)&\le m\log \frac{e^c+1}{e^c-1} \Vert X\Vert _{2, {{\widehat{\sigma }}}}^2 + \frac{c}{\sinh (\beta _{\min }/2)}\langle X, {{\widehat{{\mathcal {L}}}}}^*(X)\rangle _{{{\widehat{\sigma }}}} \\&\quad + \frac{1}{\alpha _{2, \min }}\sum _\ell \big \langle X_\ell , {{\widehat{{\mathcal {L}}}}}^*(X_\ell )\big \rangle _{{{\widehat{\sigma }}}}\\&=m \log \frac{e^c+1}{e^c-1} \Vert X\Vert _{2, {{\widehat{\sigma }}}}^2 +\Big (\frac{c}{\sinh (\beta _{\min }/2)} + \frac{1}{\alpha _{2, \min }} \Big )\big \langle X, {{\widehat{{\mathcal {L}}}}}^*(X)\big \rangle _{{{\widehat{\sigma }}}}. \end{aligned}$$

Next, by Proposition 7 the generator ${{\widehat{{\mathcal {L}}}}}^*$ has the spectral gap $\sinh (\beta _{\min }/2)$. Thus, by [39, Theorem 4.2] the above inequality implies

$$\begin{aligned} {{\textrm{Ent}}}_{2, {{\widehat{\sigma }}}}(X)&\le \bigg ( \frac{1+m\log \frac{e^c+1}{e^c-1} }{\sinh (\beta _{\min }/2)} +\frac{c}{\sinh (\beta _{\min }/2)} + \frac{1}{\alpha _{2, \min }} \bigg )\big \langle X, {{\widehat{{\mathcal {L}}}}}^*(X)\big \rangle _{{{\widehat{\sigma }}}}. \end{aligned}$$

Finally, letting $c=\log (2m+1 )$ and using $\log (1+\frac{1}{\,}m)\le \frac{1}{\,}m$ the desired inequality (44) is obtained. $\square$

5 Proof of the CMOE Conjecture

In this section, we present an alternative proof of the CMOE conjecture for single-mode phase-covariant Gaussian channels, which is first established in [15, 16].

Theorem 11

(CMOE Conjecture). For any single-mode phase-covariant Gaussian channel $\Phi$, and any quantum state $\rho$ with finite mean photon number we have

$$\begin{aligned} S\big (\Phi (\rho )\big ) \ge S\big (\Phi (\tau )\big ), \end{aligned}$$

(46)

where $\tau$ is a single-mode thermal state satisfying $S(\rho ) = S(\tau )$.

To prove this theorem, we first state a consequence of Theorem 1.

Theorem 12

Let $\{\Phi _t:\, t\ge 0\}$ denote a semigroup of single-mode phase-covariant Gaussian channels. Then, for any $\alpha > 0$ we have

$$\begin{aligned} \eta _{{{\textrm{th}}}}(\alpha )=\inf _{\rho }\, \frac{{\textrm{d}}}{{\textrm{d}}t} S\big (\Phi _t(\rho ) \big )\Big |_{t=0} + \alpha S(\rho ), \end{aligned}$$

(47)

where the infimum is taken over all quantum states $\rho$ with finite mean photon number, and $\eta _{{{\textrm{th}}}}(\alpha )$ is the infimum of the same function restricted to thermal states.

Proof of Theorem 12

Let ${\mathcal {L}}$ be the Lindbladian of the semigroup $\{\Phi _t:\, t\ge 0\}$ so that $\Phi _t=e^{-t{\mathcal {L}}}$. We have

$$\begin{aligned} \frac{{\textrm{d}}}{{\textrm{d}}t} S\big (\Phi _t(\rho ) \big )\Big |_{t=0} = \langle {\mathcal {L}}(\rho ), \log \rho \rangle . \end{aligned}$$

(48)

Then, the claim follows from Theorem 1 for $p=1$ and appropriate choices of $\nu _0, \nu _1\ge 0$ and $\omega =0$. $\square$

We need yet another technical ingredient to prove Theorem 11. For $0<x<1$, let

$$\begin{aligned} \tau _x = (1-x)\sum _{n} x^n|n\rangle \langle n|, \end{aligned}$$

(49)

be a thermal state with parameter x. Then, letting $\Phi _t= e^{-t{\mathcal {L}}}$ with ${\mathcal {L}}=\nu _0{\mathcal {L}}_0+\nu _1{\mathcal {L}}_1$ and using (48), we find that the objective function in (47) for thermal states equals

$$\begin{aligned} f_\alpha (x)&:= \frac{{\textrm{d}}}{{\textrm{d}}t} S\big (\Phi _t(\tau _x) \big )\big |_{t=0} + \alpha S(\tau _x) \\&= \frac{1}{1-x}\Big (\! (\nu _1 x-\nu _0) \log x - \alpha x\log x-\alpha (1-x)\log (1-x)\! \Big ). \end{aligned}$$

This equation can also be derived by taking the limit of $p\rightarrow 1^+$ in (18).

Assume that $\nu _0>0$ and as before let $\alpha >0$. Then, we have $\lim _{x\rightarrow 0^+ }f_\alpha (x) = \lim _{x\rightarrow 1^- }f_\alpha (x) = +\infty$. This means that the minimum of $f_\alpha (x)$ is attained at roots of $f'_\alpha (x)$. We have

$$\begin{aligned} f'_\alpha (x)&= \frac{1}{1-x}\Big ( \nu _1\log x + \frac{\nu _1x-\nu _0}{x} - \alpha \log x + \alpha \log (1-x) \Big ) \\&\quad +\frac{1}{(1-x)^2} \Big ((\nu _1x-\nu _0)\log x -\alpha x\log x - \alpha (1-x)\log (1-x) \Big )\\&=\frac{1}{(1-x)^2}\bigg (\! (\nu _1-\nu _0)\log x + \frac{(1-x)(\nu _1x-\nu _0)}{x} -\alpha \log x\bigg ) \end{aligned}$$

Therefore, $f'_\alpha (x)=0$ iff $g(x)=\alpha$ where

$$\begin{aligned} g(x):= \nu _1-\nu _0 + \frac{(1-x)(\nu _1x-\nu _0)}{x\log x}. \end{aligned}$$

