Sample problem 2

Problem- suppose the hamiltonian of a rigid rotator in a 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 text'

Solution- 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$\langle l,m^{'}\rangle$ with eigen value $E=Al(l+1)\hbar ^{2}+(B^{2}+C^{2})^{1/2}m'\hbar$and,

$|l,m'\rangle$

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 areand $E_{l,m}^{0}=A\hbar ^{2}l(l+1)+Bm\hbar$ 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 $E_{n}^{(2)}=A\hbar ^{2}l(l+1)+Bm\hbar +C^{2}\sum _{l',m'\neq l,m}{\frac {\langle l^{'},m^{'}\left|L_{y}\right|l,m\rangle ^{2}}{E_{l,m}^{0}-E_{l',m'}^{0}}}$

We know that

$L_{y}={\frac {1}{2i}}(L_{+}-L_{-})$ So,

$E_{n}^{(2)}=A\hbar ^{2}l(l+1)+Bm\hbar +{\frac {C^{2}m\hbar }{2B}}$

By exact solution for B>>C we will get: $E=A\hbar ^{2}l(l+1)+Bm'\hbar +{\frac {C^{2}m'\hbar }{2B}}+...$

For $m'\rightarrow m$ the exact solution gives the same energy,