[QUE/QFT-11005] QFT-PROBLEM

$\newcommand{\matrixelement}[3]{\langle#1|#2|#3\rangle}$
$\newcommand{\PP}[2][]{\frac{\partial^2 #1}{\partial #2^2}}$
$\newcommand{\dd}[2][]{\frac{d#1}{d#2}}$
 $\newcommand{\pp}[2][]{\frac{\partial #1}{\partial #2}}$
The Feynman propagator \(\Delta_F\) for a real scalar field is defined as the vacuum expectation value of time ordered product of fields: \[ \Delta_F(x-y) = \matrixelement{0}{T\big(\phi(x)\phi(y)\big)}{0}.\] We will now show that the Feynman propagator satisfies the equation \[ (\Box + M^2) \Delta_F(x-y)= - \delta^{(4)}(x-y).\] }Since \(\phi(x)\) obeys the Klein Gordon equation, it is obvious that \(\Delta^{(\pm)}(x-y)\) satisfy the Klein Gordon equation \begin{equation} (\Box + M^2)\Delta^{(\pm)}=0 . \end{equation} The time ordered product can be written as \begin{eqnarray} T(\phi(x)\phi(y)) &=&(\phi(x)\phi(y))\theta(x_0-y_0) + (\phi(y)\phi(x))\theta(y_0-x_0) \end{eqnarray} Now we compute action of the operator \(\displaystyle\Box =\frac{\partial^2}{x^{0\,2}}- \frac{\partial^2}{\partial x^{k\,2}}\) on the time ordered product. Note that the space derivatives in \(\displaystyle\Box = \PP{{x^0}}- \PP{{x^k}}\) will act only on the fields but the time derivative will act on theta functions also. So let us compute the time derivatives using \(\displaystyle\dd[\theta(x_0-y_0)]{x_0}= \delta(x_0-y_0)\). We will get \begin{eqnarray} { \pp{x_0}T(\phi(x)\phi(y))\nonumber }\nonumber &=&\Big\{\pp[\phi(x)]{x_0}\phi(y)\theta(x_0 - y_0) + (\phi(x)\phi(y))\pp{x_0}\theta(x_0 - y_0)\Big\} - \Big\{\phi(y)\pp[\phi(x)]{x_0}\theta(y_0 - x_0) - (\phi(y)\phi(x))\pp{x_0}\theta(y_0 - x_0)\Big\}\nonumber &=& \dot{\phi}(x)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\dot{\phi}(x)\theta(y_0 - x_0) + \big[\phi(x), \phi(y)\big] \delta(x_0 - y_0)\\ &=& \big(\partial_0{\phi}(x)\big)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\big(\partial_0{\phi}(x)\big)\theta(y_0 - x_0) \end{eqnarray} The last term in \eqRef{EQ19} becomes equal time commutator of the fields and hence it is zero. Differentiating \eqRef{EQ20} once again w.r.t. \(x_0\) we will get \begin{eqnarray}\nonumber { \frac{\partial^2}{\partial x^{02}}T(\phi(x)\phi(y)) }\nonumber &=& \ddot{\phi}(x)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\ddot{\phi}(x)\theta(y_0 - x_0) +\nonumber \dot{\phi}(x)\,\phi(y)\delta(x_0 - y_0) - \phi(y)\,\dot{\phi}(x)\delta(y_0 - x_0)\\\nonumber &=& \ddot{\phi}(x)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\ddot{\phi}(x)\theta(y_0 - x_0) \\\nonumber && \qquad + \big[\dot{\phi}(x)\,,\,\phi(y) \big] \delta(x_0-y_0) . \end{eqnarray} Substituting the value of equal time commutator \[\big[\dot{\phi}(x)\,,\,\phi(y)\big]\Big|_{x_0=y_0} = -\delta^{(3)}(\vec{x}-\vec{y}).\] we get\begin{eqnarray}\nonumber {\frac{\partial^2}{\partial x^{02}}\big\{T(\phi(x)\phi(y))\big\} }\ddot{\phi}(x)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\ddot{\phi}(x)\theta(y_0 - x_0) - \delta^{(4)}(x-y). \end{eqnarray} Hence \begin{eqnarray}\nonumber {\big(\partial_0^2-\partial_k^2 + M^2\big)T(\phi(x)\phi(y))} &=& \big(\Box + M^2\big)\phi(x)\,\phi(y)\theta(x_0 - y_0)- \phi(y)\big(\Box + M^2\big)\phi(x)\theta(y_0- x_0) - \delta^{(4)}(x-y) - \delta^{(4)}(x-y) . \end{eqnarray} Remembering that the field satisfies Klein Gordon equation and taking vacuum expectation value, we get the desired answer \begin{equation} \big(\Box + M^2\big) \Delta_F(x) = - \delta^{(4)}(x-y) .\end{equation}