Take the full answer it is frommy lecture to QFT. \subsection{Boson propagator} It is prefer to write the collision between to particles in term of \textit{amplitude} of \textit{probability}. The perturbative approximation of QFT assumed that the particles propagate freely except some points, when there are emission or absorption of quanta. we write the solution of motion aquations compled as pertirbative series around of free solution of motion equations of free field. The methode uses Green's function which R.Feynman gave his probabilist interpritation of implitude. Motion equation of free boson(Kein-Gordon equation) is writing:\ \begin{equation} \left( p^{2}-m^{2}\right) \varphi(p)=0 \end{equation} Where $\varphi(p)$ is a scalar function.\ Green's function $ G(p) $, in the space of momentum is: \begin{equation} (p^{2}- m^{2})G(p)=\delta^{4}(p) \end{equation} Then $ G(p)=\frac{\delta^{4}(p)}{p^{2}-m^{2}} $, $ \delta^{4} $ is the Dirac function defined as \begin{equation} \delta^{4}(p)=\delta(p_0)\delta(p_1)\delta(p_2)\delta(p_3) \end{equation} Feynman interpritation is that this operator is as amplitude of probability that the boson propagates with quadri-momentume. Propagator = $ \frac{i}{p^{2}-m^{2}} $. In same way, feynman defined an amplitude of probability that the boson whether emitted ( or absorbed) by particle 1, and/or absorbe by particle 2 of interactions.\ The amplitude are driveded for various kinds pf interactions between various particle, the square of the magnitude of each amplitude turns out be tje probability of that particular interaction (transition) occuring. These transition amplitudes each depend on the initial real particles, the final real particles, and the virtial particles that mediate the transition. It turns out that the factor in the amplitude representing the virtual particle contribution is identical to the feynman propagator $ \Delta_F $.\ \ \begin{equation} \Delta(x,y) = \int\frac{d^{4}k}{(2\pi)^{4}} \exp^{-ik(x-y)} \frac{1}{k^{2}-m^{2}+i\varepsilon} \\ \end{equation} \section{The spenor feynman Propagator} We will follow similar steps to drive feynman propagator for Dirac fields as we did for scalar feild , similar to what we did scalars, we will, heuristical, consider the operator field $ b\psi $ to creat a virtual particle at event y, and $ \psi $ to destroy that virtual particle particle at even x. The spinor propagator incorporates this two field operators. The propagator realy corresponds to a kind of probability density function in y and x. It represents the probability density of Dirac particle appearing at y and disapiaring at x. we show that in briefly:\ for the virtual spin $ 1/2 $ particle feynman propagator were\ \begin{equation} iS_{F}(x-y) = \langle 0 \vert T\lbrace \psi(x)b\psi(y)\rbrace \vert 0 \rangle = \langle 0 \vert \lbrace \psi(x)b\psi(y)\rbrace \vert 0 \rangle \end{equation}\begin{equation} iS_{F}(x-y) = \langle 0 \vert T\lbrace \psi(x)\bar\psi(y)\rbrace \vert 0 \rangle = \langle 0 \vert \lbrace \psi(x)\bar\psi(y)\rbrace \vert 0 \rangle \end{equation} if $ t_{y} < t_{x} $ ( particle) \begin{equation} = \langle 0 \vert[\psi^{+}(x), b\psi^{-}(y) ]_{+}\vert 0 \rangle \end{equation}\begin{equation} = \langle 0 \vert[\psi^{+}(x), \bar\psi^{-}(y) ]_{+}\vert 0 \rangle \end{equation} \begin{equation} = [\psi^{+}(x), b\psi^{-}(y)]_{+}\langle 0 \vert \vert 0 \rangle = [ \psi^{+}(x), b\psi^{-}(y)]_{+} \end{equation}\begin{equation} = [\psi^{+}(x), \bar\psi^{-}(y)]_{+}\langle 0 \vert \vert 0 \rangle = [ \psi^{+}(x), \bar\psi^{-}(y)]_{+} \end{equation} \begin{equation} = iS_{\alpha\beta}^{+}(x - y) = \frac{1}{2(2\pi)^{3}}\int (slp + m)\frac{e^{ip(x - y)}}{E}d^{3}p = \frac{- i}{(2\pi)^{4}}\int_{c^{+}}\frac{(slp + m)e^{-ip(x - y)}}{p^{2} - m^{2}}d^{4}p \end{equation}\begin{equation} = iS_{\alpha\beta}^{+}(x - y) = \frac{1}{2(2\pi)^{3}}\int (\not p + m)\frac{e^{ip(x - y)}}{E}d^{3}p = \frac{- i}{(2\pi)^{4}}\int_{c^{+}}\frac{(\not p + m)e^{-ip(x - y)}}{p^{2} - m^{2}}d^{4}p \end{equation} for the virtual spin $ 1/2 $ particle feynman propagator were\ \begin{equation} iS_{F}(x-y) = \langle 0 \vert T \lbrace \psi(x)b\psi(y)\rbrace \vert 0 \rangle = - \langle 0 \vert \lbrace b\psi(x)\psi(y)\rbrace \vert 0 \rangle \end{equation}\begin{equation} iS_{F}(x-y) = \langle 0 \vert T \lbrace \psi(x)b\psi(y)\rbrace \vert 0 \rangle = - \langle 0 \vert \lbrace \bar\psi(x)\psi(y)\rbrace \vert 0 \rangle \end{equation} if $ t_{y} < t_{x} $ ( antiparticle) \begin{equation} = \langle 0 \vert[b\psi^{+}(x), \psi^{-}(y) ]_{+}\vert 0 \rangle \end{equation}\begin{equation} = \langle 0 \vert[\bar\psi^{+}(x), \psi^{-}(y) ]_{+}\vert 0 \rangle \end{equation} \begin{equation} = [b\psi^{+}(x), \psi^{-}(y)]_{+}\langle 0 \vert \vert 0 \rangle = [ b\psi^{+}(x), \psi^{-}(y)]_{+} \end{equation}\begin{equation} = [\bar\psi^{+}(x), \psi^{-}(y)]_{+}\langle 0 \vert \vert 0 \rangle = [ \bar\psi^{+}(x), \psi^{-}(y)]_{+} \end{equation} \begin{equation} = iS^{-}(x - y) = -\frac{1}{2(2\pi)^{3}}\int (slp - m)\frac{e^{ip(x - y)}}{E}d^{3}p = \frac{ i}{(2\pi)^{4}}\int_{c^{-}}\frac{(slp + m)e^{-ip(x - y)}}{p^{2} - m^{2}}d^{4}p \end{equation}\begin{equation} = iS^{-}(x - y) = -\frac{1}{2(2\pi)^{3}}\int (\not p - m)\frac{e^{ip(x - y)}}{E}d^{3}p = \frac{ i}{(2\pi)^{4}}\int_{c^{-}}\frac{(\not p + m)e^{-ip(x - y)}}{p^{2} - m^{2}}d^{4}p \end{equation} \ The two contour integrals in the last lines () and () were combined in final step to yield the single integral over real space to get the final result for \textit{\textbf{the spinor Feynman propagator }} \begin{equation} S_{F}(x - y) = \int_{-\infty}^{+\infty}\frac{ d^{4}p}{(2\pi)^{4}}\frac{(slp + m)e^{-ip(x - y)}}{p^{2} - m^{2} + i\varepsilon} \end{equation}\begin{equation} S_{F}(x - y) = \int_{-\infty}^{+\infty}\frac{ d^{4}p}{(2\pi)^{4}}\frac{(\not p + m)e^{-ip(x - y)}}{p^{2} - m^{2} + i\varepsilon} \end{equation} The momentum space form of the propagator ( its fourier transform, is \begin{equation} S_{F}(p) = \frac{slp + m}{p^{2} - m^{2} + i\varepsilon} = (slp + m)\Delta_{F}(p) \end{equation}\begin{equation} S_{F}(p) = \frac{\not p+ m}{p^{2} - m^{2} + i\varepsilon} = (\not p + m)\Delta_{F}(p) \end{equation} Observe that: \begin{equation} (slp - m)(slp + m) = \gamma^{\mu}\gamma^{\nu}p_{\mu}p_{\nu} - m^{2} = p^{2} - m^{2} \end{equation}\begin{equation} (\not p - m)(\not p + m) = \gamma^{\mu}\gamma^{\nu}p_{\mu}p_{\nu} - m^{2} = p^{2} - m^{2} \end{equation} Then we can multiply by $ (slp + m) $$ (\not p + m) $ both the numerator and the denominator in eq. and rewrite $ S_{F}(p) $ in the form, \begin{equation} S_{F}(p) = \frac{i}{slp - m} \end{equation}\begin{equation} S_{F}(p) = \frac{i}{\not p - m} \end{equation}
