Results From QFT-1
Consider QFT of scalar field.
--- We know that most of physical information is contained in
\[
G\left(x_1, x_2, \ldots x_n\right) = \frac{\left\langle \Omega\right|\, T\left(\prod_{i=1}^{N}\hat{\phi}(x)\right)\left|\Omega\right\rangle}{\left\langle \Omega \middle| \Omega \right\rangle}
\]
where, \(x = \left(\vec{x}, t\right)\)
\[
\begin{gathered}
\left|\Omega\right\rangle = \text{interacting ground state.}\\
T\left(\phi(t_1)\,\phi(t_2)\right) = \theta\left(t_1 - t_2\right)\,\phi(t_1)\,\phi(t_2) + \theta\left(t_2 - t_1\right)\,\phi(t_1)\,\phi(t_2)\\
\left(\text{i.e.\ largest time on left}\right)
\end{gathered}
\]
The reason greens fun. is important is that it contains
--- info. about masses of particles for a given theory (given by pole positions of 2 point functions in momentum space)
--- It contains info about \(S\)-matrix elements. via LSZ formalism.
Goals For QFT-2 ---
(1) One first goal in QFT-2 will be to develope new ways of calculating green's fun. ex. Path integral formalism & study properties of Green's fun.; we will also try to know why Path integral formulation? ( (1) Canonical Formulation breaks manifested lorentz invariance (we give special role to time in hamiltonian formulation i.e. Hamiltonian generates time translations not space translations); although final results are lorentz invariance, so one advantage is that we can maintain lorentz invariance in path integral formulation.
(2) Another reason to use path integral formulation is that, for cases of interaction term in lagrangian having time derivative than hamiltonian might become complicated \(\left(H = p\dot{q} - L\right)\)
Consider 1-d QM ---
\[
L = \frac{1}{2}\,\dot{q}^{2} + \frac{\lambda}{2}\,\dot{q}^{2}\,q
\]
\[
\begin{gathered}
p = \frac{\partial L}{\partial\dot{q}} = \dot{q} + \lambda\,\dot{q}\,q\\
\Rightarrow\qquad \dot{q} = \frac{p}{1+\lambda q}
\end{gathered}
\]
\[
\begin{aligned}
H &= \dot{q}\,p - L\\
&= \frac{p^{2}}{\left(\lambda q + 1\right)} - \frac{1}{2}\,\frac{p^{2}}{\left(1+\lambda q\right)^{2}} - \frac{\lambda}{2}\,q\,\frac{p^{2}}{\left(1+\lambda q\right)^{2}}\\
&= \frac{p^{2}}{\left(1+\lambda q\right)}\left(1 - \frac{1}{2\left(1+\lambda q\right)} - \frac{q\lambda}{2\left(1+\lambda q\right)}\right)\\
&= \frac{p^{2}}{2\left(1+\lambda q\right)^{2}}\left(2 + 2\lambda q - 1 - \lambda q\right)
\end{aligned}
\]
\[
\begin{aligned}
H &= \frac{1}{2}\left(\frac{p^{2}}{1+\lambda q}\right) = \frac{p^{2}}{2}\left(1 + q\lambda\right)^{-1}\\
&= \frac{p^{2}}{2}\left(1 - q\lambda + \frac{q^{2}\lambda^{2}}{2} - \frac{\lambda^{3}\,q^{3}}{3!} + \cdots\right)
\end{aligned}
\]
Note that lagrangian only contains one term of '\(\lambda\)' whereas hamiltonian has multiple orders of \(\lambda\).
(3) Path integral approach is well suited for non-abelian gauge theories. (non-abelian gauge theories describe strong, weak & electromagnetic interactions, i.e. everything except gravity).
We can also develope path integral approach for time dependent lagrangian but our lagrangian for universe is time independent upto great accuracy.
Path Integral in QM ---
Lets start for simple example ---
\[
\begin{gathered}
H = \frac{p^{2}}{2m} + V(q)\\
L = \frac{1}{2}\,m\,\dot{q}^{2} - V(q)
\end{gathered}
\]
we define Action \((S)\) \(= \displaystyle\int dt\; L\left(q(t), \dot{q}(t)\right)\)
Now consider quantum theory ---
--- We will use Heisenberg picture (time independent states & time dependent operator)
\[
\Psi_{H} = \Psi_{S}(0)\;,\qquad O_{H}(0) = O_{S}
\]
--- set \(\hbar = 1\)
--- Operators evolve via
\[
\begin{aligned}
\hat{q}(t) &= e^{iHt}\;\hat{q}_{H}(0)\;e^{-iHt}\\
\hat{q}_{H}(t) &= e^{iHt}\;\hat{q}_{S}\;e^{-iHt}
\end{aligned}
\]
Now consider two states (eigen states of \(\hat{q}\), labelled by e.v. \(q'\) & \(q''\)). \(\left|q'\right\rangle\) & \(\left|q''\right\rangle\) respectively.
Quantity we are interested in is \(K\left(q', t';\ q'', t''\right)\)
\[
\begin{aligned}
K\left(q', t';\ q'', t''\right) &= \left\langle q''\right|\, e^{-iH\left(t''-t'\right)}\left|q'\right\rangle\\
&= \left\langle q''\right|\, \underline{e^{-iH t''}}\;\underline{e^{iH t'}}\left|q'\right\rangle
\end{aligned}
\]
ex :
\[
\begin{aligned}
\hat{q}(t')\,\underline{e^{iHt'}\left|q'\right\rangle} &= e^{iHt'}\,\hat{q}\,e^{-iHt'}\,e^{iHt'}\left|q'\right\rangle\\
&= e^{iHt'}\;\hat{q}\left|q'\right\rangle\\
&= q'\;e^{iHt'}\left|q'\right\rangle
\end{aligned}
\]
So \(e^{iHt'}\left|q'\right\rangle\) is eigen state of \(\hat{q}(t')\) with e.v. \(q'\).
Similarly,
\[
\left\langle q''\right|\,e^{-iHt''}\,\hat{q}(t'') = q''\left\langle q''\right| e^{-iHt''}
\]
So,
\(K\) is overlap between eigen states of \(\hat{q}(t')\) & \(\hat{q}(t'')\) with e.v. \(q'\) & \(q''\) respectively.
Claim ---
\[
K\left(q', t';\ q'', t''\right) = \int\left[\mathcal{D}q\right]\, e^{iS}
\qquad\text{for}\quad t'' > t'
\]
\definecolor{ForestGreen}{rgb}{0.13,0.55,0.13}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,-0.3) -- (0,3) node[left]{$q$};
\draw[->] (-0.5,0) -- (5.6,0) node[right]{$t$};
\node[left] at (0,2.4) {$q''$};
\node[left] at (0,1.0) {$q'$};
\draw[dash dot] (0,1.0) -- (1.4,1.0);
\draw[dash dot] (0,2.4) -- (4.2,2.4);
\draw[dash dot] (1.4,0) node[below]{$t'$} -- (1.4,1.0);
\draw[dash dot] (4.2,0) node[below]{$t''$} -- (4.2,2.4);
\draw[color=ForestGreen] (1.4,1.0) .. controls (2.2,2.6) and (3.2,1.6) .. (4.2,2.4);
\draw[color=ForestGreen] (1.4,1.0) .. controls (2.4,0.4) and (3.4,2.9) .. (4.2,2.4);
\draw[color=ForestGreen] (1.4,1.0) .. controls (2.0,1.8) and (3.6,0.6) .. (4.2,2.4);
\draw[color=ForestGreen] (1.4,1.0) .. controls (2.6,1.9) and (3.3,2.2) .. (4.2,2.4);
\node[right, text width=4.5cm] at (6,1.5) {above exp.\ is integral over all paths.};
\end{tikzpicture}
We think of path as descrete collection of points. Lets take \(N\) as some large integer.
\[
\text{We define}\qquad \Delta = \frac{t''-t'}{N} \quad\Rightarrow\;\text{small number}
\]
\[
\begin{aligned}
\tau_{k} &= t' + (k-1)\,\Delta \qquad k = 1, 2, 3\ldots N+1.\\
\tau_{1} &= t'\\
\tau_{2} &= t' + \Delta\\
\tau_{3} &= t' + 2\Delta\\
&\;\;\vdots\\
\tau_{N+1} &= t' + N\Delta = t''
\end{aligned}
\]
\definecolor{ForestGreen}{rgb}{0.13,0.55,0.13}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,-0.3) -- (0,3) node[left]{$q$};
\draw[->] (-0.5,0) -- (5.6,0) node[right]{$t$};
\node[left] at (0,2.4) {$q''$};
\node[left] at (0,1.0) {$q'$};
\draw[dash dot] (0,1.0) -- (1.2,1.0);
\draw[dash dot] (0,2.4) -- (3.9,2.4);
\draw[dash dot] (1.2,0) -- (1.2,1.0);
\draw[dash dot] (3.9,0) -- (3.9,2.4);
\node[below] at (1.2,-0.05) {$\tau_1 = t'$};
\node[below] at (3.9,-0.05) {$t''$};
\node[above right,font=\scriptsize,color=gray] at (1.35,0.05) {$\tau_2\,\tau_3$};
\draw[color=gray] (1.35,0.06) -- (1.35,0.16); \draw[color=gray] (1.55,0.06) -- (1.55,0.16);
\draw[color=gray] (1.75,0.06) -- (1.75,0.16); \draw[color=gray] (1.95,0.06) -- (1.95,0.16);
\draw[color=gray] (2.15,0.06) -- (2.15,0.16); \draw[color=gray] (2.35,0.06) -- (2.35,0.16);
\draw[color=gray] (2.55,0.06) -- (2.55,0.16); \draw[color=gray] (2.75,0.06) -- (2.75,0.16);
\node[above,font=\scriptsize] at (3.45,0.35) {$\tau_{N+1} = t''$};
\draw[->,font=\scriptsize] (3.45,0.3) -- (3.85,0.05);
\draw[color=ForestGreen] (1.2,1.0) .. controls (2.0,2.6) and (3.0,1.6) .. (3.9,2.4);
\draw[color=ForestGreen] (1.2,1.0) .. controls (2.2,0.4) and (3.2,2.9) .. (3.9,2.4);
\draw[color=ForestGreen] (1.2,1.0) .. controls (1.8,1.8) and (3.4,0.6) .. (3.9,2.4);
\end{tikzpicture}
We define
\[
\begin{gathered}
q_{i} = q\left(\tau_{i}\right)\ \text{ for a given path}\\
q_{1} = q\left(\tau_{1}\right) = q(t') = q'\\
q_{2} = q\left(\tau_{2}\right)\\
\vdots\\
q_{N+1} = q\left(\tau_{N+1}\right) = q(t'') = q''
\end{gathered}
\]
So our path is collection of points \(\left(q_1, q_2 \ldots q_{N+1}\right)\) (or) \(\left(q', \ldots\ldots, q''\right)\)
So action
\[
\begin{aligned}
(S) &= \int_{t'}^{t''} dt\,\left(\frac{1}{2}\,\dot{q}^{2} - V(q)\right)\\
&= \Delta\,\sum_{k=1}^{N}\left(\frac{1}{2}\left(\frac{q_{k+1}-q_{k}}{\Delta}\right)^{2} - V\left(q_{k}\right)\right)
\end{aligned}
\]
So
\[
\begin{gathered}
K\left(q', t';\ q'', t''\right) = \lim_{N\to\infty}\;\mathcal{N}\!\int dq_2\, dq_3\ldots dq_N\; e^{iS}\\
\hookrightarrow\;\text{does not depend on } q' \,\&\, q''.\quad\left(\text{fixed for all case}\right).
\end{gathered}
\]
\[
\begin{aligned}
K\left(q', t';\ q'', t''\right) &= \left\langle q''\right| e^{-iH\left(t''-t'\right)}\left|q'\right\rangle\\
&= \left\langle q''\right| e^{-iH N\Delta}\left|q'\right\rangle\\
&= \left\langle q''\right| \underbrace{e^{-iH\Delta}}_{\downarrow}\, e^{-iH\Delta}\, e^{-iH\Delta}\cdots \underbrace{e^{-iH\Delta}}_{\uparrow}\left|q'\right\rangle
\end{aligned}
\]
\[
\int dq_N \left|q_N\right\rangle\left\langle q_N\right|
\qquad\qquad
\begin{aligned}
&\text{insert completeness relation.}\\
&\int dq_2 \left|q_2\right\rangle\left\langle q_2\right|
\end{aligned}
\]
\[
= \int dq_2\, dq_3\ldots dq_N\;\prod_{k=1}^{N}\;\underline{\left\langle q_{k+1}\right| e^{-iH\Delta}\left|q_{k}\right\rangle}
\qquad
\left(\underline{q_{N+1} = q''}\ \&\ q_{1} = q'\right)
\]
Now our goal is to evaluate \(\left\langle q_{k+1}\right| e^{-iH\Delta}\left|q_{k}\right\rangle\)
\[
= \left\langle q_{k+1}\right|\, e^{-i\left(\frac{\hat{p}^{2}}{2m} + V(\hat{q})\right)\Delta}\left|q_{k}\right\rangle
\qquad
\left|
\begin{aligned}
&e^{A+B} = e^{A}\,e^{B}\,e^{-\frac{1}{2}\left[B,A\right]}\cdots\\
&\text{for our case } \left[B,A\right] \sim \Delta^{2}
\end{aligned}
\right.
