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Taking this paper (Zinn-Justin: Six-Vertex, Loop and Tiling Modles: Integrability and combinatorics) as reference (chapter 1), I would like to ask a question. First of all some fixed points:

  1. The usual quantization condition (in a siutable signature): $$\{ψ(z),ψ(w)\} = δ(z-w);\tag{1}$$
  2. As (1.1), flippin a sign: $$ ψ(z)=∑_{k∈\mathbb{Z}+\frac{1}{2}}ψ_kz^{-k-\frac{1}{2}}, \qquad\qquad\qquad ψ^*(w)=∑_{k∈\mathbb{Z}+\frac{1}{2}}ψ^*_kw^{k-\frac{1}{2}};\tag{2} $$
  3. Then the anticommutation relation: $$ \{ψ_r,ψ^*_s \}= δ_{r,s}.\tag{3} $$

PROBLEM

I would like to get $(1)$ using $(2)$ and $(3)$ but

\begin{equation} \begin{aligned} \{ψ(z),ψ^*(w)\} &= ∑_{r∈\mathbb{Z}+\frac{1}{2}}∑_{s∈\mathbb{Z}+\frac{1}{2}}z^{-k-\frac{1}{2}}w^{k-\frac{1}{2}}\{ψ_r,ψ^*_s\}\\ &= ∑_{r∈\mathbb{Z}+\frac{1}{2}}z^{-r-\frac{1}{2}}w^{r-\frac{1}{2}}\\ &= w^{-1}∑_{n∈\mathbb{Z}}\left(\frac{w}{z}\right)^n \end{aligned} \end{equation}

in this last passage I set $r+\frac{1}{2}=n$. Now the problem is to evaluate that series. Of course if, naively

\begin{equation} \begin{aligned} ∑_{n∈\mathbb{Z}}\left(\frac{w}{z}\right)^n &= ∑_{n=0}^∞ \left(\frac{w}{z}\right)^n + ∑_{n=0}^{∞}\left(\frac{z}{w} \right) -1\\ &= \frac{1}{1-\frac{w}{z}} + \frac{1}{1-\frac{z}{w}} -1\\ &= 0, \end{aligned} \end{equation}

this apparent paradox, I think, is because of the radius of convergence of the series: the first converge if $|w|<|z|$, while the second if $|z|<|w|$.

QUESTION

How can I solve this paradox and get (with some $δ$-representation and analytic continuation) the result $(1)$?

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  • $\begingroup$ You are done, taking into account the singularity of the series for $w=z$. $\endgroup$ – Jon Jun 8 '17 at 13:19
  • $\begingroup$ How can I take into account it? Where does the delta come from? Then there is that $w^{-1}$ over all... $\endgroup$ – MaPo Jun 8 '17 at 13:21
  • $\begingroup$ You have just proven that the series is zero when $w\ne z$. Just note that when $w=z$ is infinity and this is nothing else than the definition of the Dirac's delta. $\endgroup$ – Jon Jun 8 '17 at 13:41
  • $\begingroup$ First of all the series never converges. Then the definition of delta is not just a function that is $∞$ when its argument is zero. Then when $z=w$ I get $w^{-1}∑_{n}1$, which regularization do you mean? $\endgroup$ – MaPo Jun 8 '17 at 13:46
  • $\begingroup$ I assume $\sum_n 1=\infty$ taking for granted your evaluation for $w\ne z$. Anyway, the best approach to check if the behavior is that of a delta is by a test function and integration. $\endgroup$ – Jon Jun 8 '17 at 14:40
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Lat us consider the series $$ \Sigma(z,w)=w^{-1}\sum_{n\in\mathbb{Z}}\left(\frac{w}{z}\right)^n $$ and take a test function $f(z)$. One has $$ \int_{-\infty}^\infty\Sigma(z,w)f(z)dz= w^{-1}\sum_{n\in\mathbb{Z}}w^n\int_{-\infty}^\infty\frac{f(z)}{z^n}dz. $$ Moving to the complex domain and choosing a proper path, we recognize here the coefficients of a Laurent series and, indeed, our integral is just proportional to $f(w)$. So, we can identify $\Sigma(z,w)$ with $\delta(z-w)$, neglecting a possible multiplication constant.

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  • $\begingroup$ Is that $w^{-1}$ that worries me! $\endgroup$ – MaPo Jun 9 '17 at 11:07
  • $\begingroup$ No, indeed, as the coefficients of the Laurent series are given with $z^{-n-1}$. Check en.wikipedia.org/wiki/Laurent_series. $\endgroup$ – Jon Jun 9 '17 at 11:26
  • $\begingroup$ Yeah, of course since I have to pick the residue I need an extra power, you're right. I'm still a bit perplex about the question of convergence $\endgroup$ – MaPo Jun 9 '17 at 12:14
  • $\begingroup$ Indeed, about convergence I have been somewhat cavalier but consider that I am working with distributions and that $f(z)$ must belong to a set of well-behaved test functions. $\endgroup$ – Jon Jun 9 '17 at 12:17
  • $\begingroup$ Now I'm thinking about the contour: maybe there is a pole at infinity... $\endgroup$ – MaPo Jun 9 '17 at 12:24

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