# Finding the free action in momentum space

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.

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

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.