# Interpretation of basic free boson propagator (Euclidean action)

In section 2.3.4 of Di Francesco, Mathieu, Senechal's Conformal Field Theory (and I would guess in many other introductory discussions of QFT) one finds a simple sample computation of a propagator

$$K(x,y) := \langle \varphi(x)\varphi(y)\rangle.$$

In the example, the fields $\varphi:\mathbf{R}^2\to \widehat{u}(1)$ are taken to obey the Euclidean action of a free boson of mass $m$

$$S = \frac{1}{2}\int d^2x \{\partial_\mu \varphi\partial^\mu\varphi + m^2 \varphi^2\}.$$

[I'm assuming I won't ruffle any feathers talking about the fields as operator-valued functions rather than operator-valued distributions. By $\widehat{u}(1)$ I mean the affine Heisenberg algebra (see Conformal Field Theory section 14.4.4).]

In particular, the authors compute

$$K(x,y) = -\frac{1}{2\pi} \ln(\|x-y\|), \quad \textrm{if }\ m=0$$ $$K(x,y) = \frac{1}{2\pi} K_0(m \|x-y\|), \quad \textrm{if }\ m>0$$

where $K_0$ is a modified Bessel function.

I find I am at a loss for how to interpret this result.

For simplicity, let's focus on the case $m=0$. Based on the last few paragraphs of Luboš's answer to this question, it seems $K(x,y) = -\ln(\|x-y\|)/2\pi$ might be interpreted in one of two ways:

1) As something that "knows about the correlation of $\varphi(x)$ and $\varphi(y)$." In this case we would conclude that:

• When $\|x-y\|<1$, it looks like $\varphi(x)$ and $\varphi(y)$ "have the same sign." (What could that mean in this context?)
• When $\|x-y\|>1$, it looks like $\varphi(x)$ and $\varphi(y)$ "have opposite signs." (Ditto)
• When $\|x-y\|=1$, it looks like $\varphi(x)$ and $\varphi(y)$ are independent of each other. (What would it mean for $\varphi(x)$ not to be correlated with $\varphi(y)$ only when the distance between the points is 1?)

2) As a probability amplitude for a particle to go from $x$ to $y$. This perspective doesn't seem to apply here: $K<0$ when $\|x-y\|>1$, and in any event $|K|\to\infty$ as $\|x-y\|\to\infty$. I mention it since this is the interpretation Luboš specifically gives in relation to correlation of operators.

Neither of these possibilities, as I have understood them, looks correct to me.

Any help in clearing up my confusion would be most appreciated. If it is possible to see the above propagators manifested in concrete data (perhaps results gathered in a lab (condensed matter?) or from computer simulations (Ising model?)), I think I would such a treatment most satisfying, but that may be asking too much. I don't know.

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You should be careful with the free boson at $m = 0$. There is a symmetry $\phi \rightarrow \phi + const$ and an associated zero energy mode. This is, I believe, the source of the strange behavior of the correlation functions. If you look at $<\partial \phi \partial \phi>$ you see it is more reasonably behaved, as are the vertex operators. –  BebopButUnsteady Aug 4 '11 at 4:29
To expand upon Bebop's excellent comment, one gets bad behavior of the $\phi$ field in dimensions 1 and 2 but from 3 onwards things are fine. When working in 2 dimensions one is better off considering the gradient field $\nabla \phi$. –  Marek Aug 4 '11 at 9:06
@Marek - Thank you for your comments! –  Dan Kneezel Aug 5 '11 at 5:58

This is a confusing point, but BebopButUnsteady gets at the main issue. Consider higher dimensions. Then the correlation function

${1\over k^2 + m^2} = \int e^{-\tau(k^2 + m^2)} d\tau$

which is the Schwinger representation, the propagating particle interpretation. Each $\tau$ contribution to the propagator is the Fourier transform of a Gaussian of width $\sqrt{\tau}$, which is the probability of a random walk to go from x to y in time $\tau$. This interpretation is correct in all dimensions, and it is the "probability amplitude to go from x to y" interpretation in real time, because the Schwinger proper time Brownian motion path-integral continues to a quantum mechanical free particle path integral amplitude.

The Schwinger representation proves that the Euclidean G(x) is positive, and that $G(x)\rightarrow 0$ at large x, because the random walk probability distribution is positive and long-time gaussians are decayed. For short times, there is always a predictable blow up at short distances, which is the same as the power-law of the massless propagator. For the four dimensional massless propagator $1\over2\pi^2 (x-y)^2$.

