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.
cyberdeathreaper
Messages
46
Reaction score
0
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?
 
Physics news on Phys.org
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.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top