What are the energy levels and wavefunctions of a 3D SHO potential well?

[sachi]

We have a particle moving in a 3-D potential well V=1/2*m*(omega^2)*(r^2). we use separation of variables in cartesian coords to show that the energy levels are:

E(Nx,Ny,Nz)=hbar*omega(3/2 + Nx + Ny + Nz)

where Nx,Ny,Nz are integers greater than or equal to 1.

Therefore we can say that the ground state is unique, the first excited state has 3 fold degeneracy, and the second excited state has 6 fold degeneracy etc.

We now have to rewrite the wavefunctions of the 1st and 2nd excited states in terms of the 1d oscillator wavefunctions. We then have to rewrite the wavefunctions again using spherical harmonics and functions of r. We then find the L^2 and Lz eigenvalues of our wavefunctions.

For the 1st excited state this is easy, but I'm struggling with the 2nd one.
I know that we have 6 different wavefunctions. N.b the wavefunctions for the 1-d oscillator are as follows:

phi0(x)=A0*exp(-(x^2)/2(a^2))
phi1(x)=A1*x*exp(-(x^2)/2(a^2))
phi2(x)=A2*(2(x^2)-(a^2))*exp(-(x^2)/2(a^2))

n.b A0,A1,A2 are constants and a is a constant involving w, hbar etc.

Therefore our six wavefuntions for the 3-D case in the second excited state will be

phi(r)=k*(2(x^2)-(a^2))*exp(-(r^2)/2(a^2))
phi(r)=k*(2(y^2)-(a^2))*exp(-(r^2)/2(a^2))
phi(r)=k*(2(z^2)-(a^2))*exp(-(r^2)/2(a^2))
phi(r)=k*xy*exp(-(r^2)/2(a^2))
phi(r)=k*yz*exp(-(r^2)/2(a^2))
phi(r)=k*xz*exp(-(r^2)/2(a^2))

where k is a constant.
I'm finding it really difficult to express these functions as linear combinations of the spherical harmonics. thanks very much for your help.

Sachi

 

[Physics Monkey]

Hi sachi,

What you need to work is essentially products of cartesian components $$x, y, z$$ in terms in spherical harmonics $$Y^m_\ell$$, yes? This is most easily accomplished if you first write the spherical harmonics in terms of r and the various cartesian components. For example, $$Y^1_1 = -\frac{1}{2} \sqrt{\frac{3}{2 \pi}} \sin{\theta} e^{i \phi} = - \frac{1}{2} \sqrt{\frac{3}{2 \pi}}\frac{x+ i y}{r}$$ and $$Y^0_1 = \frac{1}{2} \sqrt{\frac{3}{\pi}} \cos{\theta} = \frac{1}{2} \sqrt{\frac{3}{\pi}}\frac{z}{r}$$. Clearly then z is proportional to $$r Y^0_1$$, and if you wanted x or y then you would take linear combinations of $$r Y^1_1$$ and $$r Y^{-1}_1$$. Now, if you go to $$\ell = 2$$ and work things out you will find that the $$Y^m_2$$ generally contain products of cartesian components. This is what you want. If you wanted $$z^2$$ you would need to make use of $$Y^0_2$$ and so forth. You're task is then reduced just to doing a little linear algebra. You can work out cartesian forms of the spherical harmonics yourself, or you can find several of them here http://mathworld.wolfram.com/SphericalHarmonic.html .

 

[sachi]

hi, thanks very much for the hint. I'm a little confused, do we need to consider any of the Ylm's with l greater than 2 (I'm pretty sure our lecturer said that this was the case). If so, then I can't see how to get a combination of say, xz on its own. using Y21 we can we get an xz product, but we also get a yz product, which is not present in any of the other spherical harmonics and hence can't be subtracted. the only one I can see how to do is the z^2 one,which would just require Y20

Sachi

 

[Physics Monkey]

Hi sachi,

In this case you don't need spherical harmonics higher than $$\ell = 2$$ because you have at most two cartesian components appearing together. To obtain something like $$x z$$, try looking at $$Y^{-1}_2$$ and $$Y^1_2$$. Do you notice anything? In particular, can you figure out a way to get rid of the offending $$y z$$ terms?

 

[Gokul43201]

If so, then I can't see how to get a combination of say, xz on its own. using Y21 we can we get an xz product, but we also get a yz product, which is not present in any of the other spherical harmonics and hence can't be subtracted.

The bold part is not true. Notice, from definition, that $Y_l^m$ and $Y_l^{-m}$ are complex conjugates of each other.

 

[sachi]

thanks very much for the hints everybody. the point about conjugation was the crucial one (it seems so obvious now!)