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\title{Exercise 3}
\author{Michael Goerz}
\date{May 9, 2005}

\begin{document}

\maketitle

%\begin{abstract}
%\end{abstract}

\subsubsection*{Problem 8}

\textbf{Premise:}
The operator $e^A$ is defined as
\begin{eqnarray}
e^A := \sum \limits_{n=1}^{\infty} \frac{A^n}{n!}
\end{eqnarray}
The $x$-operator is
\begin{eqnarray}
x = i\hbar \frac{\partial}{\partial p}
\end{eqnarray}

\textbf{Assertion:}
\begin{eqnarray}
e^{ipa/\hbar} \, x \, e^{-i p a/\hbar}=x+a \label{eq81}
\end{eqnarray}

\textbf{Proof:}
We apply the left hand side of eq. (\ref{eq81}) to a function $f(p)$:
\begin{eqnarray}
e^{ipa/\hbar} \, x \, e^{-i p a/\hbar} \, f(p) &=& e^{ipa/\hbar} \, i\hbar \frac{\partial}{\partial p} \, e^{-i p a/\hbar + \ln(f(p))} \nonumber\\
&=& e^{ipa/\hbar} \, i\hbar \left[-i \,a/\hbar + \frac{1}{f} f' \right] \, e^{-i p a/\hbar + \ln(f(p))} \nonumber\\
&=& e^{ipa/\hbar} \, \left[a + i\hbar \frac{1}{f} f' \right] \, f \; e^{-i p a/\hbar} \nonumber\\
&=& (a+x) f(p)
\end{eqnarray}
\begin{flushright}
$\blacksquare$
\end{flushright}


\subsubsection*{Problem 9}
The potential energy $V$ is given as
\begin{eqnarray}
V(x)=-\lambda \delta(x);\qquad \lambda > 0
\end{eqnarray}
This implies the following Schr\"odinger equation for the problem:
\begin{eqnarray}
\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \Psi(x) - \lambda \delta(x) \Psi(x) = E \Psi(x) \label{eq9.1}
\end{eqnarray}
We know that $\Psi(x)$ is continuous and $\Psi(x)$ is discontinuous at $x=0$. By integrating the Schr\"odinger equation around $x=0$, we find a condition for the discontinuity.
\begin{eqnarray}
\frac{-\hbar^2}{2m} \int\limits_{-\epsilon}^{+\epsilon} \Psi''(x) \, dx - \lambda \int\limits_{-\epsilon}^{+\epsilon} \delta(x) \Psi(x) \, dx &=& E \int\limits_{-\epsilon}^{+\epsilon} \Psi(x) \,dx; \qquad \epsilon \rightarrow 0 \nonumber\\
\frac{-\hbar^2}{2m} \left( \lim\limits_{\epsilon \rightarrow 0}(\Psi'(\epsilon) - \lim\limits_{\epsilon \rightarrow 0}(\Psi'(-\epsilon)\right) - \lambda \Psi(0) &=& 0  \label{eq9.2}
\end{eqnarray}

For $\Psi(x)$, we choose the following ansatz:
\begin{eqnarray}
\Psi(x) &=& \left\lbrace \begin{array}{lc}
 c_1 \, e^{\kappa x}; & x < 0 \\
 c_2 \, e^{-\kappa x}; & x > 0
\end{array}\right.
\end{eqnarray}
From the condition that $\Psi(x)$ must be continuous, we see that $c_1=c_2$. We can then calculate the energy by inserting $\Psi(x)$ into eq. (\ref{eq9.1}), getting
\begin{eqnarray}
E=-\frac{\hbar^2 \kappa^2}{2m} \label{eq9.3}
\end{eqnarray}
Finally, we have to calculate $\kappa$, which can be done using the discontinuity of $\Psi'(\pm 0)$. Thus, we insert $\Psi(\pm 0)$ into eq. (\ref{eq9.2}):
\begin{eqnarray}
\frac{-\hbar^2}{2 m} (\kappa + \kappa) - \lambda = 0 \nonumber\\
\Leftrightarrow \kappa = \frac{-\lambda m}{\hbar^2} \label{eq9.4}
\end{eqnarray}
From eq. (\ref{eq9.3}) and (\ref{eq9.4}) we now find the energy as
\begin{eqnarray}
E=\frac{-\hbar^2 \lambda^2 m^2}{2m\hbar^4}=-\frac{\lambda^2m}{2\hbar^2}
\end{eqnarray}

