- #1

McCoy13

- 74

- 0

**First order correction to particle-in-box eigenstates for Dirac perturbation**

## Homework Statement

Calculate the first three nonzero terms in the expansion of the correction to the ground state [itex]\psi^{1}_{1}[/itex] for a Dirac delta perturbation of strength alpha at a/2 (box from 0 to a).

## Homework Equations

[itex]\psi^{1}_{n} = \sum_{m\neqn} \frac{\left\langle\psi^{0}_{m}\right|H'\left|\psi^{0}_{n}\right\rangle}{E^{0}_{n}-E^{0}_{m}}\psi^{0}_{m}[/tex]

## The Attempt at a Solution

I started with m=2 and tried to computer the integral [tex]\left\langle\psi^{0}_{m}\right|H'\left|\psi^{0}_{n}\right\rangle[/tex] by parts.

Calculations:

[tex]\frac{2}{a}\alpha\int^{a}_{0}sin(\frac{2 \pi x}{a})sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})dx[/tex]

integration by parts

[tex]\frac{2}{a}\alpha([sin(\frac{2 \pi x}{a})sin(\frac{\pi}{2})]^{a}_{0}-\frac{2\pi}{a}\int^{a}_{0}cos(\frac{2 \pi x}{a})dx)[/tex]

This clearly equals 0. Also, this equation indicates that the correction will be 0 for any m even.

Trying m=3.

[tex]\frac{2}{a}\alpha\int^{a}_{0}sin(\frac{\pi x 3}{a})sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})dx[/tex]

integration by parts

[tex]\frac{2}{a}\alpha([sin(\frac{\pi x 3}{a})sin(\frac{\pi}{2})]^{a}_{0}-\frac{3\pi}{a}\int^{a}_{0}cos(\frac{3 \pi x}{a})dx)[/tex]

However, this also is 0, and will be for any m odd.

This clearly cannot be the correct result.

Last edited: