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\markboth{Homework \#7, Phys623, Spring 1998, Prof.~Yakovenko}
{Homework \#7, Phys623, Spring 1998, Prof.~Yakovenko}
\begin{document}
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\noindent
\begin{minipage}[t]{3.5in}
{\bf Homework \#7} --- Phys623 --- Spring 1998 \\
{\bf Deadline: 5 p.m., Monday, March 30, 1998.} \\
Return homework in class, by e-mail, or \\
put in the box on the door of Phys 2314.
\end{minipage}
\hfill
\begin{minipage}[t]{2.9in}
\raggedleft
Victor Yakovenko, Assistant Professor \\
Office: Physics 2314 \\
Phone: (301)--405--6151 \\
E-mail: yakovenk@physics.umd.edu
\end{minipage}
\medskip
\centerline{\bf Do not forget to write your name and the homework
number!}
\centerline{Equation numbers with the period, like (3.25), refer to the
equations of Schwabl.}
\centerline{Equation numbers without period, like (5), refer to
the equations of this homework.}
\begin{center}
\section*{Molecules (Chapter 15)}
\end{center}
\noindent
{\em Schwabl's Problem 15.2 can be solved using the so-called
elliptical coordinates (see p.\ 475 of the book by Baym). I do not ask
you to take those integrals.}
\bigskip
\begin{enumerate}
\item {\bf [5 points]} Schwabl's Problem 15.1.
\underline{Directions:} With the wave function suggested in
Schwabl, repeat the derivation of Ch.\ 15.3. You don't need to
take integrals, because they can be expressed in terms of the
integrals given in Ch.\ 15.3 by rescaling coordinate $r$ to
absorb $Z^*$. Obtain a new expression for (15.24) and minimize
it in $Z^*$ and $R$. If the minimization turns out to be
difficult analytically, do it numerically if you can
(\emph{Mathematica} ?).
\item {\bf [5 points]} Schwabl's Problem 15.3.
\underline{Directions:} This problem discusses a molecule that
consists of one electron and two nuclei of the nuclear charge
$Z=2$. Repeat the derivation of Ch.\ 15.3 in this case. Obtain a
new expression for (15.24) and show that it does not have a
minimum as function of $R$. You may need to plot this function
numerically (\emph{Mathematica} ?). Strictly speaking, the
variational method cannot prove the \emph{absence} of a bound
state, but it gives some indication.
\item {\bf [2 points]} Schwabl's Problem 15.4.
\underline{Directions:} Don't do calculations in this problem;
just sketch the energy vs $a$ qualitatively.
\item {\bf [5 points]} Schwabl's Problem 15.5.
\item {\bf [5 points]} Schwabl's Problem 15.7.
\underline{Directions:} Follow Ch.\ 15.5. Appendix D of Schwabl
may be useful. This potential is called the \emph{Lennard-Jones
6--12 potential} and is widely used in molecular physics. The
power 6 represents the van der Waals attraction, and the power
12 is selected for mathematical convenience.
\item {\bf [7 points]} Calculate the energy of the van der Waals
interaction between two hydrogen atoms in the ground states
(Ch.\ 15.6) using the variational method. Try the following two
variational functions:
\begin{eqnarray}
\psi_\alpha({\bf r}_1,{\bf r}_2)&=&
C\psi_0(r_1)\psi_0(r_2)(1+\alpha z_1z_2), \\
\psi_\alpha({\bf r}_1,{\bf r}_2)&=&
C\psi_0(r_1)\psi_0(r_2)[1+\alpha (x_1x_2+y_1y_2-2z_1z_2)],
\label{variational}
\end{eqnarray}
where $\psi_0(r)$ is the ground-state wave function of the
hydrogen atom, $C$ is a normalization constant, $\alpha$ is a
variational parameter. Solution of Problem 5 of Homework 6 and
the equation before Eq.\ (15.51) may be useful.
\item Thus far we considered molecules that consist of two atoms. Now
let us consider a molecule that consists of an infinite number of
atoms spaced periodically at a distance $d$: a crystal. Let us
consider one-dimensional arrangement for simplicity. The
Hamiltonian of a particle is
\begin{equation}
\hat{H}=\frac{\hat{p}^2}{2\mu} + \sum_{l=-\infty}^{\infty} V_l(x),
\end{equation}
where $l$ is an integer, and $V_l(x)=V_0(x-ld)$ is the potential
of one atom displaced to a distance $ld$. As a specific example,
we will consider the Dirac comb:
\begin{equation}
\hat{H}_D=\frac{\hat{p}^2}{2\mu}-\lambda \sum_{l=-\infty}^{\infty}\delta(x-ld),
\end{equation}
where $\delta(x)$ is the Dirac delta-function.
