Properties of the Poincaré Constant

In this appendix, we state some basic properties of the Poincaré constant \(\lambda _\rho \).

[Style1 Style2]Proposition 5

For any bipartite state \(\rho _\), we have \(\lambda _\ge \lambda _}\) where \(\rho _1\) is the marginal state on the first subsystem.

Proof

Apply the Poincaré inequality for \(\rho _\) on operators \(X_1\) that act only on the first subsystem. \(\square \)

Any Gaussian unitary U can be decomposed into single-mode squeezers, phase shifters and two-mode beam splitters [24]. The latter two classes, namely phase shifters and two-mode beam splitters, generate the space of passive transformations that have the property that their action commutes with the photon number operator \(H_m=\sum _^m }_j^\dagger }_j\). Passive transformations can be characterized in terms of their action on annihilation operators. Indeed, for any passive unitary U there is an \(m\times m\) unitary \((u_)\) such that

$$\begin U^\dagger }_j U = \sum _^m u_ }_. \end$$

(57)

We now prove a more interesting property of the Poincaré constant.

[Style1 Style2]Proposition 6

For any passive transformation U, we have \(\lambda _}=\lambda _\rho \). More generally, if \(\lambda _\rho >0\), then for any Gaussian unitary U we have

$$\begin \lambda _}>0. \end$$

Proof

Let U be passive, and suppose we want to verify the Poincaré inequality for \(U \rho U^\) on operator Y. Let \(X=U^\dagger Y U\). Then, using (57) and the fact that \((u_)\) is unitary we have

$$\begin \Vert \partial Y \Vert _^2&= \sum _^m \Big (\big \Vert [}_j, Y] \big \Vert _^2 + \big \Vert [}^\dagger _j, Y] \big \Vert _^2\Big )\\&= \sum _^m \Big (\big \Vert [U^\dagger }_j U, X] \big \Vert _^2 + \big \Vert [U^\dagger }^\dagger _j U, X] \big \Vert _^2\Big )\\&= \sum _^m \bigg ( \bigg \Vert \sum _^m u_[ }_ , X] \bigg \Vert _^2 + \bigg \Vert \sum _^m }}_[ }_^\dagger , X] \bigg \Vert _^2\bigg )\\&= \sum _^m \Big ( \big \Vert [ }_ , X] \big \Vert _^2 + \big \Vert [ }_^\dagger , X] \big \Vert _^2\Big )\\&= \Vert \partial X\Vert _^2. \end$$

It is also easily verified that \(\Vert Y\Vert ^2_ = \Vert X\Vert ^2_\) and \(\textrm(\rho X) = \textrm(U \rho U^\dagger Y)\). Therefore, \(\lambda _\rho \Vert X\Vert ^2_\le \Vert \partial X\Vert _^2\) implies \(\lambda _\rho \Vert Y\Vert ^2_\le \Vert \partial Y \Vert _^2 \) and vice versa.

To prove the second part, we only need to handle the tensor product of single-mode squeezing unitaries which take the form \(U = e^ \sum _^m r_j(}_^2 + (}_^)^2)}\). We claim that

$$\begin \lambda _} \ge \frac \lambda _, \end$$

where \(c=2 \max _j \big (\cosh ^2(r_j) + \sinh ^2(r_j)\big )\). To this end, we use

$$\begin U }_ U^\dagger = \cosh (r_j) }_ + \sinh (r_j) }_^. \end$$

Then, for \(X=U^\dagger Y U\) we have

$$\begin \Vert \partial X \Vert _^2&= \Vert \partial X \Vert _^2\\&= \sum _^m \Big ( \big \Vert [ }_ , X] \big \Vert _^2 + \big \Vert [ }_^\dagger , X] \big \Vert _^2\Big )\\&= \sum _^m \Big ( \big \Vert [ U}_ U^\dagger , Y] \big \Vert _^2 + \big \Vert [ U}_^\dagger U^\dagger , Y] \big \Vert _^2\Big )\\&= \sum _^m \Big ( \big \Vert \cosh (r_j) [}_ , Y] + \sinh (r_j) [}_^, Y]\big \Vert _^ \\&\quad + \big \Vert \cosh (r_j) [}^\dagger _ , Y] + \sinh (r_j) [}_, Y]\big \Vert _^ \Big ) \\&\le 2\sum _^m \big (\cosh ^2(r_j) + \sinh ^2(r_j)\big ) \Big ( \big \Vert [}_ , Y] \big \Vert _^ + \big \Vert [}^\dagger _ , Y] \big \Vert _^ \Big ) \\&\le c \Vert \partial Y \Vert _^2. \end$$

We also have \(\Vert Y\Vert _ = \Vert X\Vert _\). Then, the desired inequality follows. \(\square \)

The following corollary shows that the Poincaré constant does not decrease under convolution.

