3D Harmonic Oscillator - Eigenfunctions and Eigenvalues

[physicalchemishard]

Homework Statement

Due to the radial symmetry of the Hamiltonian, H=-(ħ²/2m)∇²+k(x^2+y^2+z^2)/2
it should be possible to express stationary solutions to schrodinger's wave equation as eigenfunctions of the angular momentum operators L² and L_z, where
L²=-ħ²[1/sin(Θ)(d/dΘ)(sin(Θ)d/dΘ+1/sin²(Θ)d²/dΦ²
L_z=-iħd/dΦ

for this radially symmetric system, what are the three lowest energy stationary states that are eigenfunctions of L² and L_z, with eigenvalues of 0 and 0, respectively?

Homework Equations

in a previous problem we solved for the 3D stationary states of the harmonic oscillator using separation of variables and got that Ψ_nx,ny,nz(x,y,z)=ψ_nx(x)ψ_ny(y)ψ_nz(z)
also relevant is the solution to the 1D harmonic oscillator: ψ(y)=N_nH_n(y)e^(-1/2∝y2

The Attempt at a Solution

Based off of a hint given by one of my GSI's, I was told to begin by taking some of the solutions for the lowest energy stationary states and then convert these solutions to polar coordinates, and then try to manipulate them to look like one the spherical harmonics functions. Following this hint I took the ground state solution, which turned out to be : Ψ₀₀₀=N₀³e^(1/2)r2, which can be manipulated to look like the first spherical harmonic function, and will have eigenvalues of 0 and 0 when both of the given angular momentum operators are applied, but I'm at a loss as far as finding the other two stationary states, given that all of the other stationary states that I have tried have led to nonzero eigenvalues for one of the two operators. Any help or hints would be appreciated.

 

[drvrm]

Hint:

pl.see the following and try to work out the steps -or you can consult any textbook!
3D isotropic harmonic oscillator
The potential of a 3D isotropic harmonic oscillator is

In this article it is shown that an N-dimensional isotropic harmonic oscillator has the energies
E(n) = (h/2Pi) ( n+N/2)
i.e., n is a non-negative integral number; ω is the (same) fundamental frequency of the N modes of the oscillator. In this case N = 3, so that the radial Schrödinger equation becomes,

(T +V) Psi(r, theta) = E(n) Psi(r,theta)
REFERENCE;.
<https://en.wikipedia.org/wiki/Parti...c_potential#3D_isotropic_harmonic_oscillator>

if there will be problems in understanding one can discuss.

 

[vela]

Based off of a hint given by one of my GSI's, I was told to begin by taking some of the solutions for the lowest energy stationary states and then convert these solutions to polar coordinates, and then try to manipulate them to look like one the spherical harmonics functions. Following this hint I took the ground state solution, which turned out to be : Ψ₀₀₀=N₀³e^(1/2)r2, which can be manipulated to look like the first spherical harmonic function, and will have eigenvalues of 0 and 0 when both of the given angular momentum operators are applied

Considering $Y_0^0 = 1$, there's not much to manipulate here. ;)

but I'm at a loss as far as finding the other two stationary states, given that all of the other stationary states that I have tried have led to nonzero eigenvalues for one of the two operators. Any help or hints would be appreciated.

Why are non-zero eigenvalues a problem?

I think it would be easier if you converted the spherical harmonics to cartesian coordinates. Consider $Y_1^{-1} = \sqrt{\frac{3}{8\pi}}\sin\theta e^{-i\phi}$. If you multiply this function by $r$ and expand the complex exponential, you get
$$r Y_1^{-1} = \sqrt{\frac{3}{8\pi}}(r \sin\theta \cos\phi - i r \sin \theta \sin \phi) = \sqrt{\frac{3}{8\pi}} (x - i y).$$

 

[physicalchemishard]

Considering $Y_0^0 = 1$, there's not much to manipulate here. ;)Why are non-zero eigenvalues a problem?

I think it would be easier if you converted the spherical harmonics to cartesian coordinates. Consider $Y_1^{-1} = \sqrt{\frac{3}{8\pi}}\sin\theta e^{-i\phi}$. If you multiply this function by $r$ and expand the complex exponential, you get
$$r Y_1^{-1} = \sqrt{\frac{3}{8\pi}}(r \sin\theta \cos\phi - i r \sin \theta \sin \phi) = \sqrt{\frac{3}{8\pi}} (x - i y).$$

the problem statement itself says "what are the three lowest energy stationary states that are eigenfunctions of L²and L_z, with eigenvalues of 0 and 0, respectively?"

I thought that implied we needed to find 3 different sets of eigenfunctions of the angular momentum operators with eigenvalues of 0 and 0 using the harmonic oscillator eigenfunctions. The next part of the problem asks to write these functions as linear combinations of products of the harmonic oscillator eigenfunctions. My confusion lies in the fact that it seems like the only spherical harmonic that will give a zero eigenvalue for L² is $Y_0^0$. Maybe I've misinterpreted the problem somehow?

 

[vela]

Oh, I missed that. Your confusion stems from the fact that the problem statement is weird. You're right. The only spherical harmonic involved here is $Y_0^0$. If there are three different eigenfunctions, it'll be because there are three different radial functions you can combine with $Y_0^0$. I haven't done this problem in a long time so I might be misremembering the solution, but I'm pretty sure there aren't three radial functions for $l=0$. If you come to the same conclusion, you might ask your professor if there's a typo in the problem.

 

[physicalchemishard]

Oh, I missed that. Your confusion stems from the fact that the problem statement is weird. You're right. The only spherical harmonic involved here is $Y_0^0$. If there are three different eigenfunctions, it'll be because there are three different radial functions you can combine with $Y_0^0$. I haven't done this problem in a long time so I might be misremembering the solution, but I'm pretty sure there aren't three radial functions for $l=0$. If you come to the same conclusion, you might ask your professor if there's a typo in the problem.[/QU

Just an update. I asked one of the people who wrote the Pset about this question and he said that the other solutions are formed from linear combinations of degenerate states of the harmonic oscillator eigenfunctions. Another solution would be Ψ=ψ_2,0,0(x,y,z)+ψ_0,2,0(x,y,z)+ψ_0,0,2 (x,y,z) since that would allow the x^2+y^2+z^2 term that results to be substituted with an r^2 (and the equation would then have no φ or θ dependence). The third solution will be generated in the same way, but I'm having trouble figuring out what it would be. I guess I just have to guess and check a bunch of degenerate states and see if they can be manipulated to look like $Y_0^0$.

 

[vela]

Oh, of course! For some reason I was thinking you had to have n=0.

You should try expanding $(x^2+y^2+z^2)^2$.