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## 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 [itex]H=AL^2+BL_z+CL_y[/itex], if the term \that is quadratic in the field is neglected. Obtain the exact energy eigenvalues and eigenfunctions of the hamiltonian."

## Homework Equations

[tex]L^2Y_l^m(\theta,\phi)=l(l+1)\hbar^2 Y_l^m(\theta,\phi)[/tex]

[tex]L_ZY_l^m(\theta,\phi)=m\hbarY_l^m(\theta,\phi)[/tex]

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

[tex]L_+ Y_l^m(\theta,\phi)=\sqrt{l(l+1)-m(m+1)}\hbar Y_l^{m+1}(\theta,\phi)[/tex]

[tex]L_- Y_l^m(\theta,\phi)=\sqrt{l(l+1)-m(m-1)}\hbar Y_l^{m-1}(\theta,\phi)[/tex]

## The Attempt at a Solution

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

[tex]E=Al(l+1)\hbar^2+Bm\hbar[/tex]

But what is the eigenvalue for the last part (of [itex]L_Y[/itex])? Is it just

[tex]

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

\right]

[/tex]

?

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

[tex]Y_l^{m+1}(\theta,\phi)[/tex]

and

[tex]Y_l^{m-1}(\theta,\phi)[/tex]

?

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

IHateMayonnaise

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