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\heading{E6070: Perturbation Theory for a ring in an Electric Field} 
\auname{Jacob Hanouna and Sarit Nagar}


{\bf The problem:}

\Dn

A particle with mass M and charge $e$ is placed in a one dimensional ring with radius $R$.

A homogenous electric field $\mathcal{E}$ is being induced parallel to the ring's plane.

\Dn

(1) Write the Hamiltonian $\mathcal{H}(\theta,p)$ of the system.

(2) What are the symmetries the system has without/with the perturbation.

\text{(3) Write the matrical presentation of the Hamiltonian with the suitable basis for the perturbated system.}

(4) Calculate the ground state energy $n=0$ up to the second order in the perturbation.

(5) Calculate the excited states energies $n>1$ up to the second order in the perturbation.


\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1) 
The Hamiltonian is: 

\[
\mathcal{\hat{H}}
= \frac{\hat{p}^2}{2m}+\hat{V}(\theta)
= \frac{\hat{p}^2}{2m}+e\mathcal{E}R\cos\theta
\]

%%%%%%%%%%%%%%%
 

\Dn

(2)
\text{Without the perturbation the Hamiltonian has symmetry with respect to rotations and reflections.}
In the presence of the electric field, only the reflection remains.


%%%%%%%%%%%%%%
\Dn

(3)
We will use the basis that complies with the reflection symmetry:
\begin{equation*}
|n=0\rangle \mapsto \frac1{\sqrt{2 \pi R}} \longmapsto \frac1{\sqrt{2 \pi}}		
\end {equation}
\begin{equation*}
|n,+\rangle \mapsto \frac1{\sqrt{\pi R}}\cos(k_nx) \longmapsto \frac1{\sqrt{\pi}}\cos(n\theta)
\end {equation}
\begin{equation*}
|n,-\rangle \mapsto \frac1{\sqrt{\pi R}}\sin(k_nx) \longmapsto \frac1{\sqrt{\pi}}\sin(n\theta)
\end {equation*}

\quad\newline Notice that describing the basis using the relation $x=R\theta$ changes the normalization as mentioned on the right side.\newline
There is a degeneracy between the even states and the odd states, though they are not "coupled" to each other.
Therefore, the Hamiltonian will be of the form:

\[
\mathcal{H}\mapsto
\begin{pmatrix}
\varepsilon_0&&&&&&& \\
&\varepsilon_1&&&&&& \\
&&\varepsilon_2&&&&& \\
&&&\ddots&&&& \\
&&&&&\varepsilon_1&& \\
&&&&&&\varepsilon_2& \\
&&&&&&&\ddots \\
\end{pmatrix}
+
\begin{pmatrix}
V_{00}&V^{\tiny(+)}_{01}&V^{\tiny(+)}_{02}& \cdots &V^{\tiny(-)}_{01}&V^{\tiny(-)}_{02}& \cdots \\
V^{\tiny(+)}_{10}&V^{\tiny(+)}_{11}&V^{\tiny(+)}_{12}& \cdots &0&0&\cdots& \\
V^{\tiny(+)}_{20}&V^{\tiny(+)}_{21}&V^{\tiny(+)}_{22}& \cdots &0&0&\cdots& \\
\vdots & \vdots &\vdots & \ddots & \vdots & \vdots & \ddots \\
V^{\tiny(-)}_{10}&0&0&\cdots &V^{\tiny(-)}_{11} &V^{\tiny(-)}_{12}& \cdots \\
V^{\tiny(-)}_{20}&0&0&\cdots &V^{\tiny(-)}_{21} &V^{\tiny(-)}_{22}& \cdots \\
\vdots &\vdots & \vdots & \ddots& \vdots & \vdots & \ddots \\
\end{pmatrix}
\]

where $\varepsilon_n = \frac{k^2_n}{2M}=\frac{n^2}{2MR^2}$.

