# 有限位勢壘

## 定義

${\displaystyle \psi _{L}\,\!}$ ：壘左邊，${\displaystyle x<0\,\!}$（壘外區域），
${\displaystyle \psi _{C}\,\!}$ ：壘內，${\displaystyle 0（壘內區域），
${\displaystyle \psi _{R}\,\!}$ ：壘右邊，${\displaystyle a（壘外區域）。

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

## 導引

${\displaystyle \psi _{L}(x)=A_{r}e^{ik_{0}x}+A_{l}e^{-ik_{0}x}\,\!}$ ，　壘左邊， ${\displaystyle x<0\,\!}$(2)
${\displaystyle \psi _{C}(x)=B_{r}e^{ik_{1}x}+B_{l}e^{-ik_{1}x}\,\!}$ ，　壘內， ${\displaystyle 0(3)
${\displaystyle \psi _{R}(x)=C_{r}e^{ik_{0}x}+C_{l}e^{-ik_{0}x}\,\!}$ ，　壘右邊，${\displaystyle a(4)

${\displaystyle k_{0}={\sqrt {2mE/\hbar ^{2}}}\,\!}$ ，　壘左邊與壘右邊，(5)
${\displaystyle k_{1}={\sqrt {2m(E-V_{0})/\hbar ^{2}}}\,\!}$ ，　壘內。(6)

${\displaystyle \psi _{L}(0)=\psi _{C}(0)\,\!}$
${\displaystyle {\frac {d}{dx}}\psi _{L}(0)={\frac {d}{dx}}\psi _{C}(0)\,\!}$
${\displaystyle \psi _{C}(a)=\psi _{R}(a)\,\!}$
${\displaystyle {\frac {d}{dx}}\psi _{C}(a)={\frac {d}{dx}}\psi _{R}(a)\,\!}$

${\displaystyle A_{r}+A_{l}=B_{r}+B_{l}\,\!}$(7)
${\displaystyle ik_{0}(A_{r}-A_{l})=ik_{1}(B_{r}-B_{l})\,\!}$(8)
${\displaystyle B_{r}e^{ik_{1}a}+B_{l}e^{-ik_{1}a}=C_{r}e^{ik_{0}a}+C_{l}e^{-ik_{0}a}\,\!}$(9)
${\displaystyle ik_{1}(B_{r}e^{ik_{1}a}-B_{l}e^{-ik_{1}a})=ik_{0}(C_{r}e^{ik_{0}a}-C_{l}e^{-ik_{0}a})\,\!}$(10)

${\displaystyle B_{r}={\frac {k_{0}+k_{1}}{2k_{1}}}-{\frac {k_{0}-k_{1}}{2k_{1}}}\ r\,\!}$(11)
${\displaystyle B_{l}=-{\frac {k_{0}-k_{1}}{2k_{1}}}+{\frac {k_{0}+k_{1}}{2k_{1}}}\ r\,\!}$(12)
${\displaystyle B_{l}=-{\frac {k_{0}-k_{1}}{k_{0}+k_{1}}}e^{i2k_{1}a}\ B_{r}\,\!}$(13)

${\displaystyle r={\frac {(k_{0}^{2}-k_{1}^{2})\sin(k_{1}a)}{(k_{0}^{2}+k_{1}^{2})\sin(k_{1}a)+i2k_{0}k_{1}\cos(k_{1}a)}}\,\!}$(14)

{\displaystyle {\begin{aligned}t&={\frac {2k_{1}}{k_{0}+k_{1}}}e^{-i(k_{0}-k_{1})a}B_{r}\\&={\frac {2k_{1}}{k_{0}+k_{1}}}e^{-i(k_{0}-k_{1})a}\left({\frac {k_{0}+k_{1}}{2k_{1}}}-{\frac {k_{0}-k_{1}}{2k_{1}}}r\right)\\&={\frac {i2k_{0}k_{1}e^{-ik_{0}a}}{(k_{0}^{2}+k_{1}^{2})\sin(k_{1}a)+i2k_{0}k_{1}\cos(k_{1}a)}}\\\end{aligned}}\,\!}

