Correlation Inequalities for Schrödinger Operators

Abstract

The purpose of the present paper is to analyze correlation structures of the ground states of the Schrödinger operator. We construct Griffiths inequalities for the ground state expectations by applying operator-theoretic correlation inequalities. As an example of such an application, we study the ground state properties of Schrödinger operators.

Appendix A: General Theory of Correlation Inequalities

In this appendix, we will review some basic results concerning the operator inequalities introduced in Section 3. Almost all results are taken from the author’s previous works [24, 26,27,28,29,30,31].

Proposition A.1

Let$\{A_{n}\}_{n=1}^{\infty }\subseteq {\mathscr{B}}(\mathfrak {H})$and let$A\in {\mathscr{B}}(\mathfrak {H})$. Suppose that A_nconverges to Ain the weak operator topology. If$A_{n}\unrhd 0$w.r.t.$\mathfrak {P}$for all$n\in \mathbb {N}$, then$A\unrhd 0$w.r.t.$\mathfrak {P}$.

Proof

By Remark 3.7 (i), 〈ξ|A_nη〉 ≥ 0 for all $\xi , \eta \in \mathfrak {P}$. Thus, $\displaystyle \langle \xi |A\eta \rangle =\lim _{n\to \infty }\langle \xi |A_{n}\eta \rangle \ge 0$ for all $\xi , \eta \in \mathfrak {P}$. By Remark 3.7 (i) again, we conclude that $A\unrhd 0$ w.r.t. $\mathfrak {P}$. □

Proposition A.2

Let Abe a self-adjoint positive operator on$\mathfrak {H}$. Assume that$e^{-\beta A} \unrhd 0$w.r.t.$\mathfrak {P}$for all β ≥ 0. Assume that$E=\inf \sigma (A)$is an eigenvalue of A. Then there exists a nonzero vector$\xi \in \ker (A-E)$such that ξ ≥ 0 w.r.t.$\mathfrak {P}$.

Proof

Let $\eta \in \mathfrak {H}$. By Theorem 3.4, we can express η as η = η_R + iη_I with $\eta _{R}, \eta _{I} \in \mathfrak {H}_{\mathbb {R}}$. Now, we define an antilinear involution J by Jη = η_R − iη_I. Clearly,

$$ \eta_{R}=\frac{1}{2}(\eta+J\eta),\ \ \ \eta_{I}=\frac{1}{2i} (\eta-J\eta). $$

(A.1)

Moreover, $\mathfrak {H}_{\mathbb {R}}=\{\eta \in \mathfrak {H} | J\eta =\eta \}$. Because $e^{-\beta A} \mathfrak {P} \subseteq \mathfrak {P}$, we see that $e^{-\beta A} \mathfrak {H}_{\mathbb {R}} \subseteq \mathfrak {H}_{\mathbb {R}}$ for all β ≥ 0, see Remark 3.7 (i). Hence, for all β ≥ 0, we obtain

$$ Je^{-\beta A} =e^{-\beta A}J. $$

(A.2)

Let $\xi \in \ker (A-E)$ with ξ ≠ 0. ξ can be expressed as ξ = ξ_R + iξ_I with $\xi _{R}, \xi _{I} \in \mathfrak {H}_{\mathbb {R}}$. Because ξ ≠ 0, we have ξ_R ≠ 0 or ξ_I ≠ 0. By (A.1) and (A.2), we know that $\xi _{R}, \xi _{I}\in \ker (A-E) \cap \mathfrak {H}_{\mathbb {R}}$. Without loss of generality, we may assume that ξ_R ≠ 0. By Definition 3.2 (ii) and Theorem 3.4, we have a unique decomposition ξ_R = ξ_R,+ − ξ_R,−, where $\xi _{R, \pm } \in \mathfrak {P}$ with 〈ξ_R,+|ξ_R,−〉 = 0. Let |ξ_R| = ξ_R,+ + ξ_R,−. Because ∥ξ_R∥ = ∥|ξ_R|∥, we have

$$ e^{-\beta E} \|\xi_{R}\|^{2}=\langle \xi_{R}|e^{-\beta A} \xi_{R} \rangle \le \langle |\xi_{R}||e^{-\beta A} | \xi_{R}| \rangle \le e^{-\beta E} \|\xi_{R}\|^{2}. $$

(A.3)

Thus, $|\xi _{R}| \in \ker (A-E)$. Clearly, |ξ_R|≥ 0 w.r.t. $\mathfrak {P}$. □

Theorem A.3

Let A be a self-adjoint positive operator on$\mathfrak {H}$and$B\in {\mathscr{B}}(\mathfrak {H})$. Suppose that

• (i)$e^{-\beta A} \unrhd 0$w.r.t.$\mathfrak {P}$for all β ≥ 0;

