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:

\begin{equation} 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]. \end{equation}

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?

  • $\begingroup$ 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. $\endgroup$
    – tparker
    Commented May 23, 2016 at 7:28
  • $\begingroup$ 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? $\endgroup$
    – Wagm
    Commented May 23, 2016 at 18:11
  • $\begingroup$ I'll expand my comment into an answer to address that $\endgroup$
    – tparker
    Commented May 23, 2016 at 19:08

1 Answer 1


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.

The TLDR version is this: Fourier transforming a differential expression simply turns it into a product in momentum space of the FT of the differential operator and the FT of the function being differentiated. And products are much easier to invert (you just do division) than differential operators. Trying to "invert" a differential operator (i.e. find its Green's function) is usually very hard in real space, but in momentum space it becomes trivial: you just take the reciprocal of the Fourier-transformed differential operator. So basically yes, to FT the inverse of an operator you can just naively FT the operator and then stick that FT into the denominator.


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