問題 8-1 の解答例

Problem 8-1
Note that the amplide to go from any state $f (x)$ to another state $g (x)$ is the transition
amplitude $⟨ g | 1 | f ⟩$ as defined in Eq. (7-1).
Suppose $f (x)$ and $g (x)$ are expanded in terms of the orthogonal function $ϕ_{n} (x)$ , the solutions to the wave equation associated with the kernel $K (2, 1)$ , as discussed in Sec. 4-2. Thus

\begin{matrix} (8-23) & f (x) = \sum_{n} f_{n} ϕ_{n} (x), g (x) = \sum_{n} g_{n} ϕ_{n} (x) \end{matrix}

Using the coefficients

f_{n}

and

g_{n}

and Eq. (4-59), show that the transition amplitude can be witten as

\begin{matrix} (8-24) & \iint g^{*} (x_{2}) K (x_{2}, T; x_{1}, 0) f (x_{1}) d x_{1} d x_{2} = \sum_{n} g_{n}^{*} f_{n} e^{- i E_{n} T / ℏ} \end{matrix}

Next, suppose we choose a special pair of functions

f (x)

and

g (x)

for which the expansion on the right-hand side of Eq. (8-24) is simple. Then after obtaining the functions

f_{n}

we could get some information about the wave functions

ϕ_{n} (x)

from the expansions of Eq. (8-23). Suppose we choose the functions

f (x)

and

g (x)

in the following way

\begin{aligned} (8-25) & f (x) & = {(\frac{m ω}{π ℏ})}^{1 / 4} \exp [- \frac{m ω}{2 ℏ} (x - a)^{2}] \\ (8-26) & g (x) & = {(\frac{m ω}{π ℏ})}^{1 / 4} \exp [- \frac{m ω}{2 ℏ} (x - b)^{2}] \end{aligned}

These functions represent gaussian distributions centered about

a

and

b

, respectively.
We shall call

f_{n} = f_{n} (a)

and

g_{n} = f_{n} (b)

. Determine the transition amplitude

⟨ g | 1 | f ⟩

, where

f (x)

and

g (x)

are given by Eqs. (8-25) and (8-26), respectively, and the kernel is that for a harmonic oscillator, Eq. (8-1). Perform the integrals in Eq. (8-24) to get

\begin{matrix} (8-27) & \exp {- \frac{i ω T}{2} - \frac{m ω}{4 ℏ} (a^{2} + b^{2} - 2 a b e^{- i ω T})} = \sum_{n} g_{n}^{*} (b) f_{n} (a) e^{- i E_{n} T / ℏ} \end{matrix}

From this result show that

E_{n} = ℏ ω (n + \frac{1}{2})

and that

\begin{matrix} (8-28) & f_{n} (a) = {(\frac{m ω}{2 ℏ})}^{n / 2} \frac{a^{n}}{\sqrt{n!}} \exp (- \frac{m ω a^{2}}{4 ℏ}) \end{matrix}

Using this result in Eq. (8-23) and write for

f_{n} (x)

the form given by Eq. (8-7) considering the

H_{n} (x)

still unknown. From this derive the generating function of Eq. (8-9) for these functions

H_{n} (x)

