Determining exact solutions to a perturbed simple harmonic oscillator

[slimjim]

Homework Statement

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]

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.

 

[Dick]

Homework Statement

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]

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.

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.

 

[slimjim]

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.

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)

 

[Dick]

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)

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