How Does a Perturbation Affect Energy Levels in an Infinite Potential Well?

AI Thread Summary
The discussion focuses on calculating the first-order energy level changes for a particle in a one-dimensional infinite potential well subjected to a perturbation. The perturbation is defined as a delta function at the midpoint of the well, impacting the energy levels based on the parity of the quantum number 'n'. The first-order correction to the energy levels is derived using perturbation theory, yielding non-zero corrections for odd 'n' values and zero for even 'n' values. Participants emphasize the importance of considering the implications of the perturbation's effects on the eigenvalues. A Java applet is recommended for visualizing the effects of the perturbation on the energy levels.
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Sorry for all the questions - I tend to save them till I'm done with assignments:

Here's the question:
Consider a particle of mass 'm' in a one-dimensional infinite potential well of width 'a'
<br /> V (x) = \left\{\begin{array}{c} 0 \ \ \ if \ \ \ 0 \leq x \leq a \\ \infty \ \ \ otherwise<br />
The particle is subject to a perturbation of the form:
<br /> \omega (x) = a \omega_0 \delta \left(x - \frac{a}{2} \right)<br />
Where 'a' is a real constant with dimension of energy. Calculate the changes in the energy level of the particle in the first order of \omega_0

I just need some help starting off at this point. Can anyone suggest how to begin?
 
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What does first-order pertubation theory say?
 
Okay, I think I've got it. Does this look correct?
ANS:
I'm looking for first-order correction to the nth eigenvalue - so I need to solve this:
<br /> E_n^1 = \left&lt; \psi_n^0 | H&#039; | \psi_n^0 \right&gt;<br />
Where
<br /> \psi_n^0 (x) = \sqrt{ \frac{2}{a} } sin \left( \frac{n \pi x}{a} \right)<br />
and
<br /> H&#039; = a \omega_0 \ \delta \left( x - \frac{a}{2} \right)<br />
Substituting and solving gives:
<br /> E_n^1 = \left&lt; \psi_n^0 | H&#039; | \psi_n^0 \right&gt; = 2 \omega_0 \int_0^a sin^2 \left( \frac{n \pi x}{a} \right) \delta \left( x - \frac{a}{2} \right) dx<br />
<br /> = 2 \omega_0 sin^2 \left( \frac{n \pi}{2} \right)<br />
<br /> = \left\{\begin{array}{c} 2 \omega_0 \ \ \ if \ \ \ n = &quot;odd&quot; \\ 0 \ \ \ if \ \ \ n = &quot;even&quot; <br />
Do I "choose" only non-zero answers then, or is the array the complete answer? Thanks.
 
Looks right ! And no, you don't throw away the zero-terms. In fact, you should be asking yourself if it makes sense for the perturbation to have no effect on half the spectrum (the eigenvalues for even n).
 
Found this java applet from ZapperZ's link :

http://www.quantum-physics.polytechnique.fr/en/pages/p0204.html

It shows you the solutions to the SE for a double well. You can play with the potentials to essentially mimic your problem. Make a (infinitesimally) thin, high wall in the middle, and see what happens when you just barely increase its width : only the odd eigenvalues move, as predicted.
 
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