\]
\[
\simeq\; \left\langle q_{k+1}\right|\, e^{-i\frac{\hat{p}^{2}}{2m}\Delta}\; e^{-iV(\hat{q})\Delta}\left|q_{k}\right\rangle
\]
\(\downarrow\) Note that this is approx. equal to.
\[
\begin{gathered}
\text{As}\quad e^{A+B} = e^{A}\, e^{B}\, e^{\frac{1}{2}\left[B,A\right]}\cdots\\
\text{For our case}\quad \left[B,A\right] = \sim\Delta^{2}
\;\Rightarrow\;
e^{A+B} = e^{A}\,e^{B}\cdot e^{\Delta^{2}K} = e^{A}\,e^{B}\left(1 + \Delta^{2}K + \cdots\right)
\end{gathered}
\]
Since we have \(N\)-such products \(\prod_{k=1}^{N}\left\langle q_{k+1}\right|e^{-iH\Delta}\left|q_{k}\right\rangle\) we get,
\[
\begin{gathered}
\lim_{N\to\infty}\left(1 + \Delta^{2}K\right)^{N}
\qquad\text{where}\quad \Delta = C/N\\
\lim_{N\to\infty}\left(1 + \frac{\text{const.}}{N^{2}}\right)^{N} = \;\underline{1}
\end{gathered}
\]
\[
\begin{aligned}
&\simeq\; e^{-iV\left(q_k\right)\Delta}\,\left\langle q_{k+1}\right|\, e^{-i\frac{\hat{p}^{2}}{2m}\Delta}\cdot I\,\left|q_{k}\right\rangle
\qquad\qquad \hookrightarrow I = \int dp_{k}\left|p_{k}\right\rangle\left\langle p_{k}\right|\\
&\simeq\; e^{-iV\left(q_k\right)\Delta}\int dp\;\left\langle q_{k+1}\right| e^{-i\frac{\hat{p}^{2}}{2m}\Delta}\left|p_{k}\right\rangle\left\langle p_{k}\middle| q_{k}\right\rangle\\
&\simeq\; e^{-iV\left(q_k\right)\Delta}\int dp\; e^{-i\frac{p_{k}^{2}}{2m}\Delta}\;\left\langle q_{k+1}\middle| p_{k}\right\rangle\left\langle p_{k}\middle| q_{k}\right\rangle\\
&\simeq\; e^{-iV\left(q_k\right)\Delta}\left(\frac{1}{2\pi}\right)\int dp\; e^{-i\frac{p_{k}^{2}}{2m}\Delta}\cdot e^{i\,p_{k}\left(q_{k+1}-q_{k}\right)}
\end{aligned}
\]
using
\[
\left\langle x \middle| p \right\rangle = \frac{1}{\sqrt{2\pi}}\;e^{+i\,p\cdot x}
\]
\[
\simeq\; e^{-iV\left(q_k\right)\Delta}\left(\frac{1}{2\pi}\right)\int dp_{k}\;
e^{-\frac{i\Delta}{2m}\left(p_{k} - \frac{2m\,p_{k}\left(q_{k+1}-q_{k}\right)}{\Delta} + \frac{m}{\Delta^{2}}\left(q_{k+1}-q_{k}\right)^{2} - \frac{m}{\Delta^{2}}\left(q_{k+1}-q_{k}\right)^{2}\right)}
\]
\[
\simeq\; \frac{e^{-iV\left(q_k\right)\Delta}}{2\pi}\int dp_{k}\;
e^{-\frac{i\Delta}{2m}\left(p_{k} - m\frac{\left(q_{k+1}-q_{k}\right)}{\Delta}\right)^{2}}\;
e^{\frac{i}{2\Delta}\left(q_{k+1}-q_{k}\right)^{2}}
\]
\[
\simeq\; \frac{e^{-iV\left(q_k\right)\Delta \;+\; \frac{i}{2\Delta}\left(q_{k+1}-q_{k}\right)^{2}}}{2\pi}\;\sqrt{\frac{\pi\,2m}{i\Delta}}
\]
So,
\[
\begin{aligned}
K\left(q', t';\ q'', t''\right)
&= \int dq_2\, dq_3\ldots dq_N\;\prod_{k=1}^{N}\left(\frac{1}{2\pi}\right)
e^{-iV\left(q_k\right)\Delta + \frac{i}{2\Delta}\left(q_{k+1}-q_{k}\right)^{2}}\cdot\sqrt{\frac{2\pi m}{i\Delta}}\\
&= \int dq_2\, dq_3\ldots dq_N\left(\frac{1}{\sqrt{2\pi}}\right)^{N}
e^{+i\Delta\left(\sum_{k=1}^{N}\frac{\left(q_{k+1}-q_{k}\right)^{2}}{\Delta} - V\left(q_{k}\right)\right)}
\left(\sqrt{\frac{m}{i\Delta}}\right)^{N}
\end{aligned}
\]
\[
\boxed{\;K\left(q', t';\ q'', t''\right) = \mathcal{N}\int\left[\mathcal{D}q\right]\; e^{+iS}\;}
\qquad\qquad
\mathcal{N} = \left(\sqrt{\frac{m}{2\pi\,i\Delta}}\right)^{N}
\]
Generalisation :
Suppose we want to calculate matrix elements of kind ---
\[
\left\langle q''\right|\, e^{-iHt''}\;\underline{T\left(\hat{q}(t_1)\,\hat{q}(t_2)\,\cdots\,\hat{q}(t_n)\right)}\; e^{iHt'}\left|q'\right\rangle
\]
instead of just \(\left\langle q'', t'' \middle| q', t'\right\rangle\) or \(\left\langle q''\right| e^{-iHt''}\cdot\underline{1}\cdot e^{iHt'}\left|q'\right\rangle\)
where \(t'' > t_n > t_{n-1}\;\cdots\; > t'\)
eventually we will take limits \(t'' \longrightarrow \infty\), \(t' \longrightarrow -\infty\)
Claim is
\[
\left\langle q''\right| e^{-iHt''}\, T\left(\hat{q}(t_1)\,\hat{q}(t_2)\,\cdots\,\hat{q}(t_n)\right) e^{iHt'}\left|q'\right\rangle
\;=\; \int\left[\mathcal{D}q\right]\, e^{iS}\; q(t_1)\,q(t_2)\cdots q(t_n)
\]
Please remind that \(e^{iHt'}\left|q'\right\rangle\) is eigen state of \(\hat{q}(t')\) as it satisfies
\[
\begin{aligned}
\underline{\hat{q}(t')}\;\underline{e^{iHt'}\left|q'\right\rangle} &= \underline{e^{iHt'}\,\hat{q}\,e^{-iHt'}}\;\underline{e^{iHt'}\left|q'\right\rangle}\\
&= q'\;\underline{e^{iHt'}\left|q'\right\rangle}
\end{aligned}
\]
with eigen value \(q'\).
Similarly \(\left\langle q''\right| e^{-iHt''}\) is eigen state of \(\hat{q}(t'')\), with eigen value \(q''\).
Above claim implies that path integral automatically generates matrix elements of time ordered products of operators
Since LHS & RHS of our claim is independent of choice weather
\[
\begin{gathered}
t_1 > t_2 > t_3\;\cdots\; > t_n\\
\text{or}\qquad t_1 < t_2 < t_3 \cdots < t_{n-1} < t_n
\end{gathered}
\]
We choose \(t_n > t_{n-1} > t_{n-2}\;\cdots\; > t_2 > t_1\)
LHS :
\[
\left\langle q''\right|\, e^{-iHt''}\;\hat{q}(t_n)\,\hat{q}(t_{n-1})\,\ldots\,\hat{q}(t_1)\; e^{iHt'}\left|q'\right\rangle
\]
using
\[
\left.
\begin{aligned}
\hat{q}(t_n) &= e^{iHt_n}\,\hat{q}(0)\,e^{-iHt_n}\\
&= e^{+iHt_n}\,\hat{q}\,e^{-iHt_n}
\end{aligned}
\right\}\;\text{Heisenberg picture.}
\]
\(\Rightarrow\) LHS becomes
\[
\begin{aligned}
&\left\langle q''\right| e^{-iHt''}\, e^{iHt_n}\,\hat{q}\,e^{-iHt_n}\; e^{iHt_{n-1}}\,\hat{q}\,e^{-iHt_{n-1}}\cdots\; e^{iHt_1}\,\hat{q}\,e^{-iHt_1}\,e^{iHt'}\left|q'\right\rangle\\
=\;&\left\langle q''\right|\, e^{-iH\left(t''-t_n\right)}\;\hat{q}\;e^{-iH\left(t_n - t_{n-1}\right)}\;\cdots\cdots\;
e^{-iH\left(t_2-t_1\right)}\;\hat{q}\;e^{-iH\left(t_1-t'\right)}\left|q'\right\rangle
\end{aligned}
\]
Note that, time difference \(t_n - t_{n-1}\) may not be same for all '\(n\)'.
So let us descretise total time diff. \(t'' - t' = N\Delta\)
we define
\[
\begin{gathered}
t_1 = t' + k_1\,\Delta\\
t_2 = t' + k_2\,\Delta
\qquad
\left\{\,k_r\ \text{may not be same}\right.\\
\vdots\\
t_r = t' + k_r\,\Delta\\
\text{for some integer } k_r.
\end{gathered}
\]
Now one can argue that \(k_r\) may be some floating point number
The idea is that we devide \(t'' - t'\) in large # of small time steps.
\definecolor{ForestGreen}{rgb}{0.13,0.55,0.13}
\begin{tikzpicture}[scale=1.0]
\node[left] at (0,0) {$t''$};
\draw[thick] (0.2,0) -- (9.2,0);
\foreach \x in {0.35,0.5,0.65,0.8,0.95,1.1,1.25,1.4,1.55,1.7,1.85,2.0,2.15,2.3,2.45,2.6,2.75,2.9,3.05,3.2,3.35,3.5,3.65,3.8,3.95,4.1,4.25,4.4,4.55,4.7,4.85,5.0,5.15,5.3,5.45,5.6,5.75,5.9,6.05,6.2,6.35,6.5,6.65,6.8,6.95,7.1,7.25,7.4,7.55,7.7,7.85,8.0,8.15,8.3,8.45,8.6,8.75,8.9,9.05}
\draw[red] (\x,0) -- (\x,0.12);
\node[right] at (9.25,0) {$t'$};
\foreach \x/\l in {1.0/$t_n$,1.8/$t_{n-1}$,6.4/$t_3$,7.5/$t_2$,8.6/$t_1$}{
\draw[thick,ForestGreen] (\x,-0.15) -- (\x,0.05);
\node[below,ForestGreen] at (\x,-0.15) {\l};
}
\node[above,font=\scriptsize] at (1.8,0.2) {$\delta$};
\node[above,font=\scriptsize] at (6.4,0.2) {$\delta$};
\end{tikzpicture}
We then can shift our \(t_r\) such that it falls on one of red lines (upto an small order \(\delta\) we can shift our times).
