# 有限深方形阱

## 一維阱定義

${\displaystyle \psi =\psi _{1}\,\!}$ ：阱左邊，${\displaystyle x<-L/2\,\!}$ （阱外區域），
${\displaystyle \psi =\psi _{2}\,\!}$ ：阱內，${\displaystyle -L/2 （阱內區域），
${\displaystyle \psi =\psi _{3}\,\!}$ ：阱右邊，${\displaystyle x>L/2\,\!}$ （阱外區域）。

${\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi }{dx^{2}}}+V(x)\psi =E\psi \,\!}$ (1)

### 阱內區域

${\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi _{2}}{dx^{2}}}=E\psi _{2}\,\!}$ (2)

${\displaystyle k={\frac {\sqrt {2mE}}{\hbar }}\,\!}$ (3)

${\displaystyle {\frac {d^{2}\psi _{2}}{dx^{2}}}=-k^{2}\psi _{2}\,\!}$

${\displaystyle \psi _{2}=A\sin(kx)+B\cos(kx)\quad \,\!}$

### 阱外區域

${\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi _{1}}{dx^{2}}}=(E-V_{0})\psi _{1}\,\!}$

## 束縛態

${\displaystyle \alpha ={\frac {\sqrt {2m(V_{0}-E)}}{\hbar }}\,\!}$ (4)

${\displaystyle {\frac {d^{2}\psi _{1}}{dx^{2}}}=\alpha ^{2}\psi _{1}\,\!}$

${\displaystyle \psi _{1}=Fe^{-\alpha x}+Ge^{\alpha x}\,\!}$
${\displaystyle \psi _{3}=He^{-\alpha x}+Ie^{\alpha x}\,\!}$

### 束縛態的波函數

${\displaystyle \psi _{1}=Fe^{-\alpha x}+Ge^{\alpha x}\,\!}$ ：阱左邊，${\displaystyle x<-L/2\,\!}$ （阱外區域），
${\displaystyle \psi _{2}=A\sin(kx)+B\cos(kx)\,\!}$ ：阱內，${\displaystyle -L/2 （阱內區域），
${\displaystyle \psi _{3}=He^{-\alpha x}+Ie^{\alpha x}\,\!}$ ：阱右邊，${\displaystyle x>L/2\,\!}$ （阱外區域）。

${\displaystyle x\,\!}$ 趨向負無窮，包含${\displaystyle F\,\!}$ 的項目趨向無窮。類似地，當${\displaystyle x\,\!}$ 趨向無窮，包含${\displaystyle I\,\!}$ 的項目趨向無窮。可是，波函數在任何${\displaystyle x\,\!}$ 都必須是有限值。因此，必須設定${\displaystyle F=I=0\,\!}$ 。阱外區域的波函數變為

${\displaystyle \psi _{1}(x)=Ge^{\alpha x}\,\!}$
${\displaystyle \psi _{3}(x)=He^{-\alpha x}\,\!}$

### 奇的波函數

${\displaystyle \psi _{2}=A\sin(kx)\,\!}$
${\displaystyle G=-H\,\!}$
${\displaystyle \psi _{1}(-x)=-\psi _{3}(x),\qquad \qquad x\geq 0\,\!}$

${\displaystyle \psi _{1}(-L/2)=\psi _{2}(-L/2)\,\!}$
${\displaystyle \left.{\frac {d\psi _{1}}{dx}}\right|_{x=-L/2}=\left.{\frac {d\psi _{2}}{dx}}\right|_{x=-L/2}\,\!}$

${\displaystyle Ge^{-\alpha L/2}=-A\sin(kL/2)\,\!}$ (5)
${\displaystyle \alpha Ge^{-\alpha L/2}=kA\cos(kL/2)\,\!}$ (6)

${\displaystyle \alpha =-k\cot(kL/2)\,\!}$

${\displaystyle \alpha ^{2}={\frac {2mV_{0}}{\hbar ^{2}}}-k^{2}\,\!}$

${\displaystyle k^{2}={\frac {2mV_{0}}{\hbar ^{2}}}\sin ^{2}(kL/2)\,\!}$

### 偶的波函數

${\displaystyle \psi _{2}=A\cos(kx)\,\!}$
${\displaystyle G=H\,\!}$
${\displaystyle \psi _{1}(-x)=\psi _{3}(x),\qquad \qquad x\geq 0\,\!}$

${\displaystyle \psi _{1}(-L/2)=\psi _{2}(-L/2)\,\!}$
${\displaystyle \left.{\frac {d\psi _{1}}{dx}}\right|_{x=-L/2}=\left.{\frac {d\psi _{2}}{dx}}\right|_{x=-L/2}\,\!}$

${\displaystyle Ge^{-\alpha L/2}=A\cos(kL/2)\,\!}$ (7)
${\displaystyle \alpha Ge^{-\alpha L/2}=kA\sin(kL/2)\,\!}$ (8)

${\displaystyle \alpha =k\tan(kL/2)\,\!}$

${\displaystyle \alpha ^{2}={\frac {2mV_{0}}{\hbar ^{2}}}-k^{2}\,\!}$

${\displaystyle k^{2}={\frac {2mV_{0}}{\hbar ^{2}}}\cos ^{2}(kL/2)\,\!}$

## 散射態

${\displaystyle {\frac {d^{2}\psi _{1}}{dx^{2}}}=-\kappa ^{2}\psi _{1}\,\!}$
${\displaystyle {\frac {d^{2}\psi _{3}}{dx^{2}}}=-\kappa ^{2}\psi _{3}\,\!}$

${\displaystyle \psi _{1}=C_{1}\sin(\kappa x)+D_{1}\cos(\kappa x)\,\!}$
${\displaystyle \psi _{3}=C_{3}\sin(\kappa x)+D_{3}\cos(\kappa x)\,\!}$

## 參考文獻

1. ^ Griffiths, David J. Introduction to Quantum Mechanics 2nd ed. Prentice Hall. 2005. ISBN 0-13-111892-7.