2

I have recently looked at the Wightman approach to axiomatically define a continuum QFT, using these notes [1] in particular. I am confused about where distributions appear in both the classical and quantum setting.

In a classical setting, the dynamics of the system is governed by the action, which is a map from the functional space to the real numbers, that is the action takes functions as its input.

Now if we go to the Wightman definition of the path integral. We say that the path integral can be rigorously defined as the integral over the Schwartz distribution space, $S'_\mathbb{R}, $ (space of tempered distributions) with the associated Borel measure

$$\mathcal{Z}[J] = \int _{S'_\mathbb{R}} e^{i \phi(J)} d \mu_C(\phi).$$ A physicist might write the gaussian measure, $d \mu_C(\phi)$ as $e^{-S[\phi]} \mathcal{D}\phi$ where $S[\phi]$ is the classical action we see in the classical version of the theory. In particular, the action here is the free scalar (euclidean) field action

$$S[\phi] = \int _{\mathbb{R}^4} d^4x \space\frac{1}{2}(\partial \phi)^2 + \frac{1}{2}m^2\phi^2.$$

This confuses me because the action is a functional, meaning it takes functions as its input, where as $S[\phi] $ for $\phi \in S'_\mathbb{R}$ doesn't mean anything.

I think the solution to my confusion lies somewhere in the fact that a subset of Schwartz distributions as can written in terms of any integral of a smooth function, for example $ \phi(f) = \int_ {\mathbb{R}^4} d^4x \space \phi(x)f(x)$ so when we write objects like $ \phi(x)$, we really mean $ \phi(f)$ for all $f \in S_\mathbb{R}$ and writing $ \phi(x)$ is just abuse of notation. If this is the case, I am not sure what the classical action looks like as a function of distributions (since the multiplication of distributions is not well defined) also I am not sure if the objects we deal with in classical field theory are fields or really distributions.

[1] https://www-m5.ma.tum.de/foswiki/pub/M5/Allgemeines/WojciechDybalski/Notes-QFT41.pdf

Qmechanic
  • 201,751
ColourConfined
  • 766

1 Answers1

2

This question deals with a very similar confusion about what is a "path" and what its dual in the context of a proper mathematical formulation of the path integral.

The confusion comes down to this: The argument to the measure $\mathrm{d}\mu_C(\phi)$ lives in the dual, but the arguments $\phi$ to the action $S[\phi]$ should be Schwartz functions. The problem is just notational - as I say in my answer to the linked question, what you actually show is that a covariance operator $C$ on the Schwartz functions induces the Gaußian measure $\mathrm{d}\mu_C$ on their dual. When you apply the folklore that "physicists just write $\mathrm{d}\mu_C[\phi]$ as $\mathrm{e}^{-S[\phi]}\mathcal{D}\phi$", you need to keep in mind this passage to the dual - the use of $\phi$ in both cases is just notationally confused. The action - built directly by application of the covariance operator - acts on the Schwartz functions, the measure induced by the same covariance operator acts on the dual.

Just don't use the physics notation $\mathrm{e}^{-S[\phi]}\mathcal{D}\phi$ for the induced measure if you find this too confusing. No one is actually claiming that you can apply the action/covariance operator to arbitrary distributions.

ACuriousMind
  • 124,833
  • So if we consider the a correlation function of the form $C^{n}(x_1,x_2,...,x_n)=\int_{S'\mathbb{R}} \phi(x_1)....\phi(x_n) d\mu_C(\phi)$ then $\phi(x_1) \in S'\mathbb{R}$ in the integrand but the $\phi$ in the measure is $\in S_\mathbb{R}$? In what sense is the $\phi$ in the integrand dual to the $\phi$ in the measure? I ask because I was under the impression that $S_\mathbb{R} \subset S'_\mathbb{R}$ so the space of distributions is larger? – ColourConfined Dec 28 '21 at 12:16
  • Also what happens when we try to perturb the gaussian measure? We do this by writing a more general measure $d\mu_I(\phi)= e^{-\int d^4x \mathcal{L}_I(\phi)} d\mu_C(\phi)$. Is the $\phi$ in the interacting lagrangian a test function or a distribution? I ask because when we expand as a tayor series this because an infinite series of correlators written in terms of the free measure which are distributions, it seems that upon tayor expanding these test functions magically become distributions. – ColourConfined Dec 28 '21 at 12:16
  • @SheldonCooper As I say: The measure $\mathrm{d}\mu_C$ is a measure on the space of tempered distributions, i.e. the $\phi$ in $\mathrm{d}\mu_C(\phi)$ is a distribution. The rest of your questions regarding what the n-point function means rigorously is really a separate question (spoiler: the answer involves Wick ordering and renormalization), and should be answered by any math text on path integrals. I recommend reading chapter 6 and 8 of Glimm/Jaffe's Quantum Physics (or the entire book, in fact, if the only thing you're currently reading to get into this topic are the notes you linked). – ACuriousMind Dec 28 '21 at 13:04
  • @SheldonCooper: Your formula for $d\mu_I$ is not what one really does. One replaces $d\mu_{C}$ by an approximation $d\mu_{C,\Lambda}$ which is a cutoff Gaussian measure obtained from $d\mu_{C}$ by suppressing Fourier modes $\ge \Lambda$. This results in a measure supported on smooth fields $\phi$. Only then one multiplies by the exponential with the interaction $\mathcal{L}I$. Finally one removes the cutoff and hopes for the convergence of the full measure $d\mu{I,\Lambda}$. For that one needs the the "constants" inside $\mathcal{L}_I$ to depend on $\Lambda$, i.e., you need renormalization. – Abdelmalek Abdesselam Dec 28 '21 at 21:53