# Why do correlation functions diverge in field theory?

A correlation between two random variables is generally defined as $$\text{corr}(X,Y)=\frac{\text{cov}(X,Y)}{\sigma_X \sigma_Y}$$ In particular this says that $\text{corr}(X,X)=1$. However, in physics we often look at correlation between fields like for example $$\left< \phi(x)\phi(y) \right> = \frac{1}{|x-y|^{2\Delta}}$$ in conformal field theory. Thinking of $x$ and $y$ as labels for correlated random variables we should have $$\left< \phi(x)\phi(x) \right> = 1 \, ,$$ but it actually diverges.

I have been told that this has to do with us being in some continuum limit, with renormalization and with principal values of distributions but I can't quite piece it together.

What I am ultimately interested in is recovering the moments of the probability distribution $P(\phi)$ at some point, say $x=0$. It seems like it should be possible to extract this information from the correlation functions but I don't quite know how.

• The covariance depends upon products between $\sigma_{x}$ and $\sigma_{y}$ and themselves, so it should not diverge if normalized to the product of $\sigma_{x}$ and $\sigma_{y}$. I think it would be meaningless to have a correlation that diverged, would it not? What would that even mean? – honeste_vivere Oct 19 '17 at 18:52
• Oh come on, @sammygerbil, this posting this on the math site is sentencing it to death. – DanielSank Oct 19 '17 at 19:37
• @sammygerbil it is absolutely not a math question - he is using reasoning in mathematics to question a result in physics. – JamalS Oct 19 '17 at 19:56

Note that $\langle\cdots\rangle$ is just the statistical mechanics notation for the expectation denoted by $\mathbb{E}\cdots$ in probability textbooks. So, in principle, $$\langle\phi(x)\phi(y)\rangle=\mathbb{E}\phi(x)\phi(y)$$ which for centered random variables (i.e., assuming $\mathbb{E}\phi(x)=\mathbb{E}\phi(y)=0$) should be equal to the covariance ${\rm Cov}(\phi(x),\phi(y))$ rather than what statisticians would call the coefficient of correlation, i.e., ${\rm Cov}(\phi(x),\phi(y))/\sqrt{(\mathbb{E}\phi^2(x)\times \mathbb{E}\phi^2(y))}$.

Now the main problem related to your question is that $\phi(x)$ is not a random variable. This is because the field $(\phi(x))_{x\in\mathbb{R}^d}$ is not a function but a Schwartz distribution which in general cannot be evaluated at a point. What can be evaluated are local averages $$\phi(f)=\int_{\mathbb{R}^d} \phi(x)f(x)\ d^dx"$$ for smooth test functions $f$, e.g., $C^{\infty}$ with compact support or in Schwartz space $S(\mathbb{R}^d)$. Take for instance the Euclidean 2d Ising CFT. Then $\phi$ is a random distribution, i.e., a random variable $\omega\mapsto \phi[\omega]$ with values in $S'(\mathbb{R}^2)$ on some probability space $(\Omega,\mathcal{F},\mathbb{P})$. The maps $\Omega\rightarrow \mathbb{R}$, $\omega\mapsto \phi[\omega](f)$ indexed by fixed test functions $f$, are honest real-valued random variables with moments of all orders. In particular, the variance $${\rm Var}(\phi(f))=\mathbb{E} \phi(f)^2={\rm Cst}\times \int_{\mathbb{R}^2}\int_{\mathbb{R}^2} \frac{f(x)f(y)}{|x-y|^{\frac{1}{4}}}\ d^2x\ d^2y$$ is finite.

There is a Kolmogorov-Chentsov Theorem for distributions with negative Hölder regularity which generalizes the usual one for positive regularity as in standard textbooks on Brownian motion. See for instance this article by Furlan and Mourrat. The blow-up on the diagonal is a signature of the fact the field $\phi$ is a distribution rather than a function.