\[
\text{Then}\qquad t_r = t_1 + k_r\,\Delta\qquad\text{where}\quad \underline{k_r \in \mathbb{I}}
\]
\[
\begin{gathered}
r = 1, 2, 3\;\cdots\; n\\
k_r = 0, 1, 2, \cdots\; N
\end{gathered}
\]
Since we have choosen \(t_n > t_{n-1} > t_{n-2}\;\cdots\; > t_1\)
\[
\begin{gathered}
\Rightarrow\qquad t_r > t_{r-1} > t_{r-2}\\
\Rightarrow\qquad k_r > k_{r-1} > k_{r-2}
\end{gathered}
\]
\[
\begin{aligned}
\Rightarrow\qquad t_n - t_{n-1} &= \left(t_1 + k_n\,\Delta\right) - \left(t_1 + k_{n-1}\,\Delta\right)\\
&= \left(k_n - k_{n-1}\right)\Delta
\end{aligned}
\]
also \(t_1 = t' + k_1\,\Delta\) \(\Rightarrow\) \(t_1 - t' = k_1\,\Delta\)
LHS becomes :
\[
\begin{aligned}
&\left\langle q''\right|\, e^{-i\hat{H}\left(t''-t_n\right)}\;\hat{q}\;e^{-i\hat{H}\left(t_n-t_{n-1}\right)}\;\hat{q}\;\cdots\cdots\;
e^{-i\hat{H}\left(t_2-t_1\right)}\;\hat{q}\;e^{-i\hat{H}\left(t_1-t'\right)}\left|q'\right\rangle\\
=\;&\left\langle q''\right|\, e^{-iH\left(N-k_n\right)\Delta}\;\hat{q}\;e^{-iH\left(k_n-k_{n-1}\right)\Delta}\;\cdots\cdots\;
e^{-iH\left(k_2-k_1\right)\Delta}\;\hat{q}\;e^{-iH\,k_1\,\Delta}\left|q'\right\rangle
\end{aligned}
\]
\[
\uparrow\quad\text{insert identity operator}\qquad
I = \int dq_n\left|q_n\right\rangle\left\langle q_n\right|
\]
\[
\left(\text{Since } \exists\ `n\text{' such } \hat{q}\text{, we insert `}n\text{' identity}\right)
\]
LHS becomes ---
\[
\begin{aligned}
=\;\left\langle q''\right|\, e^{-i\hat{H}\left(N-k_n\right)\Delta}\;\hat{q}&\left(\int dq_n\left|q_n\right\rangle\left\langle q_n\right|\right)e^{-i\hat{H}\left(k_n-k_{n-1}\right)\Delta}\;\cdots\\
&\cdots\;\hat{q}\int dq_2\left|q_2\right\rangle\left\langle q_2\right| e^{-iH\left(k_2-k_1\right)\Delta}\;\hat{q}\int dq_1\left|q_1\right\rangle\left\langle q_1\right|\,e^{-i\hat{H}\,k_1\Delta}\left|q'\right\rangle
\end{aligned}
\]
\[
\begin{aligned}
\hat{q}\left|q_n\right\rangle &= q(t_n)\left|q_n\right\rangle\\
&= q_n\left|q_n\right\rangle
\end{aligned}
\]
LHS \(\Rightarrow\)
\[
\begin{aligned}
\int dq_1\, dq_2\cdots dq_n\;
&\left\langle q''\right| e^{-i\hat{H}\left(N-k_n\right)\Delta}\left|q_n\right\rangle\left\langle q_n\right| e^{-i\hat{H}\left(k_n-k_{n-1}\right)\Delta}\left|q_{n-1}\right\rangle\\
&\cdots\cdots\;\left\langle q_2\right| e^{-i\hat{H}\left(k_2-k_1\right)\Delta}\left|q_1\right\rangle\left\langle q_1\right| e^{-i\hat{H}\,k_1\Delta}\left|q'\right\rangle\\
&\times\; q(t_1)\,q(t_2)\;\cdots\; q(t_n)
\end{aligned}
\]
\[
= \int dq_1\, dq_2\cdots dq_n\;\; q(t_1)\,q(t_2)\cdots q(t_n)\cdot
\prod_{j=0}^{n}\;\left\langle q_{j+1}\right| e^{-iH\left(k_{j+1}-k_j\right)\Delta}\left|q_j\right\rangle
\]
where,
\[
\begin{aligned}
q_{n+1} &= q''\\
q_{0} &= q'
\end{aligned}
\]
\[
\left\langle q_{j+1}\right| e^{-iH\left(k_{j+1}-k_j\right)\Delta}\left|q_j\right\rangle \;=\;?
\]
or
\[
\left\langle x\right| e^{-iH\,\alpha_j\,\Delta}\left|y\right\rangle \;=\;?
\qquad\qquad
\left\{
\begin{aligned}
\alpha_j &= k_{j+1} - k_j\\
q_j &= y\\
q_{j+1} &= x
\end{aligned}
\right.
\]
\[
\begin{aligned}
&\left\langle x\right|\, e^{-i\left(\frac{\hat{p}^{2}}{2m} + \hat{V}(x)\right)\alpha_j\,\Delta}\left|y\right\rangle\\
\sim\;\;&\left\langle x\right|\, e^{-i\frac{V(x)}{2}\,\alpha_j\Delta}\; e^{-i\frac{\hat{p}^{2}}{2m}\alpha_j\Delta}\; e^{-i\,V(x)\,\frac{\alpha_j\Delta}{2}}\left|y\right\rangle\\
\sim\;\;& e^{-i\left(\frac{V(x)+V(y)}{2}\right)\alpha_j\Delta}\;\left\langle x\right| e^{-i\frac{\hat{p}^{2}\,\alpha_j\Delta}{2m}}\; I\;\left|y\right\rangle
\qquad\qquad \hookrightarrow I = \int dP\left|p\right\rangle\left\langle p\right|\\
\sim\;\;& e^{-i\left(\frac{V(x)+V(y)}{2}\right)\alpha\Delta}\;\int dP\;\left\langle x\middle|p\right\rangle\left\langle p\middle|y\right\rangle\; e^{-\frac{i\,p^{2}\,\alpha\Delta}{2m}}\\
\sim\;\;& e^{-\frac{i}{2}\left(V(x)+V(y)\right)\alpha_j\Delta}\;\int dp\;\frac{1}{\sqrt{2\pi}}\,e^{i\,p\cdot x}\;\frac{1}{\sqrt{2\pi}}\,e^{-i\,p\,y}\;e^{-i\,p^{2}\alpha_j\Delta/2m}\\
\sim\;\;&\frac{1}{2\pi}\; e^{-\frac{i\,\alpha_j\Delta}{2}\left(V(x)+V(y)\right)}\;\int dp\;\; e^{i\left(p\cdot\left(x-y\right) - \frac{p^{2}\alpha_j\Delta}{2m}\right)}\\
\sim\;\;&\frac{1}{2\pi}\; e^{-\frac{i\,\alpha\Delta}{2}\left(V(x)+V(y)\right)}\;\int dp\;\;
e^{-\frac{i\alpha\Delta}{2m}\left(p^{2} - \frac{2m}{\alpha\Delta}\,p\cdot\left(x-y\right) + \frac{m^{2}\left(x-y\right)^{2}}{\alpha^{2}\Delta^{2}} - \frac{m^{2}\left(x-y\right)^{2}}{\alpha^{2}\Delta^{2}}\right)}\\
\sim\;\;&\frac{1}{2\pi}\; e^{-i\,\frac{\alpha\Delta}{2}\left(V(x)+V(y)\right)}\;\int dp\;\;
e^{-i\,\frac{\alpha\Delta}{2m}\left(p - \frac{\left(x-y\right)m}{\alpha\Delta}\right)^{2}}\; e^{\frac{i\,m\left(x-y\right)^{2}}{2\,\alpha\Delta}}\\
\sim\;\;&\frac{1}{2\pi}\; e^{-i\,\frac{\alpha\Delta}{2}\left(V(x)+V(y)\right) + \frac{i\,m}{2\alpha\Delta}\left(x-y\right)^{2}}\;\int dp\;\;
e^{-\frac{i\,\alpha\Delta}{2m}\left(p - \frac{\left(x-y\right)m}{\alpha\Delta}\right)^{2}}\\
\sim\;\;&\frac{1}{2\pi}\;\; e^{-i\,\frac{\alpha\Delta}{2}\left(V(x)+V(y)\right) + \frac{i\,m}{2\alpha\Delta}\left(x-y\right)^{2}}\;\;
\sqrt{\frac{\pi\,2m}{\alpha_j\,\Delta}}
\end{aligned}
\]
\(\Rightarrow\) LHS becomes,
\[
\int\mathcal{D}q\;\; q(t_1)\,q(t_2)\cdots q(t_n)\;\;
\prod_{j=0}^{n}\;\frac{1}{2\pi}\;
e^{-i\,\frac{\alpha\Delta}{2}\left(V\left(q_{j+1}\right)+V\left(q_j\right)\right) + \frac{i\,m}{2\alpha\Delta}\left(q_{j+1}-q_j\right)^{2}}\;
\sqrt{\frac{\pi\,2m}{\alpha\Delta}}
\qquad\left(\alpha = \alpha_j\right)
\]
\[
= \mathcal{N}\int\mathcal{D}q\;\; q(t_1)\,q(t_2)\cdots q(t_n)\;\;
e^{\,i\alpha\Delta\,\sum_{j=0}^{n}\,\frac{m}{2}\left(\frac{q_{j+1}-q_j}{\alpha\Delta}\right)^{2} - \left(\frac{V\left(q_{j+1}\right)+V\left(q_j\right)}{2}\right)}
\]
\[
\boxed{\displaystyle
\left\langle q'', t''\right| T\left(\prod_{i}^{n}\, q\left(t_i\right)\right)\left|q', t'\right\rangle
= \mathcal{N}\int\mathcal{D}q\;\; q(t_1)\,q(t_2)\;\cdots\; q(t_n)\;\; e^{\,iS}}
\qquad
\left\{
\begin{aligned}
\alpha\Delta &= \left(k_{j+1}-k_j\right)\Delta\\
&= t_{j+1} - t_j
\end{aligned}
\right.
\]
Let us do the integral more accurately.
\[
\int dp\;\; e^{-\frac{i\Delta\,p^{2}}{2m}}
\]
to make a sense of it we need to add positive real part to '\(i\)'.
\[
\int dp\;\; e^{-\left(i+\epsilon\right)\frac{\Delta\,p^{2}}{2m}}
\qquad\qquad
\left|
\begin{aligned}
&\text{For } i = 0\quad I = \text{Area} = \int dp\; e^{-\epsilon\,\Delta p^{2}/2m}\\
&\text{Shall give +ve answer}\\
&\text{[diagram below]}
\end{aligned}
\right.
\]
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\end{tikzpicture}
\[
\text{(or)}\qquad
I = \int dp\;\; e^{-i\left(1-i\epsilon\right)\frac{\Delta\,p^{2}}{2m}}
\]
Now apply gaussian formule
\[
\begin{aligned}
I &= \sqrt{\frac{\pi\,2m}{\left(1-i\epsilon\right)\,i\,\Delta}}\\
&= \sqrt{\frac{2m\,\pi}{\left(i+\epsilon\right)\Delta}}
\qquad\longleftarrow\;\underline{\text{Verify!}}
\end{aligned}
\]
Time ordering & anti-time ordering is giving same answer in path integral reps, which it can't be --- so we have to account for
Let us see how will we take \(\sqrt{i+\epsilon}\) in this \(\mathbb{C}\) plane. \(\hookrightarrow\) has two solutions
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If '\(i\)' were absent, we should have positive answer. Branch cut has to go in some other direction. Since we are defining analytic continuation from \(\epsilon \to i+\epsilon\) i.e. there should not be branch cut in way of \(\epsilon \to i+\epsilon\). We can take branch cut like (1) or (2) does not matter.
\[
\text{then}\qquad \sqrt{1+\epsilon}? = \lim_{\epsilon\to0}\left(i+\epsilon\right)^{1/2} = e^{\left(i\pi/2\right)/2} = e^{i\pi/4}
\]
This is same as,
\[
\begin{gathered}
\Delta \;\longrightarrow\; \Delta\left(1-i\epsilon\right)\\
\frac{1}{\Delta} \;\longrightarrow\; \frac{1}{\Delta}\left(1+i\epsilon\right)
\end{gathered}
\]
\[
\Delta = \frac{t''-t'}{N}
\]
\[
\begin{gathered}
\text{for}\qquad \Delta \;\longrightarrow\; \Delta\left(1-i\epsilon\right)\\
t \;\longrightarrow\; t\left(1-i\epsilon\right)
\qquad\hookrightarrow\;\text{complex time.}
\end{gathered}
\]
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\node[ForestGreen,above] at (-2.9,1.0) {$t_1$};
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let us write \(t = \tau\left(1-i\epsilon\right)\) where \(\tau \in \mathbb{R}\), \(t \in \mathbb{C}\)
We have final path integral (for \(\left\langle q(t'')\middle|q(t')\right\rangle\))
\[
\mathcal{N}\int dq_2\, dq_3\cdots dq_n\;
\exp\left(i\Delta\,\sum_{j=0}^{N}\left\{\frac{m}{2}\underbrace{\left(\frac{q_{j+1}-q_j}{\Delta}\right)^{2}}_{(1)} - \underbrace{V\left(q_j\right)}_{(2)}\right\}\right)
\]
here we have \(\Delta\cdot\dfrac{1}{\Delta^{2}} \sim \dfrac{1}{\Delta}\)
\[
\frac{1}{\Delta} \;\longrightarrow\; \frac{1}{\Delta}\left(1+i\epsilon\right)
\]
So extra term we get :
\[
i\left(\frac{i\epsilon}{\Delta}\right)\sum_{j=1}^{N}\frac{m}{2}\left(q_{j+1}-q_j\right)^{2} - i\sum_{j=0}^{N}\left(-i\epsilon\right)\Delta\,V\left(q_j\right)
\]
For large \(q\) :
\[
\begin{gathered}
\int dq_2\,dq_3\ldots dq_N\;\exp\left(-\frac{\epsilon}{\Delta}\,\sum_{j=0}^{N}\left(\frac{m}{2}\left(q_{j+1}-q_j\right)^{2}\right) + i^{2}\,\epsilon\sum_{j=0}^{N} V\left(q_j\right)\right)\\
\text{for}\quad q \to \infty\;;\quad \exp\left(-\epsilon\infty\right) \to 0
\end{gathered}
\]
This extra term has effect that integral supresses for very large \(q\). So This makes integral well defined; if it were purely oscilletory \(e^{iS}\) we remain in doubt if it is well defined or not.
