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