It's because $\langle \phi(x)\, \phi(y)\rangle$ is not perfectly analogous to $\operatorname{corr}(X,Y)$, but to $\operatorname{cov}(X,Y)$. Clearly, $\langle \phi(x)\,\phi(x)\rangle$ is analogous to the variance quanitity used to define $\operatorname{corr}(X,Y)$, so it would be more correct to say that $\operatorname{corr}(X,Y)$ is analogous to $$\frac{\langle \phi(x)\, \phi(y)\rangle}{\sqrt{\langle \phi^2(x) \rangle\langle \phi^2(y) \rangle}} \rightarrow 0\ \forall\ x\neq y.$$

As for why this happens, there are a number of theorems constraining the form that $\langle \phi(x)\, \phi(y)\rangle$ can take. First, $\langle \phi(x)\, \phi(y)\rangle$ is related to the Green's function (propagator) of the free part of the field theory, which ties it to the equations of motion/Lagrangian/Hamiltonian. The Hamiltonian, in particular, is required to be positive in order to ensure the theory is stable, requiring that the equations of motion can't have more than two derivatives. There a few mathematically equivalent theorems that have the same implication (Ostrogradsky's theorem, classically, and one outlined in Weinberg's Quantum Field Theory books from the quantum side). It's that limit on the number of derivatives, combined with the fact that space-time is four dimensional, that inevitably leads to the propagator containing divergences when space-time is continuous.

• But the variance at some point is a number that can be experimentally determined and it's certainly not infinite, so how do extract this information? – AlexM Oct 20 '17 at 11:26
• @AlexM This is one of those things that is tricky about taking continuum limits. The equations of motion (Klein-Gordon) take a form that is, essentially, Newton's second law. Viewed as such, a point has zero mass, and so can experience infinite displacement in the field direction in finite time. It's only when you consider averages over finite volumes that things become well behaved (more broadly, integrating against test functions). And any real measurement will always cover a finite non-zero volume. – Sean E. Lake Oct 20 '17 at 16:54
• I just want to make sure that I got this: so if I pick a finite volume $V$ and I measure the field average over that volume then this gives me a well defined random variable $\Phi$ with variance given by $\mathrm{var}(\Phi) = V^{-1} \int_V \mathrm{d}^dx \langle \phi(x) \phi(0) \rangle$ and I could also obtain higher moments this way. Right? – AlexM Oct 23 '17 at 20:53
• The experimental application of quantum field theory always involves eventually counting the expected number of particles that pop up in a bin of finite size, so there will always eventually be some way to get at that number. I don't know how renormalization interacts with that picture, for what it's worth. – Sean E. Lake Oct 23 '17 at 21:01
• If it makes any difference, I am a) thinking mostly of statistical systems for now which seem conceptually easier and b) in the language of conformal field theory where we assume that we know the exact form of n-point functions, hence maybe no renormalization is needed. – AlexM Oct 23 '17 at 21:30

As observed, the product $\phi(x)\phi(y)$ is unambiguous and finite for $x \neq y$. However, as $y\to x$, we have that $\phi(x)^2$ possesses an ultraviolet divergence.

In order to make sense of this, we must renormalise this composite operator, call it $[\phi^2].$ In order to do so, we could consider evaluating $\langle T\phi(x)\phi(y)\frac12 \phi(z)^2\rangle$ and adding appropriate counterterms.

In doing so, you will encounter terms like, $m^2 - \frac16(p_1+p_2)^2$ which if interpreted as a Feynman rule would come from a term $(m^2+\frac16 \square)\phi$. Thus, we can replace the unrenormalised operator by a renormalised operator, which by necessity will contain lower dimensional operators to reproduce the result of renormalising the Green's function.

For completeness, the result is,

$$\frac12 [\phi^2] = \left[ 1+\frac{g^2}{64\pi^3 (d-6)}\right]\frac12 \phi^2 + \frac{g\mu^{d/2-3}}{64\pi^3(d-6)}(m^2 + \frac16 \square)\phi + \dots$$

Thus in field theory the divergence of the Green's function should not be shocking as $y\to x$, the root cause is the same as for the other quantities of the theory which diverge.

• Do you think you could point me to some reference dealing with this form of renormalization in particular? – AlexM Oct 20 '17 at 11:28
• @AlexM Collins' book on renormalisation is in my opinion the best book on renormalisation in general, as it deals with it very systematically – JamalS Oct 20 '17 at 12:40