Now we have a damped integral.
let us write \(t = \tau\left(1-i\epsilon\right)\) ; \(\tau \in \mathbb{R}\)
So,
\[
\left\langle q''\right|\, e^{-i\hat{H}\tau''\left(1-i\epsilon\right)}\;
\underbrace{T\Big(\hat{q}\left(\tau_1(1-i\epsilon)\right)\,\hat{q}\left(\tau_2(1-i\epsilon)\right)\cdots\hat{q}\left(\tau_n(1-i\epsilon)\right)\Big)}\;
e^{i\hat{H}\tau'\left(1-i\epsilon\right)}\left|q'\right\rangle
\]
\[
\downarrow\; I = \sum_{s}\left|s\right\rangle\left\langle s\right|
\qquad\qquad
\hookrightarrow\; I = \sum_{r}\left|r\right\rangle\left\langle r\right|
\qquad\qquad
\left\{\left|r\right\rangle\right\}\ \text{are energy eigen states}
\]
let us insert complete set of energy eigen states ---
\[
I = \sum_{s}\left|s\right\rangle\left\langle s\right|
\]
Now term
\[
\begin{aligned}
1)\qquad \left\langle q''\right|\sum_{s}\left|s\right\rangle\left\langle s\right| e^{-i\hat{H}\tau''\left(1-i\epsilon\right)}
&= \sum_{s}\left\langle q''\middle|s\right\rangle\, e^{-i\,E_s\,\tau''\left(1-i\epsilon\right)}\left\langle s\right|\\
&= \sum_{s}\left\langle q''\middle|s\right\rangle\; e^{-i\,E_s\,\tau'' - \epsilon\,E_s\,\tau''}\left\langle s\right|
\end{aligned}
\]
\[
\begin{aligned}
2)\qquad e^{i\hat{H}\tau'\left(1-i\epsilon\right)}\sum_{r}\left|r\right\rangle\left\langle r\middle|q'\right\rangle
&= \sum_{r}\left\langle r\middle|q'\right\rangle\, e^{i\,E_r\,\tau'\left(1-i\epsilon\right)}\left|r\right\rangle\\
&= \sum_{r}\left\langle r\middle|q'\right\rangle\; e^{+i\,E_r\,\tau' + \epsilon\,E_r\,\tau'}\left|r\right\rangle
\end{aligned}
\]
Now; take limits : \(\tau'' \to \infty\), \(\tau' \to -\infty\)
under these limits term (1) & (2) only survives for \(E_r = E_s = E_{\Omega}\) i.e ground state.
\[
\left\langle q''\middle|\Omega\right\rangle\left\langle \Omega\middle|q'\right\rangle\;
e^{-i\,E_{\Omega}\,\tau'' - \epsilon\,E_{\Omega}\,\tau''}\;
e^{i\,E_{\Omega}\,\tau' + \epsilon\,E_{\Omega}\,\tau'}
\qquad
\nwarrow\;\text{ground state (vacuum)}
\]
Then
\[
\left\langle \Omega\right| T\left(\hat{q}\left(\tau_1(1-i\epsilon)\right)\,\hat{q}\left(\tau_2(1-i\epsilon)\right)\cdots\hat{q}\left(\tau_n(1-i\epsilon)\right)\right)\left|\Omega\right\rangle
\qquad
\begin{aligned}
&\downarrow\;\left\langle q\left(\tau''\to\infty\right)\right|\\
&\hookrightarrow\;\left|\hat{q}\left(\tau'\to-\infty\right)\right\rangle
\end{aligned}
\]
Normalising :
\[
\frac{\left\langle \Omega\right| T\left(\prod_{i=1}^{n}\,\hat{q}\left(\tau_i\left(1-i\epsilon\right)\right)\right)\left|\Omega\right\rangle}{\left\langle \Omega\middle|\Omega\right\rangle}
\;=\;
\frac{\displaystyle\int\left[\mathcal{D}q\right]\, e^{iS}\; q\left(\tau_1(1-i\epsilon)\right)\,q\left(\tau_2(1-i\epsilon)\right)\cdots q\left(\tau_n(1-i\epsilon)\right)}{\displaystyle\int\left[\mathcal{D}q\right]\, e^{iS}}
\]
\[
\left\{\text{in end we take } \epsilon \to 0 \text{ limit.}\right\}
\]
Since we are rotating time axis \(t \to \tau\left(1-i\epsilon\right)\) we can rotate it all the way to imaginary axis.
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i.e.
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achieved by \(t \;\longrightarrow\; e^{-i\pi/2}\,\tau\)
Problem is argument of \(q\left(\tau\,e^{-i\pi/2}\right)\) are imaginary; we will do calculation using these arguments and will rotate the final result back to where arguments are real.
i.e. we will have to rotate \(e^{+i\pi/2}\) after our final result. \(\hookrightarrow\) anticlockwise
If we find result for anti-time ordering; then this will be different.
Singularities appear when we do analytic continuation back to real axis. There are also studies on kinds of singularity which appear on analytic continuation.
Anti-time ordering gives us euclidean metric path integral by rotating \(t \to e^{i\pi/2}\,\tau\); but we won't use it; since we are dealing with perturbative QFT. Euclidean path integral provides results for non-perturbative phenomena.
The funda is same; after we found results using euclidean path integral we will rotate back to real axis. \(\left(t \to e^{-i\pi/2}\,\tau\right)\) where \(\left(\tau' = e^{i\pi/2}\,\tau\right)\).
If vacuum were degenerate; this would have been more complicated. Fortunately in most of QFT vacuum is non-degenerate.
In QM if there is degenerate vacuum; we have to be more careful.
Scalar field theory :
\[
S = \int dt\; d^{3}x\;\left(-\frac{1}{2}\,\eta^{\mu\nu}\,\partial_{\mu}\phi\,\partial_{\nu}\phi \;-\; V\left(\phi\right)\right)
\]
here, \(\eta^{\mu\nu} = \mathrm{diag}\left(-1, 1, 1, 1\right)\)
Think of space as discrete collection of points.
\[
\vec{x} \;\longrightarrow\; a\,\vec{n}
\qquad\qquad
\vec{n} = \left(n_1, n_2, n_3\right)
\]
\[
\downarrow\;\text{lattice size}
\]
\[
\text{Space}\ \text{---}\qquad
\begin{matrix}
\cdot & \cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot & \cdot
\end{matrix}
\]
using discretised space breaks lorentz invariance. we will use it in limit \(a \to 0\).
\[
\begin{aligned}
\text{Action } (S) &= \int dt\, d^{3}x\;\left(\frac{\dot{\phi}^{2}}{2} - \frac{\left(\nabla\phi\right)^{2}}{2} - V(\phi)\right)
\qquad\qquad \phi\left(x, t\right) \to \phi_{n}\left(t\right)\\
&= \int dt\;\; a^{3}\sum_{n=\left(n_1,n_2,n_3\right)}\Bigg[\left(\frac{1}{2}\,\partial_t\,\phi_{n}(t)\right)^{2}\\
&\qquad\qquad -\frac{1}{2}\left(\frac{\phi_{n_1+1,\,n_2,\,n_3}(t) - \phi_{n_1 n_2 n_3}(t)}{a}\right)^{2}\\
&\qquad\qquad -\frac{1}{2}\left(\frac{\phi_{n_1,\,n_2+1,\,n_3}(t) - \phi_{n_1 n_2 n_3}(t)}{a}\right)^{2}\\
&\qquad\qquad -\frac{1}{2}\left(\frac{\phi_{n_1,\,n_2,\,n_3+1}(t) - \phi_{n_1 n_2 n_3}(t)}{a}\right)^{2}\\
&\qquad\qquad -\; V(\phi)\;\Bigg]
\end{aligned}
\]
Exercise : check the path integral formulation extends to multiple variables.
\[
H = \sum_{i=1}^{M}\;\frac{p_i^{2}}{2m_i} + V\left(q_1, q_2 \ldots q_M\right)
\]
Discretize time with interval \(\Delta\).
\[
\text{i.e.}\qquad \int dt \;\longrightarrow\; \Delta\sum_{n_0}
\qquad\qquad
n = \left(n_0, n_1, n_2, n_3\right)
\]
here, \(n_0, n_1, n_2, n_3 \in\) Integers
\[
\begin{gathered}
\partial_t\,\phi \;\longrightarrow\; \frac{\phi_{n_0+1,\,n_1,n_2,n_3} - \phi_{n_0,\,n_1,n_2,n_3}}{\Delta}\\
\phi_{n_1 n_2 n_3}(t) \;\longrightarrow\; \phi_{n} = \phi_{n_0, n_1, n_2, n_3}
\end{gathered}
\]
The \(\Delta \to \Delta\left(1-i\epsilon\right)\) prescription is same as working in Euclidean prescription.
if \(\Delta\) is real time
\[
\Delta \;\longrightarrow\; \tau = i\,\Delta
\qquad
\left\{
\text{[diagram below]}
\right\}
\qquad\text{here } \tau \text{ is imaginary time.}
\]
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\node[right,font=\scriptsize] at (0.75,0.75) {$\tau$};
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\end{tikzpicture}
If we were in Euclidean formulation \(\left(-i\tau = \Delta\;\Rightarrow\;\tau = i\Delta\right)\)
\[
\begin{gathered}
\partial_t\,\phi \;\longrightarrow\; \partial_{\tau}\,\phi\\
\frac{\phi_{n_0+1,\,n_1 n_2 n_3} - \phi_{n_0 n_1 n_2 n_3}}{\Delta}
\;\longrightarrow\;
\frac{\phi_{n_0+1,\,n_1 n_2 n_3} - \phi_{n_0\, n_1 n_2 n_3}}{-i\,\tau}
\end{gathered}
\]
So,
\[
\left(\partial_{\tau}\phi\right)^{2} \;=\; -\left(\partial_t\,\phi\right)^{2}
\]
For \(\Delta = a\) ; space & time are symmetrical
Euclidean formulation makes it more menifest that there is some symmetry in \(x \leftrightarrow t\).
Once we take out continuum limit, we get a continuous rotation group, and then we do analytic continuation, to recover lorentz symmetry.
\[
S\left[\phi\right] = \text{Functional of } \phi(x) = \int d^{4}x\;\left(\frac{1}{2}\left(\partial_t\phi\right)^{2} - \frac{1}{2}\left(\nabla\phi\right)^{2} - V\left(\phi\left(x,t\right)\right)\right)
\]
\[
x = \left(\bar{x}, t\right)
\]
So,
\(S[\phi]\) : Functional of \(\phi\). in continuum formalism but it is a function of \(\left\{\phi_{i_1 i_2 i_3 i_4}\right\}\) in discrete form.
\[
\left\{\phi_{i}\right\} = \left\{\phi_{0000}\,,\ \phi_{0001}\,,\ \phi_{0011}\,,\ldots\ \phi_{(N-1)(N-1)(N-1)(N-1)}\right\}
\]
\[
\hookrightarrow\;\text{variables.}
\]
So we can define usual partial derivative --- \(\dfrac{\partial S}{\partial\phi_{i_1 i_2 i_3 i_4}}\)
(Def\(^{\text{n}}\)) Functional derivative \(\xrightarrow{\text{\;\;discretisation\;\;}}\)
\[
\frac{1}{a^{3}\Delta}\;\frac{\partial S}{\partial\phi_{i_0 i_1 i_2 i_3}} \;\equiv\; \frac{\delta S}{\delta\phi(x)}
\]
Claim :
For fixed \(y\), \(\phi(y)\) is a functional of \(\phi(x)\).
Since \(\phi(y)\) gives a number for a given \(\phi(x)\) at point '\(y\)'.
So, we can calculate,
\[
\text{functional derivative}\qquad
\frac{\delta\,\phi(y)}{\delta\,\phi(x)} = \delta^{4}\left(x-y\right)
\]
Proof :
Suppose : \(y = \left(\Delta\,n_0,\ a\,n_1,\ a\,n_2,\ a\,n_3\right)\)
\[
\begin{gathered}
\phi(y) = \phi_{n_0 n_1 n_2 n_3}\\
\frac{\delta\phi(y)}{\delta\phi(x)} = \frac{1}{a^{3}\Delta}\;\frac{\partial\,\phi_{n_0 n_1 n_2 n_3}}{\partial\,\phi_{m_0 m_1 m_2 m_3}}\\
= \frac{1}{a^{3}\Delta}\;\;\delta_{m_0 n_0}\,\delta_{m_1 n_1}\,\delta_{m_2 n_2}\,\delta_{m_3 n_3}
\end{gathered}
\]
Lets prove RHS to be equal to \(\delta^{4}\left(x-y\right)\).
\[
\int d^{4}x\;\delta^{4}\left(x-y\right)\,F(x) = F(y)
\qquad\text{for any } F.
\]
\(\downarrow\)
\[
\begin{gathered}
a^{3}\Delta\!\!\sum_{m_0 m_1 m_2 m_3}\;\frac{1}{a^{3}\Delta}\;\delta_{m_0 n_0}\,\delta_{m_1 n_1}\,\delta_{m_2 n_2}\,\delta_{m_3 n_3}\;F_{m_0, m_1 m_2 m_3}\\
= F_{n_0 n_1 n_2 n_3} \;=\; F(y)
\end{gathered}
\]
This has to hold for any \(F\).
