testing blank spaces

The operator \(H\Prime_I(\tau)\) is the interaction
picture operator. The right hand side the matrix element of interaction
Hamiltonian in the interaction picture. This answer can be easily shown
to be equal to the first order time dependent perturbation theory
result obtained in the Schrodinger quantum mechanics.
\begin{eqnarray}
c_{fi}^{(1)}
&=&\frac{1}{i\hbar}\int_{t_0}^{t_1}{}
{\,_S\hspace{-0.5pt}\matrixelement{f}{H{'}_S(\tau)}{i}_S\,
e^{i\omega_{fi}\tau}}
d\tau.\qquad \text{\HighLight{Verify this}}.
\end{eqnarray}

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\MetDat{Title}{\sf Green Function for Schrodinger equation}
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% \Flagged[Convert into a Lesson]
% \subsubsection*{Question}
\section{Lesson Overview}
\paragraph*{Syllabus}
Green function as vacuum expectation value of time ordered product of
Schrodinger field.
\paragraph*{Prerequisites}
Time development in quantum mechanics; Definition of Green function
for a partial differential equation.
\paragraph*{Objectives}
\begin{enumerate}
\item
To introduce the vacuum expectation value (VEV) of time ordered
product of field operators. and to relate it to the Green function of
the Schrodinger equation.
\item
To evaluate the VEV of time ordered product
using the solution of the field equations.
\item
To show that the Green
function is related to the propagator for Schrodinger equation in
quantum mechanics.
\end{enumerate}
\section{Recall and Discuss}
\subsection{Solution of Time Dependent Schrodinger Equation}
For time independent Hamiltonian \(\widehat{H}\) the solution of the
time dependent Schrodinger equation is
\begin{eqnarray}
\psi(\xbf,t) = \sum_n c_n e^{-iE_n(t-t_0)/\hbar} u_n(\xbf).
\end{eqnarray}
where \(E_n, u_n\) are the energy eigenvalues and eigenfunctions and
\(c_n\) are given in terms of the wave function \(\psi(\xbf,t_0\) at
time \(t_0\) by
\begin{equation}
c_n = \int u_n^*(\xbf) \psi(\xbf,t_0)\, d\xbf
\end{equation}
\subsection{Propagator in quantum mechanics}
The propagator \(K(\xbf,t; \xbf_0t_0)\) for Schrodinger equation is
defined as solution of the time dependent Schrodinger equation
\begin{equation}
i\hbar \pp{t} K(\xbf,t;\xbf_0,t_0)- \widehat{H}
K(\xbf,t;\xbf_0,t_0)=0
\end{equation}
subject to the initial condition
\begin{equation}
K(\xbf,t;\xbf_0, t_0)\big|_{t=t_0} = \delta^{\xbf-\xbf_0}.
\end{equation}
The Green function \(G(\xbf,t;\xbf_0,t_0)\) defined by
\begin{equation}
G(\xbf,t;\xbf_0,t_0) =\theta(t-t_0)K(\xbf, t;\xbf_0,t_0)
\end{equation}
obeys
\begin{equation}
i\hbar \pp{t} G(\xbf,t;\xbf_0,t_0)- \widehat{H}
G(\xbf,t;\xbf,t_0)=\delta(\xbf-\xbf_0).\qquad \HighLight{Verify this}
\end{equation}

\section{VEV of Time Ordered Product}

% \input{qft-que-04001}
\begin{AlphaList1}
\item We will show that
\begin{equation}
G(\xbf -\xbf\Prime, t -t\Prime ) = \matrixelement{0}{T \big(ψ\psi(\xbf,
t)\psi^\dagger (\xbf\Prime, t\Prime)\big)}{0}
\end{equation}
obeys the equation for Green function of the free particle Schrodinger
equation.
\item Using the expansion of the field operator in terms of free
particle wave it will be proved that
function \(N \exp(ik\xbf - iE_k t)\), where \(E_k =
\frac{\hbar^2k^2}{2m}\).
Obtain an explicit expression for this
time ordered product as a function of \(\xbf, t, \xbf\Prime , t\Prime\).
\item Have you seen this object before? Where?
\end{AlphaList1}

