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