[Style1 Style2]Corollary 1

For any two quantum states \(\rho _1,\rho _2\) and any \(\eta \in [0,1]\) we have

$$\begin \lambda _ \sigma } \ge \min \, \lambda _\big \}. \end$$

Proof

We note that \(\rho \boxplus _ \sigma = \textrm_2\big ( U_\eta \rho \otimes \sigma U_\eta ^\dagger \big )\), where \(U_\) is a beam splitter and is passive. Then, the bound on the Poincaré constant follows from Propositions 5 and 6. \(\square \)

In the following proposition, we aim to show that if a quantum state \(\rho \) satisfies the quantum Poincaré inequality, then all of its moments are finite.

[Style1 Style2]Proposition 7

Let \(\rho \) be a quantum state satisfying \(\lambda _\rho >0\). Then, all the moments of \(\rho \) are finite.

Proof

We need to show that for any integer \(\kappa >0\) we have \(\textrm( \rho H_m^\kappa )<+\infty \) where \(H_m= \sum _^m }_j^\dagger }_j\). This can be proved by a simple induction. Suppose that \(\textrm( \rho H_m^ )<+\infty \). Then, applying the Poincaré inequality for \(X= H_m^ - \textrm(\rho H_m^)\) we obtain \(\lambda _\rho \Vert X\Vert _^2\le \Vert \partial X\Vert _^2\). We have

$$\begin \Vert X\Vert _^2 = \Vert H_m^\kappa \Vert ^2_\rho - \textrm(\rho H_m^)^2 = \textrm(\rho H_m^) - \textrm(\rho H_m^)^2, \end$$

and

$$\begin \Vert \partial X\Vert _^2 = \Vert \partial H_m^\kappa \Vert _^2 =\sum _^m \Big ( \Vert [}_j, H_m^\kappa ] \Vert _^2 + \Vert [}^\dagger _j, H_m^\kappa ] \Vert _^2 \Big ). \end$$

Now, the point is that expanding \(H_m^\kappa \) and using the canonical commutation relations, we find that \(\Vert [}_j, H_m^\kappa ] \Vert _^2\) and \(\Vert [}^\dagger _j, H_m^\kappa ] \Vert _^2\) can be bounded in terms of \(\Vert H_m^\Vert _^2=\textrm\big (\rho H_m^\big )\). Using these in the Poincaré inequality gives \(\textrm(\rho H_m^)<+\infty \) as desired. \(\square \)

In the next proposition, we consider the Poincaré inequality for Gaussian states.

[Style1 Style2]Proposition 8

For any Gaussian state \(\rho \) that is faithful we have \(\lambda _\rho >0\).

Proof

By Proposition 6, it suffices to consider Gaussian states that are in Williamson’s form, i.e., Gaussian states of the form \(\rho =\tau _1\otimes \cdots \otimes \tau _m\) where \(\tau _j = (1- e^)e^}_j^\dagger }_j}\) is a thermal state and \(\nu _j = \frac}}\ge 1\), \(j=1, \dots , m\) are the Williamson’s eigenvalues of \(\rho \). Since \(\rho _\textrm\) is faithful, \(\nu _j>1\) for all j. In the following, we show that \(\lambda _>0\) in the case of \(m=1\). The proof for arbitrary m works similarly.

Recall that \(D_ D_w = e^(-z}} + }} w)}D_\). Therefore, using (15) for \(\tau =(1- e^)e^}^\dagger }}\) with \(\nu = \frac}}> 1\) we have

$$\begin \textrm\big (\tau D_D_w\big )&= ^(-z}} + }}w)} \textrm\big ( \tau D_ \big )\\&= e^(-z}} + }}w)} \chi _\tau (w-z)\\&= e^(-z}} + }}w)} e^ \nu |w-z |^2}\\&= \exp \Big (-\frac ( |z|^2 + |w|^2 ) + \frac}} w + \frac z}}\Big ). \end$$

Let \(r=|z|\), \(z_0 = z/r\) and \(s=|w|\), \(w_0=w/s\). Also, define

$$\begin Q_:= e^ r^2} D_. \end$$

Then, the above equation is equivalent to

$$\begin \textrm\big ( \tau Q_^\dagger Q_ \big )&= \exp \bigg ( \frac\Big ((\nu +1) }}_0 w_0 + (\nu -1) z_0 }}_0 \Big ) \bigg )\\&= \exp \bigg ( \frac\Big ((\nu +1) }}_0 w_0 + (\nu -1) \frac}}_0 w_0} \Big ) \bigg ). \end$$

