# Perturbation Theory/Harmonic Oscillator

1. Apr 20, 2010

### Void123

1. The problem statement, all variables and given/known data

I am given the hamiltonian, where $$H^{^}_{0}$$ is that of the harmonic oscillator and the perturbation is (lambda)*(h-bar)*(omega)*[(lowering operator)^2 + (raising operator)^2]. I am asked to find the ground state, second-order approx. energy value.

2. Relevant equations

Second order eigenenergy equation.

3. The attempt at a solution

I have written out the whole hamiltonian. Do I need to expand the lowering-raising operators in terms of n? I am a bit lost on what to plug in to the second order equation for the two wave functions m and n (which sandwich the Hamiltonian operator).

2. Apr 21, 2010

### kuruman

You know that the "meat" of the sandwich is the perturbing Hamiltonian,

$$H'=\hbar\omega(a_+^2+a_-^2)$$

Since you are looking for the second order correction to the ground state, one of the "breads", say the bra, should be <0|. Suppose you were to write the other "bread" (the ket) as |n>. What do you get when you operate on that with a+2 and a-2? What does the resulting ket need to be in order not to have a zero matrix element?

3. Apr 24, 2010

### Void123

What if I am dealing with degenerate perturbation theory? In the case of the harmonic oscillator, do I need to set up the matrix and find the eigenvalues or can I just take advantage of the lowering and raising operators acting on the eigenstate? Thanks.

4. Apr 25, 2010

### kuruman

What is your Hamiltonian? More to the point, is the ground state degenerate? In any case, "setting up the matrix" involves calculating matrix elements using the raising and lowering operators as I showed you. That you should do, degeneracy or no degeneracy.

5. Apr 27, 2010

### Void123

I have a question though: if the harmonic oscillator is (isotropic), how would two (lowering and raising) operators multiply each other, assuming they are from two different dimensions (e.g. x and y). Would you 'go up or down the ladder' the same way one usually would except plug in the different n-values for x and y, accordingly?

In other words if you had $$a_{x} a_{y}$$, which are both lowering operators, would you go the typical route of bumping the eigenstate by |n-1>, |n-2>, etc. but plug in $$n_{x}$$ and $$n_{y}$$?

The raising and lowering operators is still a fuzzy area with me.

Last edited: Apr 27, 2010
6. Apr 27, 2010

### kuruman

That is why I asked you to tell me what your Hamiltonian looks like. If you have a two-dimensional isotropic Hamiltonian, you have degeneracies. However, if you write it in Cartesian coordinates, you can separate variables and you can treat each dimension independently. And yes, you get two ladders, one for x and one for y.