In scalar Field Theory :
\[
Z\left[J\right] = \int\mathcal{D}\phi\;\; e^{\,i\,S\left(\phi\right) + i\int d^{4}x\; J(x)\,\phi(x)}
\]
\[
\left(\text{Functional of } J \text{ \& Not } \phi\right)\ !\ (?)
\]
\[
\left(-i\,\frac{\partial}{\partial J(x_1)}\right)\left(-i\,\frac{\partial}{\partial J(x_2)}\right)\cdots\left(-i\,\frac{\partial}{\partial J(x_n)}\right) Z\left[J\right]
= Z\left[J\right]\;\phi(x_1)\,\phi(x_2)\cdots\phi(x_n)
\]
\[
\left(\text{using}\quad \frac{d}{dx}\,e^{a(x)} = e^{a(x)}\,\frac{da}{dx}\right)
\]
So,
\[
\begin{aligned}
-i\,\frac{\partial}{\partial J(x_n)}\int\mathcal{D}\phi\; e^{\,i\,S(\phi) + i\int J(x)\phi(x)\,d^{4}x}
&= \left(-i\right)\left(i\right)\, Z\cdot\frac{\partial}{\partial J(x_n)}\int J(x)\,\phi(x)\,d^{4}x\\
&= Z\left[J\right]\cdot\int\frac{\partial}{\partial J(x_n)}\,J(x)\,\phi(x)\,d^{4}x\\
&= Z\left[J\right]\,\int\delta^{4}\left(x - x_n\right)\,\phi(x)\,d^{4}x\\
&= Z\left[J\right]\;\phi\left(x_n\right)
\end{aligned}
\]
\[
\begin{gathered}
-i\,\frac{\partial}{\partial J(x_{n-1})}\; Z\left[J\right]\,\phi(x_n)
= Z\left[J\right]\;\phi\left(x_{n-1}\right)\,\phi\left(x_n\right)\\
-i\,\frac{\partial}{\partial J(x_2)}\; Z\left[J\right]\,\phi(x_{n-1})\,\phi(x_n)
= Z\left[J\right]\;\phi\left(x_2\right)\,\phi\left(x_{n-1}\right)\,\phi\left(x_n\right)
\end{gathered}
\]
So,
\[
\boxed{\displaystyle
\left(\prod_{i=1}^{n}\left(\frac{-i\,\partial}{\partial J\left(x_i\right)}\right)\right) Z\left[J\right]
\;=\; Z\left[J\right]\;\prod_{i=1}^{n}\;\phi\left(x_i\right)}
\qquad\longleftarrow\;\text{classical fields/variables}
\]
Similarly,
\[
G_{n}\left(x_1, x_2 \ldots x_n\right) = \frac{\left\langle \Omega\right| T\Big(\overbrace{\hat{\phi}(x_1)\,\hat{\phi}(x_2)\,\ldots\,\hat{\phi}(x_n)}^{\text{operators.}}\Big)\left|\Omega\right\rangle}{\left\langle \Omega\middle|\Omega\right\rangle}
\]
using;
\[
\left\langle q''\right| e^{-iHt''}\;\underline{T\left(\hat{q}(t_1)\,\hat{q}(t_2)\cdots\hat{q}(t_n)\right)}\; e^{iHt'}\left|q'\right\rangle
= \int\left[\mathcal{D}q\right]\, e^{iS}\; q(t_1)\,q(t_2)\cdots q(t_n)
\]
\[
= \frac{1}{Z\left[0\right]}\;
\left(-i\,\frac{\partial}{\partial J(x_1)}\right)\left(-i\,\frac{\partial}{\partial J(x_2)}\right)\cdots\left(-i\,\frac{\partial}{\partial x_n}\right)
Z\left[J\right]\;\Bigg|_{J=0.}
\]
\[
\uparrow\quad\left(\text{classical variables}\right)
\]
Calculate \(Z[J]\) for free theory :
\[
Z\left[J\right] = \int\mathcal{D}\phi\;\; e^{\,i\,S\left(\phi\right) + i\int J(x)\,\phi(x)\,d^{4}x}
\qquad\text{---}(1)
\]
\[
\eta_{\mu\nu} = \left(-1, 1, 1, 1\right)
\]
\[
S\left[\phi\right]_{\text{free}} = \int d^{4}x\;\left(-\frac{1}{2}\,\partial_{\mu}\phi\,\partial^{\mu}\phi \;-\; \frac{1}{2}\,m^{2}\phi^{2}\right)
\]
Lets find \(\left(i\,S\left[\phi\right] + i\int J(x)\,\phi(x)\,d^{4}x\right)\) first then put it in eq (1).
Claim is: \(i\,S\left[\phi\right] + i\int J(x)\,\phi(x)\,d^{4}x\) is equal to
\[
\frac{i}{2}\int d^{4}x\left(-\eta^{\mu\nu}\,\partial_{\mu}\chi\,\partial_{\nu}\chi - m^{2}\chi^{2}\right)
\;-\;\frac{1}{2}\int d^{4}x\, d^{4}x'\;\Delta\left(x-x'\right)\, J(x)\,J(x')
\]
Where \(\Delta\left(x-x'\right)\) satisfies
\[
\left(\partial_x^{2} - m^{2}\right)\Delta\left(x-x'\right) = i\,\delta^{4}\left(x-x'\right)
\]
&
\[
\chi(x) = \phi(x) - i\int d^{4}x'\;\Delta\left(x, x'\right)\,J(x')
\]
RHS :
\[
\begin{aligned}
&\frac{i}{2}\int d^{4}x\left(-\partial_{\mu}\chi\,\partial^{\mu}\chi - m^{2}\chi^{2}\right)
\qquad
\left\{
\begin{aligned}
&\text{using } \int\partial_{x}\left(\chi\,\partial^{\mu}\chi\right) = \underbrace{\left(\chi\,\partial^{\mu}\chi\right)}_{\text{surface term.}}\\
&= 0 = \int\partial_{\mu}\chi\,\partial^{\mu}\chi + \int\chi\,\partial_{x}^{2}\chi\\
&\qquad\chi\big(\text{Boundary}\big) \to 0
\end{aligned}
\right.\\
=\;&\frac{i}{2}\int d^{4}x\;\left(\chi\,\partial^{2}\chi - m^{2}\chi^{2}\right)\\
=\;&\frac{i}{2}\int d^{4}x\;\;\chi\,\underline{\left(\partial_x^{2} - m^{2}\right)}\,\chi\\
=\;&\frac{i}{2}\int d^{4}x\;\;\chi(x)\left(\partial_x^{2} - m^{2}\right)\left(\phi(x) - i\int d^{4}x'\;\Delta\left(x-x'\right)J(x')\right)\\
=\;&\frac{i}{2}\int d^{4}x\;\;\chi(x)\left(\partial_x^{2}-m^{2}\right)\phi(x)
\;+\;\frac{1}{2}\int d^{4}x\,d^{4}x'\;\chi(x)\;\underline{\left(\partial_x^{2}-m^{2}\right)\Delta\left(x-x'\right)}\,J(x')\\
=\;&\frac{1}{2}\int d^{4}x\,d^{4}x'\;\chi(x)\;\delta^{4}\left(x-x'\right)\,J(x')
\;+\;\frac{i}{2}\int d^{4}x\;\chi(x)\left(\partial_x^{2}-m^{2}\right)\phi(x)\\
=\;&\frac{i}{2}\int d^{4}x\;\underset{(1)}{\chi(x)}\,\underset{(2)}{\left(\partial_x^{2}-m^{2}\right)\phi(x)}
\;+\;\frac{1}{2}\int d^{4}x\;\;\chi(x)\,J(x)
\end{aligned}
\]
using integral by parts \(\hookleftarrow\) we get
\[
\begin{aligned}
=\;&\frac{i}{2}\int d^{4}x\;\phi\left(\partial_x^{2}-m^{2}\right)\chi(x)
\;+\;\frac{1}{2}\int d^{4}x\;\left(\phi(x) - i\int d^{4}x'\;\Delta\left(x-x'\right)J(x')\right) J(x)\\
=\;&\frac{i}{2}\int d^{4}x\;\phi(x)\left(\partial_x^{2}-m^{2}\right)\left\{\phi - i\int d^{4}x'\;\Delta\left(x-x'\right)J(x')\right\}\\
&\qquad\qquad +\;\frac{1}{2}\int d^{4}x\;\underset{J(x)}{\phi(x)}
\;+\;\frac{1}{2}\int d^{4}x\,d^{4}x'\;\Delta\left(x-x'\right)J(x')\,J(x)\\
=\;&\frac{i}{2}\int d^{4}x\;\phi\left(\partial_x^{2}-m^{2}\right)\phi
\;+\;\frac{1}{2}\int d^{4}x\,d^{4}x'\;\phi(x)\;\underbrace{\left(\partial_x^{2}-m^{2}\right)\Delta\left(x-x'\right)}_{\nearrow\;i\,\delta^{4}\left(x-x'\right)}\;J(x')\\
&\qquad\qquad +\;\frac{i}{2}\int d^{4}x\;\underset{J(x)}{\phi(x)}
\;+\;\frac{1}{2}\int d^{4}x\,d^{4}x'\;\Delta\left(x-x'\right)J(x')\,J(x)\\
=\;&\frac{i}{2}\int d^{4}x\;\phi\left(\partial_x^{2}-m^{2}\right)\phi
\;+\;\frac{i}{2}\int d^{4}x\;\phi(x)\,J(x)
\;+\;\frac{i}{2}\int d^{4}x\;\phi(x)\,J(x)\\
&\qquad\qquad +\;\frac{1}{2}\int d^{4}x\,d^{4}x'\;\Delta\left(x-x'\right)\,J(x')\,J(x)
\end{aligned}
\]
Now,
\[
\begin{aligned}
&\frac{1}{2}\int d^{4}x\left(-\partial_{\mu}\phi\,\partial^{\mu}\phi - m^{2}\phi^{2}\right)
\;-\;\frac{1}{2}\int d^{4}x\,d^{4}x'\;\Delta\left(x-x'\right)J(x)\,J(x')\\
=\;&\frac{i}{2}\int d^{4}x\;\phi\left(\partial_x^{2}-m^{2}\right)\phi + i\int d^{4}x\;\phi(x)\,J(x)\\
&\qquad + \cancel{\frac{1}{2}\int d^{4}x\,d^{4}x'\;\Delta\left(x-x'\right)J(x')\,J(x)}
\;\cancel{- \frac{1}{2}\int d^{4}x\,d^{4}x'\;\Delta\left(x-x'\right)J(x')\,J(x)}\\
=\;&\frac{i}{2}\int d^{4}x\;\phi\left(\partial_x^{2}-m^{2}\right)\phi \;+\; i\int d^{4}x\;\phi(x)\,J(x)\\
=\;&i\,S\left[\phi\right] + i\int d^{4}x\;\phi(x)\,J(x)\;=\;\underline{\text{LHS}}.\qquad\text{proved!}
\end{aligned}
\]
\[
\Rightarrow\quad i\int d^{4}x\;\phi(x)\,J(x) = -\frac{1}{2}\int d^{4}x\,d^{4}x'\;\Delta\left(x-x'\right)J(x)\,J(x')
\]
Idea is now to replace \(\mathcal{D}\phi \;\longrightarrow\; \mathcal{D}\chi\)
as \(\phi\) is just a constant shift to \(\chi\).