( 解答 )　【 a 】遷移要素 $⟨ g | 1 | f ⟩$ に式 (8-10)： $K (2, 1) = \sum ϕ_{l} (x_{2}) ϕ_{l}^{*} (x_{1}) e^{- i E_{l} T / ℏ}$ と式 (8-23) を代入し, 関数 $ϕ_{n} (x)$ の規格直交性を用いるならば,
$\begin{aligned} ⟨ g | 1 | f ⟩ & = \iint d x_{2} d x_{1} g^{*} (x_{2}) K (2, 1) f (x_{1}) \\ = \iint d x_{2} d x_{1} \sum_{n} g_{n}^{*} ϕ_{n}^{*} (x_{2}) K (2, 1) \sum_{m} f_{m} ϕ_{m} (x_{1}) \\ = \int d x_{2} \sum_{n} g_{n}^{*} ϕ_{n}^{*} (x_{2}) \sum_{m} f_{m} \int d x_{1} \sum_{l} ϕ_{l} (x_{2}) ϕ_{l}^{*} (x_{1}) e^{- i E_{l} T / ℏ} ϕ_{m} (x_{1}) \\ = \int d x_{2} \sum_{n} g_{n}^{*} ϕ_{n}^{*} (x_{2}) \sum_{m} f_{m} \sum_{l} ϕ_{l} (x_{2}) e^{- i E_{l} T / ℏ} \int d x_{1} ϕ_{l}^{*} (x_{1}) ϕ_{m} (x_{1}) \\ = \int d x_{2} \sum_{n} g_{n}^{*} ϕ_{n}^{*} (x_{2}) \sum_{m} f_{m} \sum_{l} ϕ_{l} (x_{2}) e^{- i E_{l} T / ℏ} δ_{l m} \\ = \int d x_{2} \sum_{n} g_{n}^{*} ϕ_{n}^{*} (x_{2}) \sum_{m} f_{m} ϕ_{m} (x_{2}) e^{- i E_{m} T / ℏ} \\ = \sum_{n} g_{n}^{*} \sum_{m} f_{m} e^{- i E_{m} T / ℏ} \int d x_{2} ϕ_{n}^{*} (x_{2}) ϕ_{m} (x_{2}) \\ = \sum_{n} g_{n}^{*} \sum_{m} f_{m} e^{- i E_{m} T / ℏ} δ_{n m} = \sum_{n} g_{n}^{*} f_{n} e^{- i E_{n} T / ℏ} \end{aligned}$
よって,

\begin{matrix} (1) & ⟨ g | 1 | f ⟩ = \iint d x_{2} d x_{1} g^{*} (x_{2}) K (2, 1) f (x_{1}) = \sum_{n} g_{n}^{*} f_{n} e^{- i E_{n} T / ℏ} \end{matrix}

【 b 】遷移要素 $⟨ g | 1 | f ⟩$ に核の式 (8-1), 及び $f$ と $g$ の式 (8-25), 式 (8-26) を代入すると,

\begin{aligned} ⟨ g | 1 | f ⟩ = \iint d x_{2} d x_{1} g^{*} (x_{2}) K (2, 1) f (x_{1}) \\ = \iint d x_{2} d x_{1} {(\frac{m ω}{π ℏ})}^{1 / 2} \exp {- \frac{m ω}{2 ℏ} [(x_{2} - b)^{2} + (x_{1} - a)^{2}]} {(\frac{m ω}{2 π i ℏ \sin ω T})}^{1 / 2} \\ (2) & \times \exp {\frac{i m ω}{2 ℏ \sin ω T} [(x_{1}^{2} + x_{2}^{2}) \cos ω T - 2 x_{1} x_{2}]} \end{aligned}

まず D.F.Styer の解答を参照して, 変数を次のような無次元量に変換する：

\begin{aligned} α \equiv {(\frac{m ω}{2 ℏ})}^{1 / 2}, & {\bar{x}}_{1} = α x_{1} {\bar{x}}_{2} = α x_{2}, \bar{a} = α a, \bar{b} = α b \\ (3) & \Rightarrow x_{1} = \frac{{\bar{x}}_{1}}{α}, x_{2} = \frac{{\bar{x}}_{2}}{α}, a = \frac{\bar{a}}{α}, b = \frac{\bar{b}}{α} \end{aligned}

