First order perturbation question

[Bobbo Snap]

Homework Statement

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, $\psi_1^1$.

Homework 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:
$$E_n = \frac{n^2 \pi^2 \hbar^2}{2ma^2}$$

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.

 

[TSny]

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.

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

 

[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.

 

[TSny]

Looks right to me.

 

[paranormal]

Your approach for part a is correct. For part b, the sum is over all possible values of m, not just odd ones. This is because the wavefunction for the infinite square well is not just restricted to odd values of n, it can also have even values. So the first three non-zero terms in the expansion of the correction to the ground state would be:

\psi_1^1 = \frac{\langle \psi_2^0 |H'| \psi_1^0 \rangle}{(E_1^0 - E_2^0)}\psi_2^0 + \frac{\langle \psi_3^0 |H'| \psi_1^0 \rangle}{(E_1^0 - E_3^0)}\psi_3^0 + \frac{\langle \psi_4^0 |H'| \psi_1^0 \rangle}{(E_1^0 - E_4^0)}\psi_4^0

Since m can be any value, including even ones, there will be non-zero terms in the expansion. The orthogonality condition you mentioned only applies when the values of n and m are different. In this case, n=1 and m=2,3,4, so the terms will not be zero.

I hope this helps clarify things for you. If you are still having trouble, I recommend seeking help from your instructor or a tutor.