UA OPTI570 量子力学16 含时的哈密顿量与时间演化算符

这一讲讨论随时间变化的哈密顿量决定的量子系统时间演化算符的推导。考虑哈密顿量 H ^ ( t ) \hat H(t) H^(t)，

引理1 ∀ t ≠ t ′ \forall t \ne t' ∀t=t′, [ H ^ ( t ) , H ^ ( t ′ ) ] ≠ 0 [\hat H(t),\hat H(t')] \ne 0 [H^(t),H^(t′)]=0，则 [ H ^ ( t ) , ∫ t 0 t H ^ ( t ′ ) d t ′ ] ≠ 0 [\hat H(t),\int_{t_0}^t \hat H(t')dt'] \ne 0 [H^(t),∫t0tH^(t′)dt′]=0

证明直接计算
[ H ^ ( t ) , ∫ t 0 t H ^ ( t ′ ) d t ′ ] = H ^ ( t ) ∫ t 0 t H ^ ( t ′ ) d t ′ − ( ∫ t 0 t H ^ ( t ′ ) d t ′ ) H ^ ( t ) = ∫ t 0 t H ^ ( t ) H ^ ( t ′ ) d t ′ − ∫ t 0 t H ^ ( t ′ ) H ^ ( t ) d t ′ = ∫ t 0 t ( H ^ ( t ) H ^ ( t ′ ) − H ^ ( t ′ ) H ^ ( t ) ) d t ′ = ∫ t 0 t [ H ^ ( t ) , H ^ ( t ′ ) ] d t ′ \begin{aligned}[\hat H(t),\int_{t_0}^t \hat H(t')dt'] & = \hat H(t) \int_{t_0}^t \hat H(t')dt' - \left( \int_{t_0}^t \hat H(t')dt' \right) \hat H(t) \\ & = \int_{t_0}^t \hat H(t)\hat H(t')dt' - \int_{t_0}^t \hat H(t')\hat H(t) dt' \\ & = \int_{t_0}^t \left( \hat H(t)\hat H(t') - \hat H(t')\hat H(t) \right)dt' \\ & = \int_{t_0}^t [\hat H(t),\hat H(t')]dt' \end{aligned} [H^(t),∫t0tH^(t′)dt′]=H^(t)∫t0tH^(t′)dt′−(∫t0tH^(t′)dt′)H^(t)=∫t0tH^(t)H^(t′)dt′−∫t0tH^(t′)H^(t)dt′=∫t0t(H^(t)H^(t′)−H^(t′)H^(t))dt′=∫t0t[H^(t),H^(t′)]dt′

∀ t ≠ t ′ \forall t \ne t' ∀t=t′, [ H ^ ( t ) , H ^ ( t ′ ) ] ≠ 0 [\hat H(t),\hat H(t')] \ne 0 [H^(t),H^(t′)]=0则上述积分非零，所以
[ H ^ ( t ) , ∫ t 0 t H ^ ( t ′ ) d t ′ ] ≠ 0 [\hat H(t),\int_{t_0}^t \hat H(t')dt'] \ne 0 [H^(t),∫t0tH^(t′)dt′]=0

引理2 构造一个新的算符
F ^ ( t ) = − i ℏ ∫ t 0 t H ^ ( t ′ ) d t ′ \hat F(t)=-\frac{i}{\hbar}\int_{t_0}^t \hat H(t')dt' F^(t)=−ℏi∫t0tH^(t′)dt′

它满足
[ F ^ ( t ) , d d t F ^ ( t ) ] = − 1 ℏ 2 [ ∫ t 0 t H ^ ( t ′ ) d t ′ , H ^ ( t ) ] = 1 ℏ 2 [ H ^ ( t ) , ∫ t 0 t H ^ ( t ′ ) d t ′ ] [\hat F(t),\frac{d}{dt} \hat F(t)]=-\frac{1}{\hbar^2}[\int_{t_0}^t \hat H(t')dt',\hat H(t)]=\frac{1}{\hbar^2}[\hat H(t),\int_{t_0}^t \hat H(t')dt'] [F^(t),dtdF^(t)]=−ℏ21[∫t0tH^(t′)dt′,H^(t)]=ℏ21[H^(t),∫t0tH^(t′)dt′]

