I am currently working through chapter 12 of Polchinski and am confused as to how the equation $(12.3.39)$ for the (0,0) picture vertex operator arises. From the text:

The state–operator mapping in chapter 10 gave the NS–NS vertex operators as $$\delta(\gamma)\delta(\tilde\gamma) = e^{-\phi-\tilde\phi} $$(12.3.38) times a $(\frac{1}{2},\frac{1}{2})$ superconformal tensor. These are the analog of the fixed bosonic vertex operators. We have seen that the superconformal tensors are the lowest components of superfields, which do indeed correspond to the value of the superfield when $\theta$ and $\overline \theta$ are fixed at 0. Calling this tensor $\mathcal{O}$, eq. (12.3.15) gives the vertex operator integrated over $\theta$ and $\overline\theta$ as $$ \mathcal{V}^{0,0}=G_{-1/2}\tilde G_{-1/2}\cdot \mathcal{O} \tag{12.3.39}$$This operator appears without the $\delta(\gamma)\delta(\tilde\gamma)$.

Here the index $(0,0)$ denotes the picture, and $\mathcal{O}$ is part of the superfield

$$ \phi=\mathcal{O}(z) + \theta\Psi(z)$$ The equations (12.3.15) are

\begin{align} &G_{-1/2}\cdot \mathcal{O} = \Psi, \quad G_r \cdot \mathcal{O}=0,r \geq\frac{1}{2} \\& G_{-1/2}\cdot\Psi = \partial \mathcal{O}, \quad G_{1/2}\cdot \Psi = 2h\mathcal{O}, \quad G_r \cdot\Psi=0, \, r\geq\frac{3}{2} \end{align} $\mathcal{O}$ is a tensor of weight $h$, and $\Psi$ is a tensor of weight $h+\frac{1}{2}$. I do not see how these equations give the vertex operator integrated over the anticommuting variables. Any help would be appreciated, thanks.


1 Answer 1


I figured it out. $\mathcal{V}^{0}$ is the vertex operator integrated over $\theta$. Integrating $\phi$ over $\theta$ gives us $\Psi$, and we can use equations (12.3.15) to rewrite $\Psi$ as $G_{-1/2}\cdot \mathcal{O}$.

In order to get the $\tilde G_{-1/2}$ factor for the $\mathcal{V}^{0,0}$ operator we must expand $\phi$ further,

$$ \phi = \mathcal{O} +\theta\Psi + \overline \theta \overline \Psi+\theta \overline \theta W$$ There is then an additional term in the equation for $\delta\Psi$, namely $\overline\eta W$, which by the Ward identity gives the relation $\tilde G_{-1/2}\Psi=W$. So $\mathcal{V}^{0,0}=G_{-1/2} \tilde G_{-1/2}\cdot \mathcal{O}=W$, which is indeed the only term from $\phi$ that survives the integration over $d^2 \theta$.


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