\[
\begin{gathered}
\chi = \phi - i\int d^{4}x'\;\Delta\left(x-x'\right)J(x')\\
d\chi = d\phi\\
\Rightarrow\qquad \mathcal{D}\phi = \mathcal{D}\chi
\end{gathered}
\]
So,
\[
\begin{aligned}
Z\left[J\right] &= \int\mathcal{D}\chi\;\; e^{\,i\,S\left(\chi\right) + i\int J(x)\,\chi(x)\,d^{4}x}\\
&= \int\mathcal{D}\chi\;\;
e^{\,\frac{i}{2}\int d^{4}x\;\chi\left(\partial_x^{2}-m^{2}\right)\chi}\;\cdot\;
e^{-\frac{1}{2}\int d^{4}x\,d^{4}x'\;\Delta\left(x-x'\right)\,J(x)\,J(x')}
\end{aligned}
\]
\[
Z_{f}\left[J\right] = Z_{f}\left[0\right]\;\;
e^{-\frac{1}{2}\int d^{4}x\,d^{4}x'\;\Delta\left(x-x'\right)\,J(x)\,J(x')}
\]
\[
\Rightarrow\quad
\boxed{\displaystyle
\frac{Z_{f}\left[J\right]}{Z_{f}\left[0\right]} = \exp\left(-\frac{1}{2}\int d^{4}x\,d^{4}x'\;\Delta\left(x-x'\right)\,J(x)\,J(x')\right)}
\]
With this we can find free greens function ---
\[
G_{f}\left(x_1, x_2 \ldots x_n\right) =
\left(\frac{-i\,\partial}{\partial J(x_1)}\right)\left(\frac{-i\,\partial}{\partial J(x_2)}\right)\cdots\left(\frac{-i\,\partial}{\partial J(x_n)}\right)
\frac{Z_{f}\left[J\right]}{Z_{f}\left[0\right]}\;\Bigg|_{J=0.}
\]
example,
\[
\begin{aligned}
G_{f}\left(x_1 x_2\right) &=
\left(-\frac{i\,\partial}{\partial J(x_1)}\right)\left(\frac{-i\,\partial}{\partial J(x_2)}\right)
\frac{Z_{f}\left[J\right]}{Z_{f}\left[0\right]}\;\Bigg|_{J=0}\\
&= \left(\frac{-i\,\partial}{\partial J(x_1)}\right)\left(\frac{-i\,\partial}{\partial J(x_2)}\right)\cdot
e^{-\frac{1}{2}\int d^{4}x\,d^{4}x'\;\Delta\left(x-x'\right)J(x)\,J(x')}\\
&= \left(-\frac{i\,\partial}{\partial J(x_1)}\right)
e^{-\frac{1}{2}\int d^{4}x\,d^{4}x'\,J(x)\,\Delta\left(x-x'\right)J(x')}\\
&\qquad\cdot\left(\frac{i}{2}\int d^{4}x\,d^{4}x'\,J(x)\;\Delta\left(x-x'\right)\;\frac{\partial J(x')}{\partial J(x_2)}
\;+\;\frac{i}{2}\int d^{4}x\,d^{4}x'\;\frac{\partial J(x)}{\partial J(x_2)}\;\Delta\left(x-x'\right)J(x')\right)\\
&= \left(-\frac{i\,\partial}{\partial J(x_1)}\right)
e^{-\frac{1}{2}\int d^{4}x\,d^{4}x'\;\Delta\left(x-x'\right)J(x)\,J(x')}
\cdot\left(\frac{i}{2}\int d^{4}x\,d^{4}x'\,J(x)\,\Delta\left(x-x'\right)\cdot\delta^{4}\left(x'-x_2\right)\right.\\
&\qquad\qquad\left. +\;\frac{i}{2}\int d^{4}x\,d^{4}x'\,J(x')\,\Delta\left(x-x'\right)\,\delta^{4}\left(x-x_2\right)\right)\\
&= \left(-\frac{i}{\partial J(x_1)}\right)
e^{-\frac{1}{2}\int d^{4}x\,d^{4}x'\,J(x)\,\Delta\left(x-x'\right)J(x')}
\left\{\frac{i}{2}\int d^{4}x\;J(x)\,\Delta\left(x-x_2\right)
\;+\;\frac{i}{2}\int d^{4}x'\;J(x')\,\Delta\left(x_2-x'\right)\right\}\\
&= \left(\frac{i}{2}\int d^{4}x\,d^{4}x'\;\frac{\delta J(x)}{\delta J(x_1)}\;\Delta\left(x-x'\right)J(x')\right.\\
&\qquad +\left.\frac{i}{2}\int d^{4}x\,d^{4}x'\;J(x)\,\Delta\left(x-x'\right)\frac{\delta J(x')}{\delta J(x_1)}\right)
\times\left(\frac{1}{2}\int d^{4}x\;\frac{J(x)}{\Delta\left(x-x'\right)}
+ \frac{1}{2}\int d^{4}x'\;J(x')\,\Delta\left(x_2-x'\right)\right)\\
&\qquad + e^{-\frac{1}{2}\int d^{4}x\,d^{4}x'\,J(x)\,\Delta\left(x-x'\right)J(x')}
\left(\frac{1}{2}\int d^{4}x\;\delta^{4}\left(x-x_1\right)\Delta\left(x-x_2\right)\right.\\
&\qquad\qquad\left. +\frac{1}{2}\int d^{4}x'\;\delta^{4}\left(x'-x_1\right)\Delta\left(x_2-x'\right)\right)
\end{aligned}
\]
at \(J = 0\)
\[
\begin{aligned}
&= \frac{1}{2}\,\Delta\left(x_1-x_2\right) + \frac{1}{2}\,\Delta\left(x_2-x_1\right)\\
&= \Delta\left(x_1-x_2\right) \;=\; \Delta\left(x_1, x_2\right)
\end{aligned}
\]
Similarly,
\[
\begin{aligned}
G_{\text{free}}\left(x_1\, x_2\, x_3\, x_4\right) =
\;&\Delta\left(x_1\,x_2\right)\Delta\left(x_3\,x_4\right)
+ \Delta\left(x_1\,x_3\right)\Delta\left(x_2\,x_4\right)\\
&+ \Delta\left(x_1\,x_4\right)\Delta\left(x_2\,x_3\right)
\end{aligned}
\]
Similarly,
\[
\begin{gathered}
G_{\text{free}}\left(x_1\,x_2\ldots x_{2n}\right) =
\Delta\left(x_1\,x_2\right)\Delta\left(x_2\,x_3\right)\cdots\Delta\left(x_{2n-1},\,x_{2n}\right)
\;+\;\text{All permutations}\\
\text{Total \# of permutations} \;=\; \frac{2n!}{2^{n}\,n!}
\end{gathered}
\]
as
\[
G_{f}\left(x_1\,x_2\ldots x_{2n+1}\right) = 0
\quad\left(\text{using wicks contraction}\ \left\langle 0\middle| T\left(\text{odd \# of fields}\right)\middle|0\right\rangle = 0\right)
\]
(or) # of permutations shall be :
\[
\left(2n-1\right)\left(2n-3\right)\cdot 5\cdot3\cdot1 \;=\; \frac{2n!}{2^{n}\,n!}
\qquad\longleftarrow\;\text{How}\ (?)
\]
Feynman rules :
(1) For correlation function of \(2n\)-\(\phi\)'s
\definecolor{ForestGreen}{rgb}{0.13,0.55,0.13}
\begin{tikzpicture}[scale=0.95]
\node[ForestGreen] at (1.1,1.8) {$\times$};
\node[above,font=\scriptsize] at (1.15,1.9) {$1$};
\node[ForestGreen] at (2.6,2.3) {$\times$};
\node[above,font=\scriptsize] at (2.65,2.4) {$2$};
\node[ForestGreen] at (0.2,0.9) {$\times$};
\node[left,font=\scriptsize] at (0.1,0.9) {$2n$};
\node[ForestGreen] at (3.6,1.0) {$\times$};
\node[ForestGreen] at (1.15,-0.4) {$\times$};
\node[ForestGreen] at (2.2,-0.6) {$\times$};
\node[ForestGreen] at (3.5,-0.8) {$\times$};
\draw[thick,blue] (1.15,-0.4) -- (2.6,2.3);
\draw[thick,blue] (1.1,1.8) -- (3.5,-0.8);
\node[left,font=\scriptsize,blue] at (1.1,-0.7) {$\Delta\left(x_2\,x_\ell\right)$};
\node[right,font=\scriptsize,blue] at (3.5,-1.05) {$\Delta\left(x_1\,x_k\right)$};
\node[below right,font=\scriptsize] at (3.4,-0.75) {$k$};
\draw[->,blue] (2.6,1.35) -- (2.05,1.05);
\node[right,blue] at (2.7,1.4) {propagators};
\node[right,text width=5.2cm] at (5.2,1.2) {--- Join `$x$' by lines\\ --- Every `$x$' should have a line ending on it \& every line should end on `$x$' on both ends.};
\end{tikzpicture}
To get \(\Delta\left(x_1\,x_2 - x_{2n}\right)\) we find :
\[
\Delta\left(x_1\,x_2 - x_{2n}\right) =
\underbrace{\Delta\left(x_1, x_k\right)\;\Delta\left(x_2, x_\ell\right)\cdots}_{\substack{\text{`}n\text{' such terms}\\ \left(n\text{- propagators}\right)}}
\;+\;\underbrace{\sum\;\text{All permutations}}_{n}
\]
Suppose we take action ---
\[
S = -\int d^{4}x\;\left(\frac{1}{2}\,\eta^{\mu\nu}\,\partial_{\mu}\phi\,\partial_{\nu}\phi + \frac{m^{2}}{2}\,\phi^{2} + \frac{\lambda}{4!}\;\phi^{4}\right)
\]
\[
\lambda:\ \text{small parameter}
\]
Write
\[
\exp\left(\int\left(i\,\frac{\lambda}{4!}\;\phi^{4}\right)d^{4}x\right)\ \text{as}\ :\quad
\sum_{m=0}^{\infty}\;\frac{1}{m!}\left(\frac{-i\lambda}{4!}\right)^{m}\left\{\int d^{4}x\;\phi^{4}(x)\right\}^{m}
\]
\[
e^{iS} = e^{i\,S_{\text{free}} + i\,S_{\text{int}}}
= e^{i\,S_{\text{free}}}\cdot e^{i\,S_{\text{int}}}
= e^{i\,S_{\text{free}}}\;\sum_{m}\;\frac{1}{m!}\left(\frac{-i\lambda}{4!}\right)^{m}\left\{i\,S_{\text{int}}\right\}^{m}
\]
We are interested to find \(G\left(x_1\,x_2\ldots x_{2n}\right)\) in interacting theory.
\[
\begin{aligned}
\text{unnormalised}\quad
G\left(x_1\,x_2\ldots x_n\right) &= \left\langle 0\right| T\left\{\phi_0(x_1)\,\phi_0(x_2)\ldots\phi_0(x_n)\; e^{i\,S_{\text{int}}}\right\}\left|0\right\rangle\\
&= \left\langle 0\right| T\left\{\phi_0(x_1)\,\phi_0(x_2)\ldots\phi_0(x_n)
\sum_{m=0}^{\infty}\frac{1}{m!}\left(\frac{-i\lambda}{4!}\right)^{m}\left(\int d^{4}y\;\phi^{4}(y)\right)^{m}\ldots\right\}\left|0\right\rangle
\end{aligned}
\]
\[
\text{Normalised}\quad
G\left(x_1\,x_2\ldots x_n\right) =
\frac{G\left(x_1\,x_2\ldots x_n\right)}{G\left(x_1\,x_2\ldots x_n\big|_{n=0}\right)}
\;=\;\frac{G\left(x_1\,x_2\ldots x_n\right)}{G(0)}
\]
\[
G(0) = \left\langle \Omega\middle|\Omega\right\rangle = \left\langle 0\right| T\left\{\exp\left(i\,S_{\text{int}}\right)\right\}\left|0\right\rangle
\]
Feynman rules of interacting theory :
Besides the \(x\)'s for external states we have 4 extra \(x\)'s for each factor of \(\lambda\).
i.e. For each order in '\(\lambda\)' introduce a vertex
\begin{tikzpicture}[baseline=-2pt,scale=0.5]
\draw (-0.7,0.7) -- (0.7,-0.7); \draw (-0.7,-0.7) -- (0.7,0.7);
\node[circle,draw,inner sep=1.2pt] at (0,0) {};
\end{tikzpicture}
& calculate using 4 extra vertex in free case with a vertex factor of \(\left(\dfrac{-i\lambda}{4!}\right)\).
\[
\underline{\text{and}}\qquad \int d^{4}x\qquad \forall\ x.