并且
[ F ^ ( t ) , d d t F ^ ( t ) ] ≠ 0 ⇒ d d t e F ^ ( t ) ≠ d d t F ^ ( t ) e F ^ ( t ) [ F ^ ( t ) , d d t F ^ ( t ) ] = 0 ⇒ d d t e F ^ ( t ) = d d t F ^ ( t ) e F ^ ( t ) [\hat F(t),\frac{d}{dt} \hat F(t)] \ne 0 \Rightarrow \frac{d}{dt}e^{\hat F(t)} \ne \frac{d}{dt} \hat F(t)e^{\hat F(t)} \\ [\hat F(t),\frac{d}{dt} \hat F(t)] = 0 \Rightarrow \frac{d}{dt}e^{\hat F(t)} = \frac{d}{dt} \hat F(t)e^{\hat F(t)} [F^(t),dtdF^(t)]=0⇒dtdeF^(t)=dtdF^(t)eF^(t)[F^(t),dtdF^(t)]=0⇒dtdeF^(t)=dtdF^(t)eF^(t)

证明用对含参变量的积分求导的结论，
d d t F ^ ( t ) = − i ℏ H ^ ( t ) \frac{d}{dt} \hat F(t) = -\frac{i}{\hbar} \hat H(t) dtdF^(t)=−ℏiH^(t)

于是直接计算
[ F ^ ( t ) , d d t F ^ ( t ) ] = [ F ^ ( t ) , − i ℏ H ^ ( t ) ] = [ − i ℏ ∫ t 0 t H ^ ( t ′ ) d t ′ , − i ℏ H ^ ( t ) ] = − 1 ℏ 2 [ ∫ t 0 t H ^ ( t ′ ) d t ′ , H ^ ( t ) ] = 1 ℏ 2 [ H ^ ( t ) , ∫ t 0 t H ^ ( t ′ ) d t ′ ] \begin{aligned} [\hat F(t),\frac{d}{dt} \hat F(t)]& =[\hat F(t), -\frac{i}{\hbar} \hat H(t)] \\ & = [-\frac{i}{\hbar}\int_{t_0}^t \hat H(t')dt',-\frac{i}{\hbar} \hat H(t)] \\ & = -\frac{1}{\hbar^2}[\int_{t_0}^t \hat H(t')dt',\hat H(t)] \\ & =\frac{1}{\hbar^2}[\hat H(t),\int_{t_0}^t \hat H(t')dt']\end{aligned} [F^(t),dtdF^(t)]=[F^(t),−ℏiH^(t)]=[−ℏi∫t0tH^(t′)dt′,−ℏiH^(t)]=−ℏ21[∫t0tH^(t′)dt′,H^(t)]=ℏ21[H^(t),∫t0tH^(t′)dt′]

最后一个等号是对易子的性质。

接下来计算
d d t e F ^ ( t ) = d d t ∑ n = 0 + ∞ 1 n ! F ^ n ( t ) \begin{aligned} \frac{d}{dt} e^{\hat F(t)} = \frac{d}{dt} \sum_{n=0}^{+\infty} \frac{1}{n!} \hat F^n(t) \end{aligned} dtdeF^(t)=dtdn=0∑+∞n!1F^n(t)

第0项为 d d t F ^ ( t ) \frac{d}{dt} \hat F(t) dtdF^(t)；第1项为
1 2 ( d F ^ ( t ) d t F ^ ( t ) + F ^ ( t ) d F ^ ( t ) d t ) = 1 2 ( 2 d F ^ ( t ) d t F ^ ( t ) + [ F ^ ( t ) , d d t F ^ ( t ) ] ) \frac{1}{2} \left( \frac{d \hat F(t)}{dt} \hat F(t)+\hat F(t)\frac{d \hat F(t)}{dt} \right) = \frac{1}{2} \left( 2 \frac{d \hat F(t)}{dt} \hat F(t)+[\hat F(t),\frac{d}{dt} \hat F(t)] \right) 21(dtdF^(t)F^(t)+F^(t)dtdF^(t))=21(2dtdF^(t)F^(t)+[F^(t),dtdF^(t)])

如果 [ F ^ ( t ) , d d t F ^ ( t ) ] = 0 [\hat F(t),\frac{d}{dt} \hat F(t)]=0 [F^(t),dtdF^(t)]=0，则它等于 d F ^ ( t ) d t F ^ ( t ) \frac{d \hat F(t)}{dt} \hat F(t) dtdF^(t)F^(t)；以此类推，可以验证如果 [ F ^ ( t ) , d d t F ^ ( t ) ] = 0 [\hat F(t),\frac{d}{dt} \hat F(t)]=0 [F^(t),dtdF^(t)]=0，则第 n n n项等于 1 n ! d F ^ ( t ) d t F ^ n ( t ) \frac{1}{n!} \frac{d \hat F(t)}{dt} \hat F^n(t) n!1dtdF^(t)F^n(t)；而根据 [ F ^ ( t ) , d d t F ^ ( t ) ] [\hat F(t),\frac{d}{dt} \hat F(t)] [F^(t),dtdF^(t)]的表达式与引理1， [ F ^ ( t ) , d d t F ^ ( t ) ] = 0 [\hat F(t),\frac{d}{dt} \hat F(t)]=0 [F^(t),dtdF^(t)]=0等价于 ∀ t ≠ t ′ \forall t \ne t' ∀t=t′, [ H ^ ( t ) , H ^ ( t ′ ) ] = 0 [\hat H(t),\hat H(t')] = 0 [H^(t),H^(t′)]=0。