The blow up is a little hard to understand if you think of fields with a lattice action. If you take the fields to fluctuate with an action proportional to:

$S \propto \sum_{\langle x,y\rangle} (\phi(x) -\phi(y))^2$

you recover the free field propagator. But then $\langle\phi(x)\phi(x)\rangle$ is the value of $G(0)$, and if you take the continuum limit by choosing the scale of $\phi$ so as to fix $G(0)$, you get no correlations at all, because the scaling inverse-square falloff pushes the continuum $G$ to zero for small lattice spacings.

The right way to take the continuum limit is to allow $G(0)$ to blow up as $1\over\epsilon^2$, with a constant of proportionality that ensures that the free propagator renormalization constant equal to 1. Then you get a power-law falling propagator with a blow up at coincident points. This rescaling just introduces $\epsilon$ factors into $S$ to recover the usual $\nabla^2$ action. It's no surprise.

But in two dimensions, there is a surprise. The dependence of the propagator is logarithmic, so that you have a long-distance residue in the correlation function which you see as an additive constant at short distances. The two dimensional correlation function $\langle\phi(x)\phi(y)\rangle$ has a logarithmic dependence on the infrared cutoff, on the radius of the domain you define it on.

A 2-d lattice free field acquires a shift in the value of the correlation function depending on the size of the domain. To define the continuum limit, you have to rescale the field to make the coefficient of the two-point function 1, and you also need to add a constant to the answer to normalize for the limit of the box going to infinity. The convention for the scale is as in higher dimensions. The convention for the box is to set the correlation function to zero on a sphere of unit radius. You can make the zero point somewhere else with a different convention, and this just adds a constant to the log-correlation function.

To get rid of this annoying infrared sensitivity, people like to take the derivative of the scalar. I prefer to deal with it, because it is real. The actual correlation function in a box lattice adds back the infrared quantity in such a way that it pushes the correlations back to positive, and rescales them so that you get the right nearest-neighbor value. These positive definite correlations are the ones that have a statistical interpretation in terms of the probability of particle paths. After rescaling and subtraction, you get nonsense correlations, as you noticed.

The mathematical way you see this story in CFT books is in the somewhat ambiguous expression for the real-space propagator:

$G(x) = \int {d^2k\over (2\pi)^2} {e^{ikx}\over k^2 + i\epsilon}$

The small-x normalization is fixed here, as you can see because there is no large-k divergence, no short distance problem in defining the divergence, because of the oscillating exponential. But there is a serious problem at small k. If you make a k-lattice for understanding the integral, you get a divergence proportional to the log of the k-lattice spacing, which is the infrared problem. You have to subtract out this logarithm to get the usual answer.

Theories with these infrared issues are known as "logarithmic conformal field theories". The long-range problems are responsible for the Mermin-Wagner-Coleman theorem.

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Since you're looking for an interpretation of the divergence, perhaps it's worth mentioning that this logarithmic divergence can be thought of as a toy model for the sort of mathematical phenomena which lead to confinement in 4d Yang-Mills theory. The correlation functions of the field $\phi$ is well-defined in UV & IR cutoff theories, but it diverges when we remove the cutoffs, so we don't get to keep the evaluation observables $\phi \mapsto \phi(x)$ in these limits. Since you should be able to construct the Hilbert space of states by applying operators to the vacuum, this means that you don't quite get the Hilbert space you were expecting. Instead you get a Hilbert space made from the vacuum by applying products of field exponentials and field derivatives (possibly familiar as the vertex operators of the string worldsheet theory). This changes the space of energy eigenstates in a rather drastic way.

A similar thing happens to observables with non-zero color charge in QCD's infinite volume

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This is true, but the analogy is weak, because the correlator is only logarithmically growing in 2d, so that the field walks away logarithmically from its center value in a large box, while in QCD, the gauge field is completely random in a large box after only a fixed finite distance which doesn't grow as you relax the infrared regulator. While the mass of the quarks grows as you relax the regulator, it grows linearly because of the mass of the string from the quark to the edge of the box, and this isn't a good analogy to the 2d free scalar, where there is no exponential forgetting (mass gap). –  Ron Maimon May 7 '12 at 6:12