Computing the derivative of g(x), we have

$$\begin{aligned} g'(x) = \frac{\nu _0(\log x-x+1) + \nu _1(x(x-1)-x^2\log x)}{x^2\log ^2x}. \end{aligned}$$

Then, using $\log x\le x-1$ and $-\log x= \log \frac{1}{x}\le \frac{1}{x}-1$ we find that $g'(x)\le 0$ for $0<x<1$, and that g(x) is monotone decreasing in this interval. Moreover, since $\nu _0>0$, we have $\lim _{x\rightarrow 0^+} g(x)=+\infty$ and $\lim _{x\rightarrow 1^-} g(x) = 0$. This means that $g(x)\ge 0$ for all $0<x<1$, and g(x) takes any positive value in this interval exactly once. As a conclusion, $g(x)=\alpha$ has exactly one solution in (0, 1) for any $\alpha >0$. We conclude that if $\alpha >0$, the minimum of $f_\alpha (x)$ is attained at the unique root of $f'_\alpha (x)$. More importantly, for any $0<x_0<1$, there is $\alpha _{0}=g(x_0)$ such that the minimum of $f_{\alpha _{0}}(x)$ is attained at $x=x_0$.

Proof of Theorem 11

As a phase-covariant Gaussian channel, we have $\Phi =\Phi _{t_1}$ for some phase-covariant Gaussian semigroup $\{\Phi _t:\, t\ge 0\}$ and some $t_1>0$.¹³ Let ${\mathcal {L}}=\nu _0{\mathcal {L}}_0+\nu _1{\mathcal {L}}_1$ be the Lindbladian of this semigroup. In order to apply the above computations, by taking the limit of $\nu _0\rightarrow 0^+$ we assume with no loss of generality that $\nu _0>0$. The point is that if $S\big (\Phi _{t_1}(\rho )\big )\ge S(\Phi _{t_1}(\tau ))$ holds for any $\nu _0>0$, then by the continuity of von Neumann entropy [43, Lemma 18], it holds for $\nu _0=0$ as well.

Suppose that $S\big (\Phi _{t_1}(\rho )\big )< S(\Phi _{t_1}(\tau _{x_0}))$ where $\tau =\tau _{x_0}$ is the thermal state in the statement of the theorem with parameter $x_0$ as in (49). We note that by assumption $S(\Phi _{t}(\rho ))= S(\Phi _{t}(\tau _{x_0}))$ for $t=0$. Thus, we may define

$$\begin{aligned} t_0 =\sup \big \{t:\, S\big (\Phi _{t}^{\otimes m}(\rho )\big )\ge mS\big (\Phi _{t}(\tau _{x_0})\big ), t<t_1 \big \}. \end{aligned}$$

Then, by continuity we have $S(\Phi _{t_0}(\rho ))= S(\Phi _{t_0}(\tau _{x_0}))$ and

$$\begin{aligned} S(\Phi _{t}(\rho ))< S(\Phi _{t}(\tau _{x_0})),\qquad \forall t_0<t\le t_1. \end{aligned}$$

(50)

Recall that $\Phi _t(\tau _{x_0})$ is a thermal state for any $t>0$. Let $x_t$ be the parameter of this thermal state: $\tau _{x_t} = \Phi _t(\tau _{x_0})$. We note that $t\mapsto x_t$ is smooth. Then, by the above discussions, there is a smooth function $t\mapsto \alpha _t$ with $\alpha _t>0$ such that

$$\begin{aligned} \eta _{{{\textrm{th}}}}(\alpha _t)= \frac{{\textrm{d}}}{{\textrm{d}}s} S\big (\Phi _s(\tau _{x_t}) \big )\big |_{s=0} + \alpha _t S(\tau _{x_t}). \end{aligned}$$

Employing Theorem 12, for any $t_0\le t\le t_1$ we obtain

$$\begin{aligned} \frac{{\textrm{d}}}{{\textrm{d}}t} S\big (\Phi _t(\rho ) \big ) + \alpha _t S( \Phi _t(\rho )) \ge \frac{{\textrm{d}}}{{\textrm{d}}t} S(\tau _{x_t} ) + \alpha _t S(\tau _{x_t}). \end{aligned}$$

Taking the integral of both sides yields

$$\begin{aligned}&S\big (\Phi _{t_1}(\rho ) \big ) - S\big (\Phi _{t_0}(\rho ) \big ) + \int _{t_0}^{t_1}\!\alpha _t S( \Phi _t(\rho ))\,{\textrm{d}}t\\&\quad \ge S(\tau _{x_{t_1}}) - S(\tau _{x_{t_0}}) +\int _{t_0}^{t_1}\! \alpha _t S(\tau _{x_t})\, {\textrm{d}}t. \end{aligned}$$

We note that $S(\Phi _{t_0}(\rho ))= S(\tau _{x_{t_0}})$, so this inequality is in contradiction with (50). This implies that the starting inequality $S\big (\Phi _{t_1}(\rho )\big )< S(\Phi _{t_1}(\tau _{x_0}))$ does not hold. $\square$

Generalization of the CMOE conjecture to the multimode case is an open problem in general. However, generalizing the proof idea of Theorem 11 and using Theorem 3, we can prove the conjecture for a special class of multimode states that can be prepared by applying a passive Gaussian unitary $U_G$ on an arbitrary product state, $\rho =U_G \rho _{1}\otimes \rho _{2}\otimes \cdots \otimes \rho _{m} U_G^\dagger$.

6 Conclusion

In this paper, we introduced a meta log-Sobolev inequality, formalized by Theorem 1, that provides a general information theoretic framework to study phase-covariant Gaussian quantum channels. In general, this inequality and our technical proof, particularly the tensorization argument that we used in the second step, are of independent interest. However, by using this result, we have derived new optimal bounds in the context of p-log-Sobolev inequalities for the semigroup of attenuation channels.

Specifically, in Theorem 4, we explicitly computed the optimal p-log-Sobolev constant of the quantum Ornstein–Uhlenbeck semigroup for any $1\le p\le 2$. We also in Theorem 10 established a bound on the 2-log-Sobolev constant in the multimode case that scales logarithmically with the number of modes. We conjecture that the log-Sobolev constants in the multimode case match those of the single-mode case and the tensorization property holds in general. We showed in Corollary 6 that this conjecture holds assuming that the optimal multimode states are Gaussian.