すると式 (2) は次となる：

\begin{aligned} {(\frac{2}{π})}^{1 / 2} α \iint \frac{d {\bar{x}}_{2}}{α} \frac{d {\bar{x}}_{1}}{α} \exp {- α^{2} [\frac{1}{α^{2}} ({\bar{x}}_{2} - \bar{b})^{2} + \frac{1}{α^{2}} ({\bar{x}}_{1} - \bar{a})^{2}]} α {(\frac{1}{π i \sin ω T})}^{1 / 2} \\ \times \exp {\frac{i α^{2}}{\sin ω T} [\frac{1}{α^{2}} ({\bar{x}}_{1}^{2} + {\bar{x}}_{2}^{2}) \cos ω T - \frac{2}{α^{2}} {\bar{x}}_{1} {\bar{x}}_{2}]} \\ = {(\frac{2}{π^{2} i \sin ω T})}^{1 / 2} \iint d {\bar{x}}_{2} d {\bar{x}}_{1} \exp {- ({\bar{x}}_{2} - \bar{b})^{2} - ({\bar{x}}_{1} - \bar{a})^{2} + \frac{i \cos ω T}{\sin ω T} ({\bar{x}}_{1}^{2} + {\bar{x}}_{2}^{2}) - \frac{2 i}{\sin ω T} {\bar{x}}_{2} {\bar{x}}_{1}} \\ = {(\frac{2}{π^{2} i \sin ω T})}^{1 / 2} \exp {- ({\bar{a}}^{2} + {\bar{b}}^{2})} \\ \times \iint d {\bar{x}}_{2} d {\bar{x}}_{1} \exp {(- 1 + i \frac{\cos ω T}{\sin ω T}) ({\bar{x}}_{1}^{2} + {\bar{x}}_{2}^{2}) + 2 (\bar{a} {\bar{x}}_{1} + \bar{b} {\bar{x}}_{2}) - \frac{2 i}{\sin ω T} {\bar{x}}_{1} {\bar{x}}_{2}} \\ = {(\frac{2}{π^{2} i \sin ω T})}^{1 / 2} \exp {- ({\bar{a}}^{2} + {\bar{b}}^{2})} \\ (4) & \times \iint d {\bar{x}}_{2} d {\bar{x}}_{1} \exp {\frac{1}{\sin ω T} [i e^{i ω T} ({\bar{x}}_{1}^{2} + {\bar{x}}_{2}^{2}) + 2 (\bar{a} {\bar{x}}_{1} + \bar{b} {\bar{x}}_{2}) \sin ω T - 2 i {\bar{x}}_{1} {\bar{x}}_{2}]} \end{aligned}

積分を簡単に実行するために, 座標軸を

π / 4

回転して新しい変数

x_{1}^{'}, x_{2}^{'}

に変更する．これは単なる座標回転であるから, 積分範囲に変更はないし余計な因子も発生しない：

\begin{matrix} (5) & {\bar{x}}_{1} = \frac{1}{\sqrt{2}} (x_{1}^{'} - x_{2}^{'}), {\bar{x}}_{2} = \frac{1}{\sqrt{2}} (x_{1}^{'} + x_{2}^{'}) . \end{matrix}

すると, 式 (4) の積分部分は分離した2つの積分となる：

\begin{aligned} \iint d x_{1}^{'} d x_{2}^{'} \exp {\frac{1}{\sin ω T} [i e^{i ω T} (x_{1}^{' 2} + x_{2}^{' 2}) + \sqrt{2} (\bar{a} x_{1}^{'} - \bar{a} x_{2}^{'} + \bar{b} x_{1}^{'} + \bar{b} x_{2}^{'}) \sin ω T - i (x_{1}^{' 2} - x_{2}^{' 2})]} \\ = \int d x_{1}^{'} \frac{1}{\sin ω T} [e (e^{i ω T} - 1) x_{1}^{' 2} + \sqrt{2} (\bar{a} + \bar{b}) x_{1}^{'} \sin ω T] \\ (6) & \times \int d x_{2}^{'} \frac{1}{\sin ω T} [i (e^{i ω T} + 1) x_{2}^{' 2} + \sqrt{2} (- \bar{a} + \bar{b}) x_{2}^{' 2} \sin ω T] \end{aligned}

各々の積分は，Feynman-Hibbsの付録「役に立つ定積分」にある公式

\begin{matrix} (7) & \int_{- \infty}^{\infty} e^{a x^{2} + b x} d x = \sqrt{\frac{π}{- a}} \exp (- \frac{b^{2}}{4 a}) \end{matrix}