Starting with \(D_wD_ = e^ (-}} w+ z}})} D_\), by a similar computation we find that

$$\begin \textrm\big ( \tau Q_ Q_^\dagger \big )&= \exp \bigg ( \frac\Big ((\nu +1) \frac}}_0 w_0} + (\nu -1) }}_0 w_0 \Big ) \bigg ). \end$$

Putting these together, we arrive at

$$\begin \big \langle Q_, Q_\big \rangle _\tau&= \frac\exp \bigg ( \frac\Big ((\nu +1) }}_0 w_0 + (\nu -1) \frac}}_0 w_0} \Big ) \bigg ) \\&\quad \, + \frac\exp \bigg ( \frac\Big ((\nu +1) \frac}}_0 w_0} + (\nu -1) }}_0 w_0 \Big ) \bigg ). \end$$

Next, using the Taylor expansion of the exponential function we find that

$$\begin&\big \langle Q_, Q_\big \rangle _\tau \nonumber \\&\quad = \frac \sum _ \frac k! \ell !} (rs)^\bigg ( \Big ( \frac}}_0w_0}\Big )^k \big ((\nu +1) }}_0w_0\big )^\ell + \Big ( \frac}}_0w_0}\Big )^k \big ((\nu -1) }}_0w_0\big )^\ell \bigg )\nonumber \\&\quad = \sum _ c_ (rs)^m (}}_0 w_0)^, \end$$

(58)

where

$$\begin \Sigma =\big \ 2) \big \}, \end$$

and for every \((m, n)\in \Sigma \),

$$\begin c_ = \frac}(\nu +1)^} + (\nu -1)^}(\nu +1)^} } \Big (\frac\Big )! \Big (\frac\Big )! }> 0. \end$$

Recall that \(D_ =e^|z|^2} e^}^\dagger }e^}}\), which implies \(Q_ = e^(\nu -1)r^2} e^}^\dagger } e^ }}\). Therefore, once again considering the Taylor expansion of the exponential function and computing \(Q_\), we obtain a sum over the set \(\Sigma \). In fact, for any \((m,n)\in \Sigma \) there is an operator \(B_\) such that

$$\begin Q_ = \sum _ \sqrt} r^m z_0^n B_. \end$$

(59)

Then, we have

$$\begin \big \langle Q_, Q_\big \rangle _\tau&= \sum _ \sqrtc_} r^m s^ \bar_0^n w_0^ \big \langle B_, B_\big \rangle _\tau . \end$$

This holds for all real numbers r, s, and complex numbers \(z_0, w_0\) with modulus 1. Thus, comparing to (58), we find that

$$\begin \big \langle B_, B_\big \rangle _\tau = \delta _\delta _. \end$$

This means that operators \(B_\) for \((m,n)\in \Sigma \) are orthonormal with respect to the inner product \(\langle \cdot , \cdot \rangle _\tau \). On the other hand, since by Stone–von Neumann theorem the span of displacement operators is dense, the operators \(B_\) span the whole space. We conclude that \(\:\, (m,n)\in \Sigma \}\) is an orthonormal basis for the space \(\textbf_2(\tau )\). We also note that by definition \(B_ = }\), so \(B_\) is orthogonal to the identity operator for any \( (m,n)\ne (0,0)\).

Next, we use (7) to conclude that \(\big [}, Q_\big ] = rz_0 Q_\). Applying the expansion (59) on both sides of this equation implies

$$\begin \big [}, B_\big ] = \sqrt}}} B_ & \qquad (m-1, n-1)\in \Sigma ,\\ 0 & \qquad \text . \end\right. } \end$$

(60)

We also have \(\big [}^\dagger , Q_\big ] = r}}_0 Q_\) which gives

$$\begin \big [}^\dagger , B_\big ] = \sqrt}}} B_ & \qquad (m-1, n+1)\in \Sigma ,\\ 0 & \qquad \text . \end\right. } \end$$

(61)

Let \(X\in }_2(\tau )\) be an arbitrary operator satisfying \(\langle }, X\rangle _\tau = \langle B_, X\rangle _\tau =0\). Then, there are coefficients \(\alpha _\) such that

$$\begin X = \sum _ \alpha _ B_. \end$$

Since \(\big \:\, (m,n)\in \Sigma \big \}\) is orthonormal, we have

$$\begin \Vert X\Vert _\tau ^2= \sum _ |\alpha _|^2. \end$$

We use (60) and (61) to compute \(\Vert \partial X\Vert _\tau ^2 \). To this end, note that for \((0, 0)\ne (m, n)\in \Sigma \) we have \((m-1, n-1)\in \Sigma \) unless \(n=-m\). Similarly, for \((0, 0)\ne (m, n)\in \Sigma \) we have \((m-1, n+1)\in \Sigma \) unless \(n=m\). Therefore, letting \(\Sigma _-:= \\) and \(\Sigma _+ := \\), we have

$$\begin \Vert \partial X\Vert _\tau ^2 = \sum _ \frac}} |\alpha _|^2 + \sum _ \frac}} |\alpha _|^2. \end$$