Quantum log-Sobolev inequalities are also defined for $0\le p< 1$ and are related to reverse hypercontractivity inequalities [2, 10]. An interesting open problem is to compute the p-log-Sobolev constant for this range of parameter, which we leave it for future works.

The proof of our multimode log-Sobolev inequality is based on an entropic inequality in Lemma 9 which is of independent interest. Using this inequality, one might be able to prove log-Sobolev inequalities for the tensor product of other quantum Markov semigroups. We leave exploration of such applications of Lemma 9 for future works.

Our results have novel consequences even in the fully classical case, i.e., when restricting to states $\rho$ that are diagonal in the Fock basis. In particular, as pointed out in [6], restricting the quantum Ornstein–Uhlenbeck semigroup to diagonal states, we obtain a classical birth-death process whose log-Sobolev constants were not known prior to our work (except in the case of $p=1$). Moreover, our results have consequences for the classical operation of thinning on the space of integer-valued random variables, for which we refer to [18] and references therein.

We also presented an alternative proof of the CMOE conjecture in Theorem 11 in the single-mode case. This conjecture in the multimode case is left as a major open problem. Our results contribute to the resolution of Gaussian optimizer conjectures in the quantum case [18]. These conjectures state the optimality of Gaussian states for certain optimization problems regarding Gaussian channels. Theorem 1 establishes this conjecture for the quite general function $\Upsilon (\rho )$ in the single-mode case. Generalizing this theorem for multimode states can resolve the CMOE conjecture in the multimode case. To this end, ideas developed in [19] might be useful. We leave further exploration of this problem for future works.

Appendices

Generators of Phase-covariant Gaussian Channels

In this appendix, we show that any single-mode phase-covariant channel given by (11) is a member of a semigroup whose Lindbladian equals (12).

Let us consider the family of channels $\{\Phi _t:\, t\ge 0\}$ corresponding to parameters $\{(\lambda _t, \gamma _t):\, t\ge 0\}$ in (11):

$$\begin{aligned} \chi _{\Phi _t(\rho )}(\xi )=\exp \!\bigg (\!\!-\frac{1}{2} \gamma _t|\xi |^2\bigg )\chi _\rho \big (\!\sqrt{\lambda _t}\,\xi \big ). \end{aligned}$$

(51)

Suppose that $\{\Phi _t:\, t\ge 0\}$ is a semigroup which in particular means that $\Phi _0$ is the identity channel corresponding to parameters $\gamma _0=0$ and $\lambda _0=1$. Then, the generator of this semigroup is determined by

$$\begin{aligned} \chi _{{\mathcal {L}}(\rho )}(\xi )=-\frac{{\textrm{d}}}{{\textrm{d}}t} \chi _{\Phi _{t}(\rho )}(\xi )\Big |_{t=0}. \end{aligned}$$

Using (51), we compute the derivative

$$\begin{aligned} \frac{{\textrm{d}}}{{\textrm{d}}t} \chi _{\Phi _{t}(\rho )}(\xi )\Big |_{t=0}&= - \frac{1}{2}\gamma '_0\,|\xi |^2 {{\textrm{tr}}}(\rho D_\xi )+\frac{1}{2}\lambda '_0\,\Big (\xi \, {{\textrm{tr}}}\big (\rho \, {\textbf{a}}^\dagger D_\xi \big ) - {{\bar{\xi }}}\, {{\textrm{tr}}}\big (\rho \, {\textbf{a}}D_\xi \big )\Big )\\&=- \frac{1}{2}\gamma _0'\, {{\textrm{tr}}}\big (\rho \big [{\textbf{a}}, [{\textbf{a}}^\dagger ,D_\xi ]\big ]\big )+\frac{1}{2}\lambda _0' \Big ({{\textrm{tr}}}\big (\rho \, {\textbf{a}}^\dagger \big [{\textbf{a}}, D_\xi \big ]\big ) - {{\textrm{tr}}}\big (\rho \, {\textbf{a}}\big [{\textbf{a}}^\dagger , D_\xi \big ]\big ) \Big )\\&=- \frac{1}{2}\gamma _0'\, {{\textrm{tr}}}\big (\big [{\textbf{a}}^\dagger , [{\textbf{a}},\rho ]\big ] D_\xi \big ) + \frac{1}{2}\lambda '_0 \Big ({{\textrm{tr}}}\big (\big [{\textbf{a}}^\dagger , \rho \, {\textbf{a}}\big ]D_\xi -\big [{\textbf{a}}, \rho \, {\textbf{a}}^\dagger \big ]D_\xi \big )\Big )\\&=- \frac{1}{2}\gamma '_0\, \chi _{[{\textbf{a}}^\dagger , [{\textbf{a}},\rho ]]}(\xi ) + \frac{1}{2}\lambda '_0\,\chi _{[{\textbf{a}}^\dagger , \rho {\textbf{a}}] -[{\textbf{a}}, \rho {\textbf{a}}^\dagger ]}(\xi ), \end{aligned}$$

where $\lambda '_0=\frac{{\textrm{d}}\lambda _t}{{\textrm{d}}t}\big |_{t=0}$, $\gamma '_0=\frac{{\textrm{d}}\gamma _t}{{\textrm{d}}t}\big |_{t=0}$. Here, we use $\frac{{\textrm{d}}}{{\textrm{d}}\lambda } D_{\lambda \delta } = (\xi {\textbf{a}}^\dagger - {{\bar{\xi }}} {\textbf{a}}) D_{\lambda \xi }$ in the first line, and the well-known equations $[{\textbf{a}}, D_\xi ] =\xi D_\xi$ and $[{\textbf{a}}^\dagger , D_\xi ] ={{\bar{\xi }}} D_\xi$ in the second line. Replacing the characteristic functions in the above equation by their corresponding operators yields

