conformal field theory - Correlator of a single vertex operator

In any textbook on CFT vertex operators $V_\alpha(z,\bar{z})=:e^{i\alpha\phi(z,\bar{z})}:$ are introduced for the free boson field $\phi(z,\bar{z})$ and their correlation function is computed $\left\langle V_{\alpha_1}(z_1,\bar{z_1})\dots \right\rangle=\prod_{i. Also, this equation holds only if $\sum_i{\alpha_i}=0$, otherwise the correlator is zero.

Consider now the correlator of a single vertex operator $\left\langle V_{\alpha}(z,\bar{z}) \right\rangle$. On the one hand, it should vanish as failing the neutrality condition. On the other hand, its expansion is $\left\langle V_{\alpha}(z,\bar{z}) \right\rangle=\left\langle 1 \right\rangle+\sum_{n>0}\frac{(i\alpha)^n}{n!}\left\langle :\phi(z,\bar{z})^n: \right\rangle$. My understanding is that all $n>0$ terms vanish by definiton of normal ordering, but why does the $n=0$ term, which is the identity, also give zero?

Answer

I don't think it's the case that all $n>0$ terms vanish, because the mode expansion of $\phi$ has a zero mode $\phi_0$. Its expansion is

$$\begin{equation}\phi \left(z,\bar{z}\right) = \phi_0 - i\pi_0 \log\left(z\bar{z}\right) +i \sum_{n\neq 0} \frac{1}{n} \left(a_n z^{-n} + \bar{a}_n \bar{z}^{-n}\right)\end{equation}$$

Computing $\langle:\phi^n:\rangle$ for $n>0$, the only term that contributes when we take the vacuum expectation value is $\phi_0^n$. This is because $a_n$ and $\bar{a}_n$ annihilate the vacuum for $n>0$, and $\pi_0|0\rangle=0$ as well. Any cross-terms involving $a_n$ and $a_{-m}$ will be zero due to the normal ordering, as will any terms involving $\phi_0$ and $\pi_0$ (as $\pi_0$ is placed to the right).

As a result, we just get

$$\begin{equation} \langle V_\alpha \left(z\right) \rangle =\langle \sum_{n} \frac{\left(i\alpha \phi_0\right)^n}{n!} \rangle= \langle e^{i\alpha \phi_0} \rangle. \end{equation}$$ Because of the commutation relations between $\pi_0$ and $\phi_0$, $e^{i\beta \phi_0} |\alpha\rangle = |\alpha+\beta\rangle$, so the vacuum expectation value is $\langle e^{i\alpha \phi_0}\rangle = \delta_{\alpha,0}$; this is just the charge neutrality condition.

It's easier to obtain this result by using the definition of normal ordering [see e.g. Di Francesco];

$$\begin{equation}V_\alpha = \exp\left(i\alpha \phi_0 + \alpha \sum_{n>0} \frac{1}{n}\left(a_{-n}z^n + \bar{a}_{-n} \bar{z}^n\right)\right) \exp \left(\alpha \pi_0 \log\left(z\bar{z}\right) - \alpha \sum_{n>0}\frac{1}{n} \left(a_{n}z^{-n} + \bar{a}_{n} \bar{z}^{-n}\right)\right).\end{equation}$$ The last exponential acts trivially on $|0\rangle$, and the $a_{-n},\bar{a}_{-n}$ with $n>0$ map $|0\rangle$ on to its descendants, which are orthogonal to $|0\rangle$. So when taking the vacuum expectation value, the operator is just $e^{i\alpha \phi_0}$ as before.

Alternatively, one can use the Ward identities; the Ward identity for translational invariance $\partial_z \langle V_{\alpha} \left(z\right)\rangle = 0$ means the correlator is constant. The Ward identity $\left(z\partial_z + h_{\alpha}\right) \langle V_{\alpha}\left(z\right)\rangle =0$ then implies that $h_\alpha \langle V_{\alpha} =0 \rangle$: since $h_\alpha = \alpha^2/2$ is non-zero for $\alpha \neq 0$, the correlator must be zero. If $\alpha=0$, $V_{\alpha} = 1$ and the correlator is just 1.