# Fermion counting operator for open superstring

+ 2 like - 0 dislike
155 views

In Barton Zwiebach's A First Course in String Theory, in section 14.4, there's a fermion counting operator $(-1)^F$ which is supposed to give you $+1$ if the state is bosonic, or $-1$ if the state is fermionic. For a general state $\lambda$ he says

The eigenvalue of $(-1)^F$ on a state is equal to minus one times a sequence of factors of minus one, one for each fermionic oscillator that appears in the state.

This is probably a very trivial question, but is this extra minus one simply because the Neveu Schwarz ground state (the Neveu Schwarz vacuum) is chosen to be fermionic (and all states are constructed by applying appropriate creation operators on this fermionic vacuum)? Why is it conventionally chosen to be fermionic? At least the way Zwiebach has written, it looks like a choice.

This post imported from StackExchange Physics at 2015-03-10 12:56 (UTC), posted by SE-user leastaction
asked Mar 3, 2015

## 1 Answer

+ 3 like - 0 dislike

The NS ground state is choosen to be fermionic because we clearly want to call bosonic the first excited state $b^\mu_{-1/2}|0>$ which is a massless vector. Also, it you eventually want to do a GSO projection and throw away the fermionic states from the NS sector, you certainly want to throw away the vacuum because it is a tachyonic state.

answered Mar 16, 2015 by (5,110 points)

Nice answer. Maybe you can clarify something that has always confused me. Is the NS sector the one where we choose the spin structure on the circle that extends to the disc or do we choose the other one?

In the NS sector, we have by definition antiperiodic fermions: $\psi (\sigma + 2\pi)=-\psi(\sigma)$. The corresponding spin structure is thus the one corresponding to a connected double cover of the circle. This is precisely the one which extends to the disc. Indeed, antiperiodicity in $\sigma$ means half-integral modes in the Fourier expansion but going from the cylinder to the plane coordinate $z=e^{i\sigma-\tau}$, $\sqrt{d \sigma}$ becomes a constant times $\sqrt{dz}/\sqrt{z}$ and so $\psi(z)$ is a univalued function of $z$ (the same thing phrased differently: by general CFT, we know that a dimension 1/2 operator has an expansion of the form $\sum_n \psi_n /z^{n+1/2}$ and so $n$ half-integral corresponds to $z^{n+1/2}$ univalued). It is probably possible to give a direct geometric proof (think how the spin lift of the orthonormal bundle of the disk restricts to the circle) but I find the preceding argument safer (it is very easy to make an error modulo 2...)

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverflo$\varnothing$Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.