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\begin{document}
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\heading{E2122: Bound state in a site coupled to a 1-D conductor}
\auname{Dror Vinkler, Nir Yarimi}


{\bf The problem:}

\Dn


In this problem we find the energy of the bound state E of a particle in a system of one state coupled with a 1-D conductor.

The particle has a mass M. The conductor is infinite: $x\in\{-\infty, \infty\}$.

The site to which the particle is bound is near the point $x_0$.

If the site was decoupled from the conductor, the particle's bound energy was $E_0$.

In practice, the site is coupled to the conductor, and the hamiltonian is
\begin{equation*}
H = \frac{p^2}{2M} + | 0 \rangle E_0 \langle 0 | + | x_0 \rangle \lambda \langle 0 | + | 0 \rangle \lambda \langle x_0 |
\end{equation*}

The standard representation of the wave function in the $x$ basis is $\Psi \rightarrow (\psi_0, \psi(x))$.

Where the amplitudes are $\psi_0 = \langle 0 | \Psi \rangle$, $\psi(x) = \langle x | \Psi \rangle$.

With the normalization $|\psi_0|^2 + \int\limits_{-\infty}^{\infty} |\psi(x)|^2 dx = 1$.

\Dn

\begin{enumerate}[(1)]
\item Write $\langle 0 | H | \Psi \rangle$ in the standard representation.
\item Write $\langle x | H | \Psi \rangle$ in the standard representation. \\
\end{enumerate}
Write the equations $H | \Psi \rangle = E | \Psi \rangle$ in the special representation. \\
Tip: project the equation on the states $\langle 0 |$, $\langle x |$ and use the results from previous clauses.\\
Show that we get a closed equation for $\psi(x)$ with an effective potential of
\begin{equation*}
V(x;E) = u(E)\delta(x-x_0)
\end{equation*}
Notice that at this point we treat the bound energy $E$ as a known constant.

Tip: without loss of generality, you can assume the location $x_0$ is the origin.

\begin{enumerate}[(1)]
\setcounter{enumi}{2}
\item Write the expression for $u(E)$.
\item Write an equation for the bound energy $E$.
\item Solve the equation for the case $E_0 = 0$.
\end{enumerate}

\Dn\Dn

{\bf The solution:}

\Dn
\begin{enumerate}[(1)]

\item
\begin{equation*}
\langle 0 | H | \Psi \rangle = E_0 \psi_0 + \lambda \psi(x_0)
\end{equation*}

\item
\begin{equation*}
\langle x | H | \Psi \rangle = -\frac{\psi''(x)}{2M} + \delta(x-x_0) \lambda \psi_0
\end{equation*}

\item
\begin{equation*}
\langle 0 | H | \Psi \rangle = E_0 \psi_0 + \lambda \psi(0) = E \psi_0 = \langle 0 | E | \Psi \rangle
\end{equation*}
\begin{equation*}
\langle x | H | \Psi \rangle = -\frac{\psi''(x)}{2M} + \delta(x) \lambda \psi_0 = E \psi(x) = \langle x | E | \Psi \rangle
\end{equation*}
We can isolate $\psi_0$ from the first equation and substitute it in the second equation, and get
\begin{equation*}
-\frac{\psi''(x)}{2M} + \frac{\lambda^2}{E - E_0}\delta(x)\psi(0) = E \psi(x)
\end{equation*}
Therefore
\begin{equation*}
V(x;E) = \frac{\lambda^2}{E - E_0}\delta(x) = u(X) \delta(x)
\end{equation*}

\item
$\psi(x)$ will be of the form
\begin{equation*}
\psi(x) = \left\{
	               \begin{array}{ll}
            		A e^{-kx}  & \mbox{if } x \geq 0 \\
            		A e^{kx} & \mbox{if } x < 0
                   \end{array}
\right.
\end{equation*}
For $x\neq0$:
\begin{equation*}
-\frac{k^2}{2M} = E \Rightarrow k = \sqrt{-2ME}
\end{equation*}
For $x=0$:
\begin{equation*}
\psi'(0^{+}) - \psi'(0^{-}) = 2 M \frac{\lambda^2}{E - E_0} \psi(0)
\end{equation*}
Substituting $k$ with its value from the first equation, we get
\begin{equation*}
\sqrt{-2 M E} = \frac{M \lambda^2}{E_0 - E}
\end{equation*}

\item
For $E_0 = 0$:
\begin{equation*}
E = -\sqrt[3]{\frac{M \lambda^4}{2}}
\end{equation*}


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\end{enumerate}
\end{document}