$$\begin{aligned} \frac{{\textrm{d}}}{{\textrm{d}}t} \Phi _{t}(\rho )\Big |_{t=0}&=- \frac{1}{2}\gamma '_0\, \big [{\textbf{a}}^\dagger , [{\textbf{a}},\rho ]\big ]+ \frac{1}{2}\lambda '_0\Big ([{\textbf{a}}^\dagger , \rho \, {\textbf{a}}] -[{\textbf{a}}, \rho \, {\textbf{a}}^\dagger ]\Big )\\&=-\frac{1}{2}\gamma _0' \Big (\frac{1}{2}\big \{{\textbf{a}}^\dagger {\textbf{a}}, \rho \big \}+\frac{1}{2}\big \{{\textbf{a}}{\textbf{a}}^\dagger , \rho \big \}- {\textbf{a}}^\dagger \rho \,{\textbf{a}}-{\textbf{a}}\,\rho \,{\textbf{a}}^\dagger \Big )\\&\quad +\frac{1}{2}\lambda '_0\Big (\frac{1}{2}\big \{{\textbf{a}}^\dagger {\textbf{a}}, \rho \big \}-\frac{1}{2}\big \{{\textbf{a}}{\textbf{a}}^\dagger , \rho \big \}+ {\textbf{a}}^\dagger \rho \,{\textbf{a}}-{\textbf{a}}\,\rho \,{\textbf{a}}^\dagger \Big )\\&=-\frac{1}{2} \big (\gamma '_0+\lambda '_0\big )\Big (\frac{1}{2}\big \{{\textbf{a}}{\textbf{a}}^\dagger , \rho \big \}-{\textbf{a}}^\dagger \rho {\textbf{a}}\Big )-\frac{1}{2} \big (\gamma '_0-\lambda '_0\big )\Big (\frac{1}{2}\big \{{\textbf{a}}^\dagger {\textbf{a}}, \rho \big \}-{\textbf{a}}\rho {\textbf{a}}^\dagger \Big )\\&=-\frac{1}{2} \big (\gamma '_0+\lambda '_0\big ){\mathcal {L}}_{0}(\rho )-\frac{1}{2} \big (\gamma '_0-\lambda '_0\big ){\mathcal {L}}_1(\rho ), \end{aligned}$$

where we define

$$\begin{aligned} {\mathcal {L}}_{0}(\rho )=\frac{1}{2}\big \{{\textbf{a}}{\textbf{a}}^\dagger , \rho \big \}-{\textbf{a}}^\dagger \rho {\textbf{a}}\qquad \text {and}\qquad {\mathcal {L}}_1(\rho )=\frac{1}{2}\big \{{\textbf{a}}^\dagger {\textbf{a}}, \rho \big \}-{\textbf{a}}\rho {\textbf{a}}^\dagger . \end{aligned}$$

Therefore, if the channels defined by (51) form a semigroup, then the corresponding Lindbladian is equal to

$$\begin{aligned} {\mathcal {L}}(\rho )=\frac{1}{2} \big (\gamma _0'+\lambda _0'\big ){\mathcal {L}}_{0}(\rho )+\frac{1}{2} \big (\gamma _0'-\lambda _0'\big ){\mathcal {L}}_1(\rho ). \end{aligned}$$

(52)

We use this equation to obtain the generator of the three classes of phase-covariant Gaussian channels.

1.1 Attenuator Channels

The attenuator channel can be physically modeled by overlapping the input state and a thermal state $\sigma _\beta$ on a beam splitter with transmissivity $\lambda$:

$$\begin{aligned} \Phi ^{\text {att}}_\lambda (\rho )={{\textrm{tr}}}_2\Big ( U_{\text {BS},\lambda } (\rho \otimes \sigma _\beta ) U_{\text {BS},\lambda }^\dagger \Big ), \end{aligned}$$

where the 2-mode unitary operator $U_{\text {BS},\lambda }$ can be described by

$$\begin{aligned} \left\{ \begin{aligned} U_{\text {BS},\lambda }^\dagger {\textbf{a}}_1 U_{\text {BS},\lambda }&= \sqrt{\lambda }\, {\textbf{a}}_1 + \sqrt{1-{\lambda }}\, {\textbf{a}}_2,\\ U_{\text{ BS },\lambda }^\dagger {\textbf{a}}_2 U_{\text {BS},\lambda }&= -\sqrt{1-{\lambda }}\, {\textbf{a}}_1 + \sqrt{\lambda }\, {\textbf{a}}_2. \end{aligned}\right. \end{aligned}$$

(53)

Using these relations, the characteristic function of the two-mode state $U_{\text {BS},\lambda } (\rho \otimes \sigma _\beta ) U_{\text {BS},\lambda }^\dagger$ equals

$$\begin{aligned} {{\textrm{tr}}}\Big ( U_{\text {BS},\lambda } (\rho \otimes \sigma _\beta ) U_{\text {BS},\lambda }^\dagger D_{\xi _1}\otimes D_{\xi _2}\Big )&={{\textrm{tr}}}\Big ( \rho \otimes \sigma _\beta U_{\text {BS},\lambda }^\dagger (D_{\xi _1}\otimes D_{\xi _2})U_{\text {BS},\lambda }\Big )\\&={{\textrm{tr}}}\Big (\! \rho D_{\sqrt{\lambda } \xi _1 -\sqrt{1-{\lambda }}\xi _2}\Big )\,{{\textrm{tr}}}\Big (\! \sigma _\beta D_{\sqrt{1-{\lambda }}\xi _1 +\sqrt{\lambda } \xi _2}\Big ). \end{aligned}$$

Thus, the characteristic function of the output state of the attenuation channel is given by

$$\begin{aligned} \chi _{\Phi ^{\text {att}}_\lambda (\rho )}(\xi )&=\chi _\rho \big (\sqrt{\lambda }\,\xi _1 -\sqrt{1-{\lambda }}\,\xi _2\big )\,\chi _{\sigma _\beta } \big (\sqrt{1-{\lambda }}\,\xi _1 +\sqrt{\lambda } \,\xi _2\big )\Big |_{(\xi _1, \xi _2)=(\xi , 0)}\\&= \chi _{\rho }\big (\sqrt{\lambda }\, \xi \big )\, \chi _{\sigma _\beta } \big (\sqrt{1-{\lambda }}\,\xi \big )\\&= \exp \!\Big ({\!{-\frac{1}{2} \coth (\beta /2) (1-\lambda ) |\xi |^2}} \Big ) \chi _{\rho }\big (\sqrt{\lambda }\, \xi \big ), \end{aligned}$$

where we replace the characteristic function of the thermal state from (9).