薛定谔时间演化算符的公式 假设 [ H ^ ( t ) , H ^ ( t ′ ) ] = 0 , ∀ t ′ , t [\hat H(t),\hat H(t')]=0,\forall t',t [H^(t),H^(t′)]=0,∀t′,t（在这个条件下可以得到时间演化算符的解析式，如果这个条件不满足，就需要解薛定谔方程了），记 U ^ ( t 0 , t ) \hat U(t_0,t) U^(t0,t)为时间演化算符，它满足
∣ ψ ( t ) ⟩ = U ^ ( t , t 0 ) ∣ ψ ( t 0 ) ⟩ |\psi(t) \rangle = \hat U(t,t_0)|\psi(t_0) \rangle ∣ψ(t)⟩=U^(t,t0)∣ψ(t0)⟩

薛定谔方程为
i ℏ d d t ∣ ψ ( t ) ⟩ = H ^ ( t ) ∣ ψ ( t ) ⟩ i\hbar \frac{d}{dt}|\psi(t) \rangle = \hat H(t) |\psi(t) \rangle iℏdtd∣ψ(t)⟩=H^(t)∣ψ(t)⟩

代入 ∣ ψ ( t ) ⟩ |\psi(t) \rangle ∣ψ(t)⟩用时间演化算符的表示，
i ℏ d d t U ^ ( t , t 0 ) = H ^ ( t ) U ^ ( t , t 0 ) i\hbar \frac{d}{dt}\hat U(t,t_0) = \hat H(t) \hat U(t,t_0) iℏdtdU^(t,t0)=H^(t)U^(t,t0)

由此可以解出时间演化算符的公式为
U ^ ( t , t 0 ) = exp ⁡ ( − i ℏ ∫ t 0 t H ^ ( t ′ ) d t ′ ) \hat U(t,t_0) = \exp \left( - \frac{i}{\hbar} \int_{t_0}^t \hat H(t')dt' \right) U^(t,t0)=exp(−ℏi∫t0tH^(t′)dt′)

证明要验证 U ^ ( t , t 0 ) \hat U(t,t_0) U^(t,t0)满足薛定谔方程只需要将其代入验证即可，对于
U ^ ( t , t 0 ) = exp ⁡ ( − i ℏ ∫ t 0 t H ^ ( t ′ ) d t ′ ) = e F ^ ( t ) \hat U(t,t_0) = \exp \left( - \frac{i}{\hbar} \int_{t_0}^t \hat H(t')dt' \right)=e^{\hat F(t)} U^(t,t0)=exp(−ℏi∫t0tH^(t′)dt′)=eF^(t)

计算
i ℏ d d t U ^ ( t , t 0 ) = i ℏ d d t e F ^ ( t ) = i ℏ d d t F ^ ( t ) e F ^ ( t ) i\hbar \frac{d}{dt}\hat U(t,t_0) = i\hbar \frac{d}{dt}e^{\hat F(t)} = i\hbar \frac{d}{dt} \hat F(t)e^{\hat F(t)} iℏdtdU^(t,t0)=iℏdtdeF^(t)=iℏdtdF^(t)eF^(t)

在引理2的证明中，我们得到了
d d t F ^ ( t ) = − i ℏ H ^ ( t ) \frac{d}{dt} \hat F(t) = -\frac{i}{\hbar} \hat H(t) dtdF^(t)=−ℏiH^(t)

所以
i ℏ d d t F ^ ( t ) e F ^ ( t ) = H ^ ( t ) e F ^ ( t ) = H ^ ( t ) U ^ ( t , t 0 ) i\hbar \frac{d}{dt} \hat F(t)e^{\hat F(t)} = \hat H(t)e^{\hat F(t)} = \hat H(t) \hat U(t,t_0) iℏdtdF^(t)eF^(t)=H^(t)eF^(t)=H^(t)U^(t,t0)

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