# Integration by parts in Zee

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Hello, I'm just starting Zee's QFT in a Nutshell, I'm a bit confused about what he means by "integate by parts under the d4x". Can someone explain what he means by this? I understand how to obtain the Klein-Gordon equation from the free particle Lagrangian density, but not sure why he invokes integration by parts.
Thanks!

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mfb
Mentor
Well, you have a ##(\partial \varphi)(\partial \varphi)## term, but ##\varphi\partial ^2 \varphi## is more useful later.

Both arise from the derivative of ##(\varphi)(\partial \varphi)## and that should go to zero for large ##\phi##.

DarMM
DarMM
Science Advisor
Gold Member
To expand on what @mfb said.

We have:
$$\partial_{\mu}\left(\phi\partial^{\mu}\phi\right) = \partial_{\mu}\phi\partial^{\mu}\phi + \phi\partial^{2}\phi$$
Under the integral:
$$\int{\partial_{\mu}\left(\phi\partial^{\mu}\phi\right)d^{4}x} = \int{\partial_{\mu}\phi\partial^{\mu}\phi \hspace{1pt} d^{4}x} + \int{\phi\partial^{2}\phi \hspace{1pt} d^{4}x}$$
So assuming the left-hand side vanishes we'd have:
$$\int{\partial_{\mu}\phi\partial^{\mu}\phi \hspace{1pt} d^{4}x} = - \int{\phi\partial^{2}\phi \hspace{1pt} d^{4}x}$$
which is exactly the replacement made in Zee.

Since the LHS term is just a surface term it should vanish if the fields decay rapidly. They do, but proving so is a good bit beyond a text like Zee. Basically you'd have to prove a randomly selected field decays at infinity with probability ##1##.

mfb