\]
For \(m\) such vertex
\begin{tikzpicture}[baseline=-2pt,scale=0.5]
\draw (-0.7,0.7) -- (0.7,-0.7); \draw (-0.7,-0.7) -- (0.7,0.7);
\node[circle,draw,inner sep=1.2pt] at (0,0) {};
\end{tikzpicture}
we have \(\dfrac{1}{m!}\) from expansion of \(e^{i\,S_{\text{int}}}\).
or \(\left(\text{for } O\left(\lambda^{m}\right)\text{ we have } \dfrac{1}{m!}\ \text{factor.}\right)\)
Suppose we want to calculate \(G\left(x_1\,x_2\right)\) upto \(O(\lambda)\).
\[
\text{[diagram below]}
\qquad
\downarrow\;\text{one vertex with factor}\ \left(-\frac{i\lambda}{4!}\right)
\]
\begin{tikzpicture}[baseline={(0,0.3)},scale=0.8]
\node at (0,1.6) {$\times$}; \node[left,font=\scriptsize] at (-0.1,1.6) {$1$};
\node at (0.3,0.0) {$\times$}; \node[left,font=\scriptsize] at (0.2,0.0) {$2$};
\begin{scope}[shift={(1.6,0.85)},scale=0.5]
\draw (-0.7,0.7) -- (0.7,-0.7); \draw (-0.7,-0.7) -- (0.7,0.7);
\node[circle,draw,inner sep=1.2pt] at (0,0) {};
\end{scope}
\end{tikzpicture}
Possibilities are:
\begin{tikzpicture}[scale=0.9]
\node at (0,1.5) {$\times$}; \node[left] at (-0.1,1.5) {$x_1$};
\node at (0,-0.5) {$\times$}; \node[left] at (-0.1,-0.5) {$x_2$};
\draw (0,1.35) -- (0,-0.35);
\draw (1.7,0.75) circle (0.32);
\draw (1.7,1.39) circle (0.32);
\node[right] at (2.15,0.75) {$\longleftarrow$ vertex};
\node at (1.7,0.42) {$\bullet$};
\node[below] at (0.9,-1.0) {Diagram 1};
\node at (4.4,0.5) {$+$};
\begin{scope}[shift={(6.6,0.4)}]
\node at (-1.0,1.2) {$\times$}; \node[above left,font=\scriptsize] at (-1.0,1.25) {$1$};
\node at (-1.0,-0.9) {$\times$}; \node[below left,font=\scriptsize] at (-1.0,-0.9) {$2$};
\draw (-1.0,1.1) -- (0.4,0.1);
\draw (-1.0,-0.8) -- (0.4,0.1);
\draw (1.05,0.1) ellipse (0.62 and 0.34);
\node[above right,red] at (0.3,0.9) {C.f $= 4\cdot3\cdot1$};
\node[below] at (0.4,-1.0) {Diagram 2};
\end{scope}
\end{tikzpicture}
Diagram (1) : \(\phi^{4}\) theory
\[
G_1 = (\frac{-i\lambda}{4!})\int d^{4}x\;\;\Delta\left(x_1\,x_2\right)\,\Delta\left(x, x\right)\,\Delta\left(x, x\right)\;\cdot\;(3)
\qquad
\swarrow\;\text{Combinatorial factor}
\]
Diagram (2) :
\[
G_2 = \frac{-i\lambda}{4!}\int d^{4}x\;\;\Delta\left(x_1, x\right)\,\Delta\left(x_2, x\right)\,\Delta\left(x, x\right)\;\cdot\;\left(4\cdot3\cdot1\right)
\]
\[
G\left(x_1\,x_2\right) = G_1 + G_2
\]
ex :
\[
G\left(x_1 x_2 x_3 x_4\right) =
\text{[diagram below]}
= \frac{-i\lambda}{\cancel{\left(4!\right)}}\int d^{4}x\;\Delta\left(x_1, x\right)\Delta\left(x_2\,x\right)\Delta\left(x_3\,x\right)\Delta\left(x_4\,x\right)\cdot\left(\cancel{4\cdot3\cdot2\cdot1}\right)
\]
\definecolor{ForestGreen}{rgb}{0.13,0.55,0.13}
\begin{tikzpicture}[baseline={(0,0)},scale=0.6,color=ForestGreen]
\draw (-1,1) node[above left]{$x_1$} -- (1,-1) node[below right]{$x_4$};
\draw (-1,-1) node[below left]{$x_3$} -- (1,1) node[above right]{$x_2$};
\node at (0.3,0.15) {$x$};
\node at (0,0) {$\times$};
\end{tikzpicture}
\[
\swarrow\quad 4!\ \text{is designed to cancel out C.f.\ from tree level diagrams!}
\qquad
\downarrow\;\text{Combinatorial factor}
\]
denominator / Normalisation
\[
\begin{aligned}
\left\langle \Omega\middle|\Omega\right\rangle &= \left\langle 0\right| T\left\{e^{i\,S_{\text{int}}}\right\}\left|0\right\rangle\\
&= 1 + \int\left(\frac{-i\lambda}{4!}\right) d^{4}x \left\langle 0\middle|\phi^{4}(x)\middle|0\right\rangle
+ \left(\frac{-i\lambda}{4!}\right)^{2}\frac{1}{2!}\int d^{4}x\,d^{4}y\;\left\langle 0\middle|\phi^{4}(x)\,\phi^{4}(y)\middle|0\right\rangle + \cdots
\end{aligned}
\]
\[
= 1 +
\text{[diagram below]}
\;+\;
\text{[diagram below]}
\;+\;
\text{[diagram below]}
\;+\;\cdots\;\text{Bubble diagrams}
\]
\begin{tikzpicture}[baseline={(0,0.15)},scale=0.45]
\draw (0,0.35) circle (0.35); \draw (0,1.05) circle (0.35);
\end{tikzpicture}
\begin{tikzpicture}[baseline={(0,0.15)},scale=0.45]
\draw (0,0.35) circle (0.35); \draw (0,1.05) circle (0.35);
\draw (1.2,0.35) circle (0.35); \draw (1.2,1.05) circle (0.35);
\end{tikzpicture}
\begin{tikzpicture}[baseline={(0,0.15)},scale=0.45]
\draw (0,0.7) ellipse (0.55 and 0.42);
\draw (0,0.98) arc (140:40:0.42);
\end{tikzpicture}
diagrams like
\begin{tikzpicture}[baseline={(0,0.1)},scale=0.55]
\node[above,font=\scriptsize] at (0,0.7) {$x_1$};
\node at (0,0.7) {$\times$};
\node at (0,-0.35) {$\times$};
\node[below,font=\scriptsize] at (0,-0.35) {$x_2$};
\draw (0,0.6) -- (0,-0.25);
\draw (0.75,0.15) circle (0.3);
\draw (1.35,0.15) circle (0.3);
\end{tikzpicture}
should not be included as such disconnected diagrams factor as connected \(\times\) vacuum bubbles where bubbles's contribution cancels from denominator \(\left\langle \Omega\middle|\Omega\right\rangle\).
Factorisation of bubble diagrams ---
Suppose we have situation like this :
\begin{tikzpicture}[scale=0.9,color=blue]
\draw (-1.1,0.9) -- (-0.35,0.3);
\draw (-1.1,-0.9) -- (-0.35,-0.3);
\draw (1.1,0.9) -- (0.35,0.3);
\draw (1.1,-0.9) -- (0.35,-0.3);
\draw (0,0) ellipse (0.5 and 0.42);
\draw (-0.25,-0.25) -- (0.05,0.3);
\draw (-0.05,-0.3) -- (0.25,0.25);
\begin{scope}[shift={(4.6,0.1)}]
\draw (0,0) ellipse (0.55 and 0.45);
\draw (-0.3,-0.28) -- (0.0,0.4);
\draw (-0.05,-0.4) -- (0.25,0.32);
\end{scope}
\end{tikzpicture}
\[
\begin{aligned}
&\left(m\text{ -- interaction vertices}\right)\\
&\text{Feynman diagram with no bubble (as we have external legs)}
\end{aligned}
\qquad\qquad
\begin{aligned}
&\left(n\text{-vertices [all are internal]}\right)\\
&\text{vacuum Bubble (as no external line/leg)}
\end{aligned}
\]
Interpretation :
\[
\frac{1}{\left(m+n\right)!}\left(\frac{-i\lambda}{4!}\right)^{m+n}\quad\swarrow
\]
(1) Above two diagrams may appear as single feynman diagram
(2) Above diagram is product of two seperate diagrams. (connected \(\times\) bubble).
\[
=\;\frac{1}{m!}\left(\frac{-i\lambda}{4!}\right)^{m}\;\frac{1}{n!}\left(\frac{-i\lambda}{4!}\right)^{n}
\]
If interpretation (1) \(=\) interpretation (2)
Any diagram can be thought as product of \(\Big(\sum\limits_{\text{form}}\)diagram without bubble\(\Big)\) \(\times\) \(\left(1 + \text{Sum of bubbles}\right)\).
i.e.
\[
\begin{gathered}
\text{Sum of Feynman diagrams} = \left(\text{sum of connected}\right)\times\left(1 + \text{sum of bubbles}\right)\\
= \sum\nolimits_{\text{connected}}\;\cdot\;\sum\left(1 + \text{bubbles}\right)
\end{gathered}
\]
The key is to find combinatorial factors which is hard to find if we don't have diagrams. If we had diagram we can follow it to find combinatorial factor.
But exchange of vertices don't generate new diagrams
\[
\begin{gathered}
\text{So,}\qquad
1)\quad \frac{1}{\left(m+n\right)!}\left(\frac{-i\lambda}{4!}\right)^{m+n}\;\times\;\left(m+n\right)!
\qquad
\swarrow\;
\begin{aligned}
&\text{factors for exchange}\\
&\text{of vertices.}
\end{aligned}\\
2)\quad \frac{1}{m!}\left(\frac{-i\lambda}{4!}\right)^{m}\;\frac{1}{n!}\left(\frac{-i\lambda}{4!}\right)^{n}\;\times\;\left(m!\;\, n!\right)
\end{gathered}
\]
Counting (1)
of ways of choosing '\(m\)' vertices out of \(\left(m+n\right)\) is \(^{m+n}C_m\)
\[
= \frac{\left(m+n\right)!}{m!\;n!}
\qquad\qquad
\left\{{}^{n}C_r = \frac{n!}{r!\left(n-r\right)!}\right.
\]
——— \(\times\) ———
\[
\left(\partial_x^{2} - m^{2}\right)\Delta\left(x, x'\right) = i\,\delta^{4}\left(x-x'\right)
\qquad\left(\text{translation invariant}\right)
\]
\[
\begin{gathered}
x \;\longrightarrow\; x + a\\
x' \;\longrightarrow\; x' + a
\qquad\qquad
\text{eq}^{\text{n}}\text{ will look same.}
\end{gathered}
\]
So, this \(\Delta\left(x, x'\right)\) shall only depend upon \(\left|x - x'\right|\).
Fourier expansion of
\[
\Delta\left(x, x'\right) = \int\frac{d^{4}k}{\left(2\pi\right)^{4}}\;\; e^{i\,k\cdot\left(x-x'\right)}\;\;\widetilde{\Delta}(k)
\]
if we use above eq\(^{\text{n}}\)
\[
\begin{aligned}
\left(\partial_x^{2} - m^{2}\right)\Delta\left(x, x'\right)
&= \int\frac{d^{4}k}{\left(2\pi\right)^{4}}\;\left(\partial_x^{2} - m^{2}\right)\, e^{i\,k\cdot\left(x-x'\right)}\;\widetilde{\Delta}(k)\\
&= \int\frac{d^{4}k}{\left(2\pi\right)^{4}}\;\left(i^{2}k^{2} - m^{2}\right)\, e^{i\,k\cdot\left(x-x'\right)}\;\widetilde{\Delta}(k)\\
&= \int\frac{d^{4}k}{\left(2\pi\right)^{4}}\;\left(-k^{2} - m^{2}\right)\, e^{i\,k\cdot\left(x-x'\right)}\;\widetilde{\Delta}(k)\\
&= i\,\delta^{4}\left(x-x'\right)
\qquad\text{iff}:\quad
\widetilde{\Delta}(k) = \frac{i}{-k^{2}-m^{2}}
\end{aligned}
\]
\[
\widetilde{\Delta}(k) = \frac{i}{-\left(k^{0}k_{0} + k_i\,k^{i} + m^{2}\right)}
= \frac{i}{-\left(-k_0^{2} + k_i\cdot k_i + m^{2}\right)}
\]
\[
\widetilde{\Delta}(k) = \frac{i}{k_0^{2} - \vec{k}^{2} - m^{2}}
\qquad\cdots\left(\eta_{\mu\nu} = \left(-1,1,1,1\right)\right)
\]
Not defined fully at denominator \(= 0\).
So, it needs to be specified how to deal with singularity.
\[
\text{Singularity at}\qquad k_0 = \pm\sqrt{\vec{k}^{2} + m^{2}}
\]
Ambiguity \(\propto\) \(e^{i\,k\cdot\left(x-x'\right)}\) at \(k_0 = \pm\sqrt{\vec{k}^{2}+m^{2}}\) (in \(x\)-dependence)
i.e. expression for \(\Delta\left(x, x'\right)\) will change by \(e^{i\,k\cdot\left(x-x'\right)}\) at \(k_0 = \pm\sqrt{\vec{k}^{2}+m^{2}}\)
Also \(e^{i\,k\cdot\left(x-x'\right)}\) satisfies
\[
\begin{gathered}
\left(\partial_x^{2} - m^{2}\right) e^{i\,k\cdot\left(x-x'\right)} = 0\\
\underline{\text{LHS}}:\quad \left(-k^{2}-m^{2}\right)\, e^{i\,k\cdot\left(x-x'\right)} = (0)\cdot e^{i\,k\cdot\left(x-x'\right)}\\
\downarrow\\
\text{at}\ k_0 = \pm\sqrt{\vec{k}^{2}+m^{2}}
\end{gathered}
\]
So, this ambiguity is just reflecting the fact that this eq\(^{\text{n}}\)
\[
\left(\partial_x^{2}-m^{2}\right)\Delta\left(x, x'\right) = i\,\delta^{4}\left(x-x'\right)
\]
does not have unique sol\(^{\text{n}}\) (i.e. \(\Delta\left(x, x'\right)\)).
for ex: we can \(\Delta \to \Delta' = \Delta + \text{Const}\).
\[
\text{Still}\qquad \left(\partial_x^{2}-m^{2}\right)\Delta' = i\,\delta^{4}\left(x-x'\right)
\]
Greens function is not uniquely defined in minkowaski space.
