問題 6-13 の解答例

Problem 6-13
Assume $V (r, t)$ is actually independent of $t$ . Substituting for the free-particle kernel $K_{0}$ in Eq.(6-61), integrate the result over $t$ to show that

\begin{matrix} (6-62) & ψ (R_{b}, t_{b}) = e^{- (i / ℏ) E_{a} t_{b}} [e^{(i / ℏ) p_{a} \cdot R_{b}} + \frac{m}{2 π ℏ^{2}} \int d^{3} r_{c} \frac{1}{r_{b c}} e^{(i / ℏ) p r_{b c}} V (r_{c}) e^{(i / ℏ) p_{a} \cdot r_{c}}] \end{matrix}

where

r_{b c}

is the distance from the final point

b

to the variable point of integration

c

and

p

is the magnitude of the momentum of the electron.
Once more, suppose that the potential drops to

0

for distances which are short compared to either

R_{a}

R_{b}

. Show that Eq.(6-62) can be written as

\begin{matrix} (6-63) & ψ (R_{b}, t_{b}) = e^{(i / ℏ) E_{b} t_{b}} e^{(i / ℏ) p_{a} \cdot R_{b}} + f (q) \frac{e^{(i / ℏ) p R_{b}}}{R_{b}} \end{matrix}

where the scattering amplitude

f (q)

is defined in terms of

v (q)

[ see Eq.(6-39) ] as

\begin{aligned} (6-64) & f (q) & = - \frac{m}{2 π ℏ^{2}} v (q), \\ where & v (q) = \int d^{3} r e^{i q \cdot r / ℏ} V (r), q = p_{a} - p_{b} \end{aligned}

The last term of Eq.(6-63),

f (q) e^{I p R_{b} / ℏ} / R_{b}

, can be thought of as the spatial part of the scattered wave function. It has the form of a spherical wave radiating outward from the center of the scattering atom. The amplitude of this spherical wave at some particular scattering angle depends upon that angle through the function

f (q)

, which, by Eq.(6-64), varies with the momentum transfer

q

. Thus the complete wave functions for the electrons after scattering can be thought of as the sum of two terms. The first term is the plane wave of nonscattered electrons,

e^{i p_{a} \cdot R_{b} / ℏ}

, and the second term is the spherical wave of scattered electrons, as indicated in Fig.6-10. Use this point of view to derive the formula for the cross section

d σ / d Ω

Fig. 6-10.　a beam of electrons, represented by its equivalent wave, moves toward the atomic nucleus at

0

.　The strongest fraction of beam moves on undisturbed as a plane wave with momentum

p_{a}

.　A small fraction of the beam is scattered from the atomic nucleus and moves away from

O

as a spherical wave pattern.　The resulting strength of the wave of electrons (the number of electrons) at some point

b

located at

R_{b}

measured from the atomic nucleus

0

, is then made up of two parts. The first is the nonscattered beam given by the plane wave

e^{i p_{a} \cdot R_{b} / ℏ}

.　To this is added the scattered wave with the spherical form given by

e^{i p R_{b} / ℏ} / R_{b}

and multiplied by the function

f (q)

, which determines the angular dependence.　The combination of these two waves gives the spherical part of the scattered wave function.

( 解答例 )　原書では式(6-62)の [ ]が抜けている．また，式(6-64)では $f (q)$ に負符号が付いていない．しかし J.J.Sakurai の式と一致させるためにマイナス符号を付けたので注意する． Emended Edition by D.F.Styer では負符号がキチンと付いている．
式(6.61) において $V (r, t_{c}) = V (r_{c})$ とし, また $K_{0}$ として, 問題4-12 で求めた次の3次元自由粒子核

\begin{matrix} (1) & K_{0} = {(\frac{m}{2 π i ℏ (t_{b} - t_{c})})}^{3 / 2} \exp [\frac{i m (R_{b} - r_{c})^{2}}{2 ℏ (t_{b} - t_{c})}] \end{matrix}

を用いるならば, 式(6-61)から次式が得られる：

\begin{aligned} ψ (R_{b}, t_{b}) & = e^{i p_{a} \cdot R_{b} - i E_{a} t_{b} / ℏ} - \frac{i}{ℏ} \int_{0}^{t_{b}} d t_{c} \int d^{3} r_{c} K_{0} (R_{b}, t_{b}; r_{c}, t_{c}) V (r) e^{i p_{a} \cdot r / ℏ} e^{- i E_{a} t_{c} / ℏ} \\ (2) & = e^{i p_{a} \cdot R_{b} - i E_{a} t_{b} / ℏ} - \frac{i}{ℏ} \int_{0}^{t_{b}} d t_{c} \int d^{3} r_{c} {(\frac{m}{2 π i ℏ (t_{b} - t_{c})})}^{3 / 2} \exp [\frac{i m (R_{b} - r_{c})^{2}}{2 ℏ (t_{b} - t_{c})}] V (r_{c}) e^{i p_{a} \cdot r_{c} / ℏ} e^{- i E_{a} t_{c} / ℏ} \end{aligned}

