An Electron in a Finite Well — Question 2

Question 2

The first odd bound state appears only beyond a strict depth-width threshold

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 $z_0=a\sqrt{2mV_0}/\hbar$. For an odd state use $\psi=A\sin kx$ inside and a decaying exponential outside.

1. Derive the odd-state eigenvalue condition.

2. Find the threshold for the first odd bound state and explain why equality is excluded.

Question 2 - Solution

The first odd bound state appears only beyond a strict depth-width threshold

(a) Method and calculation

$$A\sin ka=B,\quad Ak\cos ka=-\kappa B\Rightarrow\boxed{-k\cot ka=\kappa}.$$$$\boxed{-z\cot z=\sqrt{z_0^2-z^2}},\qquad\pi/2<z<\pi\ \text{on the first odd branch}.$$

(b) Interpretation and further calculation

$$w\to0^+:\ z\to\pi/2\Rightarrow\boxed{z_0>\pi/2}.$$$$z_0=\pi/2:\ \kappa=0\Rightarrow\psi_{\mathrm{outside}}=\text{constant}\Rightarrow\boxed{\text{not normalizable}}.$$

Physical check

The threshold state is not a bound state. Every nonzero symmetric attractive well has an even ground state, but it need not have an odd state.