On P&S's QFT book, chapter 9.5, the book discussed how to derive two point correlation function for dirac field using generating functional.

Start with $$ Z[\bar{\eta}, \eta]=\int \mathcal{D} \bar{\psi} \mathcal{D} \psi \exp \left[i \int d^4 x[\bar{\psi}(i \not \partial-m) \psi+\bar{\eta} \psi+\bar{\psi} \eta]\right] \tag{9.73} $$ we can shift the field as: $$\begin{aligned}\psi &\leadsto \psi-i \int d^4 y S_F(x-y) \eta(y)\\ \bar{\psi} &\leadsto \bar{\psi}-i \int d^4 y \bar{\eta}(y) S_F(x-y) \end{aligned}$$

I met following term in the derivation: $$\int d^4 x d^4 y \bar{\eta}(y)S_F(x-y)(i\not\partial-m)\psi^{\prime}(x) $$

So how to handle this term? I know that $(i\not\partial-m)S_F(x-y)=i\delta^4(x-y)$, but this time the differential operator appear on the right side of $S_F(x-y)$.

In my attempt of derivation:

(1) Should I write them in components form? (i.e. $S_F(x-y)_{ab}$)

(2) If not, in the Integration by part, how do we handle $i\not\partial-m$? Since there is gamma matrices.

Another point is that I fail to shift the field by directly using $\psi^{\prime}\equiv \psi+(i\not\partial-m)^{-1}\eta$, I summarized this question in this post.


EDIT: $\stackrel{\star}{=}$ expansion in J.G.'s excellent answer! $$\begin{aligned}&\int d^4y \overline{\eta(y)}S_F(x-y)(i\not\partial_x-m)\psi(x) \\ &=\text{Surface}-\int d^4y \overline{\eta(y)}(i\not\partial_x+m)S_F(x-y)\psi(x)\\ &=\int d^4y \overline{\eta(y)}(i\not\partial_y-m)S_F(x-y)\psi(x) \\ &=\int d^4y \overline{\eta(y)}(i\not\partial_y-m)S_F(y-x)\psi(x)\end{aligned}$$

  • $\begingroup$ General tip: Consider to only ask 1 question per post. $\endgroup$
    – Qmechanic
    Nov 26, 2022 at 8:06
  • $\begingroup$ In your edit, what do you mean by $\psi'$? My answer's strategy was to replace $\overline{\eta}S_F\not\partial_x\psi$ [sic] with $-\overline{\eta}\not\partial_xS_F\psi$ by IBP, then with $\overline{\eta}\not\partial_yS_F\psi$ since $S_F$ is a function of $x-y$. $\endgroup$
    – J.G.
    Dec 4, 2022 at 11:08
  • $\begingroup$ @J.G. Thanks! Sorry for the notation, my $\psi^{\prime}$ is same with your $\psi$! That's clear! So we don't need to consider the gamma matrix commutation problem in $\not\partial$ and $S_F$? I re-edit my post, sorry for confusion! $\endgroup$
    – Daren
    Dec 4, 2022 at 13:59
  • $\begingroup$ @J.G. Sorry for bothering, I don't totally understand the last step in my new EDIT! $S_F(x-y)=\int\frac{d^4 k}{(2\pi)^4}\frac{i(\not k+m)}{k^2-m^2+i\epsilon}e^{-ik(x-y)}$, why could $S_F(x-y)$ invariant under $x \leftrightarrow y$? $\endgroup$
    – Daren
    Dec 4, 2022 at 14:15

1 Answer 1


(1) is overkill. Write the shift as $\psi\leadsto\psi+\delta\psi$ etc. Since the operators $\delta,\,i\not\partial-m$ commute and $(i\not\partial-m)\delta\psi(x)=\eta(x)$,$$\begin{align}\delta\left(\overline{\psi}(x)(i\not\partial-m)\psi(x)\right)&=\overline{\psi}(x)(i\not\partial-m)\delta\psi(x)+\delta\overline{\psi}(x)(i\not\partial-m)\psi(x)\\&=\overline{\psi}(x)\eta(x)-i\int d^4y\overline{\eta}(y)S_F(x-y)(i\not\partial-m)\psi(x)\\&\stackrel{\star}{=}\overline{\psi}(x)\eta(x)-i\int d^4y\overline{\eta}(y)(i\not\partial-m)S_F(x-y)\psi(x)\\&=\overline{\psi}(x)\eta(x)+\overline{\eta}(x)\psi(x),\end{align}$$where $\stackrel{\star}{=}$ assumes integration by parts with a vanishing surface term. Note for (2) that $S_F(x-y)$ is invariant under $x\leftrightarrow y$, so $\not\partial_yS_F=-\not\partial_xS_F$, which cancels IBP's usual sign change. (This also addresses another question of yours.) Hence$$Z=\int\mathcal{D}\overline{\psi}\mathcal{D}\psi\exp\left[i\int d^4x[f+\delta f]\right],\,f:=\overline{\psi}(x)(i\not\partial-m)\psi(x).$$

  • $\begingroup$ Excellent answer! Thank you very much! May I ask additional two questions? (1) Could you please check my EDIT in my post about your star formula? This also require $\partial_y \overline{\eta(y)}=0$ after the IBP, so is this true? (2) Since your derivation don't write out the spinor components explicitly, and $\gamma^{\mu}$ in $\not\partial$ may not commute with $S_F(x-y)$! (in the IBP, we exchange their order) So can we ensure us this? Thanks! $\endgroup$
    – Daren
    Dec 4, 2022 at 9:07
  • $\begingroup$ If you have time, could you please see my another post(link) above EDIT! That's about spinor field replacement! But any way, you already helped me a lot, I am really appreciate your help! $\endgroup$
    – Daren
    Dec 4, 2022 at 9:11
  • $\begingroup$ Sorry, I need un-accept your answer! Since I don’t understand the first step before IBP. So in the integration of $d^4 y$, how can we handle $\partial_x$? Thanks! $\endgroup$
    – Daren
    Dec 4, 2022 at 15:41

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