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

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


2 Answers 2


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.

  • 1
    $\begingroup$ I completely agree with what you wrote, but I might add (and this is just me being very pedantic) that in general when you do a field redefinition the path integral will change in the sense that you will pick up a jacobian factor from the measure. It shouldn't matter in the Fourier transform case since the field redefinition is linear in the fields. Your answer is great, I just wanted to add that in case someone wanted to generalize what you wrote to a more complicated case. $\endgroup$
    – Andrew
    Nov 5, 2015 at 20:27
  • $\begingroup$ @Andrew: I thought about including that (though I wasn't sure how to find the Jacobian; I like your insight that ir doesn't matter since the transformation is linear), but I decided not to do it since I didn't think that was really what the question was about. $\endgroup$
    – Javier
    Nov 5, 2015 at 20:31
  • $\begingroup$ That is fair enough. I don't actually think you should edit your answer (definitely a +1 from me for clarity), it's mostly I was feeling pedantic so I thought it might be worth a comment. Incidentally I suspect the determinant is actually one if you normalize things properly--if you discretize everything (which the path integral implicitly tells you to do) then the Fourier transform is a unitary transformation on the fields (although. It's a moot point since the determinant is definitely field independent). $\endgroup$
    – Andrew
    Nov 5, 2015 at 20:40
  • $\begingroup$ @Javier The Jacobian you can find it by writing the $\phi(x)$ as the Fourier transform of $\tilde\phi(p)$ and deriving functionally w.r.t. say $\tilde\phi(q)$. You get only $(2\pi)$-factors times a phase factor which cancels out when you take the determinant. Then you get that the Jacobian is a constant, what I wrote. $\endgroup$
    – c.p.
    Nov 6, 2015 at 13:40
  • $\begingroup$ To your reply to the question in the answer: indeed, I messed up with notation. Still, what you wrote, doesn't address the interpretation. $\endgroup$
    – c.p.
    Nov 6, 2015 at 13:42

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


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