\subsubsection*{Details}
\begin{itemize}
\item[(a)]
The field operator satisfies the equation
$$\displaystyle{i\hbar{\partial\over\partial t}\psi(\xbf,t) =
-{\hbar^2\over2m}\nabla^2\psi(\xbf,t)}$$

The function \(G(\xbf,t;\xbf\Prime,t\Prime)\) is
\begin{align*}
G(\xbf,t;\xbf\Prime, t\Prime) =&\langle
0|T\big(\psi(\xbf,t)\psi^\dagger(\xbf\Prime,t\Prime)\big)|0\rangle\\
=&
\langle0|\psi(\xbf,t)\psi^\dagger(\xbf\Prime,
t\Prime)|0\rangle\theta(t-t\Prime)
+\langle0|\psi^\dagger(\xbf\Prime
t\Prime)\psi(\xbf,t)|0\rangle\theta(t\Prime-t)
\end{align*}
Therefore,
\begin{eqnarray*}
\lefteqn{i\hbar{\partial\over\partial t}G(\xbf,t;\xbf\Prime,t\Prime)}\\
&=&
\langle0|i\hbar{\partial\over\partial t}
\psi(\xbf,t)\psi^{\dagger}(\xbf\Prime,
t\Prime)|0\rangle\theta(t-t\Prime)
+i\hbar\langle0|\psi(\xbf,t)\psi^\dagger(\xbf\Prime,t\Prime)|0\rangle
\delta(t-t\Prime)\\
&&+i\hbar \langle0|\psi^{\dagger}(\xbf\Prime,
t\Prime){\partial\over\partial t} \psi(\xbf,t)
|0\rangle\theta(t\Prime-t)-i\hbar\langle0|\psi^\dagger(\xbf\Prime,
t\Prime)\psi(\xbf,t)|0\rangle\delta(t\Prime-t)\\
&& \qquad\qquad
\mbox{\HighLight{we have used
\(\quad\dd{t}\theta(t-t\Prime)=\delta(t-t\Prime)=
-\dd{t} \theta(t\Prime-t)\) }} \\
%
&=&-{\hbar^2\over2m}\nabla^2\langle0|\psi(\xbf,t)\psi^\dagger(\xbf\Prime
,
t\Prime)|0\rangle\theta(t-t\Prime)-{\hbar^2\over2m}\nabla^2\langle0|
\psi^\dagger(\xbf\Prime, t\Prime)
\psi(\xbf,t)|0\rangle\theta(t\Prime-t)\\
&&+i\hbar\langle0|\big[\psi(\xbf,t),\psi^\dagger(\xbf\Prime,
t\Prime)\big]|0\rangle\delta(t-t\Prime)
\end{eqnarray*}
The last term becomes equal time commutator due to $\delta(t-t\Prime)$
function.\\
Therefore, we get
\begin{align*}
i\hbar{\partial\over\partial
t}G(\xbf,t;\xbf\Prime,t\Prime)+{\hbar^2\over2m}\nabla^2G(\xbf,t,
\xbf\Prime, t\Prime) =
i\hbar\delta(t-t\Prime)\delta(\xbf-\xbf^{\,\prime})
\end{align*}
\item[(b)] The field $\psi(\xbf,t)$ obeys free particle equation
therefore we
write it as a superposition
\begin{align*}
\psi(\xbf,t)= \iiint^{\infty}_{-\infty} {d^3k\over(2\pi)^{3/2}}
\exp(i\kbf\cdot\vec{\xbf}-iE_kt/\hbar)\, a(\kbf)
\end{align*}
of free particle solutions
\(\displaystyle {1\over(2\pi)^{3/2}}
\exp(i\kbf\cdot\xbf-iE_kt/\hbar)\).
Now
\begin{eqnarray*}
\psi(\xbf,t)\psi^\dagger(\xbf\Prime,t\Prime)&=&\iiint_{-\infty}^\infty{
d^3k\over(2\pi)^{3/2}} e^{i\kbf\cdot\xbf-iE_kt/\hbar} a(\kbf)\times
\iiint_{-\infty}^\infty{d^3q\over(2\pi)^{3/2}}
e^{-i\qbf\cdot\xbf\Prime +i E_qt\Prime/\hbar} a^\dagger(\qbf)
\end{eqnarray*}
Therefore, the vacuum expectation value of the time ordered product
becomes
\begin{align*}