この第2項目を

I

とし, これだけを考える．

(R_{b} - r_{c}) = r_{b c}

とおくと次となる：

\begin{aligned} I & = - \frac{i}{ℏ} {(\frac{m}{2 π i ℏ})}^{3 / 2} \int_{0}^{t_{b}} d t_{c} \int d^{3} r_{c} (t_{b} - t_{c})^{- 3 / 2} \exp [\frac{i m r_{b c}^{2}}{2 ℏ (t_{b} - t_{c})}] V (r_{c}) e^{i p_{a} \cdot r_{c} / ℏ} e^{- i E_{a} t_{c} / ℏ} \\ (3) & = - \frac{m}{2 π ℏ^{2}} {(\frac{m}{2 π i ℏ})}^{1 / 2} \int d^{3} r_{c} e^{i p_{a} \cdot r_{c} / ℏ} V (r_{c}) \int_{0}^{t_{b}} d t_{c} (t_{b} - t_{c})^{- 3 / 2} \exp [\frac{i m r_{b c}^{2}}{2 ℏ (t_{b} - t_{c})}] e^{- i E_{a} t_{c} / ℏ} \end{aligned}

さらにこの内の時間積分

I_{1}

だけを取り出して考える：

\begin{aligned} I_{1} & = \int_{0}^{t_{b}} d t_{c} (t_{b} - t_{c})^{- 3 / 2} \exp [\frac{i m r_{b c}^{2}}{2 ℏ (t_{b} - t_{c})}] e^{i E_{a} (t_{b} - t_{c} - t_{b}) / ℏ} \\ (4) & = e^{- i E_{a} t_{b} / ℏ} \int_{0}^{t_{b}} d t_{c} (t_{b} - t_{c})^{- 3 / 2} \exp [\frac{i m r_{b c}^{2}}{2 ℏ (t_{b} - t_{c})} + \frac{i E_{a}}{ℏ} (t_{b} - t_{c})] \end{aligned}

ここで

(t_{b} - t_{c})^{- 1 / 2} = y

と置くと，

(t_{b} - t_{c})^{- 3 / 2} d t = 2 d y

で積分範囲は

[1 / \sqrt{t_{b}}, \infty]

であるが

t_{b}

は十分に大きな時間であるとするならば

1 / \sqrt{t_{b}} ≃ 0

と近似してよかろう．すると,

\begin{matrix} (5) & I_{1} ≃ e^{- i E_{a} t_{b} / ℏ} 2 \int_{0}^{\infty} d y \exp [\frac{i m r_{b c}^{2}}{2 ℏ} y^{2} + i \frac{E_{a}}{ℏ} \frac{1}{y^{2}}] \end{matrix}

巻末の「役に立つ積分」から次の公式を利用する：

\begin{matrix} (A-3) & \int_{0}^{\infty} \exp {\frac{i a}{x^{2}} + i b x^{2}} d x = \sqrt{\frac{i π}{4 b}} \exp [i 2 \sqrt{a b}] \end{matrix}

この場合,

a = E_{a} / ℏ

，

b = m r_{b c}^{2} / 2 ℏ

である．また

E_{a} = p^{2} / 2 m

を用いる. すると積分

I_{1}

は次となる：

\begin{matrix} (6) & I_{1} = 2 e^{- i E_{a} t_{b} / ℏ} \sqrt{\frac{i π}{4} \frac{2 ℏ}{m r_{b c}^{2}}} \exp {i 2 \sqrt{\frac{m E_{a} r_{b c}^{2}}{2 ℏ^{2}}}} = 2 e^{- i E_{a} t_{b} / ℏ} \frac{1}{r_{b c}} \sqrt{\frac{i π ℏ}{2 m}} \exp [\frac{i p r_{b c}}{ℏ}] \end{matrix}

従って式(3)の

I

は次となる：

\begin{aligned} I & = - \frac{m}{2 π ℏ^{2}} {(\frac{m}{2 π i ℏ})}^{1 / 2} \int d^{3} r_{c} e^{i p_{a} \cdot r_{c} / ℏ} V (r_{c}) 2 e^{- i E_{a} t_{b} / ℏ} \frac{1}{r_{b c}} {(\frac{i π ℏ}{2 m})}^{1 / 2} \exp [\frac{i p r_{b c}}{ℏ}] \\ (7) & = - e^{- i E_{a} t_{b} / ℏ} \frac{m}{2 π ℏ^{2}} \int d^{3} r_{c} \frac{e^{i p r_{b c} / ℏ}}{r_{b c}} V (r_{c}) e^{i p_{a} \cdot r_{c} / ℏ} \end{aligned}

