Golden–Thompson’s Inequality for Deformed Exponentials

Abstract

Deformed logarithms and their inverse functions, the deformed exponentials, are important tools in the theory of non-additive entropies and non-extensive statistical mechanics. We formulate and prove counterparts of Golden–Thompson’s trace inequality for $q$-exponentials with parameter $q$ in the interval $[1,3]$.

1 Introduction and Main Result

Tsallis [7] generalised in 1988 the standard Bolzmann–Gibbs entropy to a non-extensive quantity $S_q$ depending on a parameter $q.$ In the quantum version it is given by

$$\begin{aligned} S_q(\rho )=\frac{1-\mathrm{Tr}~\rho ^q}{q-1}\qquad q\ne 1, \end{aligned}$$

where $\rho$ is a density matrix. It has the property that $S_q(\rho )\rightarrow S(\rho )$ for $q\rightarrow 1,$ where $S(\rho )=-\mathrm{Tr}~\rho \log \rho$ is the von Neumann entropy. The Tsallis entropy may be written on a similar form

$$\begin{aligned} S_q(\rho )=-\mathrm{Tr}~\rho \log _q(\rho ), \end{aligned}$$

where the deformed logarithm $\log _q$ is given by

$$\begin{aligned} \log _q x=\int _1^x t^{q-2}\,dt = \left\{ \begin{array}{ll} \displaystyle \frac{x^{q-1}-1}{q-1}&{}\quad q> 1\\ \log x &{}\quad q=1 \end{array}\right. \end{aligned}$$

for $x>0.$ The deformed logarithm is also denoted the $q$-logarithm. The inverse function $\exp _q$ is called the $q$-exponential and is given by

$$\begin{aligned} \exp _q(x)=\left( x(q-1)+1\right) ^{1/(q-1)}\qquad \text {for}\quad x>\frac{-1}{q-1}\,. \end{aligned}$$

The $q$-logarithm and the $q$-exponential functions converge, respectively, to the logarithmic and the exponential functions for $q\rightarrow 1.$

The aim of this article is to generalise Golden–Thompson’s trace inequality [2, 6] to deformed exponentials. The main result is the following:

Theorem 1.1

Let $A$ and $B$ be positive definite matrices.

• (i) If $1\le q<2$ then

  $$\begin{aligned} \mathrm{Tr}~\exp _q(A+B)\le \mathrm{Tr}~\exp _q(A)^{2-q}\bigl (A(q-1) + \exp _q B\big ). \end{aligned}$$

• (ii) If $2\le q \le 3$ then

  $$\begin{aligned} \mathrm{Tr}~\exp _q(A+B)\ge \mathrm{Tr}~\exp _q(A)^{2-q}\bigl (A(q-1) + \exp _q B\big ). \end{aligned}$$

Notice that for $q=1$ we recover Golden–Thomson’s trace inequality

$$\begin{aligned} \mathrm{Tr}~\exp (A+B)\le \mathrm{Tr}~\exp (A)\exp (B). \end{aligned}$$

This inequality is valid for arbitrary self-adjoint matrices $A$ and $B.$ However, it is sufficient to know the inequality for positive definite matrices, since the general form follows by multiplication with positive numbers.

2 Preliminaries

We collect a few well-known results that we are going to use in the proof of the main theorem.

The $q$-logarithm is a bijection of the positive half-line onto the open interval $(-(q-1)^{-1},\infty ),$ and the $q$-exponential is consequently a bijection of the interval $(-(q-1)^{-1},\infty )$ onto the positive half-line. For $q>1$ we may thus safely apply both the $q$-logarithm and the $q$-exponential to positive definite operators. We also notice that

$$\begin{aligned} \frac{d}{dx}\log _q(x)=x^{q-2}\qquad \text {and}\qquad \frac{d}{dx}\exp _q(x)=\exp _q(x)^{2-q}\,. \end{aligned}$$

(1)

The proof of the following lemma is rather easy and may be found in [4, Lemma 5].

