# constraint on scaling dimension

How can we show that for any scalar operator $\Delta\geq1$ (where $\Delta$ is the scaling dimension)? Where can I find a reference for reading where it comes from?

-
Ok, I removed it :-/. But a strange fact is that the results one obtaines by the method I explained for scalars, spinors, etc are the same as the authors of this paper obtain as scaling dimensions for the same quantities ... So I suspect that in the paper they are not talking about the "real" scaling dimensions either and it is a coincidence that the results agree...? So what is the difference between the real "scaling dimensions" and the "enginiering dimensions" I mentioned? – Dilaton Nov 2 '12 at 16:43
The scaling dimension defined by $\Delta$ where $\phi(x)\rightarrow\lambda^{-\Delta}\phi(\lambda x)$ and also connected to the "engineering dimension" by an addition of the anomalous dimension. They are the same when you don't have interactions. – Yair Nov 2 '12 at 16:47

This is a consequence of the Lehman spectral representation for a physical scalar operator. The two point function of this operator (the expected value of the operator with it's conjugate) can be written as an integral over propagators:

$$\langle \bar{\phi}(p)\phi(p') \rangle = (2\pi)^d\delta^d(p-p') \int_0^\infty {\rho(s)\over p^2 - s + i\epsilon} ds$$

Where each propagator falls off as ${1\over x^2}$ at short distances in 4 dimensions, and $\rho(s)>0$ for all s (because of Hilbert space positivity--- this is the norm of a state, namely $||\phi(p)|0\rangle||$).

A superposition of positive propagators falling off as ${1\over p^2}$ with positive coefficients annot produce a falloff at large p which is faster than ${1\over p^2}$. This means that the asymptotic scale dimension of the scalar operator can't be less than 1 in 4 dimensions, it can't be less then 1/2 in 3 dimensions, and it can't be negative in 2 dimensions.

This is not exactly mathematically true, because you can engineer a spectral weight which is growing near s=0 as a power law, to produce faster than 1/p^2 falloff. But it is physically true anyway, because such a growth requires an infinite number of particle species at p=0, which is inconsistent with the usual idea that a quantum field theory has a finite number of elementary fields, with a finite thermal entropy. The way to understand this is that superposing any finite tower of particles with positive spectral weights always leads to 1/p^2 falloff or slower, and a ${1\over x^a}$ propagator with $a\le 2$

The Kallen-Lehman spectral representation is a standard field theory result, it is found in most standard textbooks. The original paper is reprinted in Schwinger's reprint volume "Quantum Electrodynamics".

-
Can you give an example of $\rho(s)$ which implies a faster falloff than 1/p^2? – Yair Nov 2 '12 at 19:58
@Yair: $\rho(s)= s^{\alpha}$ gives $(|p|^2)^\alpha$ by dimensional analysis (or rescaling the integral to dimensionless form), and for $\alpha<-1$ gives a falloff faster than $1/p^2$. This is unphysical, because the total integral of $\rho$ is divergent near s=0, meaning there is an infinite number of particles at long distances. – Ron Maimon Nov 2 '12 at 20:14
Thanks for you answer Ron. I must say I do expect this "trick" to appear in some textbook but so far none of the textbook I saw mentioned that. Do you know one? – Yair Nov 2 '12 at 21:27
@Yair: It's not a trick--- it's a famous result, the positivity of the spectral weight. It is used all over the 1950s-1970s literature, but it's lost its sparkle today. I learned it from Sidney Coleman's lectures, it's in Peskin&Schroeder somewhere according to google. – Ron Maimon Nov 2 '12 at 22:04