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.

Leave a Reply

Your email address will not be published. Required fields are marked *