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

AI Thread Summary
The discussion revolves around solving the quantum harmonic oscillator (QHO) with a sinusoidal perturbing potential, VoSIN(BX). The initial approach involved a Taylor expansion of the potential and rewriting the position operator, but this proved unproductive. Participants suggested using perturbation theory to find the ground state energy and eigenket, emphasizing that first-order corrections can be calculated directly through integrals without needing a Taylor series. The original poster expressed confusion about handling the trigonometric function in the perturbation and was advised to rewrite it in exponential form. The conversation concluded with optimism about simplifying the problem using the suggested approach.
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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



H_{o} = \frac{P^{2}}{2m} + \frac{1}{2}m \varpi^{2}
H = H_{o} + Asin(BX)
E^{(o)}_{n}=\hbar\varpi(n+\frac{1}{2}) 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:
X = \sqrt{\frac{\hbar}{2m\varpi}}(a^{\dagger}+a)

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:

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

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!
 
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Have you thought about using perturbation theory? I am guessing V_0 is small since you mention it is a perturbative potential. So are you supposed to find the ground state to first order in V_0? Or 2nd order?
 
nickjer said:
Have you thought about using perturbation theory? I am guessing V_0 is small since you mention it is a perturbative potential. So are you supposed to find the ground state to first order in V_0? 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:

E_n^{(1)}=\langle n^{(0)}|V|n^{(0)} \rangle

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

E_n^{(1)}=\langle n^{(0)}|V|n^{(0)} \rangle

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!
 
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