Angular momentum eigenfunctions

[IHateMayonnaise]

Homework Statement

This is problem 18.1 from Merzbacher.

"The hamiltonian of a rigid rotator in a magnetic field perpendicular to the x-axis is of the form $H=AL^2+BL_z+CL_y$, if the term \that is quadratic in the field is neglected. Obtain the exact energy eigenvalues and eigenfunctions of the hamiltonian."

Homework Equations

$$L^2Y_l^m(\theta,\phi)=l(l+1)\hbar^2 Y_l^m(\theta,\phi)$$

$$L_ZY_l^m(\theta,\phi)=m\hbarY_l^m(\theta,\phi)$$

$$L_Y Y_l^m(\theta,\phi)=\frac{1}{2\imath}(L_+-L_-)Y_l^m(\theta,\phi)$$

$$L_+ Y_l^m(\theta,\phi)=\sqrt{l(l+1)-m(m+1)}\hbar Y_l^{m+1}(\theta,\phi)$$

$$L_- Y_l^m(\theta,\phi)=\sqrt{l(l+1)-m(m-1)}\hbar Y_l^{m-1}(\theta,\phi)$$

The Attempt at a Solution

For the first two parts of the hamiltonian the answer is easy:

$$E=Al(l+1)\hbar^2+Bm\hbar$$

But what is the eigenvalue for the last part (of $L_Y$)? Is it just

$$C\left[\frac{\sqrt{l(l+1)-m(m+1)}\hbar-\sqrt{l(l+1)-m(m-1)}\hbar}{2\imath} \right]$$

?

And for the eigenfunctions: for the first two parts of H it is just the spherical harmonics, but for the last part is it both

$$Y_l^{m+1}(\theta,\phi)$$

and

$$Y_l^{m-1}(\theta,\phi)$$

?

Many days without sleep. THis is an easy question I know, please don't judge me I am sleep deprived :)

IHateMayonnaise

 

[gabbagabbahey]

When you act on the state $Y_{\ell}{}^{m}(\theta,\phi)$ with the Hamiltonian, do you get a constant multiple of $Y_{\ell}{}^{m}(\theta,\phi)$ ? If not, it isn't an eigenfunction of the Hamiltonian.

 

[IHateMayonnaise]

When you act on the state $Y_{\ell}{}^{m}(\theta,\phi)$ with the Hamiltonian, do you get a constant multiple of $Y_{\ell}{}^{m}(\theta,\phi)$ ? If not, it isn't an eigenfunction of the Hamiltonian.

Right! This is what confused me. So I guess it cannot be an eigenfunction, and the only one is the regular old spherical harmonic.

And the eigenvalue is just

$$C\left[\frac{\sqrt{l(l+1)-m(m+1)}\hbar-\sqrt{l(l+1)-m(m-1)}\hbar}{2\imath} \right]$$

?

Of course this could be simplified..