Sample problem 2: Difference between revisions
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'' | '''''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) ''''' | ||
we rotate the system in the direction which is in the Z' axis, thus, <math>H=AL^2+(B^2+ | '''<math>H=AL^2+BL_{z}+CL_{y}</math>''' | ||
we can have The eigen state | |||
'''''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, <math>H=AL^2+(B^2+C^2)L_{z}^{'}</math> where the angel between Z and Z' can be written | |||
we can have The eigen state<math>\langle l,m^{'}\rangle</math> | |||
with eigen value <math>E=Al(l+1)\hbar^{2}+(B^2+C^2)^{1/2}m'\hbar</math>and, | |||
<math> |l,m'\rangle</math> | |||
If<math>B\gg C </math>, | |||
<math>H=AL^2+(B^2+C^2)L_{z}^{'}</math>should be considered as none pertubative Hamiltonian, and <math>CL_{y}</math> behaves as pertubative term. So the none pertubative eigen value and eigen states areand <math>E_{l,m}^{0}=A\hbar^2l(l+1)+Bm\hbar</math> | |||
and first order corrections to the eigenstates of a given Hamiltonian is zero because of <math>L_{y}</math> so the second order correction will be written in the following form | |||
<math>E_{n}^{(2)}=A\hbar^2l(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}}</math> | |||
We know that | |||
<math>L_{y}=\frac{1}{2i}(L_{+}-L_{-})</math> | |||
So, | |||
<math>E_{n}^{(2)}=A\hbar^2l(l+1)+Bm\hbar+\frac{C^2m \hbar}{2B} </math> | |||
By exact solution for B>>C we will get: | |||
<math>E=A\hbar^2l(l+1)+Bm'\hbar+\frac{C^2m' \hbar}{2B}+...</math> | |||
For <math>m'\rightarrow m</math> the exact solution gives the same energy, | |||
Latest revision as of 21:52, 30 April 2010
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)
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'
Solution- 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 \langle 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 |l,m'\rangle}
IfFailed 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 B\gg C }
,
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}^{'}}
should be considered as none pertubative Hamiltonian, 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 CL_{y}}
behaves as pertubative term. So the none pertubative eigen value and eigen states areand 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_{l,m}^{0}=A\hbar^2l(l+1)+Bm\hbar}
and first order corrections to the eigenstates of a given Hamiltonian is zero because of 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 L_{y}}
so the second order correction will be written in the following form
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+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
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 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,