# 2D Harmonic Oscillator Quantum numbers

1. Jan 20, 2016

1. The problem statement, all variables and given/known data
In the exercise, we solved the 2D Harmonic Oscillator in kartesian (x,y) and polar (r,φ) coordinates.
We found out that both have the same energy levels, but they look very different, when I plot them.
What am I missing? The polar solution seems more like it.

2. Relevant equations

3. The attempt at a solution
Two examples:

If needed, I can give you more details about the formulas or plots!

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2. Jan 20, 2016

### blue_leaf77

What are the corresponding quantum numbers $(n_x,n_y)$ in the case of Cartesian and $(n,|l|)$ in the case of polar of the picture you plot there?

3. Jan 20, 2016

They correspond to the formulas above.
$n_{x}$ and $n_{y}$ are the quantum numbers from the 1D solution.
In the polar case, $n_{r}$ corresponds to the energy and $l$ to the angular momentum.

Last edited: Jan 20, 2016
4. Jan 20, 2016

### blue_leaf77

What I asked is what values of $(n_x,n_y)$ you used for plotting

and what values of $(n,l)$ for
?

5. Jan 20, 2016

Oh, sry.. (1,1) in both cases!
They look the same for the ground state.

6. Jan 20, 2016

### blue_leaf77

That's the first mistake you made. Polar and Cartesian representations of 2D harmonic oscillator are different sets of bases, the one in polar coordinate is invariant under rotation while the one in Cartesian is not. This means, you should not assume that two pair of quantum numbers in polar and Cartesian representations having identical numerical values to be identically equal to each other. However, they are indeed connected if the energy eigenvalue are the same in both representation, in particular one eigenstate in one representation will be a linear combination of the eigenstates in the other representation. See the table in the last page in this link http://www.physics.rutgers.edu/~steves/501/Lectures_Final/Lec17_Polar Coordinates.pdf.
You can find there too that in polar coordinate the energy eigevalue is given by
$$E_n = \hbar\omega(n+1)$$
while in Cartesian coordinate by
$$E_{n_x,n_y} = \hbar\omega(n_x+n_y+1)$$
Here you can see that for the same energy in polar and Cartesian coordinates, the quantum numbers are related by $n=n_x+n_y$. Using your program, try plotting the real part of $(n,l) = (1,1)$ for polar and $(n_x,n_y) = (1,0)$ in Cartesian.

7. Jan 20, 2016

Actually, it is $n=n_{x}+n_{y}=2n_{r}+l$.
But I tried many different combinations, they behave very different in general.
(The polar solution has circular symmetry and the cartesian has mirroring symmetry)

8. Jan 20, 2016

### blue_leaf77

Indeed, they should be different. You can't find any pair of eigenfunctions in polar and in Cartesian which look identical as an eigenstate in one coordinate is a linear combination of eigenstates in the other coordinates, as I have mentioned in my previous comment. The only identical pair is the ground state because this energy level has no degeneracy. In fact, the eigenstate in polar is complex on the other hand the one in Cartesian is always real, no way to compare them one-on-one.
I have also pointed out this point in my previous comment. The polar representation is also eigenstate of angular momentum operator, which means it must indeed have circular symmetry. The Cartesian representation is eigenstate of $x$ and $y$ parity operators, so they will have mirror symmetry as you already observed.

9. Jan 20, 2016