An Electron in a Finite Well — Question 1

Question 1

Even finite-well states are selected by matching logarithmic derivatives

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 even state use $\psi=A\cos kx$ inside and $\psi=Be^{-\kappa(x-a)}$ for $x>a$. The mass is the same everywhere.

1. Match the wave function and derivative at a and eliminate the amplitudes.

2. Express the eigenvalue condition in z and $z_0$, and identify the branch containing the ground state.

Question 1 - Solution

Even finite-well states are selected by matching logarithmic derivatives

(a) Method and calculation

$$A\cos ka=B,\qquad-Ak\sin ka=-\kappa B.$$$$\boxed{k\tan ka=\kappa}.$$

(b) Interpretation and further calculation

$$k^2+\kappa^2=2mV_0/\hbar^2\Rightarrow\boxed{z\tan z=\sqrt{z_0^2-z^2}}.$$$$\boxed{0<z<\min(z_0,\pi/2)}\quad\text{for the nodeless ground branch}.$$

Physical check

A finite wall does not require the wave function to vanish at its edge. Both the value and derivative are continuous for a finite step with constant mass.