\[
V_{00}=\langle n=0|V(\theta)|n=0\rangle = 0
\end{equation*}
\begin{equation*}
V^{\tiny{(+)}}_{0m}=\langle n=0|V(\theta)|m,+\rangle= \frac{e\mathcal{E}R}{\sqrt2\pi}
\int_0^{2\pi} \cos(\theta)\cos(m\theta) d\theta = \frac{e\mathcal{E}R}{\sqrt2} \delta_{m,1}
\]

\begin{align*}
V^{\tiny{(+)}}_{nm}&=\langle n,+|V(\theta)|m,+\rangle
=\frac{e\mathcal{E}R}{\pi} \int_0^{2\pi} \cos(n\theta)\cos(\theta)\cos(m\theta) d\theta \\
&=\frac{e\mathcal{E}R}{4\pi} \left[ \int_0^{2\pi}  \left[ \cos((n-m-1)\theta)+\cos((n-m+1)\theta)\right] d\theta \quad + \\
& \quad + \int_0^{2\pi} \left[ \cos((n+m-1)\theta) + \cos((n+m+1)\theta) \right] d\theta  \right]\\
&=\frac{e\mathcal{E}R}{2} \left[ \delta_{n,m+1} +\delta_{n,m-1} +\delta_{n,-m+1} + \delta_{n,-m-1} \right]
=\frac{e\mathcal{E}R}{2} \left[ \delta_{n,m+1} +\delta_{n,m-1} \right]
\end{aling*}

The last transition is due to the fact that $n,m>0$.

\begin{equation*}
V^{\tiny{(-)}}_{0m}=\langle n=0|V(\theta)|m,-\rangle= \frac{e\mathcal{E}R}{\sqrt2\pi}
\int_0^{2\pi} \cos(\theta)\sin(m\theta) d\theta = 0
\end{equation*}
\begin{equation*}
V^{\tiny{(-)}}_{nm}=\langle n,-|V(\theta)|m,-\rangle= \frac{e\mathcal{E}R}{\pi}
\int_0^{2\pi} \sin(n\theta)\cos(\theta)\sin(m\theta) d\theta
=\frac{e\mathcal{E}R}{2} \left[ \delta_{n,m+1} +\delta_{n,m-1} \right]
\end{equation*}


V^{\tiny{(+)}}_{n,m}=\frac{e\mathcal{E}R}{2}
\begin{pmatrix}
0 & \sqrt2 & 0 & 0 & \cdots \\
\sqrt2 & 0 & 1 & 0 & \cdots \\
0 & 1 & 0 & 1 & \cdots \\ 
0 & 0 & 1 & 0 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots \\
\end{pmatrix}
\qquad \qquad \qquad
V^{\tiny{(-)}}_{n,m}=\frac{e\mathcal{E}R}{2}
\begin{pmatrix}
0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 1 & 0 & \cdots \\
0 & 1 & 0 & 1 & \cdots \\ 
0 & 0 & 1 & 0 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots \\
\end{pmatrix}


%%%%%%%%%%%%%
\Dn
(4)
Notice that $E^{[0]}_n=\varepsilon_n$ and $E^{[1]}_n=V_{nn}=0$
Therefore, the ground state energy up to the second order in the perturbation is simply
\begin{equation*}
E_0=E^{[2]}_0=\displaystyle\sum_{m\not=0} \frac{{|V_{0m}|}^2}{0-\frac{m^2}{2MR^2}}=-2Me^2\mathcal{E}^2R^4
\end{equation*}

%%%%%%%%%%%%%
\Dn

(5)
For the excited states $n,m>1$ we have $V^{\tiny{(+)}}_{n,m}=V^{\tiny{(-)}}_{n,m}$

\begin{equation*}
E^{[2]}_n=\displaystyle\sum_{m\not=n} \frac{{|V_{nm}|}^2}{\varepsilon_n-\varepsilon_m}
=\frac{Me^2\mathcal{E}^2R^4}{2} \left[ \frac1{n^2-(n-1)^2} + \frac1{n^2-(n+1)^2} \right]
=\frac{Me^2\mathcal{E}^2R^4}{4n^2-1}
\end{equation*}
Therefore the energy up to second correction is:
\begin{equation*}
E_n=\frac{n^2}{2MR^2}+\frac{Me^2\mathcal{E}^2R^4}{4n^2-1}
\end{equation*}


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