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In Schwartz's QFT textbook Section 3.5, the Lagrangian for the graviton $$\mathcal{L}=-\frac{1}{2}h\Box h+\frac{1}{3}\lambda h^3+Jh$$ with EOM $\Box h-\lambda h^2-J=0$ is perturbatively expanded in $h$ to yield: $$h(x)=\int d^4y\delta^4(x-y)h(y)=-\int d^4y[\Box_y\Pi(x,y)]h(y)=-\int d^4y\Pi(x,y)\Box_yh(y)$$

with the note from Schwartz "where we have integrated by parts in the last step."

The first minus sign I understand with the definition $\Box_x\Pi(x,y)=-\delta^4(x-y)$, but I don't understand why there's not a second minus sign between the last and second-to-last terms.

My impression was that in QFT, we treat terms at the boundary of spacetime as null so that $$\int_U A\partial_\mu B=\int_{\partial U} AB-\int_U B\partial_\mu A=-\int_U B\partial_\mu A.$$

However in this case, wouldn't that mean that

$$-\int d^4y[\Box_y\Pi(x,y)]h(y)=-\int d^4y(\Box_y[\Pi(x,y)h(y)]-\Pi(x,y)\Box_y[h(y)])=\int d^4y\Pi(x,y)\Box_yh(y)~?$$

The last term is definitely supposed to have a minus sign, as Schwartz cites the same equation again with minus sign in (3.84). What am I missing here?

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Figured this out as I was typing the question...

The d'Alembertian $\Box_y=\partial_y\partial_y$, so to 'swap' it with an adjacent element, it requires two uses of integration by parts.

That is, $\int_U A\Box B=\int_U A\partial_\mu\partial_\mu B=\int_{\partial U} A\partial_\mu B-\int_U\partial_\mu A\partial_\mu B=-\int_{\partial U}\partial_\mu (AB)-(-\int_U\partial_\mu\partial_\mu(A)B)=\int_UB\Box A$

The last two terms in the expansion of $h(x)$ then have the same sign.

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    $\begingroup$ Hey, thank you so much for the question and answer, I had the same problem and you solved nicely. Thanks! $\endgroup$
    – Janne
    Commented Mar 2, 2022 at 10:19

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