This is from chapter 14.7.1 in Schwartz's QFT book. I am trying to derive contact terms starting from field redefinition $\phi\rightarrow\phi(x)+\epsilon(x)$. For the 1-point function we have from definition of generating functionals, $$ ⟨\hat{\phi}(x)⟩ \ = \ -i \frac{1}{Z[J=0]} \frac{\partial{Z[J]}}{\partial J(x)} \ \text{at J=0} \ = \ \frac{1}{Z[J=0]} \int \mathcal{D}\phi e^{i\int d^{4}y(-\frac{1}{2}\phi\Box_{y}\phi)}\phi(x)$$ then we proceed to shift the field like stated above and end up with $$⟨\hat{\phi}(x)⟩ \ = \frac{1}{Z[J=0]} \int \mathcal{D}\phi e^{i\int d^{4}y(-\frac{1}{2}(\phi+\epsilon)\Box_{y}(\phi+\epsilon))}[\phi(x)+\epsilon(x)]$$ which I expand out to $$⟨\hat{\phi}(x)⟩ \ = \frac{1}{Z[J=0]} \int \mathcal{D}\phi e^{i\int d^{4}y(-\frac{1}{2}\phi\Box_{y}\phi)}e^{i\int d^{4}z(-\frac{1}{2}[\phi\Box_{z}\epsilon + \epsilon\Box_{z}\phi])}[\phi(x)+\epsilon(x)]$$ which can then be written to first order in $\epsilon$ as $$ \frac{1}{Z[J=0]} \int \mathcal{D}\phi e^{iS[\phi]}\{ \frac{-i}{2}(\phi(x)+\epsilon(x)) + \frac{-i}{2}\phi(x)\int d^{4}z (\phi \Box_{z}\epsilon +\epsilon\Box_{z}\phi) + \frac{i}{2} \epsilon(x) \int d^{4}z (\phi \Box_{z}\epsilon +\epsilon\Box_{z}\phi)\} .$$ This is where Schwartz states "integrated by parts to combine $\epsilon\Box\phi \ \text{and} \ \phi\Box\epsilon$ terms" and arrives at eq. (14.109) which is $$ ⟨\hat{\phi}(x)⟩ \ = \frac{1}{Z[J=0]} \int \mathcal{D}\phi e^{i\int d^{4}y(-\frac{1}{2}\phi\Box_{y}\phi)} \{\phi(x) + \ \epsilon(x) -i\phi(x)\int d^{4}z \ \epsilon(z)\Box_{z} \phi(z)\} . $$ and for the life of me I can't figure out why I have extra factors of $i/2$ and how actually the path integral works out with the d'Alembertian operator. Any help would be greatly appreciated.


1 Answer 1


Let us clarify these two issues first. The expansion of the exponential is: $$e^x = 1 + x + \mathcal{O}(x^2)\tag{1}\label{eq:exponential}$$ Second let us remind ourselves about integration by parts: $$\int \mathrm{d}^4\, x\; g(x)\square f(x) = \int \mathrm{d}^4\,x\; g(x)\partial_\mu\partial^\mu f = \int_{\partial\mathbb{R}^4} g(x)\partial^\mu f(x) \mathrm{d}\mathbf{n}_\mu -\int \mathrm{d}^4x\; \partial_\mu g(x)\partial^\mu f(x)\tag{2}\label{eq:intParts}$$ As it is common in this topic, fields are assume to vanish at infinity fast enough that boundary terms don't contribute, meaning the first term of the RHS is neglected. Now we can use \eqref{eq:intParts} two times, to place all the differential operators to act on $\phi$ instead of $\epsilon$, so that: $$-\frac{1}{2}\int\mathrm{d}^4 x\;\phi(x)\square\epsilon(x) = \frac{1}{2}\int\mathrm{d}^4x\;\partial_\mu\epsilon(x)\partial^\mu\phi(x) =-\frac{1}{2}\int\mathrm{d}^4x\;\epsilon(x)\square\phi(x).\tag{3}\label{eq:epsParts}$$

Starting with the third equation of the OP's question can use first equation \eqref{eq:exponential} to deal with the second factor: $$\langle \hat{\phi}(x)\rangle = \frac{1}{Z[0]}\int\mathcal{D}\phi e^{iS[\phi]}\left(1-\frac{i}{2}\int\mathrm{d}^4z\;(\phi(z)\square\epsilon(z)+\epsilon(z)\square\phi(z)) +\mathcal{O}(\epsilon^2)\right)\left[\phi(x) + \epsilon(x) \right] $$ where just the exponential having the $\epsilon$ field was expanded and $S$ stands for the kinetic term action of $\phi$. Now we can use the integration by parts as shown in \eqref{eq:epsParts}: $$\langle \hat{\phi}(x)\rangle = \frac{1}{Z[0]}\int\mathcal{D}\phi e^{iS[\phi]}\left(1-i\int\mathrm{d}^4z\;\epsilon(z)\square\phi(z)+\mathcal{O}(\epsilon^2)\right)\left[\phi(x) + \epsilon(x) \right]$$ Keeping only terms up to linear order in $\epsilon$ one obtains Schwartz's result: $$\langle \hat{\phi}(x)\rangle = \frac{1}{Z[0]}\int\mathcal{D}\phi e^{iS[\phi]}\left(\phi(x) + \epsilon(x) -i\phi(x)\int\mathrm{d}^4z\;\epsilon(z)\square\phi(z)\right)$$

  • $\begingroup$ Thank you very much. I would like to learn more about integrals in equation 2, could you provide me with a reference to look at about methods used in eq. (2) please? $\endgroup$
    – Blitz
    Apr 29, 2020 at 20:20
  • $\begingroup$ That is just the multivariate version of integration by parts. You can arrive to that expression by considering the integral of the divergence of the product, using the chain rule for derivatives and the Stoke's theorem. $\endgroup$
    – ohneVal
    Apr 30, 2020 at 8:05
  • $\begingroup$ Wow, that is data that I was looking for. Thanks.~ $\endgroup$
    – Plantation
    Oct 18, 2023 at 1:45

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