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?


1 Answer 1


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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.