# 【量子力学】三次元のシュレディンガー方程式を変数分離で解く

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## 解法

### 変数分離

この粒子の時間に依存しないシュレーディンガー方程式 $$\left(-\frac{\hbar^2}{2m}\nabla^2\right)\Psi = E\Psi \tag{1}$$

を解いて求めます。($$\Psi(x,y,z)$$は波動関数,$$E$$はエネルギー固有値,箱の中を考えるのでV = 0)

まず、 $$\Psi = \psi_x(x)\psi_y(y)\psi_z(z)\tag{2}$$ と変数分離を行います。

(1)式に(2)式を代入すると、 $$-\frac{\hbar^2}{2m}\left(\frac{\partial^2 \psi_x(x)}{\partial x^2}\psi_y(y)\psi_z(z) + \psi_x(x)\frac{\partial^2 \psi_y(y)}{\partial y^2}\psi_z(z) + \psi_x(x)\psi_y(y)\frac{\partial^2 \psi_z(z)}{\partial z^2}\right)\\ = E\psi_x(x)\psi_y(y)\psi_z(z)$$ なので $$-\frac{\hbar^2}{2m}\left(\frac{1}{\psi_x(x)}\frac{\partial^2 \psi_x(x)}{\partial x^2}+ \frac{1}{\psi_y(y)}\frac{\partial^2 \psi_y(y)}{\partial y^2}+ \frac{1}{\psi_z(z)}\frac{\partial^2 \psi_z(z)}{\partial z^2}\right) = E\tag{3}$$ となります。

ここで、$$\frac{1}{\psi_x(x)}\frac{\partial^2 \psi_x(x)}{\partial x^2}$$に注目して

(3)式を $$-\frac{\hbar^2}{2m}\left(\frac{1}{\psi_x(x)}\frac{\partial^2 \psi_x(x)}{\partial x^2}\right) = \frac{\hbar^2}{2m}\left(\frac{1}{\psi_y(y)}\frac{\partial^2 \psi_y(y)}{\partial y^2} + \frac{1}{\psi_z(z)}\frac{\partial^2 \psi_z(z)}{\partial z^2}\right) + E \tag{4}$$ と変形します。

ここで(4)式の左辺に注目すると、$$x$$のみの関数、右辺に注目すると$$y,\;z$$のみの関数なので、

$$-\frac{\hbar^2}{2m}\left(\frac{1}{\psi_x(x)}\frac{\partial^2 \psi_x(x)}{\partial x^2}\right) = e_x(定数)$$ と置くことができます。

これを変形すると、 $$\frac{\partial^2 \psi_x(x)}{\partial x^2} = -\frac{2me_x}{\hbar^2}\psi_x(x)$$ となり、これを解くと、

$$\psi_x(x) = A\cos kx + B\sin kx\;\;\;\left(ただし、k = \sqrt{\frac{2me_x}{\hbar^2}},\;A,Bは定数\right) \tag{5}$$ となります。

### 境界条件

また、境界条件として

$$\begin{eqnarray} \begin{cases} \psi_x(0) = 0 \\ \psi_x(a) = 0 \end{cases} \end{eqnarray}$$ を考えると、(5)式から、

$$\begin{eqnarray} \begin{cases} \psi_x(0) = A = 0\\ \psi_x(a) = A\cos ka + B\sin ka = 0 \end{cases} \end{eqnarray}$$ より、 $$B\sin ka = 0$$ が求まります。

これを満たす$$ka$$は $$ka = l\pi\;\;\;(l = 1,2,3,\cdots)$$ なので、

$$k = \frac{l\pi}{a} \tag{6}$$

となります。

#### 固有エネルギー

このことから、固有エネルギー$$e_x$$を求めることができます。

(5)式で$$k = \sqrt{\frac{2me_x}{\hbar^2}}$$だったので, $$k = \frac{l\pi}{a} = \sqrt{\frac{2me_x}{\hbar^2}}$$ となります。中辺と右辺を変形すると、

$$e_x = \frac{\hbar^2l^2\pi^2}{2ma^2}\tag{7}$$ となり、エネルギー固有値を求めることができました。

### 固有関数

$$\psi_x(x) = B\sin \left(\frac{l\pi}{a}\right)x$$ となりました。

$$\psi_y(y),\;\psi_z(z)$$も同様にして

## 参考

.katex{font-size:1.1em;}.katex-display+.katex-display{margin-top:10px;}シュレーディンガー方程式を解くという問題は数学的に難しく、解析的な関数として解が得られるような問題は非常に限られている（それ以外は数値的に、あるいは近似的

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