﻿ 无穷维Hamilton算子辛自伴延拓的存在性与唯一性研究

# 无穷维Hamilton算子辛自伴延拓的存在性与唯一性研究Research on Existence and Uniqueness of Symplectic Self-Adjoint Extension of Infinite Dimensional Hamiltonian Operator

Abstract:
In this paper, the symplectic self-adjoint extension problem of infinite dimensional Hamiltonian operator is studied. By using the method of space decomposition, the condition of dimensional Hamiltonian Infinite operator exists symplectic self-adjoint extension is given, and the conditions of symplectic self-adjoint extension which is unique are given.

1. 引言

$H=\left(\begin{array}{cc}A& B\\ C& -{A}^{*}\end{array}\right)$

$H=\left(\begin{array}{cc}A& 0\\ 0& -{A}^{*}\end{array}\right)$

$X={l}_{2},D\left(A\right)=\left\{a\in X:对某个N\in Ν,\underset{m=0}{\overset{n}{\sum }}{a}_{m}=0,当n>N时,{a}_{n}=0\right\}$

$a\in D\left(A\right),Aa\in X$ 定义为

${\left(Aa\right)}_{n}=i\left[\underset{m=0}{\overset{n-1}{\sum }}{a}_{m}+\underset{m=0}{\overset{n}{\sum }}{a}_{m}\right]$

$D\left(T\right)=\left\{x\left(t\right):x\left(t\right)\in AC\left[0,1\right],x\left(0\right)=x\left(1\right)=0\right\}$

$D\left({T}_{\alpha }\right)=\left\{x\left(t\right):x\left(t\right)\in AC\left[0,1\right],x\left(0\right)=\alpha x\left(1\right)\right\}$

2. 无穷维Hamilton算子辛自伴延拓的存在性

2.1. 预备知识

$\left(Hx,u\right)=-\left(Hx,{J}^{2}u\right)=\left(x,JHJu\right)$

$JHJ\subset {H}^{*}$ 时，对于任意 $x,y\in D\left(H\right)$

$\left(Hx,Jy\right)=\left(x,{H}^{*}Jy\right)=\left(x,JHJJy\right)=-\left(x,JHy\right)$

$\left(Hx,y\right)=-\left(Hx,{J}^{2}y\right)=\left(x,JHJy\right)=\left(x,{H}^{*}y\right)=-\left({J}^{2}x,{H}^{*}y\right)=\left(J{H}^{*}Jx,y\right)$

$H\subset J{H}^{*}J$ 时，对于任意 $x\in D\left(JHJ\right)$$Jx\in D\left(H\right)$ ，所以 $Jx\in D\left(J{H}^{*}J\right)$$x\in D\left({H}^{*}\right)$ ，且

$JHJx=J\left(J{H}^{*}J\right)Jx={H}^{*}x$

${\left(x,y\right)}^{*}=\left(x,y\right)+\left(J{H}^{*}Jx,J{H}^{*}Jy\right),x,y\in D\left(J{H}^{*}J\right)$

$0={\left(x,y\right)}^{*}=\left(x,y\right)+\left(J{H}^{*}Jx,J{H}^{*}Jy\right)=\left(x,y\right)+\left(Hx,J{H}^{*}Jy\right)$

${\left(x,z\right)}^{*}=\left(x,z\right)+\left(J{H}^{*}Jx,J{H}^{*}Jz\right)=\left(x,z\right)+\left(x,{H}^{*}J{H}^{*}Jz\right)=\left(x,z\right)+\left(x,-z\right)=0$

$\left(Hx,Jy\right)=-\left(x,JHy\right)⇔\left(J{H}^{*}Jx,Jy\right)=-\left(x,{J}^{2}{H}^{*}Jy\right)⇔\left({H}^{*}Jx,y\right)=\left(x,{H}^{*}Jy\right)$