## 解答分析

### ${\displaystyle E

${\displaystyle \kappa _{1}={\sqrt {2m(V_{0}-E)/\hbar ^{2}}}=k_{1}/i\,\!}$(15)

${\displaystyle t={\cfrac {2k_{0}\kappa _{1}e^{-ik_{0}a}}{-i(k_{0}^{2}-\kappa _{1}^{2})\sinh(\kappa _{1}a)+2k_{0}\kappa _{1}\cosh(\kappa _{1}a)}}\,\!}$

${\displaystyle T=|t|^{2}={\cfrac {4k_{0}^{2}\kappa _{1}^{2}}{(k_{0}^{2}-\kappa _{1}^{2})^{2}\sinh ^{2}(\kappa _{1}a)+4k_{0}^{2}\kappa _{1}^{2}}}\,\!}$

${\displaystyle T={\cfrac {1}{1+{\cfrac {\sinh ^{2}(\kappa _{1}a)}{4\ {\cfrac {E}{V_{0}}}\left(1-{\cfrac {E}{V_{0}}}\right)}}}}\,\!}$

### ${\displaystyle E>V_{0}}$

${\displaystyle T=|t|^{2}={\cfrac {1}{1+{\cfrac {\sin ^{2}(k_{1}a)}{4\ {\cfrac {E}{V_{0}}}({\cfrac {E}{V_{0}}}-1)}}}}\,\!}$

${\displaystyle R=|r|^{2}=1-T\,\!}$

${\displaystyle k_{1}a\,\!}$${\displaystyle \pi \,\!}$ 的整數倍的時候，粒子的透射機率等於 1 。粒子肯定地可以透射到位勢壘的另外一邊，就好像位勢壘完全不存在一樣。這現象可以在電子對於惰性氣體散射實驗中觀測得到，又稱為冉紹耳-湯森德效應Ramsauer-Townsend effect）。

### ${\displaystyle E=V_{0}}$

${\displaystyle B_{1}+B_{2}x=\psi _{C}(x)\,\!}$(16)

${\displaystyle A_{r}+A_{l}=B_{1}\,\!}$
${\displaystyle ik_{0}(A_{r}-A_{l})=B_{2}\,\!}$
${\displaystyle B_{1}+B_{2}a=C_{r}e^{ik_{0}a}+C_{l}e^{-ik_{0}a}\,\!}$
${\displaystyle B_{2}=ik_{0}(C_{r}e^{ik_{0}a}-C_{l}e^{-ik_{0}a})\,\!}$

${\displaystyle T={\cfrac {1}{1+mV_{0}a^{2}/2\hbar ^{2}}}\,\!}$
${\displaystyle R={\cfrac {mV_{0}a^{2}/2\hbar ^{2}}{1+mV_{0}a^{2}/2\hbar ^{2}}}\,\!}$

## 應用

Delta 位勢壘是另外一種很重要的位勢壘，可以視為一種特別的有限位勢壘。壘內位勢為狄拉克 Delta 函數${\displaystyle V(x)={\frac {\lambda }{m}}\delta (x)}$ ，壘外位勢為 0 的位勢壘。所有在此條目導引出來的結果，都能夠應用於 Delta 位勢壘，只需要保持 ${\displaystyle V_{0}a={\frac {\lambda ^{2}}{m^{2}}}\,\!}$ 不變，而同時取極限 ${\displaystyle V_{0}\to \infty ,\quad a\to 0\,\!}$

## 參考文獻

• Griffiths, David J. Introduction to Quantum Mechanics (2nd ed.). Prentice Hall. 2004. ISBN 0-13-111892-7.
• Claude Cohen-Tannoudji, Bernard Diu et Frank Laloë. Mécanique quantique, vol. I et II. Paris: Collection Enseignement des sciences (Hermann). 1977. ISBN 2-7056-5767-3.