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

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

Proof

By (ii) and Proposition A.1,

$$ e^{\beta B}=\sum\limits_{n\ge 0} \underbrace{\frac{\beta^{n}}{n!}}_{\ge 0} \underbrace{B^{n}}_{\unrhd 0} \unrhd 0\ \ \text{w.r.t. } \mathfrak{P} \text{ for all } \beta \ge 0. $$

(A.4)

Hence, by (i) and Proposition 3.8 (ii),

$$ \Big(\underbrace{e^{-\beta A/n}}_{\unrhd 0} \underbrace{e^{\beta B/n}}_{\unrhd 0} \Big)^{n} \unrhd 0\ \ \ \text{w.r.t. } \mathfrak{P} \text{ for all } \beta \ge 0. $$

(A.5)

Using the Trotter–Kato product formula(e.g., [35, Theorem S. 21]) and Proposition A.1, we arrive at the desired assertion. □

Theorem A.4

Let A, B be self-adjoint positive operators on$\mathfrak {H}$. Assume that B = A − C with$C\in {\mathscr{B}}(\mathfrak {H})$. Suppose that

• (i)$e^{-\beta A}\unrhd 0$w.r.t.$\mathfrak {P}$for all β ≥ 0;

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

Then we have$e^{-\beta B }\unrhd e^{-\beta A}$w.r.t.$\mathfrak {P}$for all β ≥ 0.

Proof

By the Duhamel formula, we have the norm-convergent expansion

$$ \begin{array}{@{}rcl@{}} e^{-\beta B}&=&\sum\limits_{n=0}^{\infty}D_{n}(\beta), \end{array} $$

(A.6)

$$ \begin{array}{@{}rcl@{}} D_{n}(\beta)&=&{\int}_{S_{n}(\beta)} e^{-s_{1} A}C e^{-s_{2} A}C{\cdots} e^{-s_{n} A} C e^{-(\beta-{\sum}_{j=1}^{n}s_{j})A}, \end{array} $$

(A.7)

where ${\int \limits }_{S_{n}(\beta )}={\int \limits }_{0}^{\beta }ds_{1}{\int \limits }_{0}^{\beta -s_{1}}ds_{2}{\cdots } {\int \limits }_{0}^{\beta -{\sum }_{j=1}^{n-1}s_{j}} ds_{n}$ and D₀(β) = e^−βA. Since $C \unrhd 0$ and $e^{-t A}\unrhd 0$ w.r.t. $\mathfrak {P}$ for all t ≥ 0, it holds that, by Proposition 3.8 (ii),

$$ \underbrace{ e^{-s_{1} A} }_{\unrhd 0} \underbrace{C}_{\unrhd 0} \underbrace{ e^{-s_{2} A} }_{\unrhd 0}{\cdots} \underbrace{e^{-s_{n} A}}_{\unrhd 0} \underbrace{C}_{\unrhd 0} \underbrace{e^{-(\beta-{\sum}_{j=1}^{n}s_{j})A}}_{\unrhd 0} \unrhd 0 $$

(A.8)

provided that $s_{1} \ge 0, \dots , s_{n}\ge 0$ and β − s₁ −⋯ − s_n ≥ 0. Thus, by Proposition A.1, we obtain $D_{n}(\beta )\unrhd 0$ w.r.t. $\mathfrak {P}$ for all n ≥ 0. Accordingly, by (A.6) and Proposition A.1 again, we have $e^{-\beta B}\unrhd D_{n=0}(\beta )=e^{-\beta A}$ w.r.t. $\mathfrak {P}$ for all β ≥ 0. □

Remark A.5

By (i), there exists a unique $\xi \in \mathfrak {H}$ such that ξ > 0 w.r.t. $\mathfrak {P}$ and P_A = |ξ〉〈ξ|. Of course, ξ satisfies $A\xi =\inf \sigma (A)\xi$. ♢

Theorem A.6

Let A be a self-adjoint positive operator on$\mathfrak {H}$, and let$B\in {\mathscr{B}}(\mathfrak {H})$. Suppose the following:

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

• (ii)B is ergodic w.r.t.$\mathfrak {P}$.

Then, $e^{-\beta (A-B)} \rhd 0$w.r.t. $\mathfrak {P}$for all β > 0.