$〈x,y〉=\left({H}^{*}Jx,y\right)-\left(x,{H}^{*}Jy\right)=\left(J{H}^{*}Jx,Jy\right)-\left(Hx,Jy\right)=\left(Hx,Jy\right)-\left(Hx,Jy\right)=0$

$x\in D\left(J{H}^{*}J\right)$ ，且对于任意的 $y\in D\left(J{H}^{*}J\right)$ ，有 $〈x,y〉=0$

$\left(x,{H}^{*}Jy\right)=\left({H}^{*}Jx,y\right)=\left(J{H}^{*}Jx,Jy\right)$

$x\in D\left(H\right)$$Hx=J{H}^{*}Jx$

$x={x}_{1}+{x}_{2},{x}_{1}\in D\left(H\right),{x}_{2}\in Ker\left({H}^{*}J{H}^{*}J+I\right)$

${x}_{2}\in K$ ，所以 $D\left(B\right)\subset D\left(H\right){\oplus }^{*}K$ ，即 $D\left(B\right)=D\left(H\right){\oplus }^{*}K$

K的辛共轭子空间。若 $K\subset {K}^{*}$ ，则称K为辛对称的。若 $K={K}^{*}$ ，则称K为辛自伴的。

(i) B为辛对称 $⇔$ K为辛对称；

(ii) B为辛自伴 $⇔$ K为辛自伴。

$〈x+u,y+v〉=〈x,y〉+〈x,u〉+〈u,y〉+〈u,v〉=0$

(ii)当B为辛自伴时， $K\subset {K}^{*}$ ，所以对于任意 $v\in {K}^{*}$$u\in K$ ，有 $〈u,v〉=0$ 。对任意 $x+u\in D\left(B\right)$$x\in D\left(H\right)$ ，由B是辛自伴可知 $〈x+u,v〉=0$ ，即

$\left({H}^{*}J\left(x+u\right),v\right)=\left(x+u,{H}^{*}Jv\right)⇔\left(J{H}^{*}J\left(x+u\right),Jv\right)=\left(x+u,{H}^{*}Jv\right)⇔\left(B\left(x+u\right),Jv\right)=\left(x+u,{H}^{*}Jv\right)$

${H}^{*}Jv={B}^{*}Jv$$v\in D\left(J{B}^{*}J\right)=D\left(B\right)=D\left(H\right){\oplus }^{*}K$ ，因此 $v\in K$ ，所以 $K={K}^{*}$ ，即K为辛自伴。

$\left(B\left(x+u\right),Jz\right)=\left(x+u,{B}^{*}Jz\right)=\left(J\left(x+u\right),J{B}^{*}Jz\right)=\left(J\left(x+u\right),J{H}^{*}Jz\right)=\left(x+u,{H}^{*}Jz\right)$

2.2. 主要结果及证明

$〈{k}_{1}+\alpha {x}_{0},{k}_{2}+\beta {x}_{0}〉=〈{k}_{1},{k}_{2}〉+\alpha 〈{x}_{0},{k}_{2}〉+\stackrel{¯}{\beta }〈{k}_{1},{x}_{0}〉+\alpha \stackrel{¯}{\beta }〈{x}_{0},{x}_{0}〉=0$

3. 无穷维Hamilton算子辛自伴延拓的唯一性

$\left(Jx,Jy\right)=\left(x,y\right),{J}^{2}=-I$ 。当无穷维Hamilton算子H是闭算子时，JH是闭的辛对称算子，故在一定条件下存在辛自伴延拓，本节得出了辛自伴延拓唯一的条件。

3.1. 预备知识

3.2. 主要结果及证明

$B+A{\left(C-iI\right)}^{-1}{A}^{*}$ 是自伴算子时，H存在唯一的辛自伴延拓。

$H=RTL+\left(\begin{array}{cc}0& 0\\ iI& 0\end{array}\right)$

${H}^{*}={\left(RTL\right)}^{*}+{\left(\begin{array}{cc}0& 0\\ iI& 0\end{array}\right)}^{*}=\left(\begin{array}{cc}{A}^{*}& C\\ B& -A\end{array}\right)$

