盒子
盒子
文章目录
  1. 可观测量
    1. 动量算符
    2. Hermite 共轭
    3. 算符

量子物理笔记 - 11:可观测量与算符

可观测量

可观测量 $Q(x,p)$ 的期望(要求为实数)为:
$$
\begin{aligned}
& \langle Q \rangle=\int \psi^* \hat{Q}\psi dx=\langle\psi|\hat{Q}\psi\rangle \in \mathbb{R} \\
\Rightarrow
& \langle Q \rangle=\langle Q \rangle^* \\
\Rightarrow
& \langle\psi|\hat{Q}\psi\rangle=\langle\hat{Q}\psi|\psi\rangle
\end{aligned}
$$
可以看到,可观测量的算符有一些特殊性质(Hermite 共轭),即:
$$
\hat{Q}=\hat{Q}^\dagger
$$

动量算符

动量算符为:
$$
\hat{p}=-i\hbar \nabla
$$
考虑 $f,g\in L^2$,内积为:
$$
\begin{aligned}
\langle f|\hat{p}g\rangle
&=-i\hbar\int_ {-\infty}^{\infty}f^* \nabla gdx \\
&=-i\hbar\left(\underbrace{(f^* g)|_ {-\infty}^{\infty}}_ {0}-\int_ {-\infty}^{\infty}g\nabla f^* dx\right) \\
&=\int_ {-\infty}^{\infty}i\hbar \nabla f^* gdx \\
&=\int_ {-\infty}^{\infty}(-i\hbar \nabla f)^* gdx \\
&=\langle \hat{p}f|g\rangle
\end{aligned}
$$
即动量算符是 Hermite 共轭的。

Hermite 共轭

实际上,若对于任意 $h \in L^2$,都有:
$$
\langle h|\hat{Q}h\rangle=\langle \hat{Q}h|h \rangle
$$
则算符 $\hat{Q}$ 对任意 $f,g \in L^2$ 都是 Hermite 共轭的。

考虑构造 $h=f+cg$,则:
$$
\begin{aligned}
\langle h|\hat{Q}h \rangle
&=\langle f+cg | \hat{Q}f+c\hat{Q}g\rangle \\
&=\langle f|\hat{Q}f \rangle+c^* \langle g|\hat{Q}f \rangle
+c\langle f|\hat{Q}g\rangle+c^* c\langle g|\hat{Q}g \rangle, \quad c^* c=|c|^2 \\
&=\langle\hat{Q}h|h\rangle\\
&=\langle \hat{Q}f+c\hat{Q}g|f+cg\rangle \\
&=\langle \hat{Q}f|f\rangle+c^* \langle\hat{Q}g|f\rangle
+c\langle\hat{Q}f|g\rangle+c^* c\langle\hat{Q}g|g\rangle,\quad c^* c=|c|^2 \\
&=\langle f|\hat{Q}f\rangle+c^* \langle\hat{Q}g|f\rangle
+c\langle\hat{Q}f|g\rangle+c^* c\langle g|\hat{Q}g\rangle
\end{aligned}
$$
分别取 $c=1,i$,得:
$$
\begin{aligned}
& \begin{cases}
\langle f|\hat{Q}g \rangle+\langle g|\hat{Q}f \rangle
=\langle \hat{Q}f|g \rangle+\langle \hat{Q}g|f \rangle, \quad & c=1 \\
\langle f|\hat{Q}g \rangle-\langle g|\hat{Q}f \rangle
=\langle \hat{Q}f|g \rangle-\langle \hat{Q}g|f \rangle, \quad & c=i \\
\end{cases} \\
\Rightarrow
& \langle f|\hat{Q} g\rangle=\langle\hat{Q}f|g\rangle
\end{aligned}
$$

算符

算符 $\hat{Q}$ 的伴随算符是 $\hat{Q}^\dagger$,即:
$$
\langle f|\hat{Q}g\rangle=\langle\hat{Q}^\dagger f|g \rangle
$$
Hermite 算符是自伴算符,即:
$$
\hat{Q}=\hat{Q}^\dagger
$$
考虑算符乘积的伴随,即:
$$
\begin{aligned}
\langle f|\hat{Q}\hat{R} g \rangle
&=\langle \hat{Q}^\dagger f|\hat{R}g\rangle \\
&=\langle \hat{R}^\dagger \hat{Q}^\dagger f | g \rangle \\
&=\langle (\hat{Q}\hat{R})^\dagger f|g \rangle
\end{aligned}
$$
即:
$$
(\hat{Q}\hat{R})^\dagger=\hat{R}^\dagger\hat{Q}^\dagger
$$
进一步的,若 $\hat{Q}\hat{R}$ 是 Hermition 的,则:
$$
\begin{aligned}
& \hat{Q}\hat{R}=\hat{R}^\dagger\hat{Q}^\dagger=\hat{R}\hat{Q} \\
\Rightarrow
& [\hat{Q},\hat{R}]=0
\end{aligned}
$$
即 $\hat{Q},\hat{R}$ 是对易的。

一些常见的伴随算符:
$$
\begin{aligned}
& (f(x))^\dagger=f^* (x) \\
& \left(\frac{d}{dx}\right)^\dagger=-\frac{d}{dx} \\
& a_ {+}^\dagger=a_ {-}
\end{aligned}
$$

反厄密算符等于它的负的厄密共轭:
$$
\hat{Q}^\dagger=-\hat{Q}
$$
(1)反厄密算符的期望值是虚数。
$$
\begin{aligned}
\langle Q \rangle
&=\langle\psi|\hat{Q}\psi\rangle \\
&=\langle\hat{Q}^\dagger\psi|\psi\rangle \\
&=-\langle\hat{Q}\psi|\psi\rangle \\
&=-(\langle\psi|\hat{Q}\psi\rangle)^* \\
&=-(\langle Q \rangle)^*
\end{aligned}
$$
因此:
$$
\Re(\langle Q \rangle)=\frac{1}{2}\Re(\langle Q \rangle+\langle Q \rangle^* )=0
$$
所以 $\langle Q \rangle $ 是虚数。

(2)两个厄密算符的对易子是反厄密的。
$$
\begin{aligned}
\left[\hat{P},\hat{Q}\right]^\dagger
&=(\hat{P}\hat{Q})^\dagger-(\hat{Q}\hat{P})^\dagger \\
&=\hat{Q}^\dagger\hat{P}^\dagger-\hat{P}^\dagger\hat{Q}^\dagger \\
&=\hat{Q}\hat{P}-\hat{P}\hat{Q} \\
&=-[\hat{P},\hat{Q}]
\end{aligned}
$$
因此厄米算符的对易子是虚数。

(3)两个反厄密算符的对易子是反厄密的。
$$
\begin{aligned}
\left[\hat{P},\hat{Q}\right]^\dagger
&=(\hat{P}\hat{Q})^\dagger-(\hat{Q}\hat{P})^\dagger \\
&=\hat{Q}^\dagger\hat{P}^\dagger-\hat{P}^\dagger\hat{Q}^\dagger \\
&=(-\hat{Q})(-\hat{P})-(-\hat{P})(-\hat{Q}) \\
&=-[\hat{P},\hat{Q}]
\end{aligned}
$$
因此反厄米算符的对易子是虚数。

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