Wave Functions of a Trapped Electron — Question 5

Question 5

Reflection parity determines which position matrix elements must vanish

Define $y=x-L/2$ and $x_{mn}=\int_0^L u_m(x)x u_n(x)dx$. The infinite well occupies $0<x<L$; wave functions vanish outside. Its normalized modes are $u_n(x)=\sqrt{2/L}\sin(n\pi x/L)$ and energies $E_n=n^2E_1$. The reflection parity is $u_n(L-x)=(-1)^{n+1}u_n(x)$.

1. Show that off-diagonal $x_{mn}$ vanishes for m and n of the same parity.

2. Evaluate $x_{12}$ using $2\sin a\sin b=\cos(a-b)-\cos(a+b)$.

Question 5 - Solution

Reflection parity determines which position matrix elements must vanish

(a) Method and calculation

$$m\ne n:\quad x_{mn}=\int_0^L u_myu_n\,dx+\frac L2\int_0^L u_mu_n\,dx=\int_0^L u_myu_n\,dx.$$$$m,n\text{ same parity}:\ u_myu_n\text{ odd about }L/2\Rightarrow\boxed{x_{mn}=0}.$$

(b) Interpretation and further calculation

$$x_{12}=\frac1L\int_0^Lx[\cos(\pi x/L)-\cos(3\pi x/L)]dx.$$$$\boxed{x_{12}=-\frac{2L}{\pi^2}+\frac{2L}{9\pi^2}=-\frac{16L}{9\pi^2}}.$$

Physical check

The sign depends on the chosen phases of the eigenfunctions. A transition probability involves the squared magnitude, which is independent of that convention.