$C+{A}^{*}{\left(B-iI\right)}^{-1}A$ 是自伴算子时，H存在唯一的辛自伴延拓。

(i) $D\left(B\right)\subset D\left({A}^{*}\right)$$D\left(C\right)\subset D\left(A\right)$$‖{A}^{*}{B}^{-1}A{C}^{-1}‖<1$

(ii) $D\left(B\right)\subset D\left({A}^{*}\right)$$D\left(C\right)\subset D\left(A\right)$$‖A{C}^{-1}{A}^{*}{B}^{-1}‖<1$ 时，则H的辛自伴延拓唯一。

$A{x}_{n}^{\left(1\right)}+B{x}_{n}^{\left(2\right)}\to 0,C{x}_{n}^{\left(1\right)}-{A}^{*}{x}_{n}^{\left(2\right)}\to 0$

$\left(C{x}_{n}^{\left(1\right)},{x}_{n}^{\left(1\right)}\right)+\left(B{x}_{n}^{\left(2\right)},{x}_{n}^{\left(2\right)}\right)\to 0$

$\left(B{x}_{n}^{\left(2\right)},{x}_{n}^{\left(2\right)}\right)\to 0,\left(C{x}_{n}^{\left(1\right)},{x}_{n}^{\left(1\right)}\right)\to 0$

${B}^{\frac{1}{2}}{x}_{n}^{\left(2\right)}\to 0,{C}^{\frac{1}{2}}{x}_{n}^{\left(1\right)}\to 0$

$0\in \rho \left(B\right)\cap \rho \left(C\right)$ 时， ${B}^{-1}$${C}^{-1}$ 的平方根算子分别记为 ${B}^{-\frac{1}{2}}$${C}^{-\frac{1}{2}}$

${B}^{\frac{1}{2}}\left({B}^{\frac{1}{2}}{x}_{n}^{\left(2\right)}\right)={x}_{n}^{\left(2\right)}\to 0,{C}^{\frac{1}{2}}\left({C}^{\frac{1}{2}}{x}_{n}^{\left(1\right)}\right)={x}_{n}^{\left(1\right)}\to 0$

(ⅰ)当 $D\left(B\right)\subset D\left({A}^{*}\right)$$D\left(C\right)\subset D\left(A\right)$$‖{A}^{*}{B}^{-1}A{C}^{-1}‖<1$ 时， $\left(I+{A}^{*}{B}^{-1}A{C}^{-1}\right)$ 可逆，于是对于任意

$\left(\begin{array}{c}f\\ g\end{array}\right)\in X×X$ ，取

$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}{C}^{-1}{\left(I+{A}^{*}{B}^{-1}A{C}^{-1}\right)}^{-1}\left(g+{A}^{*}{B}^{-1}f\right)\\ {B}^{*}f-{B}^{-1}A{C}^{-1}{\left(I+{A}^{*}{B}^{-1}A{C}^{-1}\right)}^{-1}\left(g+{A}^{*}{B}^{-1}f\right)\end{array}\right)$ ，则有 $\left(\begin{array}{cc}A& B\\ C& -{A}^{*}\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}f\\ g\end{array}\right)$

(i) $D\left(A\right)\subset D\left(C\right)$$D\left({A}^{*}\right)\subset D\left(B\right)$${A}^{-1}B,{\left({A}^{*}\right)}^{-1}C$ 是有界线性算子， $‖C{A}^{-1}B{\left({A}^{*}\right)}^{-1}‖<1$

(ii) $D\left(B\right)\subset D\left({A}^{*}\right)$$D\left(C\right)\subset D\left(A\right)$${A}^{-1}B,{\left({A}^{*}\right)}^{-1}C$ 是有界线性算子， $‖B{\left({A}^{*}\right)}^{-1}C{A}^{-1}‖<1$ 时，则H的辛自伴延拓唯一。

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