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- Thread starter nrqed
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In summary, there is a rule known as Wick's Rule for computing an OPE when there are several fields in a Conformal Field Theory (CFT). This rule involves taking all possible pairwise contractions, followed by combinations of two pairwise contractions, and so on. However, this rule only works for free fields and is not applicable to CFTs where interactions cannot be treated as a perturbation. Additionally, the OPE is only valid within a certain radius of convergence, and it may be necessary to take into account regular terms when evaluating the OPE of multiple fields.

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xepma

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

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

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

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.

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Ben Niehoff

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[tex]\mathcal{R}abcd = (ab)cd + (ac)bd + (ad)bc + (bc)ad + (bd)ac + (cd)ab + (ab)(cd) + (ac)(bd) + (ad)(bc)[/tex]

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.

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xepma

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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.

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xepma said:

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

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

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

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.

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.

Thanks for your input!

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DrFaustus

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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...

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That's a very interesting point you made there!xepma said: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 [itex]T(z_2)[/itex] and [itex]\Phi(z_3)[/itex] contains the second descendant of [itex]\Phi(z_3)[/itex], call it [itex]\Phi^{(-2)}(z_1)[/itex]. The most singular term in the OPE of [itex]T(z_3)[/itex] with this field is (I looked it up in DiFrancesco)

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

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

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.

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

[tex] \frac{c}{2} \frac{\Phi(z_3)}{(z_1-z_2)^4 } [/tex]

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

But wait! That's actually the same, isn't? When we do the OPE of [itex] T(z_2) \Phi(z_3) [/itex],if we get a regular term of the form [itex]\Phi^{-2}(z_3) [/itex] we may as well repace it by [itex] \Phi^{-2}(z_2) [/itex]. And if we use that to do the OPE with the remaining [itex] T(z_1) [/itex], 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 [itex] z_2-z_3 [/itex], we have

[tex] \frac{1}{(z_1-z_3)^4 } \simeq \frac{1}{(z_1-z_2)^4} [/tex]

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

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Hope someone may clear things up a bit!

- #9

Haelfix

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xepma said: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.

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

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blechman

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nrqed said:

Hope someone may clear things up a bit!

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!

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blechman

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nrqed said:...

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

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

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blechman said: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!

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!

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Count Iblis

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Count Iblis said:

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

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

Touché!

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

[tex] \oint_w dx \frac{1}{(x-w)^{n-1}} T(z) T(x) \phi(w) [/tex]

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

[tex] \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) [/tex]

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

[tex] T(x) \phi(w) = \sum_l (x-w)^{l-2} \phi^{-l}(w) [/tex]

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

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

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

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blechman

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That is:

[tex]\oint \frac{dz}{(z-x)^n}=0[/tex]

for any integer [itex]n\neq 1[/itex]

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blechman said:

That is:

[tex]\oint \frac{dz}{(z-x)^n}=0[/tex]

for any integer [itex]n\neq 1[/itex]

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.

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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.

OPEs, or operator product expansions, are mathematical tools used in conformal field theory to describe how the correlation functions of operators behave as the distance between them approaches zero. They allow us to understand the behavior of operators in a conformally invariant theory.

OPEs are useful because they allow us to calculate correlation functions of operators in a conformally invariant theory that would otherwise be very difficult to compute. They also provide a way to classify and organize operators in a theory, making it easier to study and understand the properties of the theory.

OPEs are closely related to conformal symmetry, as they are a consequence of the conformal symmetry of a theory. In fact, conformal symmetry is required for OPEs to exist. OPEs also provide a way to understand and analyze the behavior of operators under conformal transformations.

Yes, OPEs can be used in non-conformal field theories, but only in certain cases. OPEs are only valid in theories that have a well-defined conformal limit, meaning that the theory approaches a conformal theory in a certain limit. In these cases, OPEs can still be used to understand the behavior of operators, but they may not have the same usefulness as they do in conformal field theories.

There are some limitations to using OPEs in conformal field theory. For example, they are not always valid for non-local operators or in theories with a large number of fields. Additionally, there are certain situations where OPEs may not be well-defined, such as in theories with logarithmic operators. However, in general, OPEs are a powerful tool for understanding conformal field theories and their behavior.

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