Take the full answer it is frommy lecture to QFT. \subsection{Boson propagator} It is prefer to write the collision between to particles in term of \textit{amplitude} of \textit{probability}. The perturbative approximation of QFT assumed that the particles propagate freely except some points, when there are emission or absorption of quanta. we write the solution of motion aquations compled as pertirbative series around of free solution of motion equations of free field. The methode uses Green's function which R.Feynman gave his probabilist interpritation of implitude. Motion equation of free boson(Kein-Gordon equation) is writing:\ \begin{equation} \left( p^{2}-m^{2}\right) \varphi(p)=0 \end{equation} Where $\varphi(p)$ is a scalar function.\ Green's function $ G(p) $, in the space of momentum is: \begin{equation} (p^{2}- m^{2})G(p)=\delta^{4}(p) \end{equation} Then $ G(p)=\frac{\delta^{4}(p)}{p^{2}-m^{2}} $, $ \delta^{4} $ is the Dirac function defined as \begin{equation} \delta^{4}(p)=\delta(p_0)\delta(p_1)\delta(p_2)\delta(p_3) \end{equation} Feynman interpritation is that this operator is as amplitude of probability that the boson propagates with quadri-momentume. Propagator = $ \frac{i}{p^{2}-m^{2}} $. In same way, feynman defined an amplitude of probability that the boson whether emitted ( or absorbed) by particle 1, and/or absorbe by particle 2 of interactions.\ The amplitude are driveded for various kinds pf interactions between various particle, the square of the magnitude of each amplitude turns out be tje probability of that particular interaction (transition) occuring. These transition amplitudes each depend on the initial real particles, the final real particles, and the virtial particles that mediate the transition. It turns out that the factor in the amplitude representing the virtual particle contribution is identical to the feynman propagator $ \Delta_F $.\ \ \begin{equation} \Delta(x,y) = \int\frac{d^{4}k}{(2\pi)^{4}} \exp^{-ik(x-y)} \frac{1}{k^{2}-m^{2}+i\varepsilon} \\ \end{equation} \section{The spenor feynman Propagator} We will follow similar steps to drive feynman propagator for Dirac fields as we did for scalar feild , similar to what we did scalars, we will, heuristical, consider the operator field $ b\psi $ to creat a virtual particle at event y, and $ \psi $ to destroy that virtual particle particle at even x. The spinor propagator incorporates this two field operators. The propagator realy corresponds to a kind of probability density function in y and x. It represents the probability density of Dirac particle appearing at y and disapiaring at x. we show that in briefly:\ for the virtual spin $ 1/2 $ particle feynman propagator were\ \begin{equation} iS_{F}(x-y) = \langle 0 \vert T\lbrace \psi(x)b\psi(y)\rbrace \vert 0 \rangle = \langle 0 \vert \lbrace \psi(x)b\psi(y)\rbrace \vert 0 \rangle \end{equation} if $ t_{y} < t_{x} $ ( particle) \begin{equation} = \langle 0 \vert[\psi^{+}(x), b\psi^{-}(y) ]_{+}\vert 0 \rangle \end{equation} \begin{equation} = [\psi^{+}(x), b\psi^{-}(y)]_{+}\langle 0 \vert \vert 0 \rangle = [ \psi^{+}(x), b\psi^{-}(y)]_{+} \end{equation} \begin{equation} = iS_{\alpha\beta}^{+}(x - y) = \frac{1}{2(2\pi)^{3}}\int (slp + m)\frac{e^{ip(x - y)}}{E}d^{3}p = \frac{- i}{(2\pi)^{4}}\int_{c^{+}}\frac{(slp + m)e^{-ip(x - y)}}{p^{2} - m^{2}}d^{4}p \end{equation} for the virtual spin $ 1/2 $ particle feynman propagator were\ \begin{equation} iS_{F}(x-y) = \langle 0 \vert T \lbrace \psi(x)b\psi(y)\rbrace \vert 0 \rangle = - \langle 0 \vert \lbrace b\psi(x)\psi(y)\rbrace \vert 0 \rangle \end{equation} if $ t_{y} < t_{x} $ ( antiparticle) \begin{equation} = \langle 0 \vert[b\psi^{+}(x), \psi^{-}(y) ]_{+}\vert 0 \rangle \end{equation} \begin{equation} = [b\psi^{+}(x), \psi^{-}(y)]_{+}\langle 0 \vert \vert 0 \rangle = [ b\psi^{+}(x), \psi^{-}(y)]_{+} \end{equation} \begin{equation} = iS^{-}(x - y) = -\frac{1}{2(2\pi)^{3}}\int (slp - m)\frac{e^{ip(x - y)}}{E}d^{3}p = \frac{ i}{(2\pi)^{4}}\int_{c^{-}}\frac{(slp + m)e^{-ip(x - y)}}{p^{2} - m^{2}}d^{4}p \end{equation} \ The two contour