态与表象

如果体系的状态使用坐标 $\left(x,y,z\right)$ 的函数来描述, 即描写状态的波函数是坐标的函数. 但是这种表述方式并不唯一, 波函数也可以选用其他变量的函数.

态和力学量的具体表现方式称为表象.

Ehrenfest关系

波函数与力学量算符无法被观测, 观测到的是力学量的各种取值的概率分布与均值, 以及其随时间的演化. 这是对运动规律进行不同图像描述的依据.

由 $i\hbar\frac{\partial}{\partial{t}}f=Hf$ , 力学量 $\hat{A}$ 的均值为 $\langle{\hat{A}}\rangle=\langle{f}\vert{A}\vert{f}\rangle$. 所以有

$$\begin{equation} \begin{split} \frac{d}{dt}\langle{A}\rangle & = \langle{\frac{df}{dt}}\vert{A}\vert{f}\rangle + \langle{f}\vert{\frac{dA}{dt}}\vert{f}\rangle + \langle{f}\vert{A}\vert{\frac{df}{dt}}\rangle \\ & = -\frac{1}{i\hbar}\langle{Hf}\vert{A}\vert{f}\rangle + \langle{\dot{A}}\rangle + \frac{1}{i\hbar}\langle{f}\vert{A}\vert{Hf}\rangle \\ & = -\frac{1}{i\hbar}\langle{f}\vert{HA}\vert{f}\rangle + \frac{1}{i\hbar}\langle{f}\vert{AH}\vert{f}\rangle + \langle{\dot{A}}\rangle \\ & = \frac{1}{i\hbar}\langle{\left[A,H\right]}\rangle + \langle{\dot{A}}\rangle \\ \end{split} \end{equation}$$

如果 $A$ 不显含时间, 有

$$\begin{equation} \frac{d}{dt}\langle{A}\rangle = \frac{1}{i\hbar}\langle{\left[A,H\right]}\rangle \end{equation}$$

上式为 Ehrenfest关系.

若 $A$ 不显含时间 $\frac{\partial A}{\partial t}=0$, 且如果 $A$ 与 $H$ 对易 $\left[A,H\right]=0$, 则 $\frac{d}{dt}\langle{A}\rangle=0$. 算符的均值不随时间变化, 属于含时不变量.

由于 $A$ 与 $H$ 对易(同时对角化), 他们具有同一套特征向量(完全集) $\left\{\phi_n\vert n=1,2,\cdots\right\}$, $H\phi_n=\epsilon_n\phi_n$, $A\phi_n=A_n\phi_n$. 系统的状态由该完全集展开

$$\begin{equation} \begin{split} f(t) & = \sum_n \vert{\phi_n}\rangle\langle{\phi_n}\vert f(t)\rangle \\ \vert{a_n(t)}\vert^2 & = a^\ast_n(t) a_n(t) \\ \frac{d}{dt}\vert{a_n(t)}\vert^2 & = \frac{d}{dt}a^\ast_n a_n = \frac{da^\ast_n}{dt}a_n + a^\ast_n\frac{da_n}{dt} \\ & = \frac{d}{dt}\langle{\phi_n}\vert{f}\rangle^\ast\langle{\phi_n}\vert{f}\rangle + \langle{\phi_n}\vert{f}\rangle^\ast\frac{d}{dt}\langle{\phi_n}\vert{f}\rangle \\ & = \frac{d}{dt}\langle{f}\vert{\phi_n}\rangle\langle{\phi_n}\vert{f}\rangle + \langle{f}\vert{\phi_n}\rangle\frac{d}{dt}\langle{\phi_n}\vert{f}\rangle \\ & = \langle{\frac{df}{dt}}\vert{\phi_n}\rangle \langle{\phi_n}\vert{f}\rangle + \langle{f}\vert{\phi_n}\rangle \langle{\phi_n}\vert{\frac{df}{dt}}\rangle \\ & = \langle{\frac{1}{i\hbar}Hf}\vert{\phi_n}\rangle \langle{\phi_n}\vert{f}\rangle + \langle{f}\vert{\phi_n}\rangle \langle{\phi_n}\vert{\frac{1}{i\hbar}Hf}\rangle \\ & = -\frac{1}{i\hbar}\langle{f}\vert{H\phi_n}\rangle \langle{\phi_n}\vert{f}\rangle + \langle{f}\vert{\phi_n}\rangle \frac{1}{i\hbar}\langle{\phi_n}\vert{Hf}\rangle \\ & = -\frac{1}{i\hbar}\epsilon_n \langle{f}\vert{\phi_n}\rangle \langle{\phi_n}\vert{f}\rangle + \langle{f}\vert{\phi_n}\rangle \frac{1}{i\hbar}\epsilon^\ast_n\langle{\phi_n}\vert{f}\rangle \\ & = \frac{1}{i\hbar}\epsilon_n \left(\vert{a_n}\vert^2 - \vert{a_n}\vert^2\right) \\ & = 0 \\ \end{split} \end{equation}$$

