Sample problem 2

Suppose the Hamiltonian of a rigid rotator in the magnetic field perpendicular to the axis is of the form(Merzbacher 1970, problem 17-1)

$H=AL^{2}+BL_{z}+CL_{y}$

"if terms quadratic in the field are neglected. Assuming B, use Pertubation to the lowest nonvanishing order to get approximate energy eigenvalues

we rotate the system in the direction which is in the Z' axis, thus, $H=AL^{2}+(B^{2}+C^{2})L_{z}^{'}$ where the angel between Z and Z' can be written we can have The eigen state${\displaystyle }$ with eigen value $E=Al(l+1)\hbar ^{2}+(B^{2}+C^{2})^{1/2}m'\hbar$

If$B\gg C$, $H=AL^{2}+(B^{2}+C^{2})L_{z}^{'}$should be considered as none pertubative Hamiltonian, and $CL_{y}$ behaves as pertubative term. So the none pertubative eigen value and eigen states are$Insertformulahere$and $E_{l,m}^{0}=A\hbar ^{2}l(l+1)+Bm\hbar$ and first order corrections to the eigenstates of a given Hamiltonian is zero because of $L_{y}$ so the second order correction will be written in the following form $E_{n}^{(2)}=A\hbar ^{2}l(l+1)+Bm\hbar +C^{2}\sum _{l',m'\neq l,m}{\frac {\left|\right|}{E_{l,m}^{0}-E_{l',m'}^{0}}}$