An Electron in a Finite Well — Question 7

Question 7

A design window for exactly two bound states has one open and one closed endpoint

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 $z_0=a\sqrt{2mV_0}/\hbar$. Successive new bound states appear when $z_0$ exceeds $\pi/2,\pi,3\pi/2,\ldots$; a threshold solution itself is not normalizable. Take $V_0=1.00\,\mathrm{eV}$ and $\hbar^2/(2m_e)=.0381\,\mathrm{eV\,nm^2}$.

1. Find the allowed interval of $z_0$ for exactly two spatial bound states.

2. Convert it into an interval for the full well width $L=2a$.

Question 7 - Solution

A design window for exactly two bound states has one open and one closed endpoint

(a) Method and calculation

$$\text{Second state bound}:z_0>\pi/2,\quad\text{third absent}:z_0\le\pi.$$$$\boxed{\pi/2<z_0\le\pi}.$$

(b) Interpretation and further calculation

$$L=2z_0\sqrt{\frac{\hbar^2}{2m_eV_0}}.$$$$\boxed{.613214\,\mathrm{nm}<L\le1.22643\,\mathrm{nm}}.$$

Physical check

The upper endpoint still has only two normalizable bound states. Spin is excluded from this count; each spatial state is counted once.