An Electron in a Finite Well — Question 6

Question 6

A shallow one-dimensional well has a weakly bound state with quadratic binding energy

A symmetric finite well has $V=0$ for $|x|<a$ and $V=V_0>0$ outside. Bound energies satisfy $0<E<V_0$. Define $k=\sqrt{2mE}/\hbar$, $\kappa=\sqrt{2m(V_0-E)}/\hbar$, $z=ka$, $w=\kappa a$ and $z_0=a\sqrt{2mV_0}/\hbar$. Consider $z_0\ll1$ and the even ground state $w=z\tan z$, $z^2+w^2=z_0^2$. Define binding energy $B=V_0-E$.

1. Find the leading approximations for z and w.

2. Derive the leading binding energy in terms of m, $V_0$ and a.

Question 6 - Solution

A shallow one-dimensional well has a weakly bound state with quadratic binding energy

(a) Method and calculation

$$\tan z=z+O(z^3)\Rightarrow w\simeq z^2,\quad z^2+z^4\simeq z_0^2.$$$$\boxed{z\simeq z_0,\qquad w\simeq z_0^2}.$$

(b) Interpretation and further calculation

$$B=\frac{\hbar^2w^2}{2ma^2}\simeq\frac{\hbar^2z_0^4}{2ma^2}.$$$$\boxed{B\simeq\frac{2mV_0^2a^2}{\hbar^2}=V_0z_0^2},\quad\boxed{E/V_0\simeq1-z_0^2}.$$

Physical check

The binding energy is positive for every nonzero shallow well, but becomes very small. The exterior decay length diverges as the well disappears.