を当てはめて実行出来る．また, 式 (8-11) から

\begin{matrix} (8) & \frac{2 i \sin ω T}{e^{2 i ω T} - 1} = \frac{e^{i ω T} (1 - e^{- 2 i ω T})}{e^{2 i ω T} (1 - e^{- 2 i ω T})} = e^{- i ω T} \end{matrix}

も利用する．すると式 (4) すなわち式 (2) は結局次となる：

\begin{aligned} {(\frac{2}{π^{2} i \sin ω T})}^{1 / 2} \exp {- ({\bar{a}}^{2} + {\bar{b}}^{2})} \frac{π \sin ω T}{i (1 - e^{i ω T})} \exp {\frac{(\bar{a} + \bar{b})^{2} \sin ω T}{i (1 - e^{i ω T})}} \\ \times {(\frac{π \sin ω T}{- i (1 + e^{i ω T})})}^{1 / 2} \exp {- \frac{(a - b)^{2} \sin ω T}{i (1 + e^{i ω T})}} \\ = {(\frac{2}{π^{2} i \sin ω T})}^{1 / 2} \exp {- ({\bar{a}}^{2} + {\bar{b}}^{2})} \frac{π \sin ω T}{(1 - e^{2 i ω T})^{1 / 2}} \exp {i \sin ω T [\frac{({\bar{a}}^{2} + {\bar{b}}^{2}) e^{i ω T} + 2 \bar{a} \bar{b}}{e^{2 i ω T} - 1}]} \\ = {(\frac{2 i \sin ω T}{e^{2 i ω T} - 1})}^{1 / 2} \exp {- ({\bar{a}}^{2} + {\bar{b}}^{2})} \exp {\frac{i \sin ω T}{e^{2 i ω T} - 1} [({\bar{a}}^{2} + {\bar{b}}^{2}) e^{i ω T} + 2 \bar{a} \bar{b}]} \\ = \exp {- i \frac{ω T}{2}} \exp {- ({\bar{a}}^{2} + {\bar{b}}^{2})} \exp {\frac{e^{- i ω T}}{2} [({\bar{a}}^{2} + {\bar{b}}^{2}) e^{i ω T} + 2 \bar{a} \bar{b}]} \\ (9) & = \exp {- \frac{i ω T}{2} - \frac{m ω}{4 ℏ} (a^{2} + b^{2} - 2 a b e^{- i ω T})} \end{aligned}

よって式 (8-24) は次となる：

\begin{matrix} (10) & ⟨ g | 1 | f ⟩ = \exp {- \frac{i ω T}{2} - \frac{m ω}{4 ℏ} (a^{2} + b^{2} - 2 a b e^{- i ω T})} = \sum_{n} g_{n}^{*} (b) f_{n} (a) e^{- i E_{n} T / ℏ} \end{matrix}

この問題に一致する議論が Schiff の § 13 (p.87) で為されている．そこの結果をここの問題に合わせて表わすならば,

\begin{aligned} ψ (x_{2}, T) & = \int_{- \infty}^{\infty} d x_{1} K (x_{2}, T; x_{1}, t_{0}) f (x_{1}, t_{0}) \\ (11) & = {(\frac{m ω}{π ℏ})}^{1 / 4} \exp {- α^{2} x_{2}^{2} - \frac{1}{2} α^{2} a^{2} - \frac{i ω T}{2} - \frac{1}{2} α^{2} a^{2} e^{- 2 i ω T} + 2 α^{2} x a e^{- i ω T}} \end{aligned}

