# Delta位勢壘

## 定義

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

${\displaystyle V(x)=\lambda \delta (x)\,\!}$

## 導引

${\displaystyle \psi _{L}(x)=A_{r}e^{ikx}+A_{l}e^{-ikx}\quad x<0\,\!}$
${\displaystyle \psi _{R}(x)=B_{r}e^{ikx}+B_{l}e^{-ikx}\quad x>0\,\!}$

${\displaystyle \psi _{L}=\psi _{R}\,\!}$
${\displaystyle {\frac {d}{dx}}\psi _{L}={\frac {d}{dx}}\psi _{R}-{\frac {2m\lambda }{\hbar ^{2}}}\psi _{R}\,\!}$

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

${\displaystyle E\int _{-\epsilon }^{\epsilon }\psi \,dx\approx E\cdot 2\epsilon \cdot \psi (0)\,\!}$ (2)

${\displaystyle \epsilon \to 0\,\!}$ 的極限，這項目往著0去。

${\displaystyle -{\frac {\hbar ^{2}}{2m}}\left({\frac {d\psi _{R}}{dx}}{\bigg |}_{\epsilon }-{\frac {d\psi _{L}}{dx}}{\bigg |}_{-\epsilon }\right)+\lambda \int _{-\epsilon }^{\epsilon }\delta (x)\psi \,dx=0\,\!}$ (3)

${\displaystyle \int _{-\epsilon }^{\epsilon }\delta (x)\psi \,dx=\psi _{R}(0)\,\!}$ (4)

${\displaystyle \lim _{\epsilon \to 0}{\frac {d\psi _{L}}{dx}}{\bigg |}_{-\epsilon }={\frac {d\psi _{L}}{dx}}{\bigg |}_{0}\,\!}$ (5)
${\displaystyle \lim _{\epsilon \to 0}{\frac {d\psi _{R}}{dx}}{\bigg |}_{\epsilon }={\frac {d\psi _{R}}{dx}}{\bigg |}_{0}\,\!}$ (6)

${\displaystyle {\frac {d\psi _{L}}{dx}}={\frac {d\psi _{R}}{dx}}-{\frac {2m\lambda }{\hbar ^{2}}}\psi _{R}\,\!}$

${\displaystyle A_{r}+A_{l}=B_{r}+B_{l}\,\!}$
${\displaystyle ik(A_{r}-A_{l}-B_{r}+B_{l})=-{\frac {2m\lambda }{\hbar ^{2}}}(B_{r}+B_{l})\,\!}$

### 反射與透射

${\displaystyle r={\cfrac {1}{{\cfrac {i\hbar ^{2}k}{m\lambda }}-1}}\,\!}$
${\displaystyle t={\cfrac {1}{{\cfrac {im\lambda }{\hbar ^{2}k}}+1}}\,\!}$

${\displaystyle R=|r|^{2}={\cfrac {1}{1+{\cfrac {\hbar ^{4}k^{2}}{m^{2}\lambda ^{2}}}}}={\cfrac {1}{1+{\cfrac {2\hbar ^{2}E}{m\lambda ^{2}}}}}\,\!}$

${\displaystyle T=|t|^{2}=1-R={\cfrac {1}{1+{\cfrac {m^{2}\lambda ^{2}}{\hbar ^{4}k^{2}}}}}={\cfrac {1}{1+{\cfrac {m\lambda ^{2}}{2\hbar ^{2}E}}}}\,\!}$

• 由於模型的對稱性，假若，粒子從右邊入射，我們也會得到同樣的答案。
• 很奇異地，給予同樣的能量、質量、與狄拉克Delta函數的強度，Delta位勢壘與Delta位勢阱有同樣的反射係數與透射係數。