2D Harmonic Oscillator Quantum numbers

Tabasko633 · Jan 20, 2016

Homework Statement

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.

Homework Equations

$Equation for the 2D harmonic oscillator wavefunction in polar coordinates: Psi(r, phi) = N r^|l| e^(-r^2/(2 r0^2)) L_{n_r}^{|l|}(r^2/r0^2) e^(i l phi), with N a normalisation constant and L a Laguerre polynomial.$

Equation for the 2D harmonic oscillator wavefunction Psi(x,y) as a product of Hermite-function factors in x and y, with normalisation 1/(sqrt(pi) x0), Hermite polynomials H_nx(x/x0) and H_ny(y/x0), and Gaussian factor exp(-(x^2+y^2)/(2x0^2)).

The Attempt at a Solution

Two examples:

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

blue_leaf77 · Jan 20, 2016

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?

Tabasko633 · Jan 20, 2016

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

blue_leaf77 · Jan 20, 2016

Tabasko633 said:

They correspond to the formulas above.
nx and ny are the quantum numbers from the 1D solution.
In the polar case, n corresponds to the energy and l to the angular momentum.

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

Two side-by-side Mathematica plots of a 2D harmonic oscillator state on x and y from -5 to 5: left, Re[psi] kartesian with four lobes of alternating sign; right, |psi|^2 kartesian with four bright lobes on a dark background.

and what values of ##(n,l)## for

Two side-by-side polar plots from a Mathematica manipulate interface with n and l sliders: Re[ψ] polar shows a lobed blue and tan pattern, and |ψ|² polar shows concentric rings around the origin, axes from -5 to 5.

?

Tabasko633 · Jan 20, 2016

Oh, sry.. (1,1) in both cases!
They look the same for the ground state.
Should I upload the file?

blue_leaf77 · Jan 20, 2016

Tabasko633 said:

Oh, sry.. (1,1) in both cases!

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%20Coordinates.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.

Tabasko633 · Jan 20, 2016

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

blue_leaf77 · Jan 20, 2016

Tabasko633 said:

But I tried many different combinations, they behave very different in general.

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.

Tabasko633 said:

(The polar solution has circular symmetry and the cartesian has mirroring symmetry)

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.

Tabasko633 · Jan 20, 2016

blue_leaf77 said:

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.

I guess it took two comments for me to understand, but now I got it.
In order to connect these two solutions, I will try to find the linear combination!
Thank you a lot for your help, blue_leaf77

2D Harmonic Oscillator Quantum numbers

Homework Statement

Homework Equations

The Attempt at a Solution

Discussion

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