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

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SUMMARY

The discussion focuses on solving the Quantum Harmonic Oscillator (QHO) with a sinusoidal perturbing potential of the form V₀sin(BX). The primary method suggested for finding the ground state energy and eigenket |g> involves applying perturbation theory, specifically calculating the first-order correction using the integral E_n^{(1)} = ⟨n^{(0)}|V|n^{(0)}⟩ with the ground state wavefunctions of the simple harmonic oscillator. The use of Taylor expansion for the potential was deemed inappropriate, and rewriting the sinusoidal function in terms of exponentials was recommended. Participants confirmed that the integral may yield zero, necessitating a second-order approximation.

PREREQUISITES
  • Understanding of Quantum Mechanics principles, particularly the Quantum Harmonic Oscillator.
  • Familiarity with perturbation theory and its applications in quantum systems.
  • Knowledge of operator algebra, specifically creation and annihilation operators.
  • Ability to perform integrals involving wavefunctions and potential energy terms.
NEXT STEPS
  • Study the application of first-order perturbation theory in quantum mechanics.
  • Learn how to express trigonometric functions in terms of exponential functions for quantum calculations.
  • Explore the implications of second-order perturbation theory and its calculations.
  • Review the properties of ground state wavefunctions for the Quantum Harmonic Oscillator.
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Students and researchers in quantum mechanics, particularly those working on perturbation theory and the Quantum Harmonic Oscillator. This discussion is beneficial for anyone seeking to deepen their understanding of quantum perturbative techniques.

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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!
 
Last edited:
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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.
 
Last edited:
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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