この結果式(7)より, 式(2)の

ψ (R_{b}, t_{b})

は最終的に次のように書くことが出来る：

\begin{aligned} ψ (R_{b}, t_{b}) & = e^{i p_{a} \cdot R_{b} - i E_{a} t_{b} / ℏ} - e^{- i E_{a} t_{b} / ℏ} \frac{m}{2 π ℏ^{2}} \int d^{3} r_{c} \frac{e^{i p r_{b c} / ℏ}}{r_{b c}} V (r_{c}) e^{i p_{a} \cdot r_{c} / ℏ} \\ (8) & = e^{- i E_{a} t_{b} / ℏ} [e^{i p_{a} \cdot R_{b} / ℏ} - \frac{m}{2 π ℏ^{2}} \int d^{3} r_{c} \frac{e^{i p r_{b c} / ℏ}}{r_{b c}} V (r_{c}) e^{i p_{a} \cdot r_{c} / ℏ}] \end{aligned}

次にポテンシャルが影響を与える領域

r_{c}

が

R_{a}, R_{b}

に比べて非常に小さい場合を考える．このとき前と同様に，

r_{b c} ≃ R_{b} - i_{b} \cdot r_{c} ≃ R_{b}

として近似するならば,

\begin{array}{r} \frac{e^{i p r_{b c} / ℏ}}{r_{b c}} ≃ \frac{1}{R_{b} - i_{b} \cdot r_{c}} \exp [\frac{i p (R_{b} - i_{b} \cdot r_{c})}{ℏ}] ≃ \frac{1}{R_{b}} \exp [i \frac{p R_{b}}{ℏ} - i \frac{p_{b} \cdot r_{c}}{ℏ}] \end{array}

従って, 式 (8) の第 2 項は次のように書ける：

\begin{aligned} - \frac{m}{2 π ℏ^{2}} \int d^{3} r_{c} \frac{e^{i p r_{b c} / ℏ}}{r_{b c}} V (r_{c}) e^{i p_{a} \cdot r_{c} / ℏ} & ≃ - \frac{m}{2 π ℏ^{2}} \frac{e^{i p R_{b} / ℏ}}{R_{b}} \int d^{3} r_{c} e^{i p_{a} \cdot r_{c} / ℏ} e^{- i p_{b} \cdot r_{c} / ℏ} V (r_{c}) \\ = - \frac{m}{2 π ℏ^{2}} \frac{e^{i p R_{b} / ℏ}}{R_{b}} \int d^{3} r_{c} e^{i (p_{a} - p_{b}) \cdot r_{c} / ℏ} V (r_{c}) \\ (9) & = - \frac{m}{2 π ℏ^{2}} \frac{e^{i p R_{b} / ℏ}}{R_{b}} \int d^{3} r_{c} e^{i q \cdot r_{c} / ℏ} V (r_{c}) \end{aligned}

よって, 式(6-62) は次のように近似した表現式とすることが出来る：

\begin{aligned} ψ (R_{b}, t_{b}) & ≃ e^{- i E_{b} t_{b} / ℏ} [e^{i p_{a} \cdot R_{b} / ℏ} - \frac{m}{2 π ℏ^{2}} \frac{e^{i p R_{b} / ℏ}}{R_{b}} \int d^{3} r_{c} e^{i q \cdot r_{c} / ℏ} V (r_{c})] \\ = e^{- i E_{b} t_{b} / ℏ} [e^{i p_{a} \cdot R_{b} / ℏ} + \frac{e^{i p R_{b} / ℏ}}{R_{b}} f (q)], \\ (10) & where & f (q) = - \frac{m}{2 π ℏ^{2}} v (q) = - \frac{m}{2 π ℏ^{2}} \int d^{3} r_{c} e^{i q \cdot r_{c} / ℏ} V (r_{c}) \end{aligned}

最後に, 微分断面積は式(6-44) から求めることが出来る：

\begin{matrix} (11) & \frac{d σ}{d Ω} = | f (q) |^{2} = {| - \frac{m}{2 π ℏ^{2}} v (q) |}^{2} \end{matrix}

月	火	水	木	金	土	日
			1	2	3	4
5	6	7	8	9	10	11
12	13	14	15	16	17	18
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26	27	28	29	30	31