# OPEs in conformal field theory

1. Dec 3, 2009

### nrqed

When people give the rules for the operator product expansion of fields in CFT, they always give the rule for the OPE of a product of two fields. But let's say that we have three fields. To be specific, consider the OPE of $$T(z_1) T(z_2) \Phi(z_3)$$ where T is the energy-momentum tensor and $\Phi$ is a primary field, let's say. Then how do we evaluate the divergent contributions to this expression? If we do first the OPE of the two T together and then the OPE of the result with the primary field, we get a different result than if we do the OPE of $T(z_2)$ on the primary field first and then the OPE of the result with the remaining em tensor. It's easy to see that the results are different because the OPE of two T contains the central charge whereas the second way does not contain the central charge. So what is the rule to compute an OPE when there are several fields??

2. Dec 3, 2009

### xepma

I've been struggling with this question as well.

I think this is because you throw away the regular terms when you take the first OPE. Each regular term contains a field. These fields also contribute to the singular behavior if you take another OPE.

For example, the first regular term of the OPE of $T(z_2)$ and $\Phi(z_3)$ contains the second descendant of $\Phi(z_3)$, call it $\Phi^{(-2)}(z_1)$. The most singular term in the OPE of $T(z_3)$ with this field is (I looked it up in DiFrancesco)

$$T(z_3)\Phi^{(-2)}(z_1) \sim \frac{c/2}{(z_1-z_3)^4}\Phi(z_1)$$

This is precisely one of the singular terms you get if you first take the OPE of $T(z_3)$ and $T(z_2)$, and then with $\Phi^{(-2)}(z_1)$.

It looks pretty convincing, but I never found a source that confirmed it. I'm actually not entirely convinced that it is correct. The remaining regular terms in the first OPE all have have a singular behavior if you take another OPE. Hence, they all contribute to this term.

Perhaps this has something to do with the radius of convergence for the OPE? I think at some point I read that OPE of two fields is only valid up to the position of the third field. So when you take the OPE with a third field, the radius of convergence of the OPE shrinks to zero. You wouldn't expect the OPE to be valid outside this radius.

3. Dec 3, 2009

### Ben Niehoff

There is a well-defined rule for this. It's called Wick's Rule. Let R denote radial-ordering, and (ab) denote the "contraction" of a and b; that is, the pairwise OPE between a and b. Then

$$\mathcal{R}abcd = (ab)cd + (ac)bd + (ad)bc + (bc)ad + (bd)ac + (cd)ab + (ab)(cd) + (ac)(bd) + (ad)(bc)$$

where any factors that remain outside () must be Taylor expanded far enough to display all singular terms (i.e., you can't throw out regular terms until the end).

The general formula is: Do all pairwise contractions, then all combinations of two pairwise contractions, then all combinations of 3 pairwise contractions, etc.

If any of your fields are Grassmann-valued, then you need to take into account however many times they must be anti-commuted to bring them adjacent to each other whenever you take contractions.

4. Dec 4, 2009

### xepma

But that only works if you have free fields. In CFT you are not working with free fields, in general. You are not even treating the interactions as a perturbation to the system, so free fields do not enter the description.

Anyway, appendix 6.B of DiFrancesco (the big yellow book) treats the idea of the Generalized Wick Theorem, and what you need to do to take the OPE of three operators. Appendix 6.C also treats something related, namely taking the OPE of three operators is not an associative procedure.

5. Dec 4, 2009

### nrqed

Thanks for the reply, Xepma. After I posted, what I thought was the answer suddenly struck me, and it is essentially what you write in your last paragraph: that we should simply do first the two fields that are the closest to one another. But then Ben's post made me wonder if I was missing something. So I am still a bit confused.

6. Dec 4, 2009

### DrFaustus

Haven't worked much on OPEs really but have read a paper or two about it. Basically it should not matter how you decide to to the expansion because of associativity of the product of fields, that is $$\big(T(z_1) T(z_2) \big) \phi(z_3) = T(z_1) \big( T(z_2) \phi(z_3) \big)$$. (It has actually been argued by Hollands and Wald that precisely this condition, and the consequent analogous condition on the OPE coefficients, should be layed as an axiom to define a QFT.)

But as the OPE itself is only valid in the limit of all the points coinciding, you should pay attention to what limit you take first in the case of three or more fields. So it might be related to how you approach the limit...

7. Dec 4, 2009

### nrqed

That's a very interesting point you made there!!

Then, when we do the OPE of the two T first (I am considering $$T(z_1)T(z_2) \Phi(z_3)$$ )we get a term of the form

$$\frac{c}{2} \frac{\Phi(z_3)}{(z_1-z_2)^4 }$$

If we do the OPE of $$T(z_2) \Phi(z_3)$$ first, we get exactly the same expression except that it is now divided by $$(z_1-z_3)^4$$ instead of $$(z_1 - z_2)^4$$, right? So it does give a different result.

But wait!! That's actually the same, isn't? When we do the OPE of $T(z_2) \Phi(z_3)$,if we get a regular term of the form $\Phi^{-2}(z_3)$ we may as well repace it by $\Phi^{-2}(z_2)$. And if we use that to do the OPE with the remaining $T(z_1)$, we get the same result as before, when we contracted the two EM tensors first! I guess the point is that in the final result, we are only listing the terms that are singular when any two of the coordinates coincide. Any regular corrections to that are dropped. Therefore, up to regular corrections proportional to $z_2-z_3$, we have

$$\frac{1}{(z_1-z_3)^4 } \simeq \frac{1}{(z_1-z_2)^4}$$

If that's it, then you have solved it, Xepma! Of course, there is something a bit scary since, as you have said, why don't we need to include ALL the terms in the OPEs, then? Maybe we have to, in principle.

