# Delta位勢阱

## 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

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

${\displaystyle E>0\,\!}$ 時，${\displaystyle \psi _{L}\,\!}$ ${\displaystyle \psi _{R}\,\!}$ 都是行進波。可是，當${\displaystyle E<0\,\!}$ 時，${\displaystyle \psi _{L}\,\!}$ ${\displaystyle \psi _{R}\,\!}$ 都隨著座標${\displaystyle x\,\!}$ 呈指數遞減或指數遞增。

${\displaystyle x=0\,\!}$ 处，邊界條件是：

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

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

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

${\displaystyle \epsilon \to 0\,\!}$ 时，该項趋向于0。

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

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

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

### 散射態

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

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

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

### 束縛態

Delta位勢阱的束縛態，在任何一個位置，波函數都是連續的；可是，除了在${\displaystyle x=0\,\!}$ 以外，在其它任何位置，波函數隨位置的導數都是連續的。

${\displaystyle \psi _{L}(x)=A_{l}e^{\kappa x}\,\!}$
${\displaystyle \psi _{R}(x)=B_{r}e^{-\kappa x}\,\!}$

${\displaystyle A_{l}=B_{r}={\sqrt {\kappa }}\,\!}$
${\displaystyle \kappa ={\frac {m\lambda }{\hbar ^{2}}}\,\!}$

Delta位勢阱只能有一個束縛態。束縛態的能量是

${\displaystyle E=-\ {\frac {\hbar ^{2}\kappa ^{2}}{2m}}=-\ {\frac {m\lambda ^{2}}{2\hbar ^{2}}}\,\!}$

${\displaystyle \psi (x)={\frac {\sqrt {m\lambda }}{\hbar }}e^{-m\lambda \mid x\mid /\hbar ^{2}}\,\!}$

Delta位勢阱是有限深方形阱的一個特別案例。在有限深位勢阱的深度${\displaystyle V_{0}\to \infty \,\!}$ 與阱寬${\displaystyle L\to 0\,\!}$ 的極限，同時保持${\displaystyle V_{0}L=\lambda \,\!}$ ，就可以從有限深位勢阱的波函數，得到Delta位勢阱的波函數。

## 雙井迪拉克Delta函數模型

Delta函數模型其實是氫原子的一維版本根據維度比例由 达德利·赫施巴赫（“Dudley R. Herschbach”）[1]團隊所研發。此 delta函數模型以雙井迪拉克Delta函數模型最有用，因其代表一維版的水分子離子。

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

${\displaystyle V(x)=-q\left[\delta (x-{\frac {R}{2}})+\lambda \delta (x+{\frac {R}{2}})\right]}$

${\displaystyle \psi (x)~=~Ae^{-d\left|x-{\frac {R}{2}}\right|}+Be^{-d\left|x+{\frac {R}{2}}\right|}}$

${\displaystyle \left|{\begin{array}{cc}q-d&qe^{-dR}\\q\lambda e^{-dR}&q\lambda -d\end{array}}\right|=0~,\qquad E=-{\frac {d^{2}}{2}}~.}$

${\displaystyle d_{\pm }(\lambda )~=~{\textstyle {\frac {1}{2}}}q(\lambda +1)\pm {\textstyle {\frac {1}{2}}}\left\{q^{2}(1+\lambda )^{2}-4\,\lambda q^{2}\lbrack 1-e^{-2d_{\pm }(\lambda )R}]\right\}^{1/2}}$

${\displaystyle d_{\pm }=q[1\pm e^{-d_{\pm }R}]}$

${\displaystyle d_{\pm }=q~+~W(\pm qRe^{-qR})/R}$

## 外部链接

1. ^ D.R Herschbach, J.S. Avery, and O. Goscinski (eds.), Dimensional Scaling in Chemical Physics, Springer, (1992). [1]
2. ^ T.C. Scott, J.F. Babb, Alexander Dalgarno and John D. Morgan III, "The Calculation of Exchange Forces: General Results and Specific Models", J. Chem. Phys., 99, pp. 2841-2854, (1993). [2]