東京大学 新領域創成科学研究科 物質系専攻 2019年8月実施 第3問

标签：

• Tokyo-University
• Creation-and-Annihilation-Operators
• Harmonic-Oscillator-Ground-State
• Harmonic-Oscillator-First-Excited-State
• First-Order-Time-Dependent-Perturbation-Transition-Probability
• Quantum-Observable-Expectation-and-Variance

Author

Miyake

Description

Kai

(1)

$$\begin{aligned} A^\dagger A &= \frac{m \omega}{2 \hbar} \left( x - \frac{ip}{m \omega} \right) \left( x + \frac{ip}{m \omega} \right) \\ &= \frac{m \omega}{2 \hbar} \left( x^2 + \frac{p^2}{m^2 \omega^2} + \frac{i}{m \omega}[x,p] \right) \\ &= \frac{m \omega}{2 \hbar} \left( x^2 + \frac{p^2}{m^2 \omega^2} - \frac{\hbar}{m \omega} \right) \\ &= \frac{1}{\hbar \omega} \left( \frac{p^2}{2 m} + \frac{1}{2} m \omega^2 x^2 \right) - \frac{1}{2} \\ &= \frac{1}{\hbar \omega} H - \frac{1}{2} \\ \therefore \ \ H &= \hbar \omega \left( A^\dagger A + \frac{1}{2} \right) \end{aligned}$$

(2)

$$\begin{aligned} H \varphi_0 &= \hbar \omega \left( A^\dagger A + \frac{1}{2} \right) \varphi_0 \\ &= \frac{1}{2} \hbar \omega \varphi_0 \\ \therefore \ \ E_0 &= \frac{1}{2} \hbar \omega \end{aligned}$$

(3)

$$\begin{aligned} \left[ A, A^\dagger \right] &= \frac{m \omega}{2 \hbar} \left[ x + \frac{ip}{m \omega}, x - \frac{ip}{m \omega} \right] \\ &= \frac{m \omega}{2 \hbar} \frac{i}{m \omega} \left\{ -[x,p] + [p,x] \right\} \\ &= \frac{i}{2 \hbar} \cdot (-2i \hbar) \\ &= 1 \\ H \varphi_1 &= \hbar \omega \left( A^\dagger A + \frac{1}{2} \right) A^\dagger \varphi_0 \\ &= \hbar \omega \left( A^\dagger (A^\dagger A + 1) + \frac{1}{2} A^\dagger \right) \varphi_0 \\ &= \frac{3}{2} \hbar \omega A^\dagger \varphi_0 \\ &= \frac{3}{2} \hbar \omega \varphi_1 \\ \therefore \ \ E_1 &= \frac{3}{2} \hbar \omega \end{aligned}$$

(4)

$\left\langle \varphi_0 \right| A \left| \varphi_0 \right\rangle = 0$ で、かつ、

$$\begin{aligned} \left\langle \varphi_{n+1} \right| A \left| \varphi_{n+1} \right\rangle &= \frac{1}{n+1} \left\langle \varphi_n \right| A A A^\dagger \left| \varphi_n \right\rangle \\ &= \frac{1}{n+1} \left\langle \varphi_n \right| A ( A^\dagger A + 1 ) \left| \varphi_n \right\rangle \\ &= \frac{1}{n+1} \left\langle \varphi_n \right| A \left( \frac{H}{\hbar \omega} - \frac{1}{2} + 1 \right) \left| \varphi_n \right\rangle \\ &= \frac{1}{n+1} \left( \frac{E_n}{\hbar \omega} + \frac{1}{2} \right) \left\langle \varphi_n \right| A \left| \varphi_n \right\rangle \end{aligned}$$

であるから、数学的帰納法により、 $n=0,1,2, \cdots$ に対して、

$$\begin{aligned} \left\langle \varphi_n \right| A \left| \varphi_n \right\rangle = 0 \end{aligned}$$

が成り立つ。

同様にして、

$$\begin{aligned} \left\langle \varphi_n \right| A^\dagger \left| \varphi_n \right\rangle = 0 \end{aligned}$$

も成り立つ。

さらに、

$$\begin{aligned} x = \sqrt{\frac{\hbar}{2m \omega}} \left( A + A^\dagger \right) \end{aligned}$$

