I am having trouble understanding how Eq (2.6) in this paper (PDF) $$Z[\mathcal{L},\mathcal{M}_{n}]\propto\langle\Phi(u,0)\tilde{\Phi}(v,0)\rangle_{\mathcal{L}^{(n)},\mathbb{R}^{2}}$$

generalizes to Eqn (2.7) $$\langle\mathcal{O}(x,y;\mbox{ sheet i })...\rangle_{\mathcal{L},\mathcal{M}_{n}}=\frac{\langle\Phi(u,0)\tilde{\Phi}(v,0)\mathcal{O}_{i}(x,y)...\rangle_{\mathcal{L}^{(n)},\mathbb{R}^{2}}}{\langle\Phi(u,0)\tilde{\Phi}(v,0)\rangle_{\mathcal{L}^{(n)},\mathbb{R}^{2}}}$$

It's quite possible that some of you with more experience in CFTs will be able to immediately answer this for me without the context, but here it is anyway: we want to evaluate the partition function on the Riemann manifold $\mathcal{M}_{n}$ which consists of $n$ flat 2D sheets joined together at the branch cut between $u$ and $v$ in the manner shown in Fig 1 in the paper. We do this by modeling the manifold as $n$ disconnected flat sheets with twist fields inserted at the branch points. It turns out that the original partition function is proportional to the correlation function of two twist fields as in Eq. 2.6.

Later on, the paper makes use of the correlation function with insertions of the stress-energy tensor, so that generalization (with the equality sign) is crucial. Your help is appreciated!

In particular, how does this NOT mean that the partition function in Eq 2.6 is not actually proportional to the two-point function but is simply equal to one? (I am replacing the $\mathcal{O}$'s in (2.7) with one to make this claim)

  • $\begingroup$ Reading my question again, I'm slightly amused by how I instinctively use the pronoun "we" instead of the more honest "they". $\endgroup$ – dbrane May 6 '12 at 0:06
  • $\begingroup$ This may be out of my depth but I'll give it a look! $\endgroup$ – David Z May 8 '12 at 22:31

I think LHS of eqn 2.7 is normalized, meaning

$\frac{1}{Z}\int\mathcal{D}\phi \mathcal{O} exp\left(-S_{E}\left[\phi\right]\right)$ evaluated on $\mathcal{M}_{n}$

If you put $\mathcal{O} =1$, you get 1.

But $Z$ itself is proportional to the correlation function of the two primary fields

ref: http://arxiv.org/abs/hep-th/0405152

sect IIIA

Hope this is useful

  • $\begingroup$ You're right! I stupidly overlooked the fact that correlation functions aren't just path integrals with operator insertions but are normalized to the partition function, at least in the treatment by these authors. $\endgroup$ – dbrane May 10 '12 at 1:49
  • $\begingroup$ @dbrane: These formulas define the action of the twist fields. I didn't realize you were only confused about the normalization. $\endgroup$ – Ron Maimon May 10 '12 at 5:05

The second formula is not a generalization of the first, it is a simple consequence of the definition of the twist fields, like the first. These formulas are just defining the formal path integral for the Riemann surface, and then the "twist field" correlation functions are essentially defined by to reproduce the correlation functions. (I will be explicit about the definition, see below).

The paper uses the notation $\tau$ and $\bar{\tau}$ for the twist fields, instead of $\Phi$ and $\bar{\Phi}$, I'll use that. These fields don't appear in the Lagrangian, you don't integrate over them, they don't have dynamics, they do not represent particles of any kind, they aren't anything physical like that at all. These are made-up fields. They act to change the path integral boundary conditions using some cut lines, and they are local operators only in that inserting them depends only on the position where you insert them, not on how you draw the cut lines between them.

To see what they are, consider (following Cardy's explanation) n-copies of the given field theory, say a theory of a single scalar $\phi$, and then the n-copies are $\phi_i$ with i from 1 to n. Since you just duplicated the theory n-times with no interaction, you have an action which is the sum of the action of each field separately. This n-fold not-mutually interacting copies have an obvious permutation symmetry, any permutation of the fields is equivalent to any other. This cyclic symmetry is a no-brainer--- the fields are all the same, they each have the same action.

Consider the subgroup of cyclic permutations which shift field $i$ to field $i+1$, and suppose we use this symmetry to define a new field theory, in which there is a special horizontal line going between two horizontally separated points A and B. When you cross this line going up, field i turns into field i+1, when you cross going down, it turns back to field i.

To say that the field turns into another field is simply saying that the fields are "discontinuously" changing when crossing this line (I put discontinuously in quotes because quantum fields are always discontinuous, but the value at one point in the path integral is dependent on the values nearby, and in this case the field you depend on changes) --- the action in the path integral makes the value of the field $\phi_1$ just below the line related to the value of $\phi_2$ just above the line, and not at all related to the value of $\phi_1$ just above the line.

You can imagine this in a simulation of the free field theory. In such a simulation, you pick a lattice site, and replace the value at the site with the average of the same field at the four neighbors, plus a random gaussian value of a certain fixed width (depending on the lattice spacing). Right below the magic line, the up neighbor you average to find the desired value of field 1 is of field number 2, and similarly for field 2, you use field 3 above the line in the average, and so on cyclically.

This takes the n-copy theory on $R^2$ to the theory on a Riemann surface (by the definition of a Riemann surface with cuts), by definition.

But now notice something nice--- the line position is completely arbitrary, so long as the endpoints stay the same. If you move the line up, so long as it has the same endpoint, you can just locally redefine the lattice values using the cyclic permutation symmetry so that the theory with this crazy shift stays exactly the same. For example, suppose we move all the positions in the middle of the line up by one lattice spacing. For all the points that were previously above, but are now below, simply move the value of $\phi_i$ in any configuration to $\phi_{i-1}$. The cut still does the same thing, the partition function is exactly the same, but the cut is in a completely different position. This is the standard thing in Riemann surfaces--- the position of the cut is arbitrary.

What this means is that the theory with the cut can be thought of as a theory with two insertions, at either end of the cut. The property of these insertions is that when you take field $\phi_i$ around the left one counterclockwise, you end up at $\phi_{i+1}$, and if you do the same thing around the right one, you end up at $\phi_{i-1}$. This defines the twist fields $\tau$ and $\bar\tau$, which are acting at A and B. What they do is produce a start and end of a branch cut, and you can link the starts and ends to each other (or to infinity) and you have the same partition function.

So to find the correlation function with twist insertions:

  • put a cut in between the $\tau$'s and $\bar\tau$'s (any which way)
  • simulate the theory with the cuts.
  • find the expected value of O.

If O=1, then you just get the partition function on the Riemann surface (the partition function of the cut, which is just 1 in the simulation definition above, since in a probabilistic simulation Z=1). If you simulate a nontrivial operator O, you find the correlation function of O in the presence of the cut. This is what Cardy's two formulas mean.

The fact that you can interpret the twist fields as local operators is very useful, and allows you to find OPE's and solve the Riemann surface problem. But the formulas you give are simply definitions, and any confusion is in the high level picture defining these. I hope this clears everything up, but it isn't saying anything more or different than Cardy, except in different words. Perhaps this will click better.


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