Shortcut :
\[
x^{0} - \left(x'\right)^{0} \;\longrightarrow\; \left(x^{0} - x'^{0}\right)\cdot\left(1 - i\epsilon\right)
\]
So,
\[
\begin{aligned}
\Delta\left(x, x'\right) &= \int\frac{d^{4}k}{\left(2\pi\right)^{4}}\;\;
e^{-i\,k^{0}\left(x^{0}-x'^{0}\right) + i\,\vec{k}\cdot\left(\vec{x}-\vec{x}'\right)}\;\;\widetilde{\Delta}(k)\\
&= \int\frac{d^{4}k}{\left(2\pi\right)^{4}}\;\;
e^{-i\,k^{0}\left(x^{0}-x'^{0}\right)\left(1-i\epsilon\right) + i\,\vec{k}\cdot\left(\vec{x}-\vec{x}'\right)}\;\;\widetilde{\Delta}(k)\\
&= \int\frac{dk^{0}}{2\pi}\;\frac{d^{3}k}{\left(2\pi\right)^{3}}\;\;
e^{-k^{0}\left(i+\epsilon\right)\left(x^{0}-x'^{0}\right)\; +\; i\,\vec{k}\cdot\left(\vec{x}-\vec{x}''\right)}\;\;\widetilde{\Delta}(k)\\
&\qquad\qquad\qquad\downarrow\quad\text{will blow up as}\ \ k^{0} \to \left(-\infty\right).
\end{aligned}
\]
So to prevent it, we also rotate \(k^{0}\)
\[
k^{0} \;\longrightarrow\; \frac{k^{0}}{1-i\epsilon} = k^{0}\left(1+i\epsilon\right) + O\left(\epsilon^{2}\right)
\qquad
\hookrightarrow\;\text{does it really prevent}\ (?)
\]
Such that
\[
\begin{aligned}
\widetilde{\Delta}(k) = \frac{i}{k'^{2} - \vec{k}^{2} - m^{2}}
\;&\longrightarrow\;
\frac{i}{k^{0\,2}\left(1+i\epsilon\right)^{2} - \vec{k}^{2} - m^{2}}\\
&= \frac{i}{k^{0\,2}\left(1+2i\epsilon\right) - \vec{k}^{2} - m^{2}}
\qquad\left(\text{neglecting } O\left(\epsilon^{2}\right)\right)\\
&= \frac{i}{\left(k^{0}\right)^{2} - \vec{k}^{2} - m^{2} + 2i\epsilon\,k^{0\,2}}\\
&= \frac{i}{\left(k^{0}\right)^{2} - \vec{k}^{2} - m^{2} + i\epsilon}
\qquad\text{redefing } \epsilon.
\end{aligned}
\]
poles are now shifted
\[
k^{0} = \left\{
\begin{aligned}
&\sqrt{\vec{k}^{2}+m^{2}} - i\epsilon\\
-&\sqrt{\vec{k}^{2}+m^{2}} + i\epsilon
\end{aligned}
\right.
\]
\definecolor{ForestGreen}{rgb}{0.13,0.55,0.13}
\begin{tikzpicture}[scale=0.9]
\draw (-3,0) -- (3,0) node[right]{$Re\left(k^{0}\right)$};
\draw (0,-1.4) -- (0,1.5) node[above left]{$Im\left(k^{0}\right)$};
\node at (-1.7,0.22) {$\times$};
\node at (1.7,-0.22) {$\times$};
\node[right,font=\scriptsize] at (0.05,0.22) {$i\epsilon$};
\node[right,font=\scriptsize] at (0.05,-0.25) {$i\epsilon$};
\draw[ForestGreen,thick] (-2.4,-0.35) -- (2.4,0.35);
\draw[->] (1.1,1.15) node[right]{contour (or equivalently) shift the poles.} -- (0.5,0.25);
\end{tikzpicture}
So, Final form of propagator becomes :
\[
\widetilde{\Delta}_{F}(k) = \frac{i}{-k^{2} - m^{2} + i\epsilon}
\qquad\text{where}\quad \epsilon > 0.
\]
In path integral formelism \(i\epsilon\) prescription ensures that we get vacuum state to calculate partition functionel.
\[
\begin{gathered}
e^{iHT}\left|q'\right\rangle \;\longrightarrow\; e^{iHT}\sum_{n}\left|n\right\rangle\left\langle n\middle|q'\right\rangle
= \sum_{n} e^{i\,E_n\,t}\;\psi_{n}\left(q'\right)\left|n\right\rangle\\
\text{if}\quad T \;\longrightarrow\; T\left(1-i\epsilon\right)\\
\searrow\quad
e^{i\,H\,T\left(1-i\epsilon\right)}\;\sum_{n}\left|n\right\rangle\left\langle n\middle|q'\right\rangle
= \sum_{n}\; e^{i\,E_n\,T\left(1-i\epsilon\right)}\;\left\langle n\middle|q'\right\rangle\;\left|n\right\rangle
\end{gathered}
\]
at \(t = -\infty\) only \(n = 0\) state survives
\[
\sum_{n}\; e^{i\,E_n\,t}\;\, e^{\epsilon\,E_n\,T}\left|n\right\rangle\left\langle n\middle|q'\right\rangle.
\qquad
\swarrow\;
\begin{aligned}
&\text{What if } E_0 \neq 0\\
&\text{does it still survive?}
\end{aligned}
\]
Also \(m^{2} \to m^{2}\left(1-i\epsilon\right)\) makes path integral well defined! else we have damped fun. \(e^{iS}\).
Momentum space feynman rules :
Idea is to calculate fourier transform of green's function.
\[
\begin{aligned}
\widetilde{G}\left(k_1\,k_2\,\cdots\,k_n\right) = \int d^{4}x_1\, d^{4}x_2\ldots d^{4}x_n\;\;
&e^{-i\,k_1\cdot x_1}\; e^{-i\,k_2\cdot x_2}\cdots e^{-i\,k_n\cdot x_n}\\
&\times\;\; G_{n}\left(x_1\,x_2\,\cdots\,x_n\right)
\end{aligned}
\]
We have
\[
\begin{gathered}
\widetilde{\phi}(k) = \int d^{4}x\;\;\phi(x)\;\, e^{-i\,k\cdot x}\\
\phi(x) = \int\frac{d^{4}k}{\left(2\pi\right)^{4}}\;\; e^{i\,k\cdot x}\;\;\widetilde{\phi}(k)
\end{gathered}
\]
In \(Z[J]\), we will replace \(\mathcal{D}\phi \to \mathcal{D}\widetilde{\phi}\) there will be some Jacobian (const. but independent of source term \(J(x)\) which will cancel out when we devide by \(Z[0]\).)
\[
\begin{aligned}
Z\left[J\right] &= \frac{1}{Z[0]}\int\left[\mathcal{D}\phi\right] e^{\,iS + i\int d^{4}x\;\phi(x)\,J(x)}\\
&= \frac{1}{Z[0]}\int\left[\mathcal{D}\widetilde{\phi}\right]\;
e^{\,iS + i\int d^{4}x\; J(x)\left(\int\frac{d^{4}k}{\left(2\pi\right)^{4}}\; e^{i\,k\cdot x}\;\widetilde{\phi}(k)\right)}
\end{aligned}
\]
also
\[
S_{\text{free}}\left[\phi\right] = \frac{1}{2}\int d^{4}x\;\left(\eta^{\mu\nu}\,\partial_{\mu}\phi\,\partial_{\nu}\phi - m^{2}\phi^{2}\right)
\]
using \(\phi(x) = \int\frac{d^{4}k}{\left(2\pi\right)^{4}}\; e^{i\,k\cdot x}\;\widetilde{\phi}(k)\)
\[
\begin{aligned}
S_{\text{free}}\left[\widetilde{\phi}\right] = \frac{1}{2}\int d^{4}x\int\frac{d^{4}k}{\left(2\pi\right)^{4}}\;\frac{d^{4}k'}{\left(2\pi\right)^{4}}
\;\Big[&\left(i\,k_{\mu}\right)\left(i\,k'_{\nu}\right)\,\widetilde{\phi}(k)\,\widetilde{\phi}(k')\;\eta^{\mu\nu}\;
e^{i\,k\cdot x}\, e^{i\,k'\cdot x}\\
&- m^{2}\;\widetilde{\phi}(k)\,\widetilde{\phi}(k')\;
e^{i\,k\cdot x}\, e^{i\,k'\cdot x}\Big]
\end{aligned}
\]
where \(\left\{d\breve{k} = \dfrac{dk}{2\pi}\right\}\)
\[
= \frac{1}{2}\int d^{4}x\int d^{4}\breve{k}\;d^{4}\breve{k}'\;\cdot\;
e^{i\,x\cdot\left(k+k'\right)}\;\;\widetilde{\phi}(k)\;\widetilde{\phi}(k')\;\left(-k^{2}-m^{2}\right)
\]
solving \(\int d^{4}x\) we get
\[
= \frac{1}{2}\int d^{4}\breve{k}\;d^{4}\breve{k}'\;\left(2\pi\right)^{4}\,\delta^{4}\left(k+k'\right)\;
\widetilde{\phi}(k)\;\widetilde{\phi}(k')\;\left(-k^{2}-m^{2}\right)
\]
solving \(k'\) integral we get
\[
S_{\text{free}}\left[\widetilde{\phi}\right] = \frac{1}{2}\int\frac{d^{4}k}{\left(2\pi\right)^{4}}\;\;
\underbrace{\widetilde{\phi}(k)\;\widetilde{\phi}(-k)}_{\downarrow\;\text{quadratic term}\ (1)}\;\;
\underset{\nearrow\;\text{Inverse of propagator}}{(-k^{2}-m^{2})}
\]
Similarly
\[
\begin{aligned}
S_{\text{int}} &= \int d^{4}x\;\, L_{\text{int}} = \int d^{4}x\;\frac{\lambda}{4!}\,\phi^{4}(x)\\
&= \frac{\lambda}{4!}\int d^{4}x\int d^{4}\breve{k}_1\, d^{4}\breve{k}_2\, d^{4}\breve{k}_3\, d^{4}\breve{k}_4\;\;
e^{i\,x\cdot\left(k_1+k_2+k_3+k_4\right)}\;\;
\widetilde{\phi}\left(k_1\right)\widetilde{\phi}\left(k_2\right)\widetilde{\phi}\left(k_3\right)\widetilde{\phi}\left(k_4\right)
\end{aligned}
\]
solving '\(x\)' integral
\[
S_{\text{int}} = \frac{\lambda}{4!}\int d^{4}\breve{k}_1\, d^{4}\breve{k}_2\, d^{4}\breve{k}_3\, d^{4}\breve{k}_4\;
\left(2\pi\right)^{4}\,\delta^{4}\left(k_1+k_2+k_3+k_4\right)\;
\widetilde{\phi}\left(k_1\right)\widetilde{\phi}\left(k_2\right)\widetilde{\phi}\left(k_3\right)\widetilde{\phi}\left(k_4\right)
\]
the Source Action becomes
\[
\begin{aligned}
i\int \underset{\downarrow\;k'}{J(x)}\;\underset{\downarrow\;k}{\phi(x)}\;d^{4}x
&= i\int d^{4}x\int d^{4}\breve{k}\,d^{4}k'\;\;\widetilde{J}(k')\;\widetilde{\phi}(k)\; e^{i\,x\cdot\left(k+k'\right)}\\
&= i\int d^{4}\breve{k}\,d^{4}\breve{k}'\;\;\widetilde{J}(k')\;\widetilde{\phi}(k)\;\left(2\pi\right)^{4}\delta^{4}\left(k+k'\right)\\
&= i\int\frac{d^{4}k}{\left(2\pi\right)^{4}}\;\;\widetilde{J}(-k)\;\widetilde{\phi}(k)
\qquad
\hookrightarrow\;\text{linear term}\ (2)
\end{aligned}
\]
(1) is quadratic term , (2) is linear term , so we complete the square.
It turns out that:
\[
\frac{Z_{f}\left[J\right]}{Z_{f}\left[0\right]} =
\exp\left(-\frac{1}{2}\int d^{4}\breve{k}_1\; d^{4}\breve{k}_2\;\;\Delta_{F}\left(k_1, k_2\right)\;\widetilde{J}\left(k_1\right)\,\widetilde{J}\left(k_2\right)\right)
\]
\[
\Delta_{F}\left(k_1\,k_2\right) = \left(2\pi\right)^{4}\,\delta^{4}\left(k_1+k_2\right)
\]
\[
\begin{aligned}
G_{F}\left(k_1\,k_2\,\cdots\,k_n\right) =
\left(-\left(2\pi\right)^{4}\, i\,\frac{\delta}{\delta\widetilde{J}\left(-k_1\right)}\right)
&\left(-\left(2\pi\right)^{4}\, i\,\frac{\delta}{\delta\widetilde{J}\left(-k_2\right)}\right)\\
\cdots\;
&\left(-\left(2\pi\right)^{4}\, i\,\frac{\delta}{\delta\widetilde{J}\left(-k_N\right)}\right)\;
\frac{Z\left[J\right]}{Z\left[0\right]}\;\Bigg|_{J=0.}
\end{aligned}
\]