# Homework Help: Time Independant Pertubation Theory - QM

1. Dec 16, 2011

### knowlewj01

1. The problem statement, all variables and given/known data

An electron is confined to a 1 dimensional infinite well $0 \leq x \leq L$
Use lowest order pertubation theory to determine the shift in the second level due to a pertubation $V(x) = -V_0 \frac{x}{L}$ where Vo is small (0.1eV).

2. Relevant equations

[1]
$E_n \approx E_n^{(0)} + V_{nn}$

[2]
$V_{nn} = \int_{-\infty}^{\infty} \psi_{n}^{(0) *} (x) V(x) \psi_{n}^{(0)} (x) dx$

the following integral may be useful:

[3]
$\int_{0}^{2\pi}\phi sin^2 \phi d\phi = \pi^2$

3. The attempt at a solution

From [1] and the known result for E2 of an infinite well
$E_2 = \frac{4\hbar^2 \pi^2}{2mL^2} - \frac{2V_0}{L^2}\int_{0}^{L} x sin^2\left(\frac{2\pi x}{L}\right) dx$

I cant see a substitution that will get it into the form in [3], anyone have any ideas?
Also, is equation [1] a general result for the time independant case for first order pertubations?

Thanks

2. Dec 16, 2011

### knowlewj01

If i were to make the substitution:

$\phi = \frac{2\pi x}{L}$

$\frac{\phi}{x} = \frac{2\pi}{L}$

does this imply that the limits of integration change from L to 2π ?

Last edited: Dec 16, 2011
3. Dec 16, 2011

### vela

Staff Emeritus
Yes. Simply plug in the limits for x to find the limits for ɸ.

4. Dec 17, 2011

### vela

Staff Emeritus
No, it isn't. It applies when the energy eigenstates of the unperturbed Hamiltonian states are non-degenerate, like in this problem. You'll have to use a different approach for the degenerate case.