Time Independant Pertubation Theory - QM

[knowlewj01]

Homework Statement

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).

Homework 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

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 can't 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 independent case for first order pertubations?

Thanks

 

[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 ?

 

[vela]

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

 

[vela]

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

Also, is equation [1] a general result for the time independent case for first order pertubations?

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.