How Do You Solve the QHO with a Sinusoidal Perturbing Potential?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 3K views
taloc
Messages
3
Reaction score
0

Homework Statement


I am tasked with solving the QHO with a sinusoidal perturbing potential of the form VoSIN(BX). I need to find the ground state energy as well as the ground state eigenket |g>.

Homework Equations



[tex]H_{o} = \frac{P^{2}}{2m} + \frac{1}{2}m \varpi^{2}[/tex]
[tex]H = H_{o} + Asin(BX)[/tex]
[tex]E^{(o)}_{n}=\hbar\varpi(n+\frac{1}{2})[/tex] Which is the unperturbed energy

The Attempt at a Solution


My first stab at this problem involved performing a Taylor expansion of the potential and the rewriting of the X operator in terms of the creation and annihilation operators:
[tex]X = \sqrt{\frac{\hbar}{2m\varpi}}(a^{\dagger}+a)[/tex]

This process was not very rewarding. I found myself with no method for determining when to terminate the expansion.

An alternative approach would be to rewrite the potential as:

[tex]Asin(BX) = \frac{A}{2i}(e^{iBX}-e^{-iBX})[/tex]

My concern with this method is that the |n> kets used in the QHO are not eigenkets of X and therefore do not play nicely with the exponentials. Do I need to perform a change of basis? Essentially creating a new set of kets composed of a linear combination of the |n> kets? Any advice would be greatly appreciated!

Cheers!
 
Last edited:
Physics news on Phys.org
Have you thought about using perturbation theory? I am guessing [itex]V_0[/itex] is small since you mention it is a perturbative potential. So are you supposed to find the ground state to first order in [itex]V_0[/itex]? Or 2nd order?
 
nickjer said:
Have you thought about using perturbation theory? I am guessing [itex]V_0[/itex] is small since you mention it is a perturbative potential. So are you supposed to find the ground state to first order in [itex]V_0[/itex]? Or 2nd order?

nickjer,

Thank you for the reply. I do need to apply pertubation theory to solve this problem. The problem I am facing involves having the X operator of the perturbing potential "locked up" inside a trigometric function. If I were to expand the trig function in a Taylor Series I could then perform the analysis for first order corrections (which vanish) and then proceed to second order corrections. I asked my instructor about the Taylor expansion method and was told this was not the appropriate path to take. He confirmed that I must rewrite the Sin function in terms of the exponentials I gave in the initial post. I am completely lost as to how I should proceed from this point on.

Cheers,
 
First order perturbation is just an integral over the perturbing potential:

[tex]E_n^{(1)}=\langle n^{(0)}|V|n^{(0)} \rangle[/tex]

So just integrate over that potential with the ground state wavefunctions for a simple harmonic oscillator. No need for Taylor series.
 
Last edited:
nickjer said:
First order perturbation is just an integral over the perturbing potential:

[tex]E_n^{(1)}=\langle n^{(0)}|V|n^{(0)} \rangle[/tex]

So just integrate over that potential with the ground state wavefunctions for a simple harmonic oscillator. No need for Taylor series.

Wow! Is that all i need to do? That is far easier than i expected. i will attempt to generate the ground state energy tomorrow morning. I was attempting to evaluate the expression in a much more general case. I suspect that the integral will be zero and require a second order approximation, but here is to hoping.

Thank you very much for your help nickjer, i will get back to you in the morning with my results.

Ceers!