Infinite square well perturbation

[kreil]

Homework Statement

A particle of mass m in the infinite square well is subjected to the perturbation H'=Vo, 0<x<L/2, H'=0 else.

(a) use first order perturbation theory to calculate the energies of the particle

(b) what are the first order corrected wave functions?

(c) if the particle is an electron, how do the frequencies emitted by the perturbed system compare with those of the unperturbed system?

Homework Equations

E_n^{(1)}=H&#039;_{nn}=&lt;E_n^{(0)}|H&#039;|E_n^{(0)}&gt;

|E_n^{(1)}&gt; = \sum_{m \notequal n} \frac{H_{mn}^&#039;}{E_n^{(0)}-E_m^{(0)}} |E_m^{(0)}&gt;

The Attempt at a Solution

I solved part (a) and got an answer of Vo/2..i.e. each energy level is shifted up by this amount.

For part (b), I got that there is no correction to the wavefunctions, since m cannot equal n and

H&#039;_{mn}=\frac{2V_0}{L} \int_0^{L/2} sin(\frac{n \pi}{L}x)sin(\frac{m \pi}{L}x) dx = \frac{V_0}{2} \delta_{mn}

did I do part (b) correctly?

also, part (c) is confusing me..does anyone have any thoughts?

 

[nickjer]

I believe you did part (b) wrong. You might want to double check your integral. For example, if n=1 and m=2 you get 4V/3\pi. As long as one of them is even and the other odd, you will get a solution that is nonzero.

 

[kreil]

oo yes you're absolutely correct, the late hour affected my thinking.

since m can be any integer not equal to n, am i allowed to just set m=n+1? that would certainly help in calculating the denominator (and I don't see how else the denominator can be calculated unless m is defined in terms of n...)

 

[nickjer]

No, because you can also have n=1 and m=4, which violates your m=n+1 rule you made up. If you know n is even, then you are just taking a sum of all odd numbers. If n is odd, then you are taking a sum of all even numbers.

 

[kreil]

so how am I to explicitly calculate the integral?

 

[kreil]

wolfram alpha gave me the following

\int_0^{L/2} sin(\frac{m \pi x}{L}) sin(\frac{n \pi x}{L}) dx = \frac{L}{2 \pi} \left( \frac{sin(\frac{\pi}{2}(m-n))}{(m-n)} - \frac{sin(\frac{\pi}{2}(m+n))}{(m+n)} \right)

 

[vela]

Use the trig identity \sin A \sin B = [\cos(A-B)-\cos(A+B)]/2.

 

[nickjer]

Also, just solve for n=odd and n=even separately.