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

  • Thread starter McCoy13
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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:

Answers and Replies

  • #2
kuruman
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It is not. What is the integral

[tex]\int^{\infty}_{-\infty}f(x) \delta(x-x_0)dx[/tex]

?
 
  • #3
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[tex]f(x_{0})[/tex]

In this case that gives:

[tex]sin(\frac{\pi}{2}) = 1[/tex]

when taking

[tex]sin(\frac{\pi x}{a})\delta(x-\frac{a}{2})[/tex]

to be the part that gets integrated in the integration by parts.
 
  • #4
kuruman
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Forget "by parts" and make your life simple. Identify

[tex]
f(x)=sin(\frac{\pi x 3}{a})sin(\frac{\pi x}{a})
[/tex]

and see what you get.
 
  • #5
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Oh. That's easy. Thanks.

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

for m = 3

[tex]\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}}[/tex]

= [tex]\frac{\frac{2}{a}\alpha sin(\frac{3 \pi}{2})sin(\frac{\pi}{2})}{E^{(0)}_{1}-E^{(0)}_{3}}}[/tex]

= [tex]\frac{-2 \alpha}{a(E^{(0)}_{1}-E^{(0)}_{3})}[/tex]

= [tex]\frac{-2 \alpha}{a(\frac{\pi^{2} h^{2}}{2ma^{2}} - \frac{9 \pi^{2} h^{2}}{2ma^{2}})}[/tex]

= [tex]\frac{ma \alpha}{2 \pi h^{2}}[/tex]

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

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