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\exnumber{7684}
\heading{Dot - Ring System} 
\auname{Eadan Valent}
 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \  Based on a solution by: Elad Ayalon and Yair Mazal
\Dn

{\bf The problem:}

\Dn

A particle of mass $M$ is confined to a site. The undisturbed bond energy is $\epsilon_0$.
\newline The site is coupled to a large ring of length $L$ at one point, so that $\left\langle\epsilon_0\left|H\right|\Psi\right\rangle=\alpha\Psi\left(x_0\right)$. 
\newline In the first question assume that the energy $\epsilon_0$ is underneath the potential floor of the ring.

\Dn

(1) Calculate the correction $\Delta$ to the bond energy, by using second order perturbation theory.

(2) Repeat the first question under the assumption that $\epsilon_0$ is above the potential floor.

(3) In the last case, calculate the decay constant $\Gamma$ of the bound state, by using Fermi's golden rule. 

(4) Find a domain of values for $\epsilon_0$, in which the previous results are unreliable.

(5) The bond energy from the first question can be found accurately, by solving a third degree equation. Write the equation.   

\Dn
\Dn
\textbf{Guidance:}
\begin{itemize}
\item It would be convenient  to work in the momentum basis $\left|k\right\rangle$ of the ring.
\item Assume that the ring is very big, so that the discrete sum can be approximated by an integral.
\item Write the final answer using the constants $\alpha,M,\epsilon_0$.
\item If it helps it is allowed, and advisable to use the notation: $v_0=\left(2\epsilon_0\over M\right)^{1\over2}$.
\end{itemize}

\Dn\Dn\Dn


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{\bf The solution:}

\Dn
For clarity reasons let us denote by $\Psi$ a general linear combination of the momentum states on the ring, and by $\psi$ the general wave function. 
\Dn

(1) 
The system is a combination of two separate systems; the undisturbed site and the states of the ring. Therefore the general wave function for the system is:
\[
\left|\psi\right\rangle=\left|\epsilon_0\right\rangle+\sum_x\Psi\left(x\right)\left|x\right\rangle
\]
\newline The previous statement can be refined by using the eigenstates of a ring of length L:
\[
\left|\psi\right\rangle=a\left|\epsilon_0\right\rangle+{1\over \sqrt{L}}\sum_x e^{ik_n x}\left|x\right\rangle
\]
 with:
\[k_n={2\pi\over L}\times n, \ \ \ E_n={2\pi^2\over m\L^2}\times n^2, \ \ \ n=0,\pm1,\pm2,...\]

\newpage
Now that the undisturbed Hamiltonian is clear, lets evaluate the disturbance.
\newline All of the momentum states of the ring are coupled to the site. The coupling occurs in a specific site in the position basis, and because we have the freedom to choose the x-axis origin, let us call the site of the coupling $x=0$. Thus:

\[\left\langle\epsilon_0\left|H\right|\Psi_k\right\rangle=\left\langle\epsilon_0\right|H\Psi^k\left(0\right)\left|0\right\rangle=\left\langle\epsilon_0\right|H {e^{ik_n\cdot 0}\over \sqrt{L}}\left|0\right\rangle={\alpha\over\sqrt{L}}\]

Therefore the coupling of each of the momentum states to the standalone state is identical:

\[V_{n,\epsilon_0}={\alpha\over\sqrt{L}}\]


Remembering $\epsilon_0 <0$ and assuming the disturbance is small: \[V_{n,\epsilon_0} << \left|\epsilon_0\right| \]  
We can now use second order perturbation theory approximation.
\[\Delta=\sum_{n=-\infty}^\infty {\left|V_{n,\epsilon_0}\right|^2\over -\left|\epsilon_0\right|-E_n}=-{\alpha^2\over L}\sum_{n=-\infty}^\infty {1\over\left|\epsilon_0\right|+{2\pi^2\over M\L^2}\ n^2}=-{M\L\alpha^2\over 2\pi^2}\int_{-\infty}^\infty{\mathrm{d}n\over {M\L^2\over 2\pi^2} \left|\epsilon_0\right|+n^2}=-{\alpha^2\over \left(2\epsilon_0\over M\right)^{1\over2}}=-{\alpha^2\over v_0}\]

\Dn
The discrete sum turned into the integral since the ring is large enough.

