Quantum Griffiths Inequalities

Abstract

We present a general framework of Griffiths inequalities for quantum systems. Our approach is based on operator inequalities associated with self-dual cones and provides a consistent viewpoint of the Griffiths inequality. As examples, we discuss the quantum Ising model, quantum rotor model, Bose–Hubbard model, and Hubbard model. We present a model-independent structure that governs the correlation inequalities.

Appendix: Fundamental Properties of Operator Inequalities Associated with Self-Dual Cones

1.1 Positivity Preserving Operators

In this appendix, we review useful operator inequalities studied in [43].

Let $\mathfrak {H}$ be a complex Hilbert space and $\mathfrak {P}$ be a self-dual cone in $\mathfrak {H}$.

Proposition A.1

Let $\{x_n\}_{n\in \mathbb {N}}$ be a CONS of $\mathfrak {H}$. Assume that $x_n\in \mathfrak {P}$ for all $n\in \mathbb {N}$. Assume that $A\unrhd 0$ w.r.t. $\mathfrak {P}$. Then we have $\mathrm {Tr}[A]\ge 0$.

Proof

Since $x_n\in \mathfrak {P}$, we see that $\langle x_n|Ax_n\rangle \ge 0$ for all $n\in \mathbb {N}$. Thus, we arrive at $\mathrm {Tr}[A]=\sum _{n=1}^{\infty }\langle x_n|Ax_n\rangle \ge 0$. $\Box$

Proposition A.2

Let $N=\dim \mathfrak {H}\in \mathbb {N}\cup \{\infty \}$. Let $\{x_n\}_{n=1}^N$ be a CONS of $\mathfrak {H}$. Assume that $x_n\in \mathfrak {P}$ for all $n\in \{1,\dots , N\}$.¹⁹ Then the following (i) and (ii) are equivalent.

• (i) $A\unrhd 0$ w.r.t. $\mathfrak {P}$.

• (ii) $A_{mn}=\langle x_m|Ax_n\rangle \ge 0$ for all $m,n \in \{1, \dots , N\}$.

Proof

(i) $\Longrightarrow$ (ii): Trivial.

(ii) $\Longrightarrow$ (i): Let $w, z\in \mathfrak {P}$. Then we can write

$$\begin{aligned} w&=\sum _{n=1}^N c_n x_n,\ \ \ c_n=\langle w|x_n\rangle ,\end{aligned}$$

(A.1)

$$\begin{aligned} z&=\sum _{n=1}^N d_n x_n,\ \ \ d_n=\langle z|x_n\rangle . \end{aligned}$$

(A.2)

Since $w,z\ge 0$ w.r.t. $\mathfrak {P}$, we see that $c_n\ge 0, d_n \ge 0$ for all $n\in \mathbb {N}$. Thus, we have

$$\begin{aligned} \langle w|Az\rangle =\sum _{m,n=1}^N c_m d_n A_{mn} \ge 0. \end{aligned}$$

(A.3)

Since $\mathfrak {P}$ is self-dual, we have $Az\ge 0$ w.r.t. $\mathfrak {P}$. Thus, we conclude that $A\unrhd 0$ w.r.t. $\mathfrak {P}$. $\square$

Proposition A.3

Assume that $A \unrhd 0$ w.r.t. $\mathfrak {P}$. Then $\mathrm {e}^{\beta A} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$.

Proof

Since $A \unrhd 0$ w.r.t. $\mathfrak {P}$, it holds that $A^n\unrhd 0$ w.r.t. $\mathfrak {P}$ for all $n\in \mathbb {N}$. Thus ,

$$\begin{aligned} \mathrm {e}^{\beta A} =\sum _{n\ge 0}\underbrace{\frac{\beta ^n}{n!}}_{\ge 0} \underbrace{A^n}_{\unrhd 0} \unrhd 0\ \ \ \text{ w.r.t. } \mathfrak {P}\,\mathrm{for\, all}\, \beta \ge 0. \end{aligned}$$

(A.4)

$\square$

Proposition A.4

Assume that $\mathrm {e}^{\beta A} \unrhd 0$ and $\mathrm {e}^{\beta B} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$. Then $\mathrm {e}^{\beta (A+B)} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$.

