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I have the free action in position space

$$S_0[J,\phi] = \int \! d^4x[\frac{1}{2} \phi (\partial_\mu \partial^\mu - m^2 + i\epsilon)\phi + \frac{\hbar}{i}J\phi].$$

Knowing that the Fourier transform of $ \phi(x) $ gives it's representation in momentum space, and likewise for $J(x)$ how do I find the free action in momentum space?

I've tried plugging the Fourier transforms into the free action, but can't seem to simplify the expression. I think I have an algebra issue.

This is how far I've got:

$$ S_0[J,\phi] = \int \! d^4x [\frac{1}{2}(\int \! \frac{d^4k}{(2\pi)^2} \tilde{\phi}(k) e^{-ik_\mu x^\mu} )(\partial_\mu \partial^\mu - m^2 + i\epsilon) (\int \! \frac{d^4k}{(2\pi)^2} \tilde{\phi}(k) e^{-ik_\mu x^\mu}) + \frac{\hbar}{i} (\int \! \frac{d^4k}{(2\pi)^2} \tilde{J}(k)e^{-ik_\mu x^\mu}) (\int \! \frac{d^4k}{(2\pi)^2} \tilde{\phi}(k) e^{-ik_\mu x^\mu})] $$

I've tried to simplify this further using a convolution and various other attempts but nothing has worked so far. Help would be very much appreciated.

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  • $\begingroup$ Please show everyone how far you got, and at what point you can't simplify further. $\endgroup$
    – kaylimekay
    Feb 3, 2021 at 5:46
  • $\begingroup$ Thanks, I've just updated it $\endgroup$
    – Ravi
    Feb 4, 2021 at 22:02

1 Answer 1

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You're integrating over $k$ twice, which is a bad idea. If you introduce a second dummy variable $q$, then the first term becomes something like

$$\int d^4x \int d^4k\int d^4q \left[ \tilde \phi(k)\tilde \phi(q)(-k^2-m^2+i\epsilon) e^{-ix\cdot (k+q)}\right]$$

Now the only thing which depends on $x$ is that exponential. Noting that $\int d^4x \ e^{-ix\cdot a}= (2\pi)^4 \delta(a)$, you should be able to take it from here.

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