Fermion propagator as derivative of scalar propagator

Question

I've seen this expression in two spacetime dimensions,

$$ \langle \bar{\psi}(x) \psi(0) \rangle = \gamma^\mu{\partial_\mu} \langle \phi(x) \phi(0) \rangle $$

The LHS is the fermion propagator, and the expectation on RHS is the scalar propagator. For 2 dimensional case, the scalar propagator is (assuming all massless)

$$ \langle \phi(x) \phi(0) \rangle = \int \frac{d^2p}{4\pi^2} \frac{1}{p^2} e^{-ipx} $$

Two questions:

1. Why the fermion propagator is derivative of scalar propagator?
2. How are the gamma matrices defined in two dimensions?

Answer 1

The free scalar and fermion propagator is $$ G_\psi(x,y) = \int \frac{d^dp}{(2\pi)^d} \frac{-i(\gamma^\mu p_\mu + m)}{ p^2 + m^2 - i \epsilon} e^{- i p \cdot ( x - y ) } $$ The scalar propagator is $$ G_\phi(x,y) = \int \frac{d^dp}{(2\pi)^d} \frac{-i}{ p^2 + m^2 - i \epsilon} e^{- i p \cdot ( x - y ) } $$ Clearly, $$ G_\psi(x,y) = ( i \gamma^\mu \partial_\mu -m)G_\phi(x,y)~. $$

PS - In any dimension, the gamma matrices are defined to satisfy $\{ \gamma^\mu , \gamma^\nu \} = - 2 \eta^{\mu\nu}$.

PPS - I am using metric signature $(-,+,+,+,\cdots)$ in this answer.

Answer 2

\subsection{Spenor feynman Propagator} 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}\\ slp = \gamma p \end{equation}

Answer 3

This answer is from my lecture of QFT That i prepared i wish you find what you want, let me know. 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} we can readily write down the 4-momentum space form of the propagator, the Fourier transform of (3-30), which will be very useful \begin{equation} \triangle_{F}(k)= \frac{1}{k^{2}-(m^{2} + i\varepsilon)} \end{equation}

Answer 4

The $ \Gamma $ matrices can be constructed recursively, d = 2 Using the Pauli matrices, $ \gamma _{0}=\sigma _{1}$, $\gamma _{1}=-i\sigma _{2}$ where the two gamma matrices can be given by $ \Gamma^{1} = $
$ \begin{pmatrix} 1& 0\\ 0 & -1 \\ \end{pmatrix} $
and $ \Gamma^{2} = $
$ \begin{pmatrix} 0 & 1\\ 1 & 0 \\ \end{pmatrix} $
you can see this document " Jeong-Hyuck Park, Lecture note on Clifford algebra "

Answer 5

Sorry your question is strange for me gamma matrices defined in two dimensions!!!! where you find this subject?

Pauli spin matrices \begin{equation} \sigma^{i} = (\sigma^{1}, \sigma^{2}, \sigma^{3}) , where:\\ \end{equation}

$ \sigma^{1} = $
$ \begin{pmatrix} 0 & 1\\ 1 & 0 \\ \end{pmatrix} $
$ \sigma^{2} = $ $ \begin{pmatrix} 0 & - i\\ i & 0\\ \end{pmatrix} $ $ \sigma^{3} = $ $ \begin{pmatrix} 1 & 0\\ 0 & - 1\\ \end{pmatrix} $ $ \sigma^{0} = $ $ \begin{pmatrix} 1 & 0\\ 0 & 1\\ \end{pmatrix} $ \ \

$ I^{0} = $ $ \begin{pmatrix} \sigma^{0} & \textbf{0}\\ \textbf{0} & \sigma^{0}\\ \end{pmatrix} $ $ \textbf{0} = $ $ \begin{pmatrix} 0 & 0\\ 0 & 0\\ \end{pmatrix} $ \ the matrices $ \gamma_{\mu} $ conscript \textit{\textbf{Dirac matrices}}, be written: $ \gamma_{\mu} $ = $ \begin{pmatrix} 0 & -i\sigma_{i}\\ i\sigma_{i} & 0\\ \end{pmatrix} $ $ i = 1, 2, 3 $ $ \gamma_{0} $ = $ \begin{pmatrix} I & 0\\ 0 & - I\\ \end{pmatrix} $ \ \ I think that you mean this solution , but it is does not mean two dimensions as i think: We have written the Dirac matrices in blocks of $ 2\times2 $ matrices, and it is natural to write similarly the four-component Dirac field as a pair of two-component fields:\

$ \Psi = $ $ \begin{pmatrix} \Psi_{L}\\ \Psi_{R}\\ \end{pmatrix} $ $ = $ $ \begin{pmatrix} \Psi_{L}\\ 0\\ \end{pmatrix} $ $ + $ $ \begin{pmatrix} 0\\ \Psi_{R}\\ \end{pmatrix} $ \ where $ \Psi_{L} $ and $ \Psi_{R} $ are, respectively, the top and bottom two components of the four-component Dirac field:\ $ \Psi_{L} = $ $ \begin{pmatrix} \psi_{1}\\ \psi_{2}\\ \end{pmatrix} $ and $ \Psi_{R} = $ $ \begin{pmatrix} \psi_{3}\\ \psi_{4}\\ \end{pmatrix} $ \ The Dirac equation (5.2) becomes:\ $ i $ $ \begin{pmatrix} \sigma^{0} & 0\\ 0 & \sigma^{0}\\ \end{pmatrix} $ $ \begin{pmatrix} \partial_{0}\Psi_{L}\\ \partial_{0}\Psi_{R}\\ \end{pmatrix} $ $ + i $ $ \begin{pmatrix} - \sigma^{i} & 0\\ 0 & \sigma^{i}\\ \end{pmatrix} $ $ \begin{pmatrix} \partial_{i}\Psi_{L}\\ \partial_{i}\Psi_{R}\\ \end{pmatrix} $ $ -m $ $ \begin{pmatrix} 0 & \sigma^{0}\\ \sigma^{0} & 0\\ \end{pmatrix} $ $ \begin{pmatrix} \Psi_{L}\\ \Psi_{R}\\ \end{pmatrix} $ $ = 0 $ \ Block multiplication then gives two coupled equations for $ \Psi_{L} $ and $ \Psi_{R} $:\ \begin{equation} i\sigma^{0}\partial_{0}\Psi_{L} - i\sigma^{i}\partial_{i}\Psi_{L} - m\Psi_{R} = 0 \end{equation} \begin{equation} i\sigma^{0}\partial_{0}\Psi_{R} + i\sigma^{i}\partial_{i}\psi_{R} - m\Psi_{L} = 0 \end{equation} the wave function now is bi-speror with possessed four components:\ $ \Psi $ = $ \begin{pmatrix} \psi_{1}\\ \psi_{2}\\ \psi_{3}\\ \psi_{4}\\ \end{pmatrix} $