# Composite operators

1. Jun 18, 2013

### geoduck

I have probably a silly question about correlation functions of composite operators. Why can't you just calculate a correlator with fields at different points x1, x2, x3, ... and then set a couple of the points equal at the end of the calculation to get the result?

e.g.,
$$\langle 0 T\phi(x_1)\phi(x_2).... 0\rangle$$

and to have a phi^2 composite operators just set x1 equal to x2 at the end of the calculation?

When you calculate $$\langle 0 T\phi(x)\phi(y).... 0\rangle$$ perturbatively at least, it seems the result is a fairly simple function of x and y. You'll get something like:

constant*eikx*eiqy

where k and q are integrated over. So just set x=y above?

2. Jun 19, 2013

### king vitamin

The problem is that when you are dealing with quantum operators, the ordering as y -> x becomes ambiguous. In your "constant," you have creation and annihilation operators which do not commute in the limit, and you need a prescription for dealing with them. It turns out that the normal ordering prescription is an unambiguous way to take the limit:

$$:\phi(x)^2: = lim_{x \rightarrow y} \{ \phi(x) \phi(y) - \langle \phi(x) \phi(y) \rangle \}$$

3. Jun 19, 2013

### geoduck

I was thinking in terms of a functional integral approach rather than an operator approach, so I wouldn't have to worry about operators and commuting.

But in terms of the operator approach, I've seen people try this:

$$<T\phi(x+\epsilon)\phi(x-\epsilon)>$$

so by arbitrary choice, the first phi is set at a later time. Then the limit ε is taken zero.

Are correlators of noncomposite operators analytic in the coordinates? It's just weird that you can't set two of the coordinates equal at the end of the calculation.

4. Jun 19, 2013

### king vitamin

Well the functional integral is equivalent to a quantum time-ordered operator. In either case, you're going to get some divergence, either in coordinate space from x=y, or in momentum space from a UV divergent momentum integral.

I think your intuition isn't right - in general one expects operator products to be divergent at the same space-time point, and this is why loops diverge (the integration diverges where the operators overlap). Recall that time-ordered correlators can be decomposed into propagators by Wick's theorem, and propagators are Green's functions of the Klein-Gordon equation. Such functions are clearly divergent at small distances (though maybe not in d=1?).