An Electron in a Finite Well — Question 3

Question 3

Numerical intersections determine the two bound energies of a moderate well

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 $z_0=a\sqrt{2mV_0}/\hbar$. Let $z_0=2.00$. Solve the even equation $z\tan z=\sqrt{4-z^2}$ and odd equation $-z\cot z=\sqrt{4-z^2}$. A calculator or numerical root finder may be used.

1. Locate the two roots on their physical branches.

2. Find both energies as fractions of $V_0$ and their binding energies relative to the continuum.

Question 3 - Solution

Numerical intersections determine the two bound energies of a moderate well

(a) Method and calculation

$$0<z_e<\pi/2:\quad\boxed{z_e=1.02986653}.$$$$\pi/2<z_o<2:\quad\boxed{z_o=1.89549427}.$$

(b) Interpretation and further calculation

$$E/V_0=z^2/z_0^2\Rightarrow\boxed{E_e/V_0=.265156,\quad E_o/V_0=.898225}.$$$$\boxed{(V_0-E_e)/V_0=.734844,\quad(V_0-E_o)/V_0=.101775}.$$

Physical check

The odd state is closer to the continuum and therefore has a longer decay length. A root across a tangent pole would not be a valid numerical solution.