Let us consider the case where the distance between the atoms,
$d$, is substantially greater that the size of the atomic wave
function $a$. For example, a single Dirac potential
$V_0(x)=-\lambda\delta(x)$ has the wave function
\begin{equation}
\psi_0(x)=\sqrt{\kappa}e^{-\kappa|x|}
\label{psi0}
\end{equation}
where $\kappa=\lambda\mu/\hbar^2$, and the energy
$E_0=-\hbar^2\kappa^2/2\mu$. We consider that case where $\kappa
d\ll1$, so the overlap of the wave functions of different atoms
is exponentially small.
Let us construct the wave function in a crystal as a linear
superposition of the eigenfunctions of individual atoms (the
so-called \emph{LCAO method} (linear combination of atomic
orbitals) or the \emph{tight-binding approximation}):
\begin{equation}
\psi(x)=\sum_n c_n \psi_n(x), \qquad {\rm where} \quad
\psi_n(x)=\psi_0(x-nd).
\label{psi}
\end{equation}
Because all wave functions $\psi_n(x)$ have the same energies
within their atoms, the coefficients $c_n$ should be determined
from some sort of a secular equation. To derive this equation,
let us substitute the sum (\ref{psi}) into the Schr\"odinger
equation
\begin{equation}
\hat{H}\psi=E\psi,
\label{H=E}
\end{equation}
and take a scalar product of (\ref{H=E}) with $\psi_m^*(x)$.
\begin{enumerate}
\item {\bf [3 points]} Show that this generates the following
equation on the coefficients $c_n$:
\begin{equation}
\sum_n \langle\psi_m|\sum_{l\neq m}V_l|\psi_n\rangle c_n
= (E-E_0)[c_m + \sum_{n\neq m}\langle\psi_m|\psi_n\rangle c_n]
\label{secular}
\end{equation}
\item {\bf [5 points]} Show that $c_n^{(k)}=e^{ikn}$ is an
eigenvector of Eq.\ (\ref{secular}) corresponding to the
energy
\begin{eqnarray}
E(k)&=&E_0+\frac{\beta+\gamma(k)}{1+\alpha(k)}, \label{E(k)} \\
\alpha(k)&=&\sum_{n\neq0} e^{ikn} \langle\psi_0|\psi_n\rangle, \label{alpha} \\
\beta&=&\langle\psi_0|\sum_{l\neq0}V_l|\psi_0\rangle, \\
\gamma(k)&=&\sum_{n\neq0} e^{ikn} \langle\psi_0|\sum_{l\neq0}V_l|\psi_n\rangle.
\label{gamma}
\end{eqnarray}
\setcounter{continue}{\value{enumii}}
\end{enumerate}
Because the atomic wave functions typically decay exponentially
(see Eq.\ (\ref{psi0})), it is sufficient to keep only the terms
with $n=\pm1$ and $l=1$ in Eqs.\
(\ref{alpha})--(\ref{gamma}). Further examination shows that,
$\alpha$ and $\beta$ can be dropped compared with the leading
term $\gamma$. So we arrive to the following equation:
\begin{equation}
E(k)=E_0 + 2t\cos(k), \qquad {\rm where} \quad
t=\langle\psi_0|V_1|\psi_1\rangle.
\end{equation}
The matrix element $t$ is called the {\em transfer integral}
between the nearest neighboring atomic sites. In this
approximation, the secular equation (\ref{secular}) has the
form:
\begin{equation}
Ec_m = E_0c_m + t (c_{m+1}+c_{m-1}),
\label{E'(k)}
\label{secular'}
\end{equation}
where the matrix element $t$ ``transfers'' the particle from
site $m$ to the neighboring sites $m\pm1$.
\begin{enumerate}
\setcounter{enumii}{\value{continue}}
\item {\bf [5 points]} For the Dirac comb potential, check the
validity of the approximations that led from Eq.\
(\ref{E(k)}) to Eq.\ (\ref{E'(k)}) and calculate the
transfer integral $t$.
\end{enumerate}
\end{enumerate}
\end{document}