First order perturbation question

1. Mar 19, 2013

Bobbo Snap

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

Suppose we put a delta function bump in the center of the infinite square well:
$$H' = \alpha \delta(x -a/2),$$
where $\alpha$ is constant.

a) Find the first order correction to the allowed energies.

b) Find the first three non-zero terms in the expansion of the correction to the ground state, $\psi_1^1$.

2. Relevant equations

First order correction to the energy:
$$E_n^1 = \langle \psi_n^0 | H' | \psi_n^0 \rangle.$$

First order correction to the wave function:
$$\psi_n^1 = \sum_{m \neq n} \frac{\langle \psi_m^0 | H' | \psi_n^0 \rangle}{(E_n^0 - E_m^0)}\psi_m^0.$$

Wavefunction for the infinite square well:
$$\psi_n(x) = \sqrt{2/a} \sin{\frac{n \pi}{a}x}$$

Energy for the infinite square well:
$$E_n = \frac{n^2 \pi^2 \hbar^2}{2ma^2}$$

3. The attempt at a solution

I think I got part a right. It worked out to $\frac{2 \alpha}{a}$ for odd n and 0 for even n.

Part b has me a little confused. In the equation for the first order correction to the wavefunction, what am I supposed to sum over? If it is m, then the first three terms are m=3, m=5, m=7 since m is not n and m must be odd. But then I have $\langle \psi_3^0 |H'|\psi_1^0\rangle, \langle \psi_5^0 |H'| \psi_1^0 \rangle, \text{ and } \langle \psi_7^0 |H'| \psi_1^0 \rangle$ in the numerators of the first three terms. But wouldn't orthogonality then make all of these terms zero so there would be no non-zero terms in the expansion? I'm not understanding how to evaluate this sum. Any help would be appreciated.

2. Mar 19, 2013

TSny

What does $\langle \psi_3^0 |H'|\psi_1^0\rangle$ look like when written out as an integral?

3. Mar 19, 2013

Bobbo Snap

$$\langle \psi_3^0|H'|\psi_1^0 \rangle = \frac{2 \alpha}{a}\int \sin{\frac{3 \pi x}{a}} \sin{\frac{\pi x}{a}} \delta(x - a/2) = \frac{2 \alpha}{a} \sin{\frac{3 \pi}{2}} \sin{\frac{\pi }{2}} = - \frac{2 \alpha}{a}$$

I worked out something like this for all three terms (m=3, m=5, m=7). Is that the correct approach? I thought orthogonality might make this go to zero.

4. Mar 19, 2013

TSny

Looks right to me.