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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.

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

Daren
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1 Answers1

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(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$$Z=\int\mathcal{D}\overline{\psi}\mathcal{D}\psi\exp\left[i\int d^4x[f+\delta f]\right],\,f:=\overline{\psi}(x)(i\not\partial-m)\psi(x).$$

J.G.
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