也就是说, 均值与均值的概率分布都不随时间变化.

Schrödinger picture

$f(t)=U(t,0)f(0)$, $i\hbar\frac{\partial}{\partial t}f(t)=Hf(t)$, then

$$\begin{equation} \begin{split} & i\hbar\frac{\partial}{\partial t}U(t)f(0) = HU(t)f(0) \\ &\left(i\hbar\frac{\partial}{\partial t}U(t) - HU(t)\right)f(0) = 0 \\ & i\hbar\frac{\partial}{\partial t}U(t) = HU(t) \\ \end{split} \end{equation}\label{a}$$

如果 $H$ 不含时, 有 $U(t)=e^{-i\frac{H}{\hbar}t}$.

如果 $H$ 含时, 有 $f(0)=U(0)f(0)\Rightarrow U(0)=1$, and $U(t) = e^{-\frac{i}{\hbar}\int_0^t H(t^\prime)dt^\prime}$.

$U^\dagger U=I$. $\vert{f(t)}\vert^2=f^\dagger(t)f(t)=\left(U(t)f(0)\right)^\dagger U(t)f(0)=f^\dagger(0)U^\dagger(t)U(t)f(0)=\vert{f(0)}\vert^2$. 也就是说 $U^\dagger=U$, 且在 $U$ 变换下函数的模长不变, 故 $U$ 是酉的.

均值 $\langle{A(t)}\rangle=\langle{f(t)}\vert{A}\vert{f(t)}\rangle$. 这里的算符 $\hat{A}$ 不含时, 均值的时间依赖性来源于描写态的波函数 $f(t)$.

$$\begin{equation} \begin{split} & \frac{d}{dt}\langle{A(t)}\rangle=\frac{d}{dt}\langle{f(t)}\vert{A}\vert{f(t)}\rangle \\ & = \langle{\frac{1}{i\hbar}Hf}\vert{A}\vert{f}\rangle + \langle{f}\vert{A}\vert{\frac{1}{i\hbar}Hf}\rangle \\ & = -\frac{1}{i\hbar}\langle{f}\vert{HA}\vert{f}\rangle + \frac{1}{i\hbar}\langle{f}\vert{AH}\vert{f}\rangle \\ & = \frac{1}{i\hbar}\langle{\left[A,H\right]}\rangle \end{split} \end{equation}$$

考虑到波函数 $f(t)=U(t)f(0)$, 算符 $\hat{A}$ 均值的含时性可以归到算符的时间依赖性上

$$\begin{equation} \begin{split} & \langle{A(t)}\rangle = \langle{f(t)}\vert{A}\vert{f(t)}\rangle \\ & = \langle{U(t)f(0)}\vert{A}\vert{U(t)f(0)}\rangle \\ & = \langle{f(0)}\vert{U^\dagger(t)AU(t)}\vert{f(0)}\rangle \\ & = \langle{f(0)}\vert{A(t)}\vert{f(0)}\rangle \\ & A(t) = U^\dagger(t)AU(t) \\ \end{split} \end{equation}$$

$$\begin{equation} \begin{split} & f_S(0) = f_H \\ & A_H(t) = U^\dagger(t)AU(t) = e^{i\frac{H}{\hbar}t}A e^{-i\frac{H}{\hbar}t} \\ & A_H(0) = A_S \end{split} \end{equation}$$

对海森堡图像下含时的算符对时间取微分得到

$$\begin{equation} \begin{split} & \frac{d}{dt}A(t) = \frac{d}{dt}U^\dagger(t)AU(t) \\ & = \frac{dU^\dagger}{dt}AU + U^\dagger A\frac{dU}{dt} \\ \end{split} \end{equation}$$

