After reading some confusing chapter of various string theory book I'm trying to construct the Tachyon vertex operator for superstring theory. I know that this is removed after GSO projection, but for the moment I would like to construct this would-be vertex. It seems very strange to me that I couldn't find a systematic and precise treatment of this argument. However it seems to me rather natural (thought I didn't understand quite well why) to put an operator of the form

$$ e^{-\phi} $$

where $\phi$ is the field used to bosonize $\beta\gamma$ system. This operator has conformal weight $h=1/2$, so it must not be the end of the story. However this starting point is confirmed by Polchinski (vol. 2 eq (10.4.22)). Now what else shoud I add? I was thinking some momentum $$ e^{ikX(0,0)} $$ as it is done for the bosonic tachyon. But this operator has conformal weight $h=1$ itself and so I would get total conformal weight $h= 1 + 1/2 = 3/2$.

Then on the book by Blumenhagen et. al. I found this sentence:

"The ground state of NS sector is thus $e^{-\phi}c(0)|0\rangle.$" Which seems to me even wrong because the conformal weight of $c$ is $h=-1$

I know that I'm very confused about these superstring vertex operator. This is because I didn't find any book in which there is a comprehensible treatment. It would be of great help if someone provide a solution to this construction.


The tachyon vertex operator is

$$ c\,e^{-\phi}\tilde{c}e^{-\tilde{\phi}}\,e^{ik.X} $$

where the $z$-frame ghost number is $(-1,-1)$, the picture number is $(-1,-1)$ and the conformal weight $(h,\tilde{h})$ is $$ h=\tilde{h}=\left(\frac{\alpha 'k^2}{4}\right)_{X}+\left(\frac{1}{2}\right)_{\phi}+\left(-1\right)_{c}=\frac{1}{2}\left(\frac{\alpha 'k^2}{2}-1\right) $$

The BRST-closed vertex operator with picture number $(-1,-1)$ and $z$-frame ghost number $(-1,-1)$ should be of the form

$$ c\,e^{-\phi}\tilde{c}e^{-\tilde{\phi}}\mathcal{V} $$

where $\mathcal{V}$ is a $(1/2,1/2)$ superconformal tensor. In particular, the total conformal weights should be $(0,0)$, since $-1+1/2+1/2=0$. This determines the value of $k^2$ to be:

$$ -m^2=k^2=\frac{2}{\alpha'} $$

a tachyon of mass $m^2=-2/\alpha'$.

If you change to the picture $(0,0)$, there will be no $e^{-\phi}e^{-\tilde{\phi}}$ insertion, and the tachyon vertex operator will be:

$$ \frac{1}{4}(\alpha ')^2 k_{\mu}k_{\nu}\,\,c\psi^{\mu}\tilde{c}\tilde{\psi}^{\nu}e^{ik.X} $$

A $(+1,+1)$ picture tachyon vertex operator will be

$$ e^{+\phi}e^{+\tilde{\phi}}\frac{1}{4}(\alpha ')^2 k_{\mu}k_{\nu}\,\,c\left(i\partial X^{\mu}+\frac{1}{2}\alpha' k_{\rho}\psi^{\rho}\psi^{\mu}\right)\tilde{c}\left(i\bar{\partial} X^{\mu}+\frac{1}{2}\alpha' k_{\sigma}\tilde{\psi}^{\sigma}\tilde{\psi}^{\nu}\right)e^{ik.X} $$

and so on. Always with $(0,0)$ conformal weight. This is not a problem because the ghost breaks unitary, so non-trivial local operators with zero conformal weight are allowed.

Note also that the world-sheet fermion number $(-1)^{F}$ is always the same since both $e^{n\phi}$ and $\psi$ carry a fermion number $-1$, assuming that $n$ is an integer (that is the case in the NS sector). The picture changing always substitute one $e^{n\phi}$ for one $\psi$ or vice versa.


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