問題 6-3 の解答例

Problem 6-3
For a free particle, Eq. (4-29) reduces to

\begin{matrix} (6-20) & - \frac{ℏ}{i} \frac{\partial}{\partial t_{b}} K_{0} (b, a) + \frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{0} (b, a) = i ℏ δ (t_{b} - t_{a}) δ (x_{b} - x_{a}) \end{matrix}

Show, from this result and Eq. (6-19), that the kernel

K_{V}

satisfies the differential equation

\begin{matrix} (6-21) & - \frac{ℏ}{i} \frac{\partial}{\partial t_{b}} K_{V} (b, a) + \frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{V} (b, a) - V (b) K_{V} (b, a) = i ℏ δ (t_{b} - t_{a}) δ (x_{b} - x_{a}) \end{matrix}

(解答) 原書及び訳本では式 (6-21) の左辺第 3 項の符号が $+$ となっているが, これは明らかにミスプリと思われるので修正しておく (校訂版では, 両辺を $i ℏ$ で割った形に書き換えてある)．
$K_{V} (b, a)$ は式 (6-19) から $d τ_{c} = d x_{c} d t_{c}$ として次であった：

\begin{matrix} (1) & K_{V} (b, a) = K_{0} (b, a) - \frac{i}{ℏ} \int d τ_{c} K_{0} (b, c) V (c) K_{V} (c, a) \end{matrix}

この式 (1) の両辺に演算子

(- ℏ / i) \partial / \partial t_{b}

を作用させ, 問題文中の式 (6-20) 及びデルタ関数の性質を用いると,

\begin{aligned} - \frac{ℏ}{i} \frac{\partial}{\partial t_{b}} K_{V} (b, a) = - \frac{ℏ}{i} \frac{\partial}{\partial t_{b}} K_{0} (b, a) - \frac{i}{ℏ} \int_{t_{a}}^{t_{b}} d t_{c} \int_{- \infty}^{\infty} d x_{c} {- \frac{ℏ}{i} \frac{\partial}{\partial t_{b}} K_{0} (b, c)} V (x_{c}, t_{c}) K_{V} (c, a) \\ = - \frac{ℏ}{i} \frac{\partial}{\partial t_{b}} K_{0} (b, a) \\ - \frac{i}{ℏ} \int_{t_{a}}^{t_{b}} d t_{c} \int_{- \infty}^{\infty} d x_{c} [- \frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{0} (b, c) + i ℏ δ (x_{b} - x_{c}) δ (t_{b} - t_{c})] V (c) K_{V} (c, a) \\ = - \frac{ℏ}{i} \frac{\partial}{\partial t_{b}} K_{0} (b, a) + \frac{i}{ℏ} \int_{t_{a}}^{t_{b}} d t_{c} \int_{- \infty}^{\infty} d x_{c} {\frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{0} (b, c)} V (c) K_{V} (c, a) \\ + \int_{t_{a}}^{t_{b}} d t_{c} \int_{- \infty}^{\infty} d x_{c} V (x_{c}, t_{c}) K_{V} (c, a) δ (t_{b} - t_{c}) δ (x_{b} - x_{c}) \\ (2) & = - \frac{ℏ}{i} \frac{\partial}{\partial t_{b}} K_{0} (b, a) + \frac{i}{ℏ} \int d θ_{c} {\frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{0} (b, c)} V (c) K_{V} (c, a) + V (b) K_{V} (b, a) \end{aligned}

次に式 (1) の両辺に演算子

(ℏ^{2} / 2 m) \partial^{2} / \partial x_{b}^{2}

を作用させると,

\begin{matrix} (3) & \frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{V} (b, a) = \frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{0} (b, a) - \frac{i}{ℏ} \int d τ_{c} {\frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{0} (b, c)} V (c) K_{V} (c, a) \end{matrix}

以上の結果式 (2) と式 (3) とを足し合わせると,

\begin{aligned} - \frac{ℏ}{i} \frac{\partial}{\partial t_{b}} K_{V} (b, a) + \frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{V} (b, a) \\ = - \frac{ℏ}{i} \frac{\partial}{\partial t_{b}} K_{0} (b, a) \\ + \frac{i}{ℏ} \int d θ_{c} {\frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{0} (b, c)} V (c) K_{V} (c, a) + V (b) K_{V} (b, a) \\ + \frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{0} (b, a) - \frac{i}{ℏ} \int d θ_{c} {\frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{0} (b, c)} V (c) K_{V} (c, a) \\ = - \frac{ℏ}{i} \frac{\partial}{\partial t_{b}} K_{0} (b, a) + \frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{0} (b, a) + V (b) K_{V} (b, a) \\ (4) & = i ℏ δ (t_{b} - t_{a}) δ (x_{b} - x_{a}) + V (b) K_{V} (b, a) \end{aligned}

ただし, 最後の式変形は問題文の式 (6-20) を利用している．以上から,

\begin{matrix} (5) & - \frac{ℏ}{i} \frac{\partial}{\partial t_{b}} K_{V} (b, a) + \frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x_{b}^{2}} K_{V} (b, a) = i ℏ δ (t_{b} - t_{a}) δ (x_{b} - x_{a}) + V (b) K_{V} (b, a) \end{matrix}

この結果式 (5) は, 問題で提示すべき式 (6-21) に等価である．

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