これを用いると, 式 (2) は次のように表される：

\begin{aligned} ⟨ g | 1 | f ⟩ & = \int d x_{2} g^{*} (x_{2}) \int d x_{1} K (2, 1) f (x_{1}) = \int d x_{2} g^{*} (x_{2}) ψ (x_{2}, T) \\ = {(\frac{m ω}{π ℏ})}^{1 / 2} \int d x_{2} \exp {- α^{2} (x_{2} - b)^{2}} \\ \times \exp {- α^{2} x_{2}^{2} - \frac{1}{2} α^{2} a^{2} - \frac{i ω T}{2} - \frac{1}{2} α^{2} a^{2} e^{- 2 i ω T} + 2 α^{2} x a e^{- i ω T}} \\ = {(\frac{m ω}{π ℏ})}^{1 / 2} \exp {- \frac{1}{2} α^{2} a^{2} - \frac{i ω T}{2} - \frac{1}{2} α^{2} a^{2} e^{- 2 i ω T}} \\ (12) & \times \int d x_{2} \exp {- α^{2} (x_{2} - b)^{2}} \exp {- α^{2} x_{2}^{2} + 2 α^{2} a e^{- i ω T} x_{2}} \end{aligned}

この積分部分を feynman&Hibbs 付録の公式

\begin{matrix} (13) & \int_{- \infty}^{\infty} e^{a (x - x_{1})^{2}} e^{b (x - x_{2})^{2}} d x = \sqrt{\frac{- π}{a + b}} \exp [\frac{a b}{a + b} (x_{1} - x_{2})^{2}] \end{matrix}

を利用して実行するならば, やはり式 (9) と同じ結果となることが分かる．

【 c 】 $f_{n} (a)$ と $g_{n} (b)$ を式 (8-28) の形であるとして, 式 (8-27) の右辺に代入すると

\begin{aligned} \sum_{n} f_{n} (a) g_{n}^{*} (b) e^{- i E_{n} T / ℏ} & = \sum_{n} {(\frac{m ω}{2 ℏ})}^{n / 2} \frac{a^{n}}{\sqrt{n!}} \exp (- \frac{m ω a^{2}}{4 ℏ}) \cdot {(\frac{m ω}{2 ℏ})}^{n / 2} \frac{b^{n}}{\sqrt{n!}} \exp (- \frac{m ω b^{2}}{4 ℏ}) \exp (- \frac{i E_{n} T}{ℏ}) \\ (14) & = \exp {- \frac{m ω}{4 ℏ} (a^{2} + b^{2})} \sum_{n} {(\frac{m ω}{2 ℏ})}^{n} \frac{a^{n} b^{n}}{n!} \exp (- \frac{i E_{n} T}{ℏ}) \end{aligned}

そこで式 (8-27) の左辺を次のように書いてみる：

\begin{aligned} (15) & \exp [- \frac{i ω T}{2} - \frac{m ω}{4 ℏ} (a^{2} + b^{2} - 2 a b e^{- i ω T})] & = \exp {- \frac{m ω}{4 ℏ} (a^{2} + b^{2})} \exp [- \frac{i ω T}{2} + \frac{m ω}{2 ℏ} a b e^{- i ω T}] \end{aligned}

以上の式 (14) と式 (15) を比較して, 次であれば良いことが分かる：

\begin{matrix} (16) & \sum_{n} {(\frac{m ω}{2 ℏ})}^{n} \frac{a^{n} b^{n}}{n!} \exp (- \frac{i E_{n} T}{ℏ}) = \exp [- \frac{i ω T}{2} + \frac{m ω}{2 ℏ} a b e^{- i ω T}] \end{matrix}

そこで

E_{n} = ℏ ω (n + \frac{1}{2})

と仮定して, これを上式の左辺に代入して見ると,

e^{x}

のテイラー展開公式：

e^{x} = \sum_{n = 0}^{\infty} \frac{1}{n!} x^{n}

を利用することで式 (16) の右辺に一致することが分かる：

\begin{aligned} \sum_{n} {(\frac{m ω}{2 ℏ})}^{n} \frac{a^{n} b^{n}}{n!} \exp (- \frac{i E_{n} T}{ℏ}) & = \sum_{n} {(\frac{m ω}{2 ℏ})}^{n} \frac{a^{n} b^{n}}{n!} \exp {- i ω T (n + \frac{1}{2})} \\ = \exp {- \frac{i ω T}{2}} \sum_{n} {(\frac{m ω}{2 ℏ})}^{n} \frac{a^{n} b^{n}}{n!} e^{- i ω T n} \\ = \exp {- \frac{i ω T}{2}} \sum_{n} \frac{1}{n!} {(\frac{m ω}{2 ℏ} a b e^{- i ω T})}^{n} \\ = \exp {- \frac{i ω T}{2}} \times \exp (\frac{m ω}{2 ℏ} a b e^{- i ω T}) \\ = \exp [- \frac{i ω T}{2} + \frac{m ω}{2 ℏ} a b e^{- i ω T}] \end{aligned}