注意到 $\eqref{a}$, 故

$$\begin{equation} \begin{split} & \frac{d}{dt}A(t) = \left(\frac{1}{i\hbar}HU\right)^\dagger AU + U^\dagger A\frac{1}{i\hbar}HU \\ & = -\frac{1}{i\hbar}U^\dagger HAU + \frac{1}{i\hbar}U^\dagger AHU \\ & \left[U,H\right] = 0 \\ & \Rightarrow \frac{dA(t)}{dt} = \frac{1}{i\hbar}\left(U^\dagger AUH - HU^\dagger AU\right) \\ & A_H(t) = U^\dagger(t) A_S U(t) \\ & \frac{dA(t)}{dt} = \frac{1}{i\hbar}\left(A_H(t)H - HA_H(t)\right) \\ & \frac{dA_H(t)}{dt} = \frac{1}{i\hbar}\left[A_H(t), H\right] \\ \end{split} \end{equation}$$

上式是海森堡运动方程. $f_H(t) = f_H(0) = f_S(0)$, $f_H(t) = e^{i\frac{H}{\hbar}t}f_S(t)$, $\frac{\partial}{\partial t}f_H(t) = \frac{iH}{\hbar}e^{i\frac{iHt}{\hbar}}f_S(t) + e^{\frac{iHt}{\hbar}}\frac{\partial f_S(t)}{\partial t} = e^{\frac{iHt}{\hbar}}\left[\frac{iH}{\hbar}f_S(t) + \frac{1}{i\hbar}Hf_S(t)\right]=0$, 即 $\frac{\partial}{\partial t}f_H(t)=0$.

Interaction picture

$H(t) = H_0 + H^\prime(t)$. Let $f_I(t) = e^{\frac{iH_0}{\hbar}t}f_S(t)$, and $f_S(t) = e^{-\frac{iH_0}{\hbar}t}f_I(t)$. The time evolution is governed by

$$\begin{equation} \begin{split} & i\hbar\frac{\partial}{\partial t}f_I(t) = i\hbar\frac{\partial} {\partial t}\left(e^{i\frac{H_0}{\hbar}t}f_S(t)\right) \\ & = i\hbar\frac{iH_0}{\hbar}e^{\frac{iH_0}{\hbar}t}f_S(t) + e^{i\frac{H_0}{\hbar}t} \left(H_0 + H^\prime(t)\right)f_S(t) \\ & = e^{i\frac{H_0}{\hbar}t}H^\prime(t)f_S(t) \\ & = e^{i\frac{H_0}{\hbar}t}H^\prime e^{-i\frac{H_0}{\hbar}t} e^{i\frac{H_0}{\hbar}t} f_S(t) \\ & = H^\prime_I(t) f_I(t) \\ \end{split} \end{equation}$$

meaning $i\hbar\frac{\partial}{\partial t}f_I(t) = H^\prime_I(t) f_I(t)$. Operator under interaction picture is $A_I(t) = e^{i\frac{H_0}{\hbar}t} A_S e^{-i\frac{H_0}{\hbar}t}$, its time evolution is

$$\begin{equation} \begin{split} & \frac{d}{dt}A_I(t) = \frac{d}{dt}\left(e^{i\frac{H_0}{\hbar}t} A_S e^{-i\frac{H_0}{\hbar}t}\right) \\ & = \frac{iH_0}{\hbar}e^{i\frac{H_0}{\hbar}t} A_S e^{-i\frac{H_0}{\hbar}t} - \frac{i}{\hbar} e^{i\frac{H_0}{\hbar}t} A_S H_0e^{-i\frac{H_0}{\hbar}t} \\ & = \frac{iH_0}{\hbar}A_I(t) - \frac{i}{\hbar} e^{i\frac{H_0}{\hbar}t} A_S e^{-i\frac{H_0}{\hbar}t}H_0 \\ & = \frac{i}{\hbar}H_0 A_I(t) - \frac{i}{\hbar} A_I(t)H_0 \\ & = \frac{i}{\hbar}\left[H_0, A_I(t)\right] \\ \end{split} \end{equation}$$

算符的时间演化由 $H_0$ 决定, 而波函数 $\vert{f}\rangle$ 的时间演化由 $H^\prime_I(t)$ 决定.

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