# 2D Harmonic Oscillator

1. Jul 13, 2005

### cyberdeathreaper

This might be another problem that our class hasn't covered material to answer yet - but I want to be sure.

The question is the following:
Find the eigenfunctions and eigenvalues of a two-dimensional isotropic harmonic oscillator.

Again, I need help simply starting.

2. Jul 13, 2005

### Staff: Mentor

Have you done the one-dimensional harmonic oscillator? Have you seen any other two- or three-dimensional setups yet? Have you used the method of separation of variables? (e.g. for the two-dimensional rectangular infinite square well)

If you've covered those topics, you should have all the tools you need. Write down the potential energy function for the two-dimensional oscillator, stick it into the two-dimensional Schrödinger equation, and separate the variables to get two one-dimensional equations.

3. Jul 13, 2005

### cyberdeathreaper

We have covered the 1D harmonic oscillator, but we haven't seen any other higher dimensional setups yet. We have also used the seperation of variables so far, just not in regards to higher-dimensions.

Just as a general question - once the equation is broken down into two 1D equations, how are the eigenvalues and eigenfunctions obtained? Is it:
$$H \Psi = E \Psi$$
with Psi being the eigenfunctions and E being the eigenvalues?

4. Jul 13, 2005

### Berislav

No, you first write the Hamiltonian operator (and use the SE) simply through operator of momentum and potential energy of the oscillator. Only now you will have a PDE, as both operators have two variables. Then use the separation of variables method.

5. Jul 13, 2005

### cyberdeathreaper

Okay, I think I've got it then. Is this correct:
$$\hat{H} = \frac{ (p_x)^2 }{2m} + \frac{ (p_y)^2 }{2m} + \frac{mw^2}{2} \left( x^2 + y^2 \right)$$
Which is broken up into components:
$$\hat{H} = \hat{H_x} + \hat{H_y}$$
Noting the 1-D harmonic oscillator gives:
$$E_x = \hbar w \left( n_x + \frac{1}{2} \right)$$
$$E_y = \hbar w \left( n_y + \frac{1}{2} \right)$$
$$\psi_n (x) = A_n (a_+)^n \psi_0 (x)$$
$$\psi_n (y) = A_n (a_+)^n \psi_0 (y)$$
and noting the seperation of variables on the wavefunction:
$$\Psi (x,y) = \Psi_x (x) \Psi_y (y)$$
Putting this together forms the results - first, using:
$$\hat{H} \Psi = E \Psi$$
and plugging in the components gives:
$$\left( H_x + H_y \right) \Psi_x \Psi_y = E \Psi_x \Psi_y$$
So the eigenvalues are:
$$\hbar w \left( n_x + \frac{1}{2} \right) + \hbar w \left( n_y + \frac{1}{2} \right)$$
Which simplifies to:
$$\hbar w \left( n + 1 \right)$$
And the eigenfunctions are simply:
$$\psi_n (x) \psi_n (y) = (A_n)^2 (a_+)^{2n} \psi_0 (x) \psi_0 (y)$$
Is this right - or do I have something wrong here? Thanks for the help.

Last edited: Jul 14, 2005
6. Jul 13, 2005

### Gokul43201

Staff Emeritus
Looks good too (though you haven't said anywhere what n is).

7. Jul 13, 2005

### cyberdeathreaper

with n = 0,1,2......
correct?

8. Jul 13, 2005

### Gokul43201

Staff Emeritus
Yes.

Also $p_x$ should be $\hat{p_x} = \frac {i \hbar}{\sqrt{2m}}\frac {\partial}{\partial x}$

9. Jul 13, 2005

### qbert

This is not quite right.

The eigenvalues are right, but there's really no need to simplify
down to n. You have a set of eigenfunctions in x and y with independent
values, and two indices.

$$\Psi_{nm} = \psi_n(x) \psi_m(y) = C_{nm} (a_+^n \psi_0(x)) (b_+^m \psi_0(y))$$

where the "b" operators are exactly analagous to the "a" operators but operate
in y instead of x. So if we want the eigenfunctions that give $p \hbar \omega$
we need combinations of n and m that add up to p. (That is, there isn't
one eigenfunction for a given "n".)

10. Jul 14, 2005

### cyberdeathreaper

I see - so technically the eigenvalues are:
$$\hbar w \left( n + m + 1 \right)$$
and the eigenfunctions are:
$$\Psi_{nm} = C_{nm} (a_+^n \psi_0(x)) (b_+^m \psi_0(y))$$
with:
$$C_{nm} = A_n A_m$$

right?

Last edited: Jul 14, 2005
11. Jul 14, 2005

### Gokul43201

Staff Emeritus
Oops ! Didn't notice that glitch.

cyber : If you are using indexes n,m for the wavefunction, use the same indexes for the eigenvalues. n is your n_x and m is your n_y.

Just also noticed in #5 that an n_y changed into an n_x when writing the eigenvalue.

Last edited: Jul 14, 2005
12. Jul 14, 2005

### cyberdeathreaper

Corrected the problems you pointed out Gokul. Other than those minor issues though, my solution proposed in #10 is correct then?

13. Jul 14, 2005

### Gokul43201

Staff Emeritus
#10 is now correct.