よって

f_{n}

が式 (8-28) の形であるならば,

E_{n}

は

E_{n} = ℏ ω (n + \frac{1}{2})

で良いことが示された．

【 d 】 $ϕ_{n} (x)$ として式 (8-7) を仮定し, 式 (8-25) と式 (8-28) を式 (8-23) に代入してみる．ただし簡単化のために変数を次のように置き換える：

\begin{matrix} (17) & β = \sqrt{\frac{m ω}{ℏ}}, \end{matrix}

すると, 式 (8-25) と式 (8-28) および式 (8-7) は次に書ける：

\begin{aligned} (18) & f (x) & = {(\frac{β}{\sqrt{π}})}^{1 / 2} \exp {- \frac{β^{2}}{2} (x - a)^{2}} \\ (19) & f_{n} & = \frac{β^{n}}{\sqrt{2^{n}}} \frac{a^{n}}{\sqrt{n!}} \exp (- \frac{β^{2}}{4} a^{2}) \\ (20) & ϕ_{n} (x) & = {(\frac{β}{\sqrt{π}})}^{1 / 2} \frac{1}{\sqrt{2^{n} n!}} H_{n} (β x) \exp (- \frac{β^{2}}{2} x^{2}) \end{aligned}

これらを式 (8-23)：

f (x) = \sum_{n} f_{n} ϕ_{n} (x)

に代入すると,

\begin{aligned} {(\frac{β}{\sqrt{π}})}^{1 / 2} \exp {- \frac{β^{2}}{2} (x - a)^{2}} = \sum_{n} \frac{β^{n}}{\sqrt{2^{n}}} \frac{a^{n}}{\sqrt{n!}} \exp (- \frac{β^{2}}{4} a^{2}) \times {(\frac{β}{\sqrt{π}})}^{1 / 2} \frac{1}{\sqrt{2^{n} n!}} H_{n} (β x) \exp (- \frac{β^{2}}{2} x^{2}), \\ \exp {- \frac{β^{2}}{2} x^{2} + β^{2} a x - \frac{β^{2}}{2} a^{2}} = \sum_{n} \frac{β^{n} a^{n}}{2^{n} n!} \exp (- \frac{β^{2}}{4} a^{2}) \exp (- \frac{β^{2}}{2} x^{2}) H_{n} (β x), \\ \exp (- \frac{β^{2}}{2} x^{2}) \exp (β^{2} a x) \exp (- \frac{β^{2}}{2} a^{2}) = \exp (- \frac{β^{2}}{4} a^{2}) \exp (- \frac{β^{2}}{2} x^{2}) \sum_{n} \frac{β^{n} a^{n}}{2^{n} n!} H_{n} (β x), \\ \exp (β^{2} a x - \frac{β^{2}}{4} a^{2}) = \sum_{n} H_{n} (β x) \frac{1}{n!} {(\frac{β}{2} a)}^{n}, \\ (21) & \Rightarrow \exp {- {(\frac{β}{2} a)}^{2} + 2 (\frac{β}{2} a) β x} = \sum_{n} H_{n} (β x) \frac{1}{n!} {(\frac{β}{2} a)}^{n} \end{aligned}

そこで, 次のような置き換えをする：

\begin{matrix} (22) & t = \frac{β}{2} a, y = β x \end{matrix}

すると, 上式は次のように書き直すことが出来る：

\exp (- t^{2} + 2 t x) = \sum_{n} H_{n} (y) \frac{t^{n}}{n!}

これはまさに関数

H_{n} (x)

に対する母関数についての式 (8-9) である．

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