Proof

Set H = A − B. We apply Fröhlich’s idea [7] and use the Duhamel expansion:

$$ \begin{array}{@{}rcl@{}} e^{-\beta H}&=&\sum\limits_{n\ge 0} \mathscr{D}_{n}(\beta), \end{array} $$

(A.9)

$$ \begin{array}{@{}rcl@{}} \mathscr{D}_{n}(\beta) &=& {\int}_{S_{n}(\beta)} e^{-s_{1} A} B e^{-s_{2} A} {\cdots} e^{-s_{n} A} B e^{-(\beta-{\sum}_{j=1}^{n} s_{j}) A}. \end{array} $$

(A.10)

In a manner similar to that used in the proof of Theorem A.4, we know that

$$ \begin{array}{@{}rcl@{}} &\mathscr{D}_{n}(\beta) \unrhd 0, \end{array} $$

(A.11)

$$ \begin{array}{@{}rcl@{}} &e^{-s_{1} A} B e^{-s_{2} A} {\cdots} e^{-s_{n} A} B e^{-(\beta-{\sum}_{j=1}^{n}s_{j}) A} \unrhd 0 \end{array} $$

(A.12)

w.r.t. $\mathfrak {P}$, provided that $s_{1} \ge 0, \dots , s_{n}\ge 0$ and β − s₁ −⋯ − s_n ≥ 0.

Let $\xi , \eta \in \mathfrak {P}\backslash \{0\}$. Since $e^{-\beta A} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all β ≥ 0, we have $e^{-\beta A} \eta \in \mathfrak {P}\backslash \{0\}$. Let β > 0 be fixed arbitrarily. Because B is ergodic w.r.t. $\mathfrak {P}$, there exists an $n\in \{0\}\cup \mathbb {N}$ such that 〈ξ|Bⁿe^−βAη〉 > 0. Now, let

$$ F(s_{1}, \dots, s_{n})=\Big\langle \xi \Big|e^{-s_{1} A} B e^{-s_{2} A} {\cdots} e^{-s_{n} A} B e^{-(\beta-{\sum}_{j=1}^{n}s_{j}) A} \eta\Big\rangle. $$

(A.13)

By (A.12), it holds that $F(s_{1}, \dots , s_{n}) \ge 0$. In addition, we have $F(0, \dots , 0)=\langle \xi |B^{n} e^{-\beta A}\eta \rangle >0$. Because $F(s_{1},\dots , s_{n})$ is continuous in $s_{1}, \dots , s_{n}$, we obtain

$$ \langle \xi|\mathscr{D}_{n}(\beta) \eta\rangle ={\int}_{S_{n}(\beta)} F(s_{1}, \dots, s_{n}) >0. $$

(A.14)

By (A.9) and (A.11), we see that $e^{-\beta H} \unrhd {\mathscr{D}}_{n}(\beta )$, which implies

$$ \langle \xi|e^{-\beta H}\eta\rangle \ge \langle \xi|\mathscr{D}_{n}(\beta)\eta\rangle>0. $$

(A.15)

Since ξ and η are in $\mathfrak {P}\backslash \{0\}$, we conclude that e^−βHη > 0 w.r.t. $\mathfrak {P}$. Since β is arbitrary, we obtain that $e^{-\beta H} \rhd 0$ w.r.t. $\mathfrak {P}$ for all β > 0. □

Theorem A.7

Let$A\in {\mathscr{B}}(\mathfrak {H})$. Assume that u > 0 w.r.t.$\mathfrak {P}$and$A\unrhd 0$w.r.t.$\mathfrak {P}$. Then, 〈u|Au〉 = 0 if and only if A = 0.

Proof

We will divide the proof into several steps.

• Step 1.Let $A\in {\mathscr{B}}(\mathfrak {H})$. IfAu = 0 for all $u\in \mathfrak {P}$, thenA = 0.

□

Proof

By Remark 3.3, each $u\in \mathfrak {H}$ can be written as u = v₁ − v₂ + i(w₁ − w₂), where $v_{1}, v_{2}, w_{1}, w_{2}\in \mathfrak {P}$ such that 〈v₁|v₂〉 = 0 and 〈w₁|w₂〉 = 0. Thus, the assumption implies that Au = 0 for all$u\in \mathfrak {H}$. □

• Step 2.Let $A\in {\mathscr{B}}(\mathfrak {H})$withA ≠ 0. Assume thatu > 0 w.r.t. $\mathfrak {P}$. If $A\unrhd 0$w.r.t. $\mathfrak {P}$, thenAu ≠ 0.

Proof

Assume that Au = 0. Then, 〈v|Au〉 = 0 for all $v\in \mathfrak {P}$, implying that 〈A^∗v|u〉 = 0. Since u > 0 and A^∗v ≥ 0 w.r.t. $\mathfrak {P}$, we conclude that A^∗v must be zero. Because v is arbitrary, A^∗ = 0 by Step 1. □

Completion of the proof.

Suppose that 〈u|Au〉 = 0. Assume that A ≠ 0. Since Au ≥ 0 and u > 0 w.r.t. $\mathfrak {P}$, Au must be zero. However, this contradicts with Step 2.