# Perturbation Theory

1. Dec 12, 2006

### holden

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

I'm given that a harmonic oscillator is in a uniform gravitational field so that the potential energy is given by: $$V(x)=\frac{1}{2}m\omega^2x^2 - mgx$$, where the second term can be treated as a perturbation. I need to show that the first order correction to the energy of a stationary state is zero, and (the part I'm having trouble with) find the new energy eigenvalues to second order.

2. Relevant equations

$$H^1_{nn} = <\psi_n|H^1\psi_n>$$
$$E_n^2 = \Sigma_{j!=n}\frac{|H_{jn}|^2}{E_n^0 - E_j^0}$$

3. The attempt at a solution

For the first part, I just took -mgx as H^1 and used the first equation above, plugging in $$a_- - a_+$$ for x, so that I got $$(a_- - a_+)\psi_n^0$$, which is zero (right?)

For the second part.. I'm not sure how to go about using the second equation. $$H_{jn} = <\psi_j^0|H\psi_n^0>$$.. but what do I use for H there? The same thing as above? And.. how do I go about doing a summation for these? I can't find any examples for the second order correction. I think En is just $$\frac{\hbar\omega}{2}$$, but Ej confuses me.

Thanks a ton for the help.

Last edited: Dec 12, 2006
2. Dec 12, 2006

### Physics Monkey

When doing perturbation theory, you should put thing doing the perturbing inside your matrix elements. Also, you have already observed that since the perturbing potential is linear in x, many simplifications occur. So before worrying about the sum, try computing the matrix elements. You might then be surprised at how easy the sum is.

Let me know if you need more help.

3. Dec 12, 2006

### holden

Thanks for the response! I guess my question is that I don't know how to calculate the matrix elements.. I gave it a shot, though.. here's what I came up with:

$$H_{jn} = <\psi_j^0|H^1\psi_n^0>$$
$$= -mg<\psi_j^0|x\psi_n^0>$$
$$= -mg<\psi_j^0|(a_-- a_+)\psi_n^0>$$
$$= mg\sqrt{(n+1)\hbar\omega}\int_{-\infty}^{\infty}\psi_j^0\psi_{n+1}^0$$

So it seems the only way this equation can give you anything is if n=j-1... so then you'd just get one for the integral. So then I guess there's nothing left to sum and you get $$\frac{m^2g^2(n+1)\hbar\omega}{E_n^0 - E_j^0}$$.. I doubt this is right, but even if it is, now I'm not sure what En and Ej are..

Last edited: Dec 12, 2006
4. Dec 12, 2006

### Physics Monkey

Ok, holden, now we've got a concrete problem we can tackle. You're actually almost right, although you've missed a few terms. So how to calculate matrix elements? What you want to do is use the properties of creation and annihilation operators to simplify all calculations.

What does a creation operator do to a state $$\psi_n$$? The answer is that $$a^+ \psi_n$$ is proportional to $$\psi_{n+1}$$. In other words, it raises the n index by one. A similar story holds for the annihilation operator, it lowers the n index by one.

Thus, if I want to calculate the matrix element $$\langle \psi_n | a^+ | \psi_m \rangle$$, for all n and m, this is very easy because n must be just m+1 otherwise the matrix element is zero. Again, a similar argument applies to the annihilation operator.

Therefore the matrix elements you have to calculate are very easy. One involves a lowering operator and the other a creation operator. Now you tell me how will j be related to n in each case and what is the value of matrix element in each case?

PS j is just another label like n which runs over all energy states (except n).

5. Dec 12, 2006

### holden

Oh, OK. I was just leaving off the annihilation operator term. I already said that j = n+1 for the case involving the creation operator, and so the matrix element is $$-mg\sqrt{(n+1)\hbar\omega}$$. For the annihilation operator case, j must equal n-1 and you get $$-mg\sqrt{n\hbar\omega}$$ for the matrix element.. but.. uh.. now what? :)

6. Dec 12, 2006

### Physics Monkey

First, you have some hbar omega's where you shouldn't. Or maybe you have your creation and annihilation operators defined so that they're not dimensionless. But regardless, you've made some error because $$\sqrt{\hbar \omega}$$ does not have the units of length. Also, check your signs; usually the position operator is defined as const. times $$a + a^+$$, so I don't know where these minus signs in your matrix elements are coming from. That's one thing to work out.

Once you've got these algebra things worked out, the other important part of second order perturbation theory is the so called energy denominator. The meaning of $$E^0_n - E^0_j$$ is the difference of the unperturbed energies associated with j and n. This difference is easily calculatable since j is either n+1 or n-1. Now the sum is easy because there are only two terms (remember that n is fixed). So fix up the algebra and put all the pieces together to see what you get.

7. Dec 12, 2006

### holden

I have in my notes that the x operator is defined as $$a_- - a_+$$.. that is wrong? I also have that $$a_-\psi_n = \sqrt{n\hbar\omega}\psi_{n-1}$$ and $$a_+\psi_n = \sqrt{(n+1)\hbar\omega}\psi_{n+1}$$, so that's where the hbar's and omegas are coming from. Suppose that could be wrong too, though.. but that's what my prof gave us =/

For the difference in En and Ej's, that's just hbar omega right? Since they differ by one energy level.

Last edited: Dec 12, 2006
8. Dec 13, 2006

### holden

still not sure where i went wrong :(