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 independant 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 independant 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.

 

[paranormal]

for any help!
Yes, equation [1] is a general result for first order perturbation theory in the time independent case. To solve this problem, you can use the fact that the integral in [2] can be simplified to [3] since the perturbation potential is linear in x. Therefore, we can rewrite [2] as:

V_{nn} = -\frac{V_0}{L}\int_{0}^{L} x sin^2\left(\frac{2\pi x}{L}\right) dx

Using the substitution u = 2\pi x/L, we can rewrite the integral as:

V_{nn} = -\frac{V_0}{L}\int_{0}^{2\pi} \frac{L}{2\pi} u sin^2(u) du

Using [3], this becomes:

V_{nn} = -\frac{V_0}{2\pi}\left(\frac{L^2}{2}\right) \pi^2 = -\frac{V_0 L^2}{4}

Substituting this into [1], we get the new energy for the second level as:

E_2 = \frac{4\hbar^2 \pi^2}{2mL^2} - \frac{V_0 L^2}{4} = \frac{4\hbar^2 \pi^2}{2mL^2} - \frac{V_0}{4}

Therefore, the shift in the second level due to the perturbation is -V_0/4, which is a small correction to the original energy.