Setting $\lambda =\lambda _t=e^{-2ct}$ with $c>0$ in the above relation, we observe that attenuator channels $\{\Phi ^{\text {att}}_{\lambda _t}:\, t\ge 0\}$ form a semigroup. Comparing the above equation with (51) also gives $\gamma _t=\coth (\beta /2) (1-\lambda _t)$. Therefore, by plugging $\lambda _0'=-2c$ and $\gamma _0'=2c\coth (\beta /2)$ into (52), we obtain the corresponding generator

$$\begin{aligned} {\mathcal {L}}(\rho )&=c \big (\!\coth (\beta /2)-1\big ){\mathcal {L}}_{0}(\rho )+c \big (\!\coth (\beta /2)+1\big ){\mathcal {L}}_1(\rho ). \end{aligned}$$

1.2 Amplifier Channels

Following a similar procedure as for attenuator channels, we can derive the generator of the semigroup of amplifier channels. An amplifier channel can be modeled by applying a two-mode squeezing unitary with squeezing parameter $\lambda \ge 1$ on the input state and a thermal state $\sigma _\beta$, and then tracing over the second mode:

$$\begin{aligned} \Phi ^{\text {amp}}_\lambda (\rho )={{\textrm{tr}}}_2\Big ( U_{\text {2S},\lambda } (\rho \otimes \sigma _\beta ) U_{\text {2S},\lambda }^\dagger \Big ). \end{aligned}$$

Here, the action of the two-mode squeezing unitary can be described by

$$\begin{aligned} \left\{ \begin{aligned} U_{2S,\lambda }^\dagger {\textbf{a}}_1 U_{2S,\lambda }&= \sqrt{\lambda }\, {\textbf{a}}_1 + \sqrt{{\lambda }-2}\, {\textbf{a}}_2^\dagger ,\\ U_{2S,\lambda }^\dagger {\textbf{a}}_2 U_{2S,\lambda }&= \sqrt{\lambda }\, {\textbf{a}}_2+\sqrt{{\lambda }-1}\, {\textbf{a}}_1^\dagger . \end{aligned}\right. \end{aligned}$$

(54)

Using these relations, we have

$$\begin{aligned} U_{2S,\lambda }^\dagger D_{\xi _1}\otimes D_{\xi _2} U_{2S,\lambda }=D_{\xi _1\sqrt{\lambda }-{\bar{\xi }}_2\sqrt{\lambda -1}}\otimes D_{\xi _2\sqrt{\lambda }-{\bar{\xi }}_1\sqrt{\lambda -1}}. \end{aligned}$$

Thus, the characteristic function of the output state is given by

$$\begin{aligned} \chi _{\Phi ^{\text {amp}}_\lambda (\rho )}(\xi )&=\chi _\rho \big (\sqrt{\lambda }\, \xi _1-\sqrt{\lambda -1}\,{\bar{\xi }}_2\big )\,\chi _{\sigma _\beta } \big (\sqrt{\lambda }\,\xi _2-\sqrt{\lambda -1}\,{\bar{\xi }}_1\big ) \Big |_{(\xi _1, \xi _2)=(\xi , 0)}\\&= \chi _{\rho }\big (\sqrt{\lambda }\,\xi \big )\, \chi _{\sigma _\beta } \big (\sqrt{{\lambda }-1}\,\xi \big )\\&= \exp \!\Big (\!{{-\frac{1}{2} \coth (\beta /2) (\lambda -1) |\xi |^2}} \Big ) \chi _{\rho }\big (\sqrt{\lambda }\,\xi \big ), \end{aligned}$$

where in the last line the characteristic function of the thermal state (9) is replaced.

Setting $\lambda =\lambda _t=e^{2ct}$ with $c>0$, we observe that amplifier channels $\{\Phi ^{\text {amp}}_{\lambda _t}:\, t\ge 0\}$ form a semigroup. Here, $\gamma _t=\coth (\beta /2) (\lambda _t-1)$. Thus, by plugging $\lambda _0'=2c$ and $\gamma _0'=2c\coth (\beta /2)$ in (52), we obtain the generator

$$\begin{aligned} {\mathcal {L}}(\rho )&=c \big (\!\coth (\beta /2)+1\big ){\mathcal {L}}_0(\rho )+c \big (\!\coth (\beta /2)-1\big ){\mathcal {L}}_{1}(\rho ). \end{aligned}$$

Note that here we get the same coefficients as those of the generator of the attenuator semigroup, but swapped.

1.3 Additive-noise Channels

The classical additive-noise Gaussian channels can be modeled by applying a displacement operator whose parameter is chosen at random according to a Gaussian probability distribution:

$$\begin{aligned} \Phi ^{\text {add}}_\gamma (\rho )=\frac{2}{\pi \gamma }\int e^{-\frac{2}{\gamma }|\alpha |^2} D_\alpha \rho D^{\dagger }_\alpha \, {\textrm{d}}^2\alpha . \end{aligned}$$

The characteristic function of the output state of this channel is given by

$$\begin{aligned} \chi _{\Phi ^{\text {add}}_\gamma (\rho )}(\xi )&=\frac{2}{\pi \gamma }\int e^{-\frac{2}{\gamma }|\alpha |^2} {{\textrm{tr}}}\big (D_\alpha \rho D^{\dagger }_\alpha D_\xi \big )\, {\textrm{d}}^2\alpha \\&=\frac{2}{\pi \gamma }\chi _\rho (\xi ) \int e^{-\frac{2}{\gamma } |\alpha |^2} e^{\xi {\bar{\alpha }}-{\bar{\xi }}\alpha }\, {\textrm{d}}^2\alpha \\&=\exp \!\Big ({-\frac{1}{2}\gamma |\xi |^2}\Big )\chi _\rho (\xi ), \end{aligned}$$

where we use $D^{\dagger }_\alpha D_\xi D_\alpha =\exp (\xi {\bar{\alpha }}-{\bar{\xi }}\alpha ) D_\xi$ in the second line.

