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.

 

[paranormal]

I would recommend using perturbation theory to solve this problem. This involves treating the perturbation as a small correction to the unperturbed system and using a series expansion to approximate the energy and wave function. This method is often used for systems that cannot be solved exactly.

To start, you can assume a perturbation parameter, such as λ, and expand the energy and wave function in terms of this parameter. Then, using the perturbation Hamiltonian, H1, you can calculate the first-order correction to the energy and wave function. This will give you an approximate solution that can be refined by including higher-order corrections.

Alternatively, you can use the variational method to approximate the ground state energy and wave function. This involves choosing a trial wave function and minimizing the expectation value of the Hamiltonian with respect to this wave function. This method can also be used to improve the initial perturbation theory solution.

In any case, it is important to keep in mind the physical meaning of the perturbation and the assumptions made in the perturbation theory approach. It is also helpful to compare the approximate solution to the exact solution of the unperturbed system to ensure the results are reasonable.