(62)

It is not hard to verify that \(\frac}} \ge \frac\) for every \((0,0)\ne (m,n)\in \Sigma \Sigma _-\) and \(\frac}}\ge \frac\) for every \((0,0)\ne (m,n)\in \Sigma \setminus \Sigma _+\). On the other hand, \(\Sigma _-\cap \Sigma _+=\emptyset \). Using these in (62) we arrive at

$$\begin \Vert \partial X\Vert _\tau ^2 \ge \frac \sum _ |\alpha _|^2 = \frac\Vert X\Vert _\tau ^2. \end$$

Hence, \(\tau \) satisfies the Poincaré inequality with constant \(\lambda _\tau \ge \frac\). \(\square \)

Score Operator for Gaussian States

In this appendix, we compute the SLD score operator for a Gaussian state \(\rho _\textrm\) that is in Williamson’s form. Such a state takes the form \(\rho _\textrm= \tau _1\otimes \cdots \otimes \tau _m\) where \(\tau _j = (1-e^) e^}_j^\dagger }_j}\) is a thermal state. We also have

$$\begin \mu _j = \Vert }_j\Vert _} = \Vert }_j\Vert _ = \frac})}. \end$$

We show that

$$\begin S_, j} = -\frac }_j. \end$$

Since \(S_, j} \!=\!\pi _}^([}_j, \rho _\textrm])\), this equality is equivalent to \([}_j, \rho _\textrm] \!=\! -\frac\pi _}^(}_j)\). Thus, we need to show that

$$\begin }_j\rho _\textrm - \rho _\textrm }_j = -\frac (\rho _\textrm }_ + }_j \rho _\textrm), \end$$

that is a simple consequence of \(}_j \rho _\textrm = e^ \rho _\textrm }_j\).

Details of the Proof of Lemma 5

We claim that for constant

$$\begin C=\frac})^2}, \end$$

we have \(Cg(x, y)\le h(x, y)\) for all \(x, y\ge 0\) where

$$\begin g(x, y) = \big (e^\sqrt - e^\sqrt\big )^2, \end$$

and

$$\begin h(x, y) = \bigg ( \sqrt}(y-x) + \frac \sqrt} \bigg )^2,\qquad \quad \mu = \frac})}. \end$$

With the change of variables \(x'=e^\sqrt, y'=e^\sqrt\) we have \(g(x, y) = (y'-x' )^2\). We also compute

$$\begin h(x, y)&= \bigg ( \sqrt}(y-x) + \frac \sqrt} \bigg )^2\\&= \frac \bigg ( (y-x) + \frac \frac \bigg )^2\\&= \frac \bigg ( \frac} y - \frac}} x \bigg )^2\\&= \frac})^2} \cdot \frac^2+e^^2} (^2-^2)^2\\&\ge \frac})^2} \cdot \frac(x'+y')^2} (^2-^2)^2\\&= \frac})^2} (y'-x')^2\\&= C g(x, y). \end$$

This proves the desired inequality.

An Extension of the Cauchy–Schwarz Inequality

The following lemma is used in the proof of Lemma 6.

[Style1 Style1]Lemma 7

Let \(\rho _1, \rho _2\) be two quantum states let \(T=\textrm_1\big ( \pi _}^(A) B \big )\) where A is an operator acting only on the first subsystem. Then we have

$$\begin \Vert T\Vert _ \le \Vert A\Vert _ \cdot \Vert B\Vert _. \end$$

Proof

We assume with no loss of generality that \(\Vert A\Vert _=1\). Then, there is an orthonormal basis \(\\) for \(}_2(\rho _1)\) including \(A_1=A\). Also, let \(\\) be an orthonormal basis for \(}_2(\rho _2)\). Now, expanding B in the tensor product basis, we have

$$\begin B = \sum _ a_ A_k\otimes C_\ell , \end$$

for some \(a_\in }\). Then, by the orthonormality of \(\, \\) we have

$$\begin \Vert T\Vert _^2 = \Big \Vert \sum _ a_ C_\ell \Big \Vert _^2 = \sum _\ell |a_|^2 \le \sum _ |a_|^2 = \Vert B\Vert ^2_. \end$$

\(\square \)