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I would be glad if someone can help me understand the argument in appendix B.1 and B.2 (page 76 to 80) of this paper.

The argument in B.1 supposedly helps understand how the authors in that paper got from equation 3.6 to 3.8 on page 18 of the same paper. These 3 equations form the crux of the calculation done in this paper and unfortunately I am unable to see this.

In the appendix B.2 they calculate certain rational functions which are completely mysterious to me. Like I can't understand what it means to say that $\frac{1}{1-x}$ is the partition function of the operator $\partial$. Similarly one can see such functions in the equations B.7, B.8 and B.10 of that paper.

Curiously these polynomials had also appeared on page 6 and 7 in this paper long before the above paper. I am completely unable to understand how the series of polynomials between equation 15 to 20 of this paper were gotten and what they mean.

I haven't seen any book ever discuss the methods being used here. I would be glad to hear of some pedagogic and expository references on the background of all this.

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up vote 3 down vote accepted

Their convention for the partition function is explained in the second paragraph before the equation (B.7) of the paper by Shiraz et al. It is $$z(x) = \sum_{operators} x^{\Delta_{operator}}$$ Note that this is just a different way of writing the usual $\mbox{Tr }\exp(-\beta H)$ if you identify $\exp(-\beta)\equiv x$ and $\Delta\equiv H$. Yes, the dimension is the same thing as the Hamiltonian (of the radial quantization) and it is often helpful to avoid exponentials and write powers of $x$ only, so therefore the exponential redefinition of $\beta$ vs $x$. The trace is the summation over the basis.

They're calculating the partition function of a whole theory, not the $\partial$ operator itself. So the partition function is the sum over operators, as described above. In this simple case, the operators are $\partial_i \partial_j \dots \phi$, i.e. arbitrary derivatives of $\phi$ by $d$ different partial derivative symbols.

The derivatives with respect to different directions commute with each other and are completely independent. So imagine $d=1$ for a while, only one direction. In that case, you have operators $$\phi, \partial \phi, \partial^2 \phi, \dots$$ and their dimensions are $$\Delta=0,1,2,\dots$$ plus the dimension of $\phi$ if it were nonzero. The partition sum is the sum of $x^\Delta$ over these operators which means $$1+x+x^2+\dots = \frac{1}{1-x}.$$ The sum is obtained as geometric series. Note that the coefficients of the Taylor expansion are simply equal to one: there is no source where you could have gotten something else.

Now, the operators in the $d$-dimensional space may be obtained by acting with some derivatives in the 1st direction; some in 2nd, and so forth, on $\phi$. So the space of operators is a tensor product of spaces from each of the $d$ directions, and the partition sum is therefore the product of the partition sums from the individual directions, i.e. the $d$-th power of $1/(1-x)$.

There are other factors multiplying the total partition function but you haven't asked about it, and I can't explain every detail in a 50-page paper you haven't asked about. But yes, the other paper you mentioned almost certainly uses the same basic insight about the geometric series.

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Thanks for your reply. The partition function as defined initially is a trace of a single operator (namely $x^H$) which is basically sum over expectation values of it over a basis. Can you kindly explain why this is the same as summing over $x$ raised to the scaling dimension of all the operators allowed in the theory? I wonder if this has anything to do with the fact that after Eucleadianizing the confomal theory what behaves as a hamiltonian is not the $0-th$ component of the momentum operator but a sum of that with the $0-$ component of the special conformal transformations. – user6818 Feb 10 '11 at 15:46
But I can't carry this argument to its logical end to prove the implied equality as $Tr x^H = \sum _{operators} x^{\Delta _ {operators}}$. May be you can give me references towards this. – user6818 Feb 10 '11 at 15:48

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