\langle0|T&\big[\psi(\xbf,t)\psi^\dagger(\xbf\Prime,t\Prime)\big]
|0\rangle\\&=
\iiint_{-\infty}^\infty {d^3k\over(2\pi)^{3/2}} \iiint^\infty_{-\infty}
{d^3q\over(2\pi)^{3/2}} e^{i(\kbf\cdot\vec{\xbf}-E_kt/\hbar)}
e^{-i(\qbf\cdot \xbf\Prime-E_qt/\hbar)}\\
&\times\big\{
\langle0|a(\kbf)a^\dagger(\qbf)|0\rangle\theta(t-t\Prime)+\langle0|a^
\dagger(\qbf) a (\kbf)|0\rangle\theta(t\Prime-t)\big\}
\end{align*}
The last term vanishes because $a|0\rangle=0$.
Also
\begin{eqnarray}\nonumber
\langle0|a(\kbf)a^\dagger(\qbf)|0\rangle\theta(t-t\Prime)
&=&\langle0|\big[a(\kbf),a^\dagger(\qbf)\big]
|0\rangle\theta(t-t\Prime)+\langle0|a^\dagger(\qbf)a(\kbf)|0\rangle
\theta(t -t\Prime)\\
&=&\delta(\kbf-\qbf)\theta(t-t\Prime)
\end{eqnarray}
Substituting in Eq.(1) and using $\delta(\kbf-\qbf)$ to do the
$\qbf$
integrals gives
\begin{align*}
\langle0|T\psi(\xbf,t)\psi^\dagger(\xbf\Prime,t\Prime)|0\rangle=
\theta(t-t\Prime)
\iiint_{-\infty}^\infty
{d^3k\over(2\pi)^3}\exp\Big[\kbf\cdot(\xbf-\xbf\Prime) -
{i\hbar^2\over2m} {(t-t\Prime)\kbf^2\over\hbar}\Big]
\end{align*}
The integrals over $\kbf$ are now Gaussian and can be done by
completing the
sequences. This gives the answer for the vacuum expectation value of the
time
ordered product as
\begin{align*}
\langle0|T\psi(\xbf,t)\psi^\dagger(\xbf\Prime,t\Prime)|0\rangle =
\theta(t-t\Prime) \left[{m\over2\pi
i\hbar(t-t\Prime)}\right]^{3/2}\exp\left({
im(\xbf-\xbf\Prime)^2\over2\hbar(t-t\Prime)}\right)
\end{align*}
\item[(c)]
This expression is just the Green function for free particle wave
function in Schrodinger quantum mechanics.
$$
\psi(\xbf,t)=\iiint d\xbf\Prime G(\xbf,t;\xbf\Prime,
t\Prime)\psi(\xbf\Prime, t\Prime)d\xbf\Prime
$$
\end{itemize}
\section{EndNotes}
\subsection{Green function}
The Green function for a partial differential equation
\begin{equation}
L G(\xbf,t;\xbf_0,t_0) = \delta(\xbf-\xbf_0)
\end{equation}
will depend on \(t-t_0\) if the operator \(L\) does not depend on time
explicitly. In this case the solution can be obtained in terms of
eigenfunctions of \(L\), as you have seen in quantum mechanics. The
Green function can be written as
\begin{equation}
G(\xbf,t;\xbf,t_0)= \theta(t-t_0))K(\xbf,t;\xbf_0,t_0)
\end{equation}
and \(K(\xbf,t;\xbf_0,t_0)\) obeys the equation
\begin{equation}
L K(\xbf,t;\xbf,t_0)=0, \qquad
\text{where }K(\xbf,t;\xbf_0,t_0)\big|_{t=t_0}=\delta(\xbf-\xbf_0)
\end{equation}
As you would realize that the corresponding objects in quantum
mechanics and field theory discussed in this lesson are just special
cases for \(L=H\).

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