Exploring the Eigenfunctions and Eigenvalues of a 2D Harmonic Oscillator

In summary, the conversation discusses finding the eigenfunctions and eigenvalues of a two-dimensional isotropic harmonic oscillator. The speaker asks if the class has covered relevant material, such as the one-dimensional harmonic oscillator and the method of separation of variables. They point out that once the Hamiltonian operator is written and separated into components, the eigenvalues can be found by plugging in the components and simplifying. The correct solution is given in the end, with the eigenvalues being \hbar w \left( n + m + 1 \right) and the eigenfunctions being \Psi_{nm} = C_{nm} (a_+^n \psi_0(x)) (b_+^m \psi_0(y)).
  • #1
cyberdeathreaper
46
0
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.
 
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  • #2
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
We have covered the 1D harmonic oscillator, but we haven't seen any other higher dimensional setups yet. We have also used the separation 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:
[tex]
H \Psi = E \Psi
[/tex]
with Psi being the eigenfunctions and E being the eigenvalues?
 
  • #4
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
Okay, I think I've got it then. Is this correct:
[tex]
\hat{H} = \frac{ (p_x)^2 }{2m} + \frac{ (p_y)^2 }{2m} + \frac{mw^2}{2} \left( x^2 + y^2 \right)
[/tex]
Which is broken up into components:
[tex]
\hat{H} = \hat{H_x} + \hat{H_y}
[/tex]
Noting the 1-D harmonic oscillator gives:
[tex]
E_x = \hbar w \left( n_x + \frac{1}{2} \right)
[/tex]
[tex]
E_y = \hbar w \left( n_y + \frac{1}{2} \right)
[/tex]
[tex]
\psi_n (x) = A_n (a_+)^n \psi_0 (x)
[/tex]
[tex]
\psi_n (y) = A_n (a_+)^n \psi_0 (y)
[/tex]
and noting the separation of variables on the wavefunction:
[tex]
\Psi (x,y) = \Psi_x (x) \Psi_y (y)
[/tex]
Putting this together forms the results - first, using:
[tex]
\hat{H} \Psi = E \Psi
[/tex]
and plugging in the components gives:
[tex]
\left( H_x + H_y \right) \Psi_x \Psi_y = E \Psi_x \Psi_y
[/tex]
So the eigenvalues are:
[tex]
\hbar w \left( n_x + \frac{1}{2} \right) + \hbar w \left( n_y + \frac{1}{2} \right)
[/tex]
Which simplifies to:
[tex]
\hbar w \left( n + 1 \right)
[/tex]
And the eigenfunctions are simply:
[tex]
\psi_n (x) \psi_n (y) = (A_n)^2 (a_+)^{2n} \psi_0 (x) \psi_0 (y)
[/tex]
Is this right - or do I have something wrong here? Thanks for the help.
 
Last edited:
  • #6
Looks good too (though you haven't said anywhere what n is).
 
  • #7
with n = 0,1,2...
correct?
 
  • #8
Yes.

Also [itex]p_x[/itex] should be [itex]\hat{p_x} = \frac {i \hbar}{\sqrt{2m}}\frac {\partial}{\partial x} [/itex]
 
  • #9
So the eigenvalues are:
[tex]
\hbar w \left( n_x + \frac{1}{2} \right) + \hbar w \left( n_x + \frac{1}{2} \right)
[/tex]
Which simplifies to:
[tex]
\hbar w \left( n + 1 \right)
[/tex]
And the eigenfunctions are simply:
[tex]
\psi_n (x) \psi_n (y) = (A_n)^2 (a_+)^{2n} \psi_0 (x) \psi_0 (y)
[/tex]

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.

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

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 [itex] p \hbar \omega [/itex]
we need combinations of n and m that add up to p. (That is, there isn't
one eigenfunction for a given "n".)
 
  • #10
I see - so technically the eigenvalues are:
[tex]
\hbar w \left( n + m + 1 \right)
[/tex]
and the eigenfunctions are:
[tex]
\Psi_{nm} = C_{nm} (a_+^n \psi_0(x)) (b_+^m \psi_0(y))
[/tex]
with:
[tex]
C_{nm} = A_n A_m
[/tex]

right?
 
Last edited:
  • #11
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:
  • #12
Corrected the problems you pointed out Gokul. Other than those minor issues though, my solution proposed in #10 is correct then?
 
  • #13
#10 is now correct.
 

1. What is a 2D Harmonic Oscillator?

A 2D Harmonic Oscillator is a physical system that exhibits harmonic motion in two dimensions. It consists of a particle moving in a two-dimensional plane, subjected to a restoring force proportional to its displacement from the equilibrium position.

2. What is the formula for the potential energy of a 2D Harmonic Oscillator?

The potential energy of a 2D Harmonic Oscillator can be expressed as U(x,y) = 1/2 kx^2 + 1/2 ky^2, where k is the spring constant and x and y are the displacements in the x and y directions, respectively.

3. How does the frequency of a 2D Harmonic Oscillator change with changes in the spring constant?

The frequency of a 2D Harmonic Oscillator is directly proportional to the square root of the spring constant. This means that as the spring constant increases, the frequency of the oscillator also increases.

4. How does the energy of a 2D Harmonic Oscillator change with changes in the amplitude of motion?

The energy of a 2D Harmonic Oscillator is directly proportional to the square of the amplitude of motion. This means that as the amplitude of motion increases, the energy of the oscillator also increases.

5. What is the relationship between the period and frequency of a 2D Harmonic Oscillator?

The period (T) and frequency (f) of a 2D Harmonic Oscillator are reciprocally related, meaning that T = 1/f. This means that as the frequency increases, the period decreases and vice versa.

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