# Delta function well as limit of finite square well

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.31.

The finite square well has the potential

$\displaystyle V(x)=\begin{cases} 0 & x<-a\\ -V_{0} & -a\le x\le a\\ 0 & x>a \end{cases} \ \ \ \ \ (1)$

where ${V_{0}}$ is a positive constant energy, and ${a}$ is a constant location on the ${x}$ axis.

The delta function potential ${V=-\alpha\delta(x)}$ can be thought of as the limit of the finite square well as ${a\rightarrow0}$ and ${V_{0}\rightarrow\infty}$ in such a way that the area of the rectangle in the well is a constant. That is, the integral of the potential is the same in both cases, so that

$\displaystyle 2aV_{0}=\alpha \ \ \ \ \ (2)$

The energies of the bound states for the even solution of the finite square well are given by

$\displaystyle \tan z=\sqrt{\frac{z_{0}^{2}}{z^{2}}-1} \ \ \ \ \ (3)$

where

 $\displaystyle z_{0}$ $\displaystyle \equiv$ $\displaystyle \frac{a}{\hbar}\sqrt{2mV_{0}}\ \ \ \ \ (4)$ $\displaystyle z$ $\displaystyle \equiv$ $\displaystyle \frac{a}{\hbar}\sqrt{2m(E+V_{0})} \ \ \ \ \ (5)$

Substituting for ${V_{0}}$ we get

 $\displaystyle z_{0}$ $\displaystyle =$ $\displaystyle \frac{1}{\hbar}\sqrt{ma\alpha}\ \ \ \ \ (6)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle \frac{1}{\hbar}\sqrt{2ma^{2}E+ma\alpha} \ \ \ \ \ (7)$

As ${a\rightarrow0}$, ${z_{0}/z\rightarrow1}$, so ${\tan z}$ becomes very small. In this limit, ${\tan z\approx z}$, so we can approximate equation 3 by

 $\displaystyle z$ $\displaystyle =$ $\displaystyle \sqrt{\frac{z_{0}^{2}}{z^{2}}-1}\ \ \ \ \ (8)$ $\displaystyle \frac{1}{\hbar}\sqrt{2ma^{2}E+ma\alpha}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\alpha}{2aE+\alpha}-1}\ \ \ \ \ (9)$ $\displaystyle \frac{1}{\hbar^{2}}\left(2ma^{2}E+ma\alpha\right)$ $\displaystyle =$ $\displaystyle \frac{\alpha}{2aE+\alpha}-1\ \ \ \ \ (10)$ $\displaystyle \left(2ma^{2}E+ma\alpha\right)\left(2aE+\alpha\right)$ $\displaystyle =$ $\displaystyle -2aE\hbar^{2} \ \ \ \ \ (11)$

If we retain only the term in ${a}$ (discarding higher powers of ${a}$), we get

 $\displaystyle ma\alpha^{2}$ $\displaystyle =$ $\displaystyle -2aE\hbar^{2}\ \ \ \ \ (12)$ $\displaystyle E$ $\displaystyle =$ $\displaystyle -\frac{m\alpha^{2}}{2\hbar^{2}} \ \ \ \ \ (13)$

This is the energy we found earlier when analyzing the delta function well.

We can do a similar analysis for the scattering states. The transmission coefficient for the finite square well is

$\displaystyle T^{-1}=1+\frac{V_{0}^{2}}{4E\left(E+V_{0}\right)}\sin^{2}\left(\frac{2a}{\hbar}\sqrt{2m\left(E+V_{0}\right)}\right) \ \ \ \ \ (14)$

For small ${a}$ we can use the approximation ${\sin x\approx x}$ and we get

$\displaystyle T^{-1}\approx1+\frac{2mV_{0}^{2}a^{2}}{\hbar^{2}E} \ \ \ \ \ (15)$

Substituting for ${V_{0}}$ gives

 $\displaystyle T^{-1}$ $\displaystyle \approx$ $\displaystyle 1+\frac{m\alpha^{2}}{2\hbar^{2}E}\ \ \ \ \ (16)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{1}{1+m\alpha^{2}/2\hbar^{2}E} \ \ \ \ \ (17)$

This is the same formula we obtained when analyzing the delta function potential directly.