【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

物理

今回は三次元のシュレディンガー方程式を変数分離で解いていきたいと思います。

三次元といっても、一次元のときとやることは変わらないので落ち着いて解いていきましょう。

スポンサーリンク

問題設定

質量\(m\)の粒子が箱の中で運動している.箱の3辺の長さは,\(x,y,z\)軸方向でそれぞれ\(a,b,c\)である。また, 箱の中のポテンシャル\(V\)はゼロとし,箱の外部では\(V= \infty\)とする。波動関数は連続で, 箱の外ではゼロであり, 箱の中で規格化されているものとする.

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

以上から\(\psi_x(x)\)は

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

\(\psi_y(y),\;\psi_z(z)\)も同様にして

$$ \psi_y(y) = C\sin \left(\frac{m\pi}{b}\right)y \\ \psi_z(z) = D\sin \left(\frac{n\pi}{c}\right)z \\(ただし、n,m = 1,2,3,\cdots ,\;C,Dは定数) $$ となります。

よって元の関数は $$ \Psi = C_0\sin \left(\frac{l\pi}{a}x\right)\sin \left(\frac{m\pi}{b}y\right)\sin \left(\frac{n\pi}{c}z\right)\\ (ただし、C_0 = BCD) $$ となります。

規格化

ここで、\(\Psi\)を規格化し、\(C_0\)を求めてみましょう。

全領域にわたって積分しますが、箱の中しか波動関数が存在できないので箱の中の領域だけで積分します。

$$ \int_{0}^{a}\int_{0}^{b}\int_{0}^{c} dxdydz\;\;|\Psi|^2 = 1 $$ なので $$ \int_{0}^{a}\int_{0}^{b}\int_{0}^{c} dxdydz\;\;C_0^2\sin^2 \left(\frac{l\pi}{a}x\right)\sin^2 \left(\frac{m\pi}{b}y\right)\sin^2 \left(\frac{n\pi}{c}z\right) = 1 $$ となります。

ここで、 $$ \int_{0}^{a}dx\;\sin^2 \left(\frac{l\pi}{a}x\right) = \int_{0}^{a}dx\;\frac{1-\cos\left(2\frac{l\pi}{a}x\right)}{2} = \frac{a}{2} $$ より、

$$ \begin{eqnarray} \int_{0}^{a}\int_{0}^{b}\int_{0}^{c} dxdydz\;\;C_0^2\sin^2 \left(\frac{l\pi}{a}x\right)\sin^2 \left(\frac{m\pi}{b}y\right)\sin^2 \left(\frac{n\pi}{c}z\right) &=& C_0^2\frac{a}{2}\frac{b}{2}\frac{c}{2}\\ &=& C_0^2\frac{abc}{8}\\ &=& 1 \end{eqnarray} $$ より、

$$ C_0 = \sqrt{\frac{8}{abc}} $$ となります。

よって、 $$ \Psi = \sqrt{\frac{8}{abc}}\sin \left(\frac{l\pi}{a}x\right)\sin \left(\frac{m\pi}{b}y\right)\sin \left(\frac{n\pi}{c}z\right) $$ となります。

固有エネルギー(全体)

最後に固有エネルギーを求めましょう。

これは各成分の固有エネルギーの和になるので

$$ E = e_x + e_y + e_z $$ なので(7)式で求めたように\(e_y,e_z\)ももとめると、 $$ e_y = \frac{\hbar^2n^2\pi^2}{2mb^2}\\ e_z = \frac{\hbar^2m^2\pi^2}{2mc^2} $$ となります。

よって $$ E = \frac{\hbar^2\pi^2}{2m}\left(\frac{l^2}{a^2} + \frac{n^2}{b^2} + \frac{m^2}{c^2}\right) $$ となります。

参考

量子力学Ⅰ/箱の中の自由粒子
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、あるいは近似的に解くが、それでも精度の高い計算には困難を伴う)。この授業では、最も簡単ないくつ

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