# Propagator from Path integral

In class we have proved something like:

$$\frac{\partial^2 Z(J,\bar{J})}{\partial J(x) \partial \bar{J}(x')}\frac{1}{Z}|_{J=\bar{J}=0}=\Delta(x-x').$$

That by introducing source terms to path integral we can simply recover Feynman propagator. But how do we actually recover Feynman propagator in practice? We have:

$$Z(J, \bar{J})=\int {\cal D} \bar{\psi} {\cal D} \psi \exp \left[ i \int \,d^4x \bar{\psi} ( i \gamma_\mu D^\mu - m ) \psi + J\psi + \bar{J}\bar{\psi} \right].$$

Giving us something like (I think):

$$\frac{\partial^2 Z(J,\bar{J})}{\partial J(x) \partial \bar{J}(x')}|_{J=0}=\int {\cal D} \bar{\psi} {\cal D} \psi( \bar{\psi}\psi)\exp \left[ i \int \,d^4x \bar{\psi} ( i \gamma_\mu D^\mu - m ) \psi \right] = \det( i \gamma_\mu D^\mu - m)( i \gamma_\mu D^\mu - m)^{-1}.$$

We divide both sides by $Z$ so determinant disappears, but how to go from the last line to old good Feynman propagator for Diracs field?

• According to the page you linked to, the Feynman propagator for Dirac fields is $\tilde{S}_F(p) = \frac{1}{\gamma_\mu p^\mu-m}$. This is the same as your result upon making the identification $p_\mu = -i D_\mu$ and taking into account the different sign convention. – tparker May 23 '16 at 7:28
• Actually Qmechanic added the link. But my expression is in position space, if I transform quantity such as $\frac{1}{( i \gamma_\mu D^\mu - m)}$ to momentum space, does the trivial expressions $\partial_\mu -> ip_\mu$ still applies? From the first impression it seems that FT of something in denominator is non-trivial? – Wagm May 23 '16 at 18:11
• I'll expand my comment into an answer to address that – tparker May 23 '16 at 19:08

You're right that the Feynman propagator for a spinor field is indeed $(i \gamma^\mu D_\mu - m)^{-1}$. The tricky part is interpreting exactly what the "inverse" means. It doesn't just mean that you invert the gamma matrices (although you do do that). The derivative operator is also being inverted. That is, if we define a function $G(y, x) := (i \gamma^\mu D_\mu - m)^{-1}$, then
$$\int d^4x\, G(y, x)\, f(x) = g(y) \implies (i \gamma^\mu D_\mu - m) g(x) = f(x).$$
It may help to think of $G(x, y)$ as being like a matrix (since it has two "indices" $x$ and $y$) and $f(x)$ as being like a vector (since it has one "index" $x$). Then the integration over $x$ is like matrix multiplication, and $G(y, x)$ is like the matrix inverse of the infinite-dimensional (not 4x4) "matrix" $(i \gamma^\mu D_\mu - m)$.
From this, it shouldn't be too hard to see that $$(i \gamma^\mu D_\mu - m)\, G(x, y) = \delta^4(x - y). \qquad \qquad (1)$$ So when we say that $G$ is the "inverse" of $(i \gamma^\mu D_\mu - m)$, we mean that it is the Green's function for the differential operator, not just the matrix inverse of the gamma matrices. We now use translational invariance to realize that $G(x, y)$ actually only depends on the single argument $x - y$. If we Fourier transform both sides, the integral over $x$ turns this into a convolution of $(i \gamma^\mu D_\mu - m)$ and $G$. The Fourier transform of a convolution is simply the product of the Fourier transforms, so the Fourier transform is easy: $$(-\gamma^\mu p_\mu - m)\, \tilde{G}(k) = 1$$ (where $\tilde{G}(k)$ is now just a 4x4 matrix, the "1" on the RHS is the 4x4 identity matrix, and the LHS just involves 4x4 matrix multiplication over the spinor indices.) So we finally get $$\tilde{G}(k) = (-\gamma^\mu p_\mu - m)^{-1},$$ where the inverse now just means 4x4 matrix inversion.