Find the ground state energy of a particle restricted to move in one dimension subject to the potential below using perturbation theory.
The box is like a normal box except it has a step in the middle of it making it have 2 wells with length a and the width of the step is b.
I'm not sure how to account for the potential since there are two wells.
My setup right now using braket notation is E = 2<Y|Y> + <Y|Vo|Y> where Y is the wavefunction Yo = (2/a)1/2 sin(nπx/a)
Can I get some feedback as to if I am headed in the right direction or how to get there if I am not?
Thank you.
That situation is treated several places online, including here:
http://www.hep.man.ac.uk/u/forshaw/BoseFermi/Double%20Well.html
Thanks for the link. It is a bit confusing though since it deals with 2 particles. I'm not really sure which functions apply to my problem.
The example deals with two wave functions for one particle, if I understand it correctly. One is symmetric and the other antisymmetric. If the partcle starts in one well it leaks through to the other well.
My quantum mechanics is very rusty so I'm afraid that is all the help i'm able to give you.
Write the potential as:
V + V'
where V is the ordinary square well and V' is the step of width b in the middle.
Then, to first order in perturbation theory, the energy increases by:
<Y|V'|Y>
This can be written as the integral:
Int |Y(x)|^2 V'(x) dx
Ok well I did the problem and came up with a really long answer. My concern is that I used the wrong limits. If the box has segments of a, b, and a, the perturbation part is going to be from a to a+b. This is what I integrate over, correct? And also, the original wavefunction that I used, the a in the denominator of the sin function and in the 2/a part should be replaced with the width of the whole box 2a+b since that is the unperturbed system. I'm just not sure of my answer because nothing really canceled.
If you put one boundary of the box at
x = 0 and the other at x = L, then you are saying that L = 2 a + b. The perturbation is nonzero over a length b near the middle (around L/2)?
The unperturbed wavefunction is:
Y(x) = sqrt(2/L)Sin(pi x/L)
What you need to do is integrate:
|Y(x)|^2 V(x)dx
where V(x) is the perturbation
This gives the correction to the ground state energy.
Right, the whole box is length 2a+b. The step is positioned right in the middle at length b with two equal lengths of a on either side of it. The height of the step is the perturbation Vo. For L I put in the length of the whole box 2a+b. I integrated from a to a+b which is the length of the perturbation. I see that you did not put an "n" in your function. Do you assume n=1 for ground state?
Yes, n = 1 for the ground state. You just add the result of the integral to the unperturbed ground state energy of
hbar^2 pi^2/(2 m L^2)
to obtain the ground state energy to first order in the pertubation.
Yep that's what I did. Thanks for your help Count Iblis and drwls.
I'm actually working on this same problem and in a desparate attempt found this online looking for help.
I'm still slightly confuse about the limits of integration and what to use for V(x). When integrating how do you handle the V(x) term?
With the limits (from a => a+b) with my Y as (2/(2a+b))^.5sin(pi*x/(2a+b)), I get a huge number of uncancelled terms that just seem to sit there. Is this correct?