Calculating Correlation of Composite Operators

[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*e^ikx*e^iqy

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

 

[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:

<br /> :\phi(x)^2: = lim_{x \rightarrow y} \{ \phi(x) \phi(y) - \langle \phi(x) \phi(y) \rangle \} <br />

 

[geoduck]

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:
<br /> :\phi(x)^2: = lim_{x \rightarrow y} \{ \phi(x) \phi(y) - \langle \phi(x) \phi(y) \rangle \} <br />

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:

&lt;T\phi(x+\epsilon)\phi(x-\epsilon)&gt;

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.

 

[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?).