Proof

Note that $\mathrm {e}^{\beta A} \mathrm {e}^{\beta B} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$. Thus, $(\mathrm {e}^{\beta A/n}\mathrm {e}^{\beta B/n})^n \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$ and $n\in \mathbb {N}$. By the Trotter–Kato product formula, we obtain the desired assertion. $\square$

The following proposition is repeatedly used in this study.

Proposition A.5

Assume the following:

• (i) $\mathrm {e}^{\beta A} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$.

• (ii) $B\unrhd 0$ w.r.t. $\mathfrak {P}$.

Then we have $\mathrm {e}^{\beta (A+B)} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$.

Proof

By (ii) and Proposition A.3, it holds that $\mathrm {e}^{\beta B} \unrhd 0$ w.r.t.  $\mathfrak {P}$  for  all $\beta \, \ge 0$. Thus, applying Proposition A.4, we conclude the assertion. $\square$

Proposition A.6

Let A be a positive self-adjoint operator. Assume that $\mathrm {e}^{-\beta A} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$. Assume that $E=\inf \mathrm {spec}(A)$ is an eigenvalue of A. Then there exists a nonzero vector $x\in \ker (A-E)$ such that $x\ge 0$ w.r.t. $\mathfrak {P}$.

Proof

Step 1 Let J be an antilinear involution given by Proposition A.7 below. Set $\mathfrak {H}_J=\{x\in \mathfrak {H}\, |\, Jx=x\}$. We will show that $\ker (A-E)\cap \mathfrak {H}_J\ne \{0\}$.

To see this, let $x\in \ker (A-E)$. Then we have the decomposition $x=\mathfrak {R}x+i \mathfrak {I}x$ with $\mathfrak {R}x=\frac{1}{2}(\mathbbm {1}+J)x$ and $\mathfrak {I}x =\frac{1}{2i}(\mathbbm {1}-J)x$. Clearly , $\mathfrak {R}x, \mathfrak {I}x\in \mathfrak {H}_J$. Since $x\ne 0$, it holds that $\mathfrak {R}x\ne 0$ or $\mathfrak {I}x\ne 0$. Since $\mathrm {e}^{-\beta A}\unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$, A commutes with J. Thus, $\mathfrak {R}x, \mathfrak {I}x\in \ker (A-E)\cap \mathfrak {H}_J$.

Step 2 Take $x\in \ker (A-E)\cap \mathfrak {H}_J$. By Proposition A.7 (iii), we have a unique decomposition $x=x_+-x_-$, where $x_{\pm }\in \mathfrak {P}$ and $\langle x_+| x_-\rangle =0$. Let $|x|=x_++x_-$. Then we have

$$\begin{aligned} \mathrm {e}^{-\beta E}\Vert x\Vert =\langle x|\mathrm {e}^{-\beta A}x\rangle \le \langle |x||\mathrm {e}^{-\beta A}|x|\rangle \le \mathrm {e}^{-\beta E}\underbrace{\Vert |x|\Vert }_{=\Vert x\Vert }. \end{aligned}$$

(A.5)

Thus, $|x|\in \ker (A-E)$. Clearly, $|x|\ge 0$ w.r.t. $\mathfrak {P}$. $\square$

Proposition A.7

A self-dual cone $\mathfrak {P}$ has the following properties:

• (i) $\mathfrak {P}\cap (-\mathfrak {P})=\{0\}$.

• (ii) There exists a unique antilinear involution J in $\mathfrak {H}$ such that $Jx=x$ for all $x\in \mathfrak {P}$.

• (iii) Each element $x\in \mathfrak {H}$ with $Jx=x$ has a unique decomposition $x=x_+-x_-$ where $x_+,x_-\in \mathfrak {P}$ and $\langle x_+| x_-\rangle =0$.

• (iv) $\mathfrak {H}$ is linearly spanned by $\mathfrak {P}$.

Proof

See, e.g., [5]. $\square$

1.2 Reflection Positive Operators

To apply Theorem 3.11, it is crucial to show that $\mathrm {e}^{-\beta H} \succeq 0$ w.r.t. $\mathscr {L}^2(\mathfrak {H})_+$ for all $\beta >0$. The following proposition is often useful in proving this condition:

Proposition A.8

Let $H_0$ be a self-adjoint operator on $\mathscr {L}^2(\mathfrak {H})$ bounded from below. Let $V \in \mathscr {B}(\mathscr {L}^2(\mathfrak {H}))$ be self-adjoint. Assume the following:

• (i) $\mathrm {e}^{-\beta H_0} \succeq 0$ w.r.t. $\mathscr {L}^2(\mathfrak {H})_+$ for all $\beta \ge 0$.