\Dn
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\Dn

(2)
Let us repeat the process, only now  $\epsilon_0 >0$:

           
\[\Delta={\alpha^2\over L}\sum_{n=-\infty}^\infty {1\over\epsilon_0-{2\pi^2\over M\L^2}\ n^2}= 
 {\alpha^2\over 2\ L\sqrt{\epsilon_0}}\int_{n=-\infty}^\infty {1\over {\sqrt{\epsilon_0}}- {\sqrt{2\pi^2\over M\L^2}}\ n}+{1\over  {\sqrt{\epsilon_0}}+{\sqrt{2\pi^2\over M\L^2}}\ n}=\]

\[ \ \ \  \  \ \  {d\over b} \times \left.\ln({|a+bn|\over |a-bn|})\right|_{-\infty}^\infty=0\]



\Dn
We know that :  $ d\int{1\over {a+bn}} \ \mathrm{d}n = {d\over b} \times \ln\left|a+bn\right|+c$
\[    \    \   \ \ \   \ \ \ \ \  \ \ \ \     ln(1) = 0\] 


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

(3)
Let us use $\delta$ for mean level spacing and not the usual $\Delta$, which we defined as the perturbation.
\newline Using the dispersion relation to find the mean level spacing, and assuming that the decay into the continuum occurs at the  energy of the standalone state ($v_E = v_0$):
\[\mathrm{d}E=v_E \mathrm{d}p\ \ \Rightarrow\ \ \delta=v_0 {\pi\over L}\]
Now according to Fermi's Golden Rule:
\[\Gamma={2\pi\over\delta}\left|V_{n,\epsilon_0}\right|^2={2\alpha^2\over v_0}\]


%%%%%%%%%%%%%
\Dn
\newpage
(4)
We shall find the values of $\epsilon_0$ for which our approximations are valid, and deduce that the results are unreliable for the exact opposite conditions.
\Dn
\newline In order for the perturbation theory to be valid, one must require:
\[\left|V_{n,\epsilon_0}\right|<<\epsilon_0\ \rightarrow \ \Delta<<\epsilon_0\]
\Dn
The requirement for the validity of Fermi's Golden Rule is obvious out of basic unit analysis:
\[\Gamma<<\epsilon_0\]  
\Dn
This can be interpreted as the bandwidth of decay being small in relation to the original energy level.  
\Dn
\\

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\Dn

(5)
The Schr\"{o}dinger equation can be written as follows:
\[\epsilon_0 |\epsilon_0\rangle +{\alpha\over\sqrt{L}} \sum_{k}\Psi_{k} = E|\epsilon_0\rangle \]
\[\epsilon_k \Psi_k + {\alpha\over\sqrt{L}} |\epsilon_0\rangle  = E\Psi_k, \ \ \ \ k=1,2,3,...\]

the $k$-th equation may be written as:
\[\Psi_k  \ \ = \ \ \frac{{\alpha\over\sqrt{L}}}{E_k-\epsilon_k}|\epsilon_0\rangle\]
Now by substituting this expression into the 0-th equation:
\[{\alpha^2\over L}\sum_k \frac{1}{E-\epsilon_k} \ \ = \ \  E-\epsilon_0\] 
This sum has been calculated in the first clause, and for an arbitrary energy $E$:
\[-{\alpha^2\over v_E}=E-\epsilon_0\]
By writing $v_E$ explicitly ($v_E=\sqrt{2E\over M}$), we find the third degree equation for the exact energy:
\[M\alpha^2=2E\cdot\left(E-\epsilon_0\right)^2\]
\Dn
\newline Additional information: When using a computer program to find the roots, we see the complex part of the solution in accordance to whats expected when using "P+Q"  formalism and finding the rezolvent and its pole:
\[E_n = \epsilon_0 + \Delta - i {\Gamma\over 2}\]  
Here the requirement of clause (4) is made clear for those familiar with this process.
\Dn 



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