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: $$ \{ψ_r,ψ^*_s \}= δ_{r,s}.\tag{3} $$


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 $|z|<|w|$.


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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.