integrals in the last lines () and () were combined in final step to yield the single integral over real space to get the final result for \textit{\textbf{the spinor Feynman propagator }} \begin{equation} S_{F}(x - y) = \int_{-\infty}^{+\infty}\frac{ d^{4}p}{(2\pi)^{4}}\frac{(slp + m)e^{-ip(x - y)}}{p^{2} - m^{2} + i\varepsilon} \end{equation} The momentum space form of the propagator ( its fourier transform, is \begin{equation} S_{F}(p) = \frac{slp + m}{p^{2} - m^{2} + i\varepsilon} = (slp + m)\Delta_{F}(p) \end{equation} Observe that: \begin{equation} (slp - m)(slp + m) = \gamma^{\mu}\gamma^{\nu}p_{\mu}p_{\nu} - m^{2} = p^{2} - m^{2} \end{equation} Then we can multiply by $ (slp + m) $ both the numerator and the denominator in eq. and rewrite $ S_{F}(p) $ in the form, \begin{equation} S_{F}(p) = \frac{i}{slp - m} \end{equation}
Take the full answer it is frommy lecture to QFT. \subsection{Boson propagator} It is prefer to write the collision between to particles in term of \textit{amplitude} of \textit{probability}. The perturbative approximation of QFT assumed that the particles propagate freely except some points, when there are emission or absorption of quanta. we write the solution of motion aquations compled as pertirbative series around of free solution of motion equations of free field. The methode uses Green's function which R.Feynman gave his probabilist interpritation of implitude. Motion equation of free boson(Kein-Gordon equation) is writing:\ \begin{equation} \left( p^{2}-m^{2}\right) \varphi(p)=0 \end{equation} Where $\varphi(p)$ is a scalar function.\ Green's function $ G(p) $, in the space of momentum is: \begin{equation} (p^{2}- m^{2})G(p)=\delta^{4}(p) \end{equation} Then $ G(p)=\frac{\delta^{4}(p)}{p^{2}-m^{2}} $, $ \delta^{4} $ is the Dirac function defined as \begin{equation} \delta^{4}(p)=\delta(p_0)\delta(p_1)\delta(p_2)\delta(p_3) \end{equation} Feynman interpritation is that this operator is as amplitude of probability that the boson propagates with quadri-momentume. Propagator = $ \frac{i}{p^{2}-m^{2}} $. In same way, feynman defined an amplitude of probability that the boson whether emitted ( or absorbed) by particle 1, and/or absorbe by particle 2 of interactions.\ The amplitude are driveded for various kinds pf interactions between various particle, the square of the magnitude of each amplitude turns out be tje probability of that particular interaction (transition) occuring. These transition amplitudes each depend on the initial real particles, the final real particles, and the virtial particles that mediate the transition. It turns out that the factor in the amplitude representing the virtual particle contribution is identical to the feynman propagator $ \Delta_F $.\ \ \begin{equation} \Delta(x,y) = \int\frac{d^{4}k}{(2\pi)^{4}} \exp^{-ik(x-y)} \frac{1}{k^{2}-m^{2}+i\varepsilon} \\ \end{equation} \section{The spenor feynman Propagator} We will follow similar steps to drive feynman propagator for Dirac fields as we did for scalar feild , similar to what we did scalars, we will, heuristical, consider the operator field $ b\psi $ to creat a virtual particle at event y, and $ \psi $ to destroy that virtual particle particle at even x. The spinor propagator incorporates this two field operators. The propagator realy corresponds to a kind of probability density function in y and x. It represents the probability density of Dirac particle appearing at y and disapiaring at x. we show that in briefly:\ for the virtual spin $ 1/2 $ particle feynman propagator were\ \begin{equation} iS_{F}(x-y) = \langle 