An Electron in a Finite Well — Question 4

Question 4

Normalizing the exterior tails gives a nonzero probability beyond the walls

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 $z_0=a\sqrt{2mV_0}/\hbar$. An even state has $ka=\kappa a=\pi/4$ and $\psi=A\cos kx$ inside. Outside, $\psi=A\cos(ka)e^{-\kappa(|x|-a)}$. Take A positive.

1. Normalize the state including both exterior tails.

2. Find the total probability outside the well.

Question 4 - Solution

Normalizing the exterior tails gives a nonzero probability beyond the walls

(a) Method and calculation

$$1=A^2\left[a+\frac{\sin2ka}{2k}+\frac{\cos^2ka}{\kappa}\right].$$$$ka=\kappa a=\pi/4:\quad\boxed{A^2=\frac1{a(1+4/\pi)}}.$$

(b) Interpretation and further calculation

$$P_{\mathrm{out}}=2A^2\cos^2ka\int_0^\infty e^{-2\kappa y}dy=\frac{A^2\cos^2ka}{\kappa}.$$$$\boxed{P_{\mathrm{out}}=\frac2{\pi+4}=.280050},\quad\boxed{P_{\mathrm{in}}=.719950}.$$

Physical check

Normalizing only the interior would overestimate all probabilities. The stated parameters satisfy the even matching equation because $\tan(\pi/4)=1$.