Lemma 2.1

Let $\varphi :\mathcal D\rightarrow \mathcal A_\text {sa}$ be a map defined in a convex cone $\mathcal D$ in a Banach space $X$ with values in the self-adjoint part of a $C^*$-algebra $\mathcal A.$ If $\varphi$ is Fréchet differentiable, convex and positively homogeneous then

$$\begin{aligned} d\varphi (x)h\le \varphi (h). \end{aligned}$$

for $x,h\in \mathcal D.$

Let $H$ be any $n\times n$ matrix. The map

$$\begin{aligned} A\rightarrow \mathrm{Tr}~\left( H^* A^p H\right) ^{1/p}, \end{aligned}$$

defined in positive definite $n\times n$ matrices, is concave for $0<p\le 1$ and convex for $1\le p\le 2,$ cf. [1, Theorem 1.1]. By a slight modification of the construction given in Remark 3.2 in the same reference, cf. also [3], we obtain that the mapping

$$\begin{aligned} (A_1,\dots ,A_k)\rightarrow \mathrm{Tr}~\left( H_1^*A_1^pH_1+\cdots +H_k^* A_k H_k\right) ^{1/p}, \end{aligned}$$

(2)

defined in $k$-tuples of positive definite $n\times n$ matrices, is concave for $0<p\le 1$ and convex for $1\le p\le 2;$ for arbitrary $n\times n$ matrices $H_1,\dots ,H_k.$

3 Deformed Trace Functions

Theorem 3.1

Let $H_1,\dots ,H_k$ be matrices with $H_1^*H_1 +\cdots + H_k^*H_k=1$ and define the function

$$\begin{aligned} \varphi (A_1,\dots ,A_k)=\mathrm{Tr}~\exp _q\left( \sum _{i=1}^k H_i^*\log _q(A_i) H_i\right) \end{aligned}$$

(3)

in $k$-tuples of positive definite matrices. Then $\varphi$ is positively homogeneous of degree one. It is concave for $1\le q\le 2$ and convex for $2\le q\le 3.$

Proof

For $q>1$ we obtain

$$\begin{aligned} \varphi (A_1,\dots , A_k)= & {} \displaystyle \mathrm{Tr}~\exp _q\left( \sum _{i=1}^k H_i^*\log _q(A_i) H_i\right) \\= & {} \displaystyle \mathrm{Tr}~\left( 1+(q-1)\sum _{i=1}^k H_i^*\log _q(A_i)H_i\right) ^{1/(q-1)}\\= & {} \displaystyle \mathrm{Tr}~\left( 1+(q-1)\sum _{i=1}^k H_i^*\frac{A_i^{q-1}-1}{q-1}H_i\right) ^{1/(q-1)}\\= & {} \displaystyle \mathrm{Tr}~\left( 1+\sum _{i=1}^k H_i^*(A_i^{q-1} -1)H_i \right) ^{1/(q-1)}\\= & {} \displaystyle \mathrm{Tr}~\left( H_1^* A_1^{q-1}H_1+\cdots +H_k^*A_k^{q-1}H_k\right) ^{1/(q-1)}. \end{aligned}$$

From this identity it follows that $\varphi$ is positively homogeneous of degree one. The concavity for $1<q\le 2$ and the convexity for $2\le q\le 3$ now follows from (2). The statement for $q=1$ follows by letting $q$ tend to one. $\square$

Corollary 3.2

Let $L$ be a positive definite matrix, and let $H_1,\dots ,H_k$ be matrices such that $H_1^*H_1 +\cdots + H_k^*H_k\le 1.$ Then the function

$$\begin{aligned} \varphi (A_1,\dots ,A_k)=\mathrm{Tr}~\exp _q\left( L+H_1^*\log _q(A_1)H_1+\cdots +H_k^*\log _q(A_k)H_k\right) , \end{aligned}$$

defined in $k$-tuples of positive definite matrices, is concave for $1\le q\le 2$ and convex for $2\le q\le 3.$

