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)

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\left l,m'\rangle}}

with eigen value Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=Al(l+1)\hbar^{2}+(B^2+C^2)^{1/2}m'\hbar} and,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{n}^{(2)}=A\hbar^2l(l+1)+Bm\hbar+\frac{C^2m \hbar}{2B} }

By exact solution for B>>C we will get: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=A\hbar^2l(l+1)+Bm'\hbar+\frac{C^2m' \hbar}{2B}+...}

For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m'\rightarrow m} the exact solution gives the same energy,