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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?

Qmechanic
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Daren
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1 Answers1

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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.
Jeanbaptiste Roux
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  • Thank you very much, that's make sense! But it seems that there is a minus sign difference! Would the $[-i(i\not\partial-m)]^{-1}$ in $k-$space equal to $\frac{i}{\not p-m}$? – Daren Nov 20 '22 at 10:33
  • It is possible that I've made a sign error, however, I don't see where right now. All I can say is that $[-i(i\not \partial-m)]^{-1}=i (i \not \partial -m)^{-1}$, hence I don't see where is my mistake... – Jeanbaptiste Roux Nov 20 '22 at 10:42
  • Oh, I see, you are right, thank you very much! I am really appreciate your help! – Daren Nov 20 '22 at 12:12
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    May I add another comment to help my understanding? Could I think $(i\not\partial_x-m)[i(i\not\partial-m){x,y}^{-1}]=i\delta(x-y)$, so we regard $[i(i\not\partial-m){x,y}^{-1}]$ as the Green function of operator $(i\not\partial_x-m)$? – Daren Nov 20 '22 at 12:57
  • @Daren Yes this absolutely the purpose of the manipulation I've done. But you have to be careful to not confuse Green's function and Feynman's propagator, there are a few differences as per Wikipedia – Jeanbaptiste Roux Nov 20 '22 at 13:31
  • I see, thank you very much, thanks for your help again! – Daren Nov 20 '22 at 14:07
  • Hello, I find that I still don't totally understand it, could you please take a look of my this new post ?https://physics.stackexchange.com/questions/737686/spinor-field-path-quantization? – Daren Nov 21 '22 at 15:12
  • Jeanbaptiste, can I interest you in making this mathematics understandable to me? https://physics.stackexchange.com/a/692833 – HolgerFiedler Dec 03 '22 at 03:56