Proof

We may without loss of generality assume $H_1^*H_1 +\cdots + H_k^*H_k< 1$ and put $H_{k+1}=\left( 1-( H_1^*H_1 +\cdots + H_k^*H_k)\right) ^{1/2}.$ We then have

$$\begin{aligned} H_1^*H_1+\cdots +H_k^*H_k+H_{k+1}^*H_{k+1}=1 \end{aligned}$$

and may use the preceding theorem to conclude that the function

$$\begin{aligned} \begin{array}{l} (A_1,\dots ,A_{k+1})\rightarrow \mathrm{Tr}~\exp _q\left( H_1^*\log _q(A_1)H_1+\cdots +H_{k+1}^*\log _q(A_{k+1})H_{k+1}\right) \end{array} \end{aligned}$$

of $k+1$ variables is concave for $1\le q\le 2$ and convex for $2\le q\le 3.$ Since $H_{k+1}$ is invertible we may choose

$$\begin{aligned} A_{k+1}=\exp _q\left( H_{k+1}^{-1}LH_{k+1}^{-1}\right) \end{aligned}$$

which makes sense since $H_{k+1}^{-1}LH_{k+1}^{-1}$ is positive definite. Concavity for $1\le q\le 2$ and convexity for $2\le q\le 3$ in the first $k$ variables of the above function then yields the result. $\square$

Setting $q=1$ we recover in particular [5, Theorem 3].

Corollary 3.3

Let $H_1,\dots ,H_k$ be matrices with $H_1^*H_1+\cdots +H_k^*H_k\le 1,$ and let $L$ be self-adjoint. The trace function

$$\begin{aligned} (A_1,\dots ,A_k)\rightarrow \mathrm{Tr}~\exp \left( L+H_1^*\log (A_1)H_1+\cdots +H_k^* \log (A_k)H_k\right) \end{aligned}$$

is concave in positive definite matrices.

Corollary 3.4

The trace function $\varphi$ defined in (3) satisfies

$$\begin{aligned} \varphi (B_1,\dots ,B_k)\le \displaystyle \mathrm{Tr}~\exp _q \left( \sum _{i=1}^k H_i^*\log _q(A_i) H_i\right) ^{2-q} \sum _{j=1}^k H_{j}^*(d\!{}\log _q (A_j)B_j) H_j \end{aligned}$$

for $1\le q\le 2$ and

$$\begin{aligned} \varphi (B_1,\dots ,B_k)\ge \displaystyle \mathrm{Tr}~\exp _q \left( \sum _{i=1}^k H_i^*\log _q(A_i) H_i\right) ^{2-q} \sum _{j=1}^k H_j^*(d\!{}\log _q(A_j)B_j) H_j \end{aligned}$$

for $2\le q\le 3,$ where $A_1,\dots ,A_k$ and $B_1,\dots ,B_k$ are positive definite matrices.

Proof

For $1\le q\le 2$ we obtain

$$\begin{aligned} d\!{}\varphi (A_1,\dots ,A_k)(B_1,\dots ,B_k)\ge \varphi (B_1,\dots ,B_k) \end{aligned}$$

by Lemma  2.1. By the chain rule for Fréchet differentiable mappings between Banach spaces we therefore obtain

$$\begin{aligned} \varphi (B_1,\dots ,B_k)\le & {} \displaystyle \sum _{j=1}^k d_j \varphi (A_1,\dots ,A_k)B_j \\= & {} \displaystyle \sum _{j=1}^k \mathrm{Tr}~d\!{}\exp _q \left( \sum _{i=1}^k H_i^*\log _q(A_i) H_i\right) H_j^*(d\!{}\log _q(A_j)B_j) H_j\\= & {} \displaystyle \sum _{j=1}^k \mathrm{Tr}~\exp _q \left( \sum _{i=1}^k H_i^*\log _q(A_i) H_i\right) ^{2-q} H_j^*(d\!{}\log _q(A_j)B_j) H_j \end{aligned}$$

where we used the identity $\mathrm{Tr}~d\!{}f(A)B=\mathrm{Tr}~f'(A)B$ valid for differentiable functions. This proves the first assertion. The result for $2\le q\le 3$ follows similarly. $\square$