0 \vert T\lbrace \psi(x)\bar\psi(y)\rbrace \vert 0 \rangle = \langle 0 \vert \lbrace \psi(x)\bar\psi(y)\rbrace \vert 0 \rangle \end{equation} if $ t_{y} < t_{x} $ ( particle) \begin{equation} = \langle 0 \vert[\psi^{+}(x), \bar\psi^{-}(y) ]_{+}\vert 0 \rangle \end{equation} \begin{equation} = [\psi^{+}(x), \bar\psi^{-}(y)]_{+}\langle 0 \vert \vert 0 \rangle = [ \psi^{+}(x), \bar\psi^{-}(y)]_{+} \end{equation} \begin{equation} = iS_{\alpha\beta}^{+}(x - y) = \frac{1}{2(2\pi)^{3}}\int (\not p + m)\frac{e^{ip(x - y)}}{E}d^{3}p = \frac{- i}{(2\pi)^{4}}\int_{c^{+}}\frac{(\not p + m)e^{-ip(x - y)}}{p^{2} - m^{2}}d^{4}p \end{equation} for the virtual spin $ 1/2 $ particle feynman propagator were\ \begin{equation} iS_{F}(x-y) = \langle 0 \vert T \lbrace \psi(x)b\psi(y)\rbrace \vert 0 \rangle = - \langle 0 \vert \lbrace \bar\psi(x)\psi(y)\rbrace \vert 0 \rangle \end{equation} if $ t_{y} < t_{x} $ ( antiparticle) \begin{equation} = \langle 0 \vert[\bar\psi^{+}(x), \psi^{-}(y) ]_{+}\vert 0 \rangle \end{equation} \begin{equation} = [\bar\psi^{+}(x), \psi^{-}(y)]_{+}\langle 0 \vert \vert 0 \rangle = [ \bar\psi^{+}(x), \psi^{-}(y)]_{+} \end{equation} \begin{equation} = iS^{-}(x - y) = -\frac{1}{2(2\pi)^{3}}\int (\not p - m)\frac{e^{ip(x - y)}}{E}d^{3}p = \frac{ i}{(2\pi)^{4}}\int_{c^{-}}\frac{(\not p + m)e^{-ip(x - y)}}{p^{2} - m^{2}}d^{4}p \end{equation} \ The two contour integrals in the last lines () and () were combined in final step to yield the single integral over real space to get the final result for \textit{\textbf{the spinor Feynman propagator }} \begin{equation} S_{F}(x - y) = \int_{-\infty}^{+\infty}\frac{ d^{4}p}{(2\pi)^{4}}\frac{(\not p + m)e^{-ip(x - y)}}{p^{2} - m^{2} + i\varepsilon} \end{equation} The momentum space form of the propagator ( its fourier transform, is \begin{equation} S_{F}(p) = \frac{\not p+ m}{p^{2} - m^{2} + i\varepsilon} = (\not p + m)\Delta_{F}(p) \end{equation} Observe that: \begin{equation} (\not p - m)(\not p + m) = \gamma^{\mu}\gamma^{\nu}p_{\mu}p_{\nu} - m^{2} = p^{2} - m^{2} \end{equation} Then we can multiply by $ (\not p + m) $ both the numerator and the denominator in eq. and rewrite $ S_{F}(p) $ in the form, \begin{equation} S_{F}(p) = \frac{i}{\not p - m} \end{equation}
Take the full answer it is frommy lecture to QFT. \subsection{Boson propagator} It is prefer to write the collision between to particles in term of \textit{amplitude} of \textit{probability}. The perturbative approximation of QFT assumed that the particles propagate freely except some points, when there are emission or absorption of quanta. we write the solution of motion aquations compled as pertirbative series around of free solution of motion equations of free field. The methode uses Green's function which R.Feynman gave his probabilist interpritation of implitude. Motion equation of free boson(Kein-Gordon equation) is writing:\ \begin{equation} \left( p^{2}-m^{2}\right) \varphi(p)=0 \end{equation} Where $\varphi(p)$ is a scalar function.\ Green's function $ G(p) $, in the space of momentum is: \begin{equation} (p^{2}- m^{2})G(p)=\delta^{4}(p) \end{equation} Then $ G(p)=\frac{\delta^{4}(p)}{p^{2}-m^{2}} $, $ \delta^{4} $ is the Dirac function defined as \begin{equation} \delta^{4}(p)=\delta(p_0)\delta(p_1)\delta(p_2)\delta(p_3) \end{equation} Feynman interpritation is that this operator is as amplitude of probability that the boson propagates with quadri-momentume. Propagator = $ \frac{i}{p^{2}-m^{2}} $. In same way, feynman defined an amplitude of probability that the boson whether emitted ( or absorbed) by particle 1, and/or absorbe by particle 2 of interactions.