Patrick

Last edited: Dec 4, 2009
8. Dec 4, 2009

### nrqed

I am still a bit confused. Some sources insit that the OPE of two operators is valid only up to the position of the closest third field, as Xepma pointed out. This seems to imply that we would need to do the OPE of the two closes fields first. On the other hand, since we are working in a conformally invariant theory, isn't the notion of distance irrelevant?
Hope someone may clear things up a bit!

9. Dec 5, 2009

### Haelfix

The procedure outlined there is the answer, or at least how I understand it to be done! Eqn 6.206 is whats needed.

10. Dec 5, 2009

### blechman

That's not quite what a CFT is: you still get position dependence. Take a look at all the formulas you wrote down, they have z's in them! Rather, a CFT doesn't have a built-in DISTANCE SCALE, so if you grow or shrink the system, then you end up with the same theory. However, relative distances do exist in a CFT, otherwise it would be a most strange theory indeed!

In other words, one can still talk about the distance between A and B, but ONLY in relation to the distance between A and C.

Hope that helps!

11. Dec 5, 2009

### blechman

Isn't that just saying that the OPE no longer converges when you hit another operator insertion?!

12. Dec 6, 2009

### nrqed

Thanks for the reply. Yes, I see what you mean: in a CFT absolute distances are not defined but relative distances are. It makes sense but let's consider three fields at z1, z2 and z3. Using a conformal transformation I can send any of these coordinates to 0, 1 and infinity. So isn't the relative distances changed if we map z1 to 0, z2 to 1 and z3 to infinity and if we map z1 to 1, z3 to 0 and z2 to infinity?

Thanks for the help!

13. Dec 6, 2009

### Count Iblis

You then have to include the products of the local scale factors exponentiated to the power of the scaling dimensions.

14. Dec 6, 2009

### nrqed

Ahh! Yes, that makes perfect sense. Thank you!

15. Dec 6, 2009

### nrqed

Touché!!

16. Dec 6, 2009

### nrqed

There is a calculation in DiFrancesco et al that ressembles closely the calculation I had in mind, but they do some things that baffle me.

The calculation is the calculation of $T(z) \phi^{(-n)} (w)$ on page 179 (section 6.6.2). Here, $\phi^{(-n)}(w) = L_{-n} \phi(w)$, i.e. a descendant field. In any case, the calculaton boils down to evaluating

$$\oint_w dx \frac{1}{(x-w)^{n-1}} T(z) T(x) \phi(w)$$

I don't write the division by 2 Pi i to save time.

They then say that the most divergent terms can be found by doing the OPE of the two em tensors first, keeping only the singular terms, so the most singular terms are given by

$$\oint_w dx ~ \frac{1}{(x-w)^{n-1}} \bigl( \frac{c/2}{(z-x)^4} + \frac{2T(x)}{(z-x)^2} + \frac{\partial T(x)}{z-x} \bigr) \phi(w)$$

I have a problem with not including the regular terms, but I will get back to this later.
Consider the second term (the 2T(x) term). They do the OPE of the em tensor with the primary field, using

$$T(x) \phi(w) = \sum_l (x-w)^{l-2} \phi^{-l}(w)$$

Here, the sum over l is over both positive and negative values of l. But then, they drop all the negative values of l! Giving them for the second term

$$2 \sum_{l=0}^\infty \oint_w dx ~ \frac{(x-w)^{l-2}}{(x-w)^{n-1} (z-x)^2} \phi^{-l}(w)$$

My question is: why do we drop all the terms wih negative values of l!? They give more singular integrands than the terms retained, so the explanation cannot be because we want to keep only the singular terms. So I am baffled as to why these terms are dropped. I hope someone can clarify this for me!!

Thanks

17. Dec 6, 2009

### blechman

I'm not looking TOO closely at your formulas, but remember that from Cauchy's Integral Theorem, it's not an issue of being more or less singular: all you are ever concerned with is the 1/x term (to the first power!). That is why you can usually get away with dropping other terms. Any other term always vanishes in the line integral.

That is:

$$\oint \frac{dz}{(z-x)^n}=0$$

for any integer $n\neq 1$

18. Dec 6, 2009

### nrqed

Yes, I agree. That's a good point. And because of this there is an upper limit to the value of l in the summation (I should have mentioned that, sorry). However, there is no lower limit on l. In the integrand, there is a piece 1/(z-x)^2 which is regular (z is outside of the contour). Therefore, we can Taylor expand this and generate a nonzero contribution for any negative power of l, no matter how small it is.

19. Dec 7, 2009

### nrqed

I realized my mistake. The field $\phi$ is a primary field, therefore the most divergent term in the OPE $T(z) \phi(w)$ is proportional to $1/(z-w)^2$. I was treating phi as a completely arbitrary field. I feel dumb! :-)

But the upshot of all this, going back to my first post in this thread, is that indeed we may do the OPE in any order we wish, but we need to keep an infinite number of terms in the expansions.