4 Proof of the Main Theorem

In order to prove Theorem 1.1 i we set $k=2$ in Corollary 3.4 and obtain

$$\begin{aligned} \varphi (B_1,B_2)\le \mathrm{Tr}~\exp _q(X)^{2-q}\left( H_1^* (d\!{}\log _q(A_1)B_1) H_1 + H_2^* (d\!{}\log _q(A_2)B_2) H_2\right) \end{aligned}$$

for $1\le q\le 2$ and positive definite matrices $A_1,A_2$ and $B_1,B_2$ where

$$\begin{aligned} X=H_1^*\log _q(A_1) H_1 + H_2^* \log _q(A_2) H_2\,. \end{aligned}$$

If we set $A_1=B_1$ and $A_2=1$ the inequality reduces to

$$\begin{aligned} \varphi (B_1,B_2)\le \mathrm{Tr}~\exp _q(H_1^* \log _q(B_1) H_1)^{2-q}\left( H_1^* B_1^{q-1} H_1 +H_2^* B_2 H_2\right) . \end{aligned}$$

We now set $H_1=\varepsilon ^{1/2}$ for $0<\varepsilon <1,$ and to fixed positive definite matrices $L_1$ and $L_2$ we choose $B_1$ and $B_2$ such that

$$\begin{aligned} \begin{array}{rl} L_1&{}=H_1^*\log _q(B_1) H_1=\varepsilon \log _q(B_1)\\ L_2&{}=H_2^*\log _q(B_2) H_2=(1-\varepsilon )\log _q(B_2). \end{array} \end{aligned}$$

It follows that

$$\begin{aligned} B_1=\exp _q\left( \varepsilon ^{-1}L_1\right) \qquad \text {and}\qquad B_2=\exp _q\left( (1-\varepsilon )^{-1}L_2\right) . \end{aligned}$$

Inserting in the inequality we obtain

$$\begin{aligned} \mathrm{Tr}~\exp _q(L_1+L_2)\le & {} \mathrm{Tr}~\exp _q(L_1)^{2-q}\left( \varepsilon \exp _q(\varepsilon ^{-1}L_1)^{q-1}+(1-\varepsilon )\exp _q((1-\varepsilon )^{-1}L_2)\right) \\= & {} \mathrm{Tr}~\exp _q(L_1)^{2-q}\left( L_1(q-1)+\varepsilon +(1-\varepsilon )\exp _q((1-\varepsilon )^{-1}L_2)\right) . \end{aligned}$$

This expression decouble $L_1$ and $L_2$ and reduces the minimisation problem over $\varepsilon$ to the commutative case. We furthermore realise that minimum is obtained by letting $\varepsilon$ tend to zero and that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} (1-\varepsilon )\exp _q\left( (1-\varepsilon )^{-1}L_2\right) =\exp _q(L_2). \end{aligned}$$

We finally replace $L_1$ and $L_2$ with $A$ and $B.$ This proves the first statement in Theorem 1.1.

The proof of the second statement is virtually identical to the proof of the first. Since now $2\le q\le 3$ the second inequality in Corollary 3.4 applies. Setting $k=2$ and applying the same substitutions as in the proof of the first statement we arrive at the inequality

$$\begin{aligned} \begin{array}{l} \mathrm{Tr}~\exp _q(L_1+L_2)\ge \mathrm{Tr}~\exp _q(L_1)^{2-q}\left( L_1(q-1)+\varepsilon +(1-\varepsilon )\exp _q((1-\varepsilon )^{-1}L_2)\right) . \end{array} \end{aligned}$$

Since $2\le q\le 3$ the function

$$\begin{aligned} \varepsilon \rightarrow \varepsilon +(1-\varepsilon )\exp _q\left( (1-\varepsilon )^{-1}L_2\right) \end{aligned}$$

is now decreasing, and we thus maximise the right hand side in the above inequality by letting $\varepsilon$ tend to zero. This proves the second statement in Theorem 1.1.