• (ii) $V\succeq 0$ w.r.t. $\mathscr {L}^2(\mathfrak {H})_+$.

Let $H=H_0-V$. We have $\mathrm {e}^{-\beta H} \succeq 0$ w.r.t. $\mathscr {L}^2(\mathfrak {H})_+$ for all $\beta \ge 0$.

Proof

Note that

$$\begin{aligned} \mathrm {e}^{\beta V}=\sum _{n\ge 0} \underbrace{\frac{\beta ^n}{n!}}_{\ge 0} \underbrace{V^n}_{\succeq 0} \succeq 0\ \ \ \text{ w.r.t. } \mathscr {L}^2(\mathfrak {H})_+. \end{aligned}$$

(A.6)

Thus, by the Trotter–Kato product formula, we obtain

$$\begin{aligned} \mathrm {e}^{-\beta H}=\mathrm {s}-\displaystyle \lim _{n\rightarrow \infty }\Big ( \underbrace{\mathrm {e}^{-\beta H_0/n}}_{\succeq 0} \underbrace{ \mathrm {e}^{\beta V/n}}_{\succeq 0} \Big )^n\succeq 0\ \ \ \text{ w.r.t. } \mathscr {L}^2(\mathfrak {H})_+ \,\mathrm{for \, all}\, \beta \ge 0, \end{aligned}$$

(A.7)

where $\mathrm {s}-\displaystyle \lim _{n\rightarrow \infty }$ means the strong limit. $\square$

Corollary A.9

Let $H_0=\mathcal {L}(A)+\mathcal {R}(A)$, where A is self-adjoint and bounded from below. Let

$$\begin{aligned} V=\sum _{j=1}^{\infty }\mathcal {L}(B_j) \mathcal {R}(B_j), \end{aligned}$$

(A.8)

where $B_j\in \mathscr {B}(\mathfrak {H})$ is self-adjoint and the right hand side of (A.8) is a weak convergent sum. Define $H=H_0-V$. Then we obtain $\mathrm {e}^{ -\beta H} \succeq 0$ w.r.t.  $\mathscr {L}^2(\mathfrak {H})_+$ for all $\beta \ge 0$.

Proof

Observe that $\mathrm {e}^{-\beta H_0}=\mathcal {L}(\mathrm {e}^{-\beta A}) \mathcal {R}(\mathrm {e}^{-\beta A})\succeq 0$ w.r.t. $\mathscr {L}^2(\mathfrak {H})_+$ for all $\beta \ge 0$. Since $V\succeq 0$ w.r.t. $\mathscr {L}^2(\mathfrak {H})_+$, we obtain the desired assertion by Proposition A.8. $\square$

The following lemma will be often useful:

Lemma A.10

Let $A_j,\ j=1, \dots , N$ be a bounded operator acting in $\mathfrak {H}$. Let $M=(M_{ij})$ be a positive semidefinite $N\times N$ matrix. Then we have

$$\begin{aligned} \sum _{i, j=1}^N M_{ij} \mathcal {L}(A_i^*) \mathcal {R}(A_j) \succeq 0\ \ \text{ w.r.t. } \mathscr {L}^2(\mathfrak {H})_+. \end{aligned}$$

(A.9)

Proof

There exists a unitary matrix U such that $M=U^* DU$, where $D=\mathrm {diag}(\lambda _j)$ is a diagonal matrix with $\lambda _j\ge 0$. Set $\tilde{A}_i=\sum _{j=1}^N U_{ij} A_j$. Then we see

$$\begin{aligned} \text{ LHS } \text{ of } \text{(A.9) }=\sum _{j=1}^N \lambda _{j} \mathcal {L}(\tilde{A}_j^*) \mathcal {R}(\tilde{A}_j) \succeq 0\ \ \text{ w.r.t. } \mathscr {L}^2(\mathfrak {H})_+. \end{aligned}$$

(A.10)

This completes the proof. $\Box$