P&S QFT derivation of (9.72) (correlation function of spinor)

Question

On P&S QFT book page 302, the book considered spinor two point correlation function derivation, begin with \begin{equation} \left\langle 0\left|T \psi\left(x_1\right) \bar{\psi}\left(x_2\right)\right| 0\right\rangle=\frac{\int \mathcal{D} \bar{\psi} \mathcal{D} \psi \exp \left[i \int d^4 x \bar{\psi}(i \not \partial-m) \psi\right] \psi\left(x_1\right) \bar{\psi}\left(x_2\right)}{\int \mathcal{D} \bar{\psi} \mathcal{D} \psi \exp \left[i \int d^4 x \bar{\psi}(i \not \partial-m) \psi\right]} \tag{A} \end{equation} then $$\left\langle 0\left|T \psi\left(x_1\right) \bar{\psi}\left(x_2\right)\right| 0\right\rangle=\frac{\text{det}[-i(i\not\partial-m)][-i(i\not\partial-m)]^{-1}_{x_1,x_2}}{\text{det}[-i(i\not\partial-m)]}=[-i(i\not\partial-m)]^{-1}_{x_1,x_2} \tag{B}$$ define Green's Function: $$(i\not\partial-m)S_F(x-y)=i\delta^4 (x-y) \tag{C}$$ multiply each side with $(i\not\partial-m)^{-1}$, then $$S_F(x-y)=-\delta^4 (x-y)[-i(i\not\partial-m)]^{-1} \tag{D}$$ while from (9.72), $$\left\langle 0\left|T \psi\left(x_1\right) \bar{\psi}\left(x_2\right)\right| 0\right\rangle=S_F\left(x_1-x_2\right)=\int \frac{d^4 k}{(2 \pi)^4} \frac{i e^{-i k \cdot\left(x_1-x_2\right)}}{\not k-m+i \epsilon} \tag{9.72}$$

So I am confused:

(1) It seems there is an additional $-\delta^4(x-y)$ in D?

(2) How does the indices $x_1$ and $x_2$ contracted?

Answer 1

Indeed, you can't just "multiply both sides" by the Green's function. You have to proceed in another manner, like so:

1. Take the Fourier transform of $(C)$: \begin{equation} (i\not\partial-m)S_F=i\delta \Longrightarrow (\not p-m)\tilde{S}_F=i \end{equation}
2. Multiply both sides by $(\not p-m)^{-1}$ and take the inverse Fourier transform: \begin{equation} \tilde{S}_F = \frac{i}{\not p-m} \Longrightarrow S_F(x-y)=\int \frac{d^4p}{(2\pi)^4} \frac{ie^{-i p(x-y)}}{\not p-m} \end{equation} And voilà.