問題 7-9 の解答例

Problem 7-9
Use this result Eq. (7-68) to show that if $S$ corresponds to a harmonic oscillator

\begin{matrix} (1) & S = \frac{m}{2} \int_{t_{1}}^{t_{2}} [{\dot{x}}^{2} - ω^{2} x^{2}] d t \end{matrix}

then

\begin{aligned} ⟨ \exp {\frac{i}{ℏ} \int_{t_{1}}^{t_{2}} f (t) x (t) d t} ⟩ \\ = ⟨ 1 ⟩ \exp {\frac{i}{ℏ} \frac{m ω}{2 \sin ω (t_{2} - t_{1})} [\frac{2 x_{2}}{m w} \int_{t_{1}}^{t_{2}} f (t) \sin ω (t - t_{1}) d t \\ + \frac{2 x_{1}}{m ω} \int_{t_{1}}^{t_{2}} f (t) \sin ω (t_{2} - t) d t \\ (2) & - \frac{2}{m^{2} ω^{2}} \int_{t_{1}}^{t_{2}} d t \int_{t_{1}}^{t} f (t) f (s) \sin ω (t_{2} - t) \sin ω (s - t_{1}) d s]} \end{aligned}

where

x_{1}

and

x_{2}

are the initial and final coordinates of the oscillator.

( 解答 ) 調和振動子の古典的経路に対する作用 $S_{c l}$ は, 問題 2-2 中の式 (2-9) で与えれている：

\begin{matrix} (3) & S_{c l} = \frac{m ω}{2 \sin ω T} [(x_{2}^{2} + x_{1}^{2}) \cos ω T - 2 x_{2} x_{1}] \end{matrix}

ただし

T = t_{2} - t_{1}

である．また, 外力が働いている調和振動子の古典的経路に対する作用

S_{c l}^{'}

は, 問題 3-11 中の式 (3-66) で与えれている：

\begin{aligned} S_{c l}^{'} = \frac{m ω}{2 \sin ω T} & [(x_{2}^{2} + x_{1}^{2}) \cos ω T - 2 x_{2} x_{1} \\ + \frac{2 x_{2}}{m ω} \int_{t_{1}}^{t_{2}} d t f (t) \sin ω (t - t_{1}) \\ + \frac{2 x_{1}}{m ω} \int_{t_{1}}^{t_{2}} d t f (t) \sin ω (t_{2} - t) \\ (4) & - \frac{2}{m^{2} ω^{2}} \int_{t_{1}}^{t_{2}} d t \int_{t_{1}}^{t} d s f (t) f (s) \sin ω (t_{2} - t) \sin ω (s - t_{1})] \end{aligned}

ただし表記を

a \to 1, b \to 2

に変更した．式 (4) から式 (3) を引き算することで

(i / ℏ) {S_{c l}^{'} - S_{c l}}

を求めると,

\begin{aligned} \frac{i}{ℏ} {S_{c l}^{'} - S_{c l}} & = \frac{i}{ℏ} \frac{m ω}{2 \sin ω T} [\frac{2 x_{2}}{m ω} \int_{t_{1}}^{t_{2}} d t f (t) \sin ω (t - t_{1}) \\ + \frac{2 x_{1}}{m ω} \int_{t_{1}}^{t_{2}} d t f (t) \sin ω (t_{2} - t) \\ (5) & - \frac{2}{m^{2} ω^{2}} \int_{t_{1}}^{t_{2}} d t \int_{t_{1}}^{t} d s f (t) f (s) \sin ω (t_{2} - t) \sin ω (s - t_{1})] \end{aligned}

この結果式 (5) を式 (7-68) に代入すれば, 題意の式 (2) が得られる：

\begin{aligned} (7-68) & ⟨ \exp {\frac{i}{ℏ} \int f (t) x (t) d t} ⟩ = ⟨ 1 ⟩ \exp {\frac{i}{ℏ} (S_{c l}^{‘} - S_{c l})} \\ = ⟨ 1 ⟩ \exp {\frac{i}{ℏ} \frac{m ω}{2 \sin ω (t_{2} - t_{1})} [\frac{2 x_{2}}{m w} \int_{t_{1}}^{t_{2}} f (t) \sin ω (t - t_{1}) d t \\ + \frac{2 x_{1}}{m ω} \int_{t_{1}}^{t_{2}} f (t) \sin ω (t_{2} - t) d t \\ (6) & - \frac{2}{m^{2} ω^{2}} \int_{t_{1}}^{t_{2}} d t \int_{t_{1}}^{t} f (t) f (s) \sin ω (t_{2} - t) \sin ω (s - t_{1}) d s]} \end{aligned}

ただし

T = t_{2} - t_{1}

である．