\subsubsection*{Problem 10}
As in Problem 9, we have singularities in the potential, but now we have one at $x=-a$ and one at $x=+a$, splitting the x-axis in three parts. At each singularity, we have to meet conditions of continuity for $\Psi(x)$. We choose the following ansatz:
\begin{eqnarray}
\Psi(x) &=& c \, e^{\kappa x}; \qquad x>a \\
\Psi(x) &=& c \, e^{- \kappa x}; \qquad x<-a \label{ansatz2}\\
\Psi(x) &=& e^{\kappa x} + e^{- \kappa x}; \qquad -a<x<a\label{ansatz3}
\end{eqnarray}
The parameter $c$ allows $\Phi(x)$ to be continuous at $x=\pm a$. For reasons of symmetry, this parameter has to be the same for both $x<-a$ and $x>a$. We can also formulate an antisymmetric ansatz $\Psi(x) = e^{\kappa x} - e^{- \kappa x}$ (which then also requires a minus in eq. (\ref{ansatz2}))
The continuity condition can be written as:
\begin{eqnarray}
c \, e^{-\kappa x} &=& e^{\kappa x} + e^{- \kappa x} \nonumber \\
\Rightarrow \quad c &=& 1 + e^{2 \kappa a}
\end{eqnarray}
As in Problem 9, $\Psi'(x)$ must be discontinuous at the singularities. We find a description of this discontinuity by integrating the Schr\"odinger equation at the specified points.
\begin{eqnarray}
-\frac{\hbar^2}{2m} \int\limits_{a-\epsilon}^{a+\epsilon} \Psi''(x) \, dx - \lambda \int\limits_{a-\epsilon}^{a+\epsilon} \delta(x-a) \Psi(x) \, dx &=& E \int\limits_{a-\epsilon}^{a+\epsilon} \Psi(x) \,dx \nonumber \\
-\frac{\hbar^2}{2m} \left( \lim\limits_{\epsilon \rightarrow 0}(\Psi'(a+\epsilon) - \lim\limits_{\epsilon \rightarrow 0}(\Psi'(a-\epsilon)\right) - \lambda \Psi(a) &=& 0 \nonumber \\
-\frac{\hbar^2}{2m} \left( -\kappa \, c \, e^{-\kappa a}- \kappa\left(e^{\kappa a}-e^{-\kappa a}\right)\right) - \lambda \Psi(a)\left(e^{\kappa a}+e^{-\kappa a}\right) &=& 0
\end{eqnarray}
After inserting $c$, we can use Mathematica to further simplify the expression:
\begin{equation}
2 \, \lambda  \cosh(2 \kappa )=\frac{e^{2 \kappa } \kappa  \hbar ^2}{m}
\end{equation}
For the antisymmetric ansatz we find from the continuity condition of $\Psi(x)$
\begin{eqnarray}
c \, e^{- \kappa x} &=& e^{\kappa x} - e^{- \kappa x} \nonumber \\
\Rightarrow \quad c &=& 1 - e^{2 k a}
\end{eqnarray}
Integrating the Schr\"odinger equation in this case yields:
\begin{eqnarray}
-\frac{\hbar^2}{2m} \left( -\kappa \, c \, e^{-\kappa a}- \kappa\left(e^{\kappa a}+e^{-\kappa a}\right)\right) - \lambda \Psi(a)\left(e^{\kappa a}-e^{-\kappa a}\right) &=& 0
\end{eqnarray}
When we insert the constant, we find the following as the transcendent equation describing the antisymmetric case:
\begin{equation}
\lambda  \sinh (2 \kappa )=\frac{\kappa  \hbar ^2 \cosh (2 \kappa )}{m}
\end{equation}
\end{document}


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