In Polchinski's exposition of the RNS formalism for the superstring (String Theory: Volume II, chapter 10), in page 8, he mentions the worldsheet fermion number operator, which he calls $F$. He then goes on to define \begin{equation} (-1)^F = e^{i \pi F}.\tag{10.2.19} \end{equation} To my best knowledge, this object is meant to commute with the worldsheet bosonic operators $X^\mu(z)$ and anticommute with the fermionic ones, $\psi^\mu(z)$ (this section of the book is focused only on the open superstring, so the antiholomorphic fermion is abscent). Since the fermions can be expanded as \begin{equation} \psi^\mu(z) = \sum_{r \in \mathbb{Z}+ \nu} \frac{\psi^\mu_r}{z^{r+ 1/2}},\tag{10.2.7} \end{equation} where $\nu =0$ in the R sector and $\nu=1/2$ in the NS sector, $e^{i \pi F}$ can only anticommute with $\psi^\mu(z)$ if it anticommutes with each one of the modes. This requirement can be written in the form \begin{equation} e^{i \pi F} \psi^\mu_r = - \psi^\mu_r e^{i \pi F} = \psi^\mu_r (-1) e^{i \pi F} = \psi^\mu_r e^{i \pi(F+1)}, \end{equation} which, upon expanding both sides, means that $$F \psi^\mu_r = \psi^\mu_r(F-1).$$ Here comes the problem: if he were to define the worldsheet fermion operator as \begin{equation} F^\prime = \sum_{r \in \mathbb{Z}+ \nu} \psi^\mu_r \psi_{-r \mu}, \end{equation} is is easy to see that indeed $$F^\prime \psi^\mu_r = \psi^\mu_r(F^\prime-1).$$ However, he defines it as \begin{equation} F = \sum_{a=0}^4 i^{\delta_{a,0}} \Sigma^{2a,2a+1},\tag{10.2.21+22} \end{equation} where \begin{equation} \Sigma^{\mu \rho} = - \frac{i}{2} \sum_{r \in \mathbb{Z}+ \nu} \left[ \psi^\mu_r , \psi^\rho_{-r} \right]\tag{10.2.20} \end{equation} are the spacetime Lorentz generators for the spinors. What is the purpose of this definition? This thing doesn´t even satisfy $[\psi^\mu_r,F]=\psi^\mu_r$, as is shown in equation (10.2.23), so I don't understand how this could be the fermion number operator.


1 Answer 1


Recall that $$\{ \psi_r^\mu , \psi_s^\nu \} = \eta^{\mu\nu} \delta_{r,-s}. \tag{1}$$ Let's start by trying your definition, \begin{equation} \begin{split} [ F' , \psi_r^\mu ] &= \sum_s [ \psi_s^\nu \psi_{-s,\nu} , \psi_r^\mu ] \\ &= \sum_s \psi_s^\nu \{ \psi_{-s,\nu} , \psi_r^\mu \} - \sum_s \{ \psi_s^\nu , \psi_r^\mu \} \psi_{-s,\nu} \\ &= \sum_s \psi_s^\nu \delta_{r,s} \delta^\mu_\nu - \sum_s \eta^{\mu\nu} \delta_{r,-s} \psi_{-s,\nu} \\ &= \psi_r^\mu - \psi_r^\mu = 0 . \end{split} \end{equation} So your definition does not work. Basically, notice that (1) implies $F' = \sum_r \psi_r^\mu \psi_{-r\mu} = \frac{D}{2} \sum_r 1$, so your $F'$ is (1) not well-defined and needs to be regulated and (2) is proportional to the identity operator and most definitely cannot act as a worldsheet fermion number operator.

The definition given in Polchinski actually works though. I suggest you work through the algebra again. It's very similar to the one shown here.

  • $\begingroup$ Thank you @Prahar. Indeed my $F^\prime$ was silly, but I still have not been able to show that Polchinski's $F$ satisfies $[F,\psi_r^\mu]=-\psi^\mu_r$. Using that $\Sigma^{2a,2a+1}=i\sum_s \psi^{2a+1}_{-s}\psi^{2a}_s$, I found \begin{align} [F,\psi^\mu_r] &= i \sum_s i^{\delta_{a0}} [ \psi^{2a+1}_{-s} \psi^{2a}_s , \psi^\mu_r ] \\ &= i \sum_s i^{\delta_{a0}} \big( \psi^{2a+1}_{-s} \{ \psi^{2a}_s , \psi^\mu_r \} - \{ \psi^{2a+1}_{-s}, \psi^\mu_r \} \psi^{2a}_s \big) \\ &= i \sum_s i^{\delta_{a0}} \big( \eta^{2a,\mu} \psi^{2a+1}_r - \eta^{2a+1,\mu} \psi^{2a}_r \big). \end{align} $\endgroup$
    – Bairrao
    Commented Sep 18, 2023 at 18:02
  • $\begingroup$ For $\mu=2b+1$ for instance, the above gives $[F,\psi^{2b+1}_r]=-i \psi^{2b}_r$. $\endgroup$
    – Bairrao
    Commented Sep 18, 2023 at 18:04
  • $\begingroup$ @Bairrao - The claim is not about $F$ at all. Rather, the statement is that $(-1)^F$ anticommutes with $\psi^\mu$. $\endgroup$
    – Prahar
    Commented Sep 18, 2023 at 18:47
  • $\begingroup$ @Bairrao - The property that you should expect for $F$ is also given in Polchinski in equation (10.2.23). First, check if you can reproduce that. Then use that to show anti-commutativity of $(-1)^F$ $\endgroup$
    – Prahar
    Commented Sep 18, 2023 at 18:54
  • 1
    $\begingroup$ That did it for me. I was assuming that the only way for $(-1)^F$ to anticommute with $\psi^\mu$ was to have $[F,\psi^\mu_r] = - \psi^\mu_r$. Thanks a lot. $\endgroup$
    – Bairrao
    Commented Sep 19, 2023 at 15:29

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