Perturbation: First order correction to particle-in-box eigenstates

[McCoy13]

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 $\psi^{1}_{1}$ 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]<h2>The Attempt at a Solution</h2><br /> I started with m=2 and tried to computer the integral $$\left\langle\psi^{0}_{m}\right|H'\left|\psi^{0}_{n}\right\rangle$$ by parts.<br /> <br /> Calculations:<br /> $$\frac{2}{a}\alpha\int^{a}_{0}sin(\frac{2 \pi x}{a})sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})dx$$<br /> integration by parts<br /> $$\frac{2}{a}\alpha([sin(\frac{2 \pi x}{a})sin(\frac{\pi}{2})]^{a}_{0}-\frac{2\pi}{a}\int^{a}_{0}cos(\frac{2 \pi x}{a})dx)$$<br /> <br /> This clearly equals 0. Also, this equation indicates that the correction will be 0 for any m even.<br /> <br /> Trying m=3.<br /> $$\frac{2}{a}\alpha\int^{a}_{0}sin(\frac{\pi x 3}{a})sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})dx$$<br /> integration by parts<br /> $$\frac{2}{a}\alpha([sin(\frac{\pi x 3}{a})sin(\frac{\pi}{2})]^{a}_{0}-\frac{3\pi}{a}\int^{a}_{0}cos(\frac{3 \pi x}{a})dx)$$<br /> <br /> However, this also is 0, and will be for any m odd.<br /> <br /> This clearly cannot be the correct result.[/itex]

 

[kuruman]

It is not. What is the integral

$$\int^{\infty}_{-\infty}f(x) \delta(x-x_0)dx$$

?

 

[McCoy13]

$$f(x_{0})$$

In this case that gives:

$$sin(\frac{\pi}{2}) = 1$$

when taking

$$sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})$$

to be the part that gets integrated in the integration by parts.

 

[kuruman]

Forget "by parts" and make your life simple. Identify

$$f(x)=sin(\frac{\pi x 3}{a})sin(\frac{\pi x}{a})$$

and see what you get.

 

[McCoy13]

Oh. That's easy. Thanks.

Just to confirm my answer, I've now got:

for m = 3

$$\frac{\frac{2}{a}\alpha\int^{a}_{0}sin(\frac{3 \pi x}{a})sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})dx}{E^{(0)}_{1}-E^{(0)}_{3}}$$

= $$\frac{\frac{2}{a}\alpha sin(\frac{3 \pi}{2})sin(\frac{\pi}{2})}{E^{(0)}_{1}-E^{(0)}_{3}}}$$

= $$\frac{-2 \alpha}{a(E^{(0)}_{1}-E^{(0)}_{3})}$$

= $$\frac{-2 \alpha}{a(\frac{\pi^{2} h^{2}}{2ma^{2}} - \frac{9 \pi^{2} h^{2}}{2ma^{2}})}$$

= $$\frac{ma \alpha}{2 \pi h^{2}}$$

Where h actually stands for h-bar and m stands for mass (rather than the index).