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\heading{E607: Perturbation theory for a ring + scatterer}
 
\auname{Katya Pupko}


{\bf The problem:}

Consider a particle with mass m and charge q, on a one dimensional ring, length L.

There is a scattere in the ring which described by $V(x)=u\cos(\frac{2\pi x}{L})$.

Calculate the second order correction to the energy in u (for any n).
\Dn\Dn

{\bf The Solution:}

\begin{verbatim}
Should be polished.
\end{verbatim}


The Hamiltonian that describes the system is:
	\[H=\frac{p^2}{2m}+V(x)
\]
The Hamiltonian has symmetry with respect to reflections.
The basis we chose to work with complies with the reflection symmetry (the even/odd states):
\[|n=0> =\sqrt{\frac{1}{L}} \]
	
\[|n_{odd}>=\sqrt{\frac{2}{L}}\sin(\frac{2\pi nx}{L})\]
\[|n_{even}>=\sqrt{\frac{2}{L}}\cos(\frac{2\pi nx}{L})\]

The perturbation does not couple the odd states to the even states.
So we focus on the perturbation's influence on the even states, and the odd states.

	\[V_{nm}=<m|V(x)|n>
\]
For the even states:
	\[V_{0,1}=V_{1,0}=\int^{0}_{L}\frac{2}{L}\cos(\frac{2\pi x}{L})u\cos(\frac{2\pi x}{L})dx=\frac{u}{\sqrt{2}}
\]
	\[V_{nm}(even)=\int^{0}_{L}\frac{2}{L}\cos(\frac{2\pi nx}{L})\cos(\frac{2\pi mx}{L})u\cos(\frac{2\pi x}{L})dx
\]
	\[V_{nm}(even)=\frac{2}{L}u\int^{0}_{L}\frac{1}{2}(\cos(\frac{2\pi x(n-m)}{L}+\cos(\frac{2\pi x(n+m)}{L}))\cos(\frac{2\pi x}{L})dx=\frac{u}{2}\delta_{|n-m|,1}
\]
For the odd states:
	\[V_{nm}(odd)=\int^{0}_{L}\frac{2}{L}\sin(\frac{2\pi nx}{L})\sin(\frac{2\pi mx}{L})u\cos(\frac{2\pi x}{L})dx
\]
	\[V_{nm}(odd)=\frac{2}{L}u\int^{0}_{L}\frac{1}{2}(\cos(\frac{2\pi x(n-m)}{L}-\cos(\frac{2\pi x(n+m)}{L}))\cos(\frac{2\pi x}{L})dx=\frac{u}{2}\delta_{|n-m|,1}
\]

The perturbation matrix for the block that belongs to the even and odd states:
\[
{V_{nm}(even)} 
=\frac{u}{2} \left( \amatrix { 0 & \sqrt{2} & 0 & ......\cr \sqrt{2} & 0 & 1 & ......\cr 0 & 1 & 0 & ......\cr 0& 0& 1  \cr .& & & .\cr .& & &} \right) 
\]
\[
{V_{nm}(odd)} 
=\frac{u}{2} \left( \amatrix { 0 & 1 & 0 & ...........\cr 1 & 0 &1 & ...........\cr 0 & 1 & 0 & ...........\cr 0& 0& 1  \cr .& & & .\cr .& & & & .} \right) 
\]

Therefore the corrections to the energy:

The first order corrections to the energy levels:
\[E^{[1]}_{n\neq0}=V_{n_{0},n_{0}}=0\]	
	
The zero order energy is given by:
\[E_{n}=\frac{2\pi^{2}n^{2}}{mL^{2}}\]

So the correction of the second order is:

For $n=0$:
	\[E^{[2]}_{0}=\sum_{l\neq 0} \frac{|<0|V|l>|^{2}}{E_{0}-E_{l}}=\sum_{l\neq 0} \frac{|V_{ol}|^{2}}{0-\frac{2\pi^{2}l^{2}}{mL^{2}}}=-\frac{mL^{2}}{4\pi^{2}}{u^2}\]	
	
For $n=1,2,3......$ :
	\[E^{[2]}_{n}=\sum_{l\neq n} \frac{|<n|V|l>|^{2}}{E_{n}-E_{l}}=\sum_{l\neq 0} \frac{|V_{nl}|^{2}}{\frac{2\pi^{2}n^{2}}{mL^{2}}-\frac{2\pi^{2}l^{2}}{mL^{2}}}=\sum_{l\neq n}\frac{1}{\frac{2\pi ^{2}n^{2}}{mL^{2}}-\frac{2\pi ^{2}l^{2}}{mL^{2}}}(\frac{u}{2})^{2}\]	
The matrix elements are all zero,except for l=n+1 and l=n-1.So the expression for the second order correction is:
	\[E^{[2]}_{n}=\frac{mL^{2}}{8\pi^{2}}(\frac{1}{2n-1}-\frac{1}{2n+1})u^{2}
\]
		
	
	
And the correction to the ground state energy, up to the second order is:
	\[E_{n}=E^{[0]}_{n}+E^{[1]}_{n}+E^{[2]}_{n}
\]
For $n=0$:
\[E_{0}=0+0-\frac{mL^{2}}{4\pi^{2}}u^2=-\frac{mL^{2}}{4\pi^{2}}u^2\]
For $n\neq0$:
	\[E_{n}=\frac{2n^{2}\pi^{2}}{mL^{2}}+0+\frac{mL^{2}}{8\pi^{2}}(\frac{1}{2n-1}-\frac{1}{2n+1})u^{2}\]
	
	\[E_{n}=\frac{2n^{2}\pi^{2}}{mL^{2}}+\frac{mL^{2}}{8\pi^{2}}(\frac{1}{2n-1}-\frac{1}{2n+1})u^{2}\]

\Dn




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