Does "sum over all paths" in the path integral imply "sum over all paths" in momentum space when one Fourier-transforms?

Question

How is the Fourier-transformed-field path integral interpreted? Is it still a "sum of all paths" in momentum space? Just that with another action?

Consider for instance the (Euclidean) partition function

$$Z=\int \mathcal{D}[\phi, \bar \phi] \mathrm{e}^{ -S[\phi,\bar \phi]}$$ of the, say, scalar complex field $\phi$? If the action is --for sensible polynomial $P$ in the fields-- $$S[\phi,\bar\phi]=\int(\bar\phi(-\nabla^2+m^2) \phi+P(\phi,\bar\phi))d^nx,$$ one gets $$S[\tilde\phi,\overline{\tilde \phi}]= \int(\overline{\tilde{\phi}}(p^2+m^2) \phi+P(\tilde\phi,\overline{\tilde\phi}))\frac{d^np}{(2\pi)^n}, $$ where $\tilde\phi$ is the Fourier transform of $\phi$. But changing $\mathcal{D}[\phi, \bar \phi]$ to $\mathcal{D}[\tilde\phi, \overline {\tilde\phi}]$ gives at most one constant factor and we are left with a

$$Z=(\mathrm{constant}) \int \mathcal{D}[\tilde\phi, \overline{\tilde \phi}] \exp\left({-\int(\overline{\tilde{\phi}}(p^2+m^2) \phi+P(\tilde\phi,\overline{\tilde\phi}))\frac{d^np}{(2\pi)^n}}\right)$$

Answer 1

I see now that your question is about the interpretation. Well, the interpretation is that you now integrate over the space of all fields in momentum space. Of course, mathematically the region of integration is still the space of functions $\mathbb{R}^4\to\mathbb{R}$ (or whatever kind of field applies) and so the meaning of $\mathcal{D}\phi$ is more or less the same, but in physics we usually don't think in those terms. We would say that the action is the same, as long as it is implied that one expression is to be used with position space fields and the other is to be used with momentum space fields.

To emphasize, while the two actions you wrote are indeed different as functionals, physically we think of them as the same, because there is a one-to-one correspondence between fields and their Fourier transforms. I don't know if there's much more to say about the issue; you just did a change of variables, like in a regular integral. This can actually be pretty useful; the partition function must be invariant when doing change of variables, since after all it is an integral. Doing a change of variables can be used to prove gauge invariance, for example.

Below is the old answer.

---

I think there's something wrong in your understanding, because writing $Z[\phi,\bar{\phi}]$ doesn't make sense. $\phi$ is the variable of integration; once you've done the integral, the result doesn't depend on $\phi$ anymore. Usually we calculate $Z[J]$ with $J$ some kind of classical current; such a dependence could come from a $J\phi$ term in the action. It's the action that is a functional of the fields, not $Z$.

This is important, because if you want to use Fourier-transformed fields you don't need to change variables. Your integral is

$$\int \mathcal{D}\phi \mathcal{D} \bar{\phi} e^{-S[\phi,\bar{\phi}]}$$

Here it's irrelevant whether you write $S$ as an integral over $x$ or over $p$; each field configuration $\phi(x)$ has a corresponding Fourier transform $\phi(p)$, and you can calculate the action with any of them.

Answer 2

I see the other answers are quite well placed. But I still think it might help going a step back and looking at Momentum space path integrals from classical paths. Here read this paper https://arxiv.org/ftp/quant-ph/papers/0403/0403005.pdf