であるから、

$$\begin{aligned} \left\langle \varphi_n \right| x \left| \varphi_n \right\rangle = 0 \end{aligned}$$

が $n=0,1,2, \cdots$ について成り立つ。

(5)

まず、

$$\begin{aligned} \left\langle \varphi_1 \right| V \left| \varphi_0 \right\rangle &= F \left\langle \varphi_0 \right| A x \left| \varphi_0 \right\rangle \\ &= F \sqrt{\frac{\hbar}{2m \omega}} \left\langle \varphi_0 \right| A \left( A + A^\dagger \right) \left| \varphi_0 \right\rangle \\ &= F \sqrt{\frac{\hbar}{2m \omega}} \left\langle \varphi_0 \right| A A^\dagger \left| \varphi_0 \right\rangle \\ &= F \sqrt{\frac{\hbar}{2m \omega}} \left\langle \varphi_0 \right| \left(A^\dagger A + 1 \right) \left| \varphi_0 \right\rangle \\ &= F \sqrt{\frac{\hbar}{2m \omega}} \end{aligned}$$

であるから、

$$\begin{aligned} c_1(t) &= \frac{F}{i \hbar} \sqrt{\frac{\hbar}{2m \omega}} \int_0^t e^{i \omega t'} dt' \\ &= \frac{F}{i \hbar} \sqrt{\frac{\hbar}{2m \omega}} \left[ \frac{e^{i \omega t'}}{i \omega} \right]_0^t \\ &= - \frac{F}{\hbar \omega} \sqrt{\frac{\hbar}{2m \omega}} \left( e^{i \omega t} - 1 \right) \\ &= \frac{F}{\sqrt{2 \hbar \omega^3 m}} \left( 1 - e^{i \omega t} \right) \end{aligned}$$

を得る。

さらに、

$$\begin{aligned} \left| e^{-i \frac{E_0}{\hbar} t} \right|^2 &= 1 \\ \left| c_1(t) e^{-i \frac{E_1}{\hbar} t} \right|^2 &= \frac{F^2}{2 \hbar \omega^3 m} \left( 1 - e^{i \omega t} \right) \left( 1 - e^{- i \omega t} \right) \\ &= \frac{F^2}{\hbar \omega^3 m} \left( 1 - \cos \omega t \right) \end{aligned}$$

であるから、時刻 $t$ において状態 $\varphi_1$ に見いだされる確率は、

$$\begin{aligned} \frac{\frac{F^2}{\hbar \omega^3 m} \left( 1 - \cos \omega t \right)} { 1 + \frac{F^2}{\hbar \omega^3 m} \left( 1 - \cos \omega t \right)} \end{aligned}$$

である。

(6)

(3), (5) より

$$\begin{aligned} \left\langle \varphi_0 \right| x \left| \varphi_0 \right\rangle &= \left\langle \varphi_1 \right| x \left| \varphi_1 \right\rangle = 0 \\ \left\langle \varphi_0 \right| x \left| \varphi_1 \right\rangle &= \left\langle \varphi_1 \right| x \left| \varphi_0 \right\rangle = \sqrt{\frac{\hbar}{2m \omega}} \end{aligned}$$

であるから、

$$\begin{aligned} \left\langle \psi \right| x \left| \psi \right\rangle &= c_1^\ast (t) e^{-i \frac{E_0-E_1}{\hbar} t} \left\langle \varphi_1 \right| x \left| \varphi_0 \right\rangle + c_1 (t) e^{-i \frac{E_1-E_0}{\hbar} t} \left\langle \varphi_0 \right| x \left| \varphi_1 \right\rangle \\ &= \frac{F}{\sqrt{2 \hbar \omega^3 m}} \left( 1 - e^{- i \omega t} \right) e^{i \omega t} \sqrt{\frac{\hbar}{2m \omega}} + \frac{F}{\sqrt{2 \hbar \omega^3 m}} \left( 1 - e^{i \omega t} \right) e^{- i \omega t} \sqrt{\frac{\hbar}{2m \omega}} \\ &= \frac{F}{2 m \omega^2} \left\{ \left( e^{i \omega t} - 1 \right) + \left( e^{- i \omega t} - 1 \right) \right\} \\ &= \frac{F}{m \omega^2} \left( \cos \omega t - 1 \right) \end{aligned}$$

を得る。