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% Equation environments
\def\bea{\begin{eqnarray}}
\def\eea{\end{eqnarray}}
\def\ben{\begin{equation}}
\def\een{\end{equation}}
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%
% Special letters 
\def\ham{{\mathcal H}}
\def\mu{{\mathcal U}}
\def\cd{c^{\dagger}}
\def\ch{(c_{H})}
\def\la{{\langle\, }}
\def\ra{{\,\rangle }}
\def\ptl{\partial}
\def\pm{\prime}
\def\d{\delta}
\def\t{\tau}


\begin{document} 

\begin{center}
Solution to Problem Set 1
\end{center}

A general time-dependent hamiltonian describing only single particle
processes is,
\ben
\ham(t) = \sum_{i,j} V_{ij}(t) \cd_i c_j.
\een
Since the hamiltonian is hermitian, the matrix $V_{ij}$ is symmetric. The 
time evolution operator is defined as,
\ben
U(t,t^{\pm}) = T {\rm exp} \left[ - \int_{t^{\pm}}^t \ham (t^{\pm \pm}) d t^{
\pm \pm} \right],
\een
where $T$ stands for time ordering operator. The matrix element of the 
evolution operator in one particle space is defined as,
\ben
\mu_{ij}(t,t^{\pm}) = \la 0 | c_i U(t,t^{\pm}) \cd_j | 0 \ra.
\een
The matrix $B(t,t^{\pm})$ is defined in terms of the matrix $V(t)$ as,
\ben
B(t,t^{\pm}) = T {\rm exp} \left[ - \int_{t^{\pm}}^t V(t^{\pm \pm}) d 
t^{\pm \pm} \right].
\een
Each term in the above exponential is to be understood as multiplication of
several matrices of the form $V_{ij}(t)$, after appropriate time ordering.

\begin{enumerate}

\item
First we examine the time evolution of $B(t,t^{\pm})$, by comparing it with
$B(t + \d t,t^{\pm})$.  We have,
\bea
B(t + \d t,t^{\pm}) &=& T {\rm exp} \left[ - \int_{t^{\pm}}^t V(t^{\pm \pm}) d
t^{\pm \pm} - \d t V(t+) \right] \nonumber \\
&=& T \sum_n \frac{(-1)^n}{n!} \left[ \int_{t^{\pm}}^t V(t^{\pm \pm}) d 
t^{\pm \pm} + \d t V(t+) \right]^n  \nonumber \\
&=& B(t,t^{\pm}) - T \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(n-1)!}
\left[ \int_{t^{\pm}}^t V(t^{\pm \pm}) d t^{\pm \pm} \right]^{n-1} \d t V(t+)
+ {\mathcal O}(\d t)^2 \nonumber \\
&=& B(t,t^{\pm}) - \d t V(t+) B(t,t^{\pm}) . 
\eea
Here $t+ = t + \epsilon$, with $\epsilon \rightarrow 0$. Due to time
ordering, in the last line of the above expression, $V(t+)$ moves to the left
of all other terms. Thus, $B(t,t^{\pm})$ obeys the differential equation,
\ben
\frac{\ptl}{\ptl t} B_{ij}(t,t^{\pm}) = - V_{ik}(t) B_{kj}(t,t^{\pm}).
\een
Similarly, it is easy to show that the time evolution operator satisfies the
differential equation,
\ben
\frac{\ptl}{\ptl t} U(t,t^{\pm}) = - \ham(t) U(t,t^{\pm})
\een
Using equations (3) and (7) we get,
\bea
\frac{\ptl}{\ptl t} \mu _{ij}(t,t^{\pm}) &=&
- \la 0 | c_i \ham(t) U(t,t^{\pm}) \cd_j | 0 \ra \nonumber \\
&=& - \sum_{k,l} \la 0 | c_i V_{kl}(t) \cd_k c_l U(t,t^{\pm}) \cd_j | 0 \ra
\nonumber \\
&=& - \sum_l V_{il}(t) \la 0 | c_l U(t,t^{\pm}) \cd_j | 0 \ra \nonumber \\
&=& - V_{il}(t) \mu_{lj}(t,t^{\pm}).
\eea
Comparing equations (6) and (8) we find that both $B_{ij}(t,t^{\pm})$ and 
$\mu _{ij}(t,t^{\pm})$ satisfy the same differential equation. Moreover, at
the initial condition, i.e. at $t= t^{\pm}$, $B_{ij}(t,t) = \mu _{ij}(t,t) =
\delta_{ij}$. Thus, 
\ben
B_{ij}(t,t^{\pm}) =  \mu _{ij}(t,t^{\pm}).
\een
This proves that the matrix $B_{ij}(t,t^{\pm})$, as defined in equation (4),
describes the evolution operator in the space of single particles.

\item
We examine the time evolution of the Heisenberg operator $\ch_i$, which is 
defined as,
\ben
\ch_i(\t) = U(0,\t) c_i U(\t,0).
\een
Using equation (7) we get,
\[
\frac{\ptl}{\ptl \t} \ch_i(\t) =  U(0,\t) \left[ \ham(\t), c_i \right] U(\t,0)
.
\]
For the quadratic hamiltonian defined in equation (1) we get,
\[
\left[ \ham(\t), c_i \right] = - \sum_l V_{il}(\t) c_l.
\]
Note that the above expression does not depend whether the particles are 
fermionic or bosonic. Using the above two expressions we get,
\ben
\frac{\ptl}{\ptl \t} \ch_i(\t) = - \sum_l V_{il}(\t) \ch_l (\t).
\een
Thus, we find that the time evolution of the annihilation operators (and 
also of the creation operators) can be entirely described in the space of
single particles. This is the essential simplification that is possible for
a hamiltonian describing one particle processes. Such simplification is not 
possible if the hamiltonian includes multi-particle processes.

Next we define the operator,
\ben
A_i(\t) = \sum_j B_{ij}(\t,0) c_j.
\een
Using equation (6), its time evolution is given by,
\bea
\frac{\ptl}{\ptl \t}A_i(\t) &=& - \sum_j V_{ik}(\t) B_{kj}(\t,0) c_j
\nonumber \\
&=& - V_{ik}(\t) A_k (\t).
\eea
Comparing equations (11) and (13) we find that both the operators $\ch_i(\t)$
and $A_i(\t)$ satisfy the same differential equation. Moreover, at initial 
time, i.e. $t=0$, $\ch_i(0) = A_i(0) = c_i$. Thus, they are the same operators
and we can write,
\ben
\ch_i(\t) = \sum_j B_{ij}(\t,0) c_j.
\een
This completes the demonstration of what we have already noted, namely, the
time evolution of the annihilation operators (and also of creation operators)
is given by the matrix of the evolution operator $U(t,t^{\pm})$ in the space
of single particles.


\end{enumerate}

\end{document}