By choosing $\gamma =2ct$ with $c>0$, the channels $\{\Phi ^{\text {add}}_t:\, t\ge 0\}$ form a semigroup. Here, $\lambda _t=1$. Thus, by inserting $\lambda _0'=0$ and $\gamma _0'=2c$ in (52) we obtain the generator

$$\begin{aligned} {\mathcal {L}}(\rho )=c{\mathcal {L}}_0(\rho )+c{\mathcal {L}}_{1}(\rho ). \end{aligned}$$

$\Upsilon (\rho )$ is Well Defined if ${{\textrm{tr}}}(\rho {\textbf{a}}^\dagger {\textbf{a}})<+\infty$

In this appendix, we show that $\Upsilon (\rho )$ given by

$$\begin{aligned} \Upsilon (\rho )&={\hat{p}}\bigg (\!\nu _0 \Big [{{\textrm{tr}}}\big (\rho {\textbf{a}}{\textbf{a}}^\dagger \big ) - {{\textrm{tr}}}\big ( \rho ^{1/p} {\textbf{a}}\rho ^{1/{\hat{p}}} {\textbf{a}}^\dagger \big ) \Big ] +\nu _1\Big [ {{\textrm{tr}}}\big (\rho {\textbf{a}}^\dagger {\textbf{a}}\big )- {{\textrm{tr}}}\big ( \rho ^{1/p} {\textbf{a}}^\dagger \rho ^{1/{\hat{p}}} {\textbf{a}}\big )\!\Big ]\!\bigg ) \\&\quad + \omega {{\textrm{tr}}}\big (\rho {\textbf{a}}^\dagger {\textbf{a}}\big ) + S(\rho ). \end{aligned}$$

is well defined for any state $\rho$ with finite mean photon number. For such a state both ${{\textrm{tr}}}(\rho {\textbf{a}}^\dagger {\textbf{a}}), {{\textrm{tr}}}(\rho {\textbf{a}}{\textbf{a}}^\dagger )$ are finite. Moreover, if $\rho = \sum _j \lambda _j |\psi _j\rangle \langle \psi _j|$ is the eigen-decomposition of $\rho$, then we have

$$\begin{aligned} {{\textrm{tr}}}\big ( \rho ^{1/p} {\textbf{a}}\rho ^{1/{\hat{p}}} {\textbf{a}}^\dagger \big )&= \sum _{i, j} \lambda _i^{1/p}\lambda _j^{1/{\hat{p}}} \big | \langle \psi _i| {\textbf{a}}|\psi _j\rangle \big |^2\\&\le \sum _{i, j} \Big ( \frac{1}{p} \lambda _i + \frac{1}{{\hat{p}}} \lambda _j \Big ) \big | \langle \psi _i| {\textbf{a}}|\psi _j\rangle \big |^2\\&= \sum _i \frac{1}{p}\lambda _i \big \Vert {\textbf{a}}^\dagger |\psi _i\rangle \big \Vert ^2 + \sum _j \frac{1}{{\hat{p}}} \lambda _j \big \Vert {\textbf{a}}|\psi _j\rangle \big \Vert ^2\\&= \frac{1}{p}{{\textrm{tr}}}\big (\rho {\textbf{a}}{\textbf{a}}^\dagger \big ) + \frac{1}{{\hat{p}}}{{\textrm{tr}}}\big (\rho {\textbf{a}}^\dagger {\textbf{a}}\big ), \end{aligned}$$

where in the second line we use Young’s inequality. Thus, ${{\textrm{tr}}}\big ( \rho ^{1/p} {\textbf{a}}\rho ^{1/{\hat{p}}} {\textbf{a}}^\dagger \big )$ and similarly ${{\textrm{tr}}}\big ( \rho ^{1/p} {\textbf{a}}^\dagger \rho ^{1/{\hat{p}}} {\textbf{a}}\big )$ are finite. Also, using the non-negativity of Umegaki’s quantum relative entropy $D(\rho \Vert \tau ) = {{\textrm{tr}}}(\rho \log \rho ) - {{\textrm{tr}}}(\rho \log \tau )$, for the thermal state $\tau = (1-1/e) e^{-{\textbf{a}}^\dagger {\textbf{a}}}$ we have

$$\begin{aligned} 0\le D(\rho \Vert \tau ) = -S(\rho ) - {{\textrm{tr}}}(\rho \log \tau ) = -S(\rho ) -\log (1-1/e) + {{\textrm{tr}}}\big (\rho {\textbf{a}}^\dagger {\textbf{a}}\big ). \end{aligned}$$

Therefore, since ${{\textrm{tr}}}\big (\rho {\textbf{a}}^\dagger {\textbf{a}}\big )$ is finite, $S(\rho )$ is also finite.

Verification of (36)

Recall that

$$\begin{aligned} \phi (x, y)&= -y^2\log y^2 -(1-y^2)\log (1-y^2) + y^2\log x^2 + (1-y^2)\log (1-x^2) \\&\quad -\frac{\big ( x^{\frac{2}{p}}-y^{\frac{2}{p}} \big )\big ( x^{\frac{2}{{\hat{p}}}}-y^{\frac{2}{{\hat{p}}} }\big )\log x^2}{\big (1-x^{\frac{2}{p}}\big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )} . \end{aligned}$$

