Sample problem 2

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'

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 stateFailed to parse (syntax error): {\displaystyle {\left l,m'\rangle}}

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

Failed to parse (syntax error): {\displaystyle \left | 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,