\ The amplitude are driveded for various kinds pf interactions between various particle, the square of the magnitude of each amplitude turns out be tje probability of that particular interaction (transition) occuring. These transition amplitudes each depend on the initial real particles, the final real particles, and the virtial particles that mediate the transition. It turns out that the factor in the amplitude representing the virtual particle contribution is identical to the feynman propagator $ \Delta_F $.\ \ \begin{equation} \Delta(x,y) = \int\frac{d^{4}k}{(2\pi)^{4}} \exp^{-ik(x-y)} \frac{1}{k^{2}-m^{2}+i\varepsilon} \\ \end{equation} \section{The spenor feynman Propagator} We will follow similar steps to drive feynman propagator for Dirac fields as we did for scalar feild , similar to what we did scalars, we will, heuristical, consider the operator field $ b\psi $ to creat a virtual particle at event y, and $ \psi $ to destroy that virtual particle particle at even x. The spinor propagator incorporates this two field operators. The propagator realy corresponds to a kind of probability density function in y and x. It represents the probability density of Dirac particle appearing at y and disapiaring at x. we show that in briefly:\ for the virtual spin $ 1/2 $ particle feynman propagator were\ \begin{equation} iS_{F}(x-y) = \langle 0 \vert T\lbrace \psi(x)b\psi(y)\rbrace \vert 0 \rangle = \langle 0 \vert \lbrace \psi(x)b\psi(y)\rbrace \vert 0 \rangle \end{equation} if $ t_{y} < t_{x} $ ( particle) \begin{equation} = \langle 0 \vert[\psi^{+}(x), b\psi^{-}(y) ]_{+}\vert 0 \rangle \end{equation} \begin{equation} = [\psi^{+}(x), b\psi^{-}(y)]_{+}\langle 0 \vert \vert 0 \rangle = [ \psi^{+}(x), b\psi^{-}(y)]_{+} \end{equation} \begin{equation} = iS_{\alpha\beta}^{+}(x - y) = \frac{1}{2(2\pi)^{3}}\int (slp + m)\frac{e^{ip(x - y)}}{E}d^{3}p = \frac{- i}{(2\pi)^{4}}\int_{c^{+}}\frac{(slp + m)e^{-ip(x - y)}}{p^{2} - m^{2}}d^{4}p \end{equation} for the virtual spin $ 1/2 $ particle feynman propagator were\ \begin{equation} iS_{F}(x-y) = \langle 0 \vert T \lbrace \psi(x)b\psi(y)\rbrace \vert 0 \rangle = - \langle 0 \vert \lbrace b\psi(x)\psi(y)\rbrace \vert 0 \rangle \end{equation} if $ t_{y} < t_{x} $ ( antiparticle) \begin{equation} = \langle 0 \vert[b\psi^{+}(x), \psi^{-}(y) ]_{+}\vert 0 \rangle \end{equation} \begin{equation} = [b\psi^{+}(x), \psi^{-}(y)]_{+}\langle 0 \vert \vert 0 \rangle = [ b\psi^{+}(x), \psi^{-}(y)]_{+} \end{equation} \begin{equation} = iS^{-}(x - y) = -\frac{1}{2(2\pi)^{3}}\int (slp - m)\frac{e^{ip(x - y)}}{E}d^{3}p = \frac{ i}{(2\pi)^{4}}\int_{c^{-}}\frac{(slp + m)e^{-ip(x - y)}}{p^{2} - m^{2}}d^{4}p \end{equation} \ The two contour integrals in the last lines () and () were combined in final step to yield the single integral over real space to get the final result for \textit{\textbf{the spinor Feynman propagator }} \begin{equation} S_{F}(x - y) = \int_{-\infty}^{+\infty}\frac{ d^{4}p}{(2\pi)^{4}}\frac{(slp + m)e^{-ip(x - y)}}{p^{2} - m^{2} + i\varepsilon} \end{equation} The momentum space form of the propagator ( its fourier transform, is \begin{equation} S_{F}(p) = \frac{slp + m}{p^{2} - m^{2} + i\varepsilon} = (slp + m)\Delta_{F}(p) \end{equation} Observe that: \begin{equation} (slp - m)(slp + m) = \gamma^{\mu}\gamma^{\nu}p_{\mu}p_{\nu} - m^{2} = p^{2} - m^{2} \end{equation} Then we can multiply by $ (slp + m) $ both the numerator and the denominator in eq. and rewrite $ S_{F}(p) $ in the form, \begin{equation} S_{F}(p) = \frac{i}{slp - m} \end{equation}