Then, computing the derivative term-by-term yields

$$\begin{aligned} \frac{{\textrm{d}}}{{\textrm{d}}x} \phi (x, y)&= \frac{2y^2}{x} - \frac{2x(1-y^2)}{1-x^2} - \frac{2\big (x^{\frac{2}{p}}-y^{\frac{2}{p}}\big )\big (x^{\frac{2}{{\hat{p}}}} -y^{\frac{2}{{\hat{p}}}}\big )}{x\big (1-x^{\frac{2}{p}}\big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )}\\&\quad - \frac{2x^{\frac{2}{p}}\big (x^{\frac{2}{{\hat{p}}}} - y^{\frac{2}{{\hat{p}}}} \big )\log x^2}{px\big (1-x^{\frac{2}{p}}\big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )} - \frac{2x^{\frac{2}{{\hat{p}}}}\big (x^{\frac{2}{p}} - y^{\frac{2}{p}}\big ) \log x^2}{{\hat{p}}x\big (1-x^{\frac{2}{p}}\big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )} \\&\quad - \frac{2x^{\frac{2}{p}}\big (x^{\frac{2}{p}}-y^{\frac{2}{p}}\big ) \big (x^{\frac{1}{{\hat{p}}}} - y^{\frac{1}{{\hat{p}}}}\big )\log x^2}{px \big (1-x^{\frac{2}{p}}\big )^2\big (1-x^{\frac{2}{{\hat{p}}}}\big )} - \frac{2x^{\frac{2}{{\hat{p}}}}\big (x^{\frac{2}{{\hat{p}}}}-y^{\frac{2}{{\hat{p}}}} \big )\big (x^{\frac{2}{p}} - y^{\frac{2}{p}}\big )\log x^2}{{\hat{p}}x \big (1-x^{\frac{2}{p}}\big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )^2}\\&= -\frac{2(x^2-y^2)}{x(1-x^2)} - \frac{2\big (x^{\frac{2}{p}} -y^{\frac{2}{p}}\big )\big (x^{\frac{2}{{\hat{p}}}}-y^{\frac{2}{{\hat{p}}}} \big )}{x\big (1-x^{\frac{2}{p}}\big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )} \\&\quad - \frac{2x^{\frac{2}{p}}\big (1-y^{\frac{2}{p}}\big ) \big (x^{\frac{2}{{\hat{p}}}} - y^{\frac{2}{{\hat{p}}}}\big )\log x^2}{px \big (1-x^{\frac{2}{p}}\big )^2\big (1-x^{\frac{2}{{\hat{p}}}}\big )} - \frac{2x^{\frac{2}{{\hat{p}}}}\big (1-y^{\frac{2}{{\hat{p}}}}\big )\big (x^{\frac{2}{p}} - y^{\frac{2}{p}}\big )\log x^2}{{\hat{p}}x\big (1-x^{\frac{2}{p}}\big ) \big (1-x^{\frac{2}{{\hat{p}}}}\big )^2}. \end{aligned}$$

Therefore, we have

$$\begin{aligned}&\frac{x}{2} \frac{{\textrm{d}}}{{\textrm{d}}x} \phi (x, y) + \frac{x^{\frac{2}{p}} \big (1-y^{\frac{2}{p}}\big )\big (x^{\frac{2}{{\hat{p}}}} - y^{\frac{2}{{\hat{p}}}} \big )}{\big (1-x^{\frac{2}{p}}\big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )} \bigg [ \frac{\log x^2}{p(1-x^{\frac{2}{p}})}+1 \bigg ]\\&\quad + \frac{x^{\frac{2}{{\hat{p}}}}\big (1-y^{\frac{2}{{\hat{p}}}}\big ) \big (x^{\frac{2}{p}} - y^{\frac{2}{p}}\big )}{\big (1-x^{\frac{2}{p}}\big ) \big (1-x^{\frac{2}{{\hat{p}}}}\big )} \bigg [ \frac{\log x^2}{{\hat{p}}(1-x^{\frac{2}{{\hat{p}}}})}+1 \bigg ] \\&\quad = -\frac{(x^2-y^2)}{(1-x^2)} \\&\qquad - \frac{\big (x^{\frac{2}{p}} -y^{\frac{2}{p}}\big )\big (x^{\frac{2}{{\hat{p}}}}-y^{\frac{2}{{\hat{p}}}}\big ) -x^{\frac{2}{p}}\big (1-y^{\frac{2}{p}}\big )\big (x^{\frac{2}{{\hat{p}}}} - y^{\frac{2}{{\hat{p}}}}\big )-x^{\frac{2}{{\hat{p}}}}\big (1-y^{\frac{2}{{\hat{p}}}} \big )\big (x^{\frac{2}{p}} - y^{\frac{2}{p}}\big )}{\big (1-x^{\frac{2}{p}} \big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )} \\&\quad = -\frac{(x^2-y^2)}{(1-x^2)} + \frac{x^2-y^2 +x^{\frac{2}{p}}y^2 - x^2y^{\frac{2}{p}}+ x^{\frac{2}{{\hat{p}}}}y^2 -x^2y^{\frac{2}{{\hat{p}}}}}{\big (1-x^{\frac{2}{p}}\big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )} \\&\quad = \frac{-(x^2-y^2)\big (1-x^{\frac{2}{p}}\big )\big (1 -x^{\frac{2}{{\hat{p}}}}\big ) +\big (1-x^2\big )\big (x^2-y^2 +x^{\frac{2}{p}}y^2 - x^2y^{\frac{2}{p}}+ x^{\frac{2}{{\hat{p}}}}y^2 -x^2y^{\frac{2}{{\hat{p}}}}\big )}{\big (1-x^2\big )\big (1-x^{\frac{2}{p}} \big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )} \\&\quad = \frac{x^2\Big ( \! \big (1- x^{\frac{2}{{\hat{p}}}}\big ) \big (1-y^{\frac{2}{{\hat{p}}}}\big ) \big (x^{\frac{2}{p}} - y^{\frac{2}{p}} \big ) + \big (1- x^{\frac{2}{p}}\big )\big (1-y^{\frac{2}{p}}\big ) \big (x^{\frac{2}{{\hat{p}}}} - y^{\frac{2}{{\hat{p}}}}\big ) \! \Big )}{\big (1-x^2\big )\big (1-x^{\frac{2}{p}}\big )\big (1-x^{\frac{2}{{\hat{p}}}}\big )}. \end{aligned}$$

This gives (36).

