# Determining exact solutions to a perturbed simple harmonic oscillator

1. Jan 21, 2013

### slimjim

1. The problem statement, all variables and given/known data
Consider as an unperturbed system H0 a simple harmonic oscillator with mass m,
spring constant k and natural frequency w = sqrt(k/m), and a perturbation H1 = k′x =
k′sqrt(hbar/2m)(a+ + a−)

Determine the exact ground state energy and wave function of the perturbed system

here a+ and a- are the ladder operators: a+/-=1/sqrt( 2hbar*m*ω)(-/+p^2 +(mωx) )

and H0=1/(2m)*(p^2 +(mωx)^2) = hbar*ω*[(a+)*(a-)+1/2]

3. The attempt at a solution

I just need a nudge in the right direction for this one. The questoin confuses me because I didnt realize exact solutions could be found for perturbed systems.

we want to solve Hψ=Eψ where H= H0 + λH1
(the unperturbed hamiltonian plus the perturbation)

I am confused about the factor of λ on H1.

griffiths text says " for the moment we'll take lambda to be a small number, later we'll crank it up to one, and H will be the true hamiltonian"

so should I take λ=1?

either way, when I write out Hψ=Eψ, i get a very nasty diff. eq. which leads me to believe there is a better method.

2. Jan 21, 2013

### Dick

Some systems can be solved exactly no matter how large the perturbation is. The potential energy in the hamiltonian is kx^2/2. The perturbation is k'x when lambda=1. You can complete the square in x to get a new hamiltonian with a shifted center and shifted energy that looks a lot like the first.

3. Jan 22, 2013

### slimjim

I did as you suggested. The potential gets shifted left in x, and decreased by a constant amount.

Coincidentally, the potential is uniformly decreased by the same value I calculated for E2, the second order perturbation in energy.

So I am thinking that the exact ground state energy would be decreased by this amount as well.

And the wave functions would simply be shifted to the left (-x) by the same amount that the potential was shifted. (the wave func. should have same shape as the wavefunction for an unperturbed simple harmonic oscillator, and must also be normalized)

4. Jan 22, 2013

### Dick

I haven't really done the perturbative calculation for a long time, but that sounds right.