Proof of $\alpha _2=\min _{p\ge 1} \alpha _p$

In this appendix, we show that for any $p\ge 1$ the following inequality holds:

$$\begin{aligned} \frac{p{\hat{p}}}{4}\big (1-e^{-\beta /p}\big )\big (1-e^{-\beta /{\hat{p}}}\big )\ge \big (1-e^{-\beta /2}\big )^2. \end{aligned}$$

First, we may restrict to $p\in [1, 2]$. Then, with the change of variable $p=\frac{2}{1+t}$ with $t\in [0,1]$ the above inequality is equivalent to

$$\begin{aligned} \big (1-e^{-\frac{\beta }{2} (1+t) }\big ) \big (1-e^{-\frac{\beta }{2} (1-t) }\big )\ge (1-t^2) \big (1-e^{-\beta /2}\big )^2. \end{aligned}$$

Starting with the left hand side, we compute

$$\begin{aligned} \big (1-e^{-\frac{\beta }{2} (1+t) }\big ) \big (1-e^{-\frac{\beta }{2} (1-t) }\big )&= 1+e^{-\beta } - e^{-\beta /2}\big (e^{\beta t/2}+e^{-\beta t/2}\big )\\&= 1+ e^{-\beta } -2 e^{-\beta /2} \sum _{k:\text { even}} \frac{1}{k!} \Big (\frac{\beta t}{2}\Big )^k\\&\ge 1+ e^{-\beta } -2 e^{-\beta /2}\bigg ( 1+ t^2\sum _{k\ge 2: \text { even}} \frac{1}{k!} \Big (\frac{\beta }{2}\Big )^k \bigg )\\&= \big (1-e^{-\beta /2}\big )^2 -t^2 e^{-\beta /2} \Big ( e^{\beta /2} + e^{-\beta /2} -2 \Big )\\&= (1-t^2) \big (1-e^{-\beta /2}\big )^2. \end{aligned}$$

Here, the inequality holds since $0\le t\le 1$. We are done.

Proof of Proposition 7

It suffices to prove the proposition for $m=1$. In this case, we drop subscript j and denote ${{\widehat{{\mathcal {L}}}}}$ and ${{\widehat{\sigma }}}$ simply by ${\mathcal {L}}$ and $\sigma$, respectively.

Using the commutation relation $[{\textbf{a}}, {\textbf{a}}^\dagger ]=1$, we find that $[{\textbf{a}}, {\textbf{q}}_z ] = \frac{z}{\sqrt{2}}$, and

$$\begin{aligned} \big [{\textbf{a}}, e^{s{\textbf{q}}_z - \frac{\coth (\beta /2)}{4} s^2}\big ] = \frac{z}{\sqrt{2}} \Big ( \frac{{\textrm{d}}}{{\textrm{d}}t} e^{st-\frac{\coth (\beta /2)}{4}s^2}\Big )\Big |_{t={\textbf{q}}_z}=\frac{z}{\sqrt{2}} s e^{s{\textbf{q}}_z - \frac{\coth (\beta /2)}{4} s^2}. \end{aligned}$$

This means that

$$\begin{aligned} e^{s{\textbf{q}}_z - \frac{\coth (\beta /2)}{4} s^2}{\textbf{a}}= \Big ({\textbf{a}}-\frac{z}{\sqrt{2}} s\Big )e^{s{\textbf{q}}_z - \frac{\coth (\beta /2)}{4} s^2}, \end{aligned}$$

and similarly

$$\begin{aligned} e^{s{\textbf{q}}_z - \frac{\coth (\beta /2)}{4} s^2}{\textbf{a}}^\dagger = \Big ({\textbf{a}}^\dagger +\frac{{{\bar{z}}}}{\sqrt{2}} s\Big )e^{s{\textbf{q}}_z - \frac{\coth (\beta /2)}{4} s^2}. \end{aligned}$$

Using these and $|z|=1$, a straightforward computation shows that

$$\begin{aligned}&{\mathcal {L}}^*(e^{s{\textbf{q}}_z - \frac{\coth (\beta /2)}{4} s^2})\\&\quad = e^{\beta /2}\left( \frac{1}{2} {\textbf{a}}^\dagger {\textbf{a}}+\frac{1}{2}\left( {\textbf{a}}^\dagger +\frac{{{\bar{z}}}}{\sqrt{2}} s\right) \left( {\textbf{a}}- \frac{z}{\sqrt{2}}s\right) - {\textbf{a}}^\dagger \left( {\textbf{a}}- \frac{z}{\sqrt{2}} s\right) \right) e^{s{\textbf{q}}_z - \frac{\coth (\beta /2)}{4} s^2}\\&\qquad + e^{-\beta /2}\left( \frac{1}{2} {\textbf{a}}{\textbf{a}}^\dagger +\frac{1}{2}\left( {\textbf{a}}- \frac{z}{\sqrt{2}}s\right) \left( {\textbf{a}}^\dagger +\frac{{{\bar{z}}}}{\sqrt{2}} s\right) - {\textbf{a}}\left( {\textbf{a}}^\dagger + \frac{{{\bar{z}}}}{\sqrt{2}} s\right) \right) e^{s{\textbf{q}}_z - \frac{\coth (\beta /2)}{4} s^2}\\&\quad = e^{\beta /2}\Big (\frac{1}{2} s{\textbf{q}}_z-\frac{1}{4} s^2 \Big )e^{s{\textbf{q}}_z - \frac{\coth (\beta /2)}{4} s^2}+e^{-\beta /2} \Big (-\frac{1}{2} s{\textbf{q}}_z-\frac{1}{4} s^2 \Big )e^{s{\textbf{q}}_z - \frac{\coth (\beta /2)}{4} s^2}\\&\quad = \frac{e^{\beta /2}-e^{-\beta /2}}{2} s\Big ( {\textbf{q}}_z -\frac{\coth (\beta /2)}{2} s \Big )e^{s{\textbf{q}}_z - \frac{\coth (\beta /2)}{4} s^2}\\&\quad = \frac{e^{\beta /2}-e^{-\beta /2}}{2} s\frac{{\textrm{d}}}{{\textrm{d}}s} e^{s{\textbf{q}}_z - \frac{\nu }{4} s^2}\\&\quad = \frac{e^{\beta /2}-e^{-\beta /2}}{2} \sum _{k=0}^\infty \frac{s^{k}}{(k-1)!}h_n({\textbf{q}}_z). \end{aligned}$$

Compared to (39), we find that

$$\begin{aligned} {\mathcal {L}}^*\big (h_k({\textbf{q}}_z)\big ) = \big (\sinh (\beta /2)k\big )\, h_k({\textbf{q}}_z). \end{aligned}$$

Finally, note that the operators $h_k({\textbf{q}}_z)$, over all $k\ge 0$ and $|z|=1$, span the whole space of polynomials of ${\textbf{a}}$ and ${\textbf{a}}^\dagger$, which is dense in $L_2(\sigma )$.