# 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?

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

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

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 $$$$\langle V_\alpha \left(z\right) \rangle =\langle \sum_{n} \frac{\left(i\alpha \phi_0\right)^n}{n!} \rangle= \langle e^{i\alpha \phi_0} \rangle.$$$$ Because of the commutation relations between $$\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]; $$$$V_\alpha = \exp\left(i\alpha \phi_0 + \alpha \sum_{n>0} \frac{1}{n}\left(a_{-n}z^n + \bar{a}_{-n} \bar{z}^n\right)\right) \exp \left(\alpha \pi_0 \log\left(z\bar{z}\right) - \alpha \sum_{n>0}\frac{1}{n} \left(a_{n}z^{-n} + \bar{a}_{n} \bar{z}^{-n}\right)\right).$$$$ The last exponential acts trivially on $$|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.

• Yep, I think this is exactly right! I first thought that even if $<:\phi^n:>$ terms are non-zero, they can't cancel the contribution of $<1>$, because they all contain some power of $\alpha$. As you point out, there is a similar computation in QM: $e^{ia\hat{p}}|x>=|x+a>$, so $<x|e^{ia\hat{p}}|x>=\delta_{a,0}$. Naively this can't be so since the null-contribution is $<x|x>=1$ (nevermind non-normalizability) while all others contain powers of $a$. Probably I don not have a good understanding of this formula, so I didn't recognize it in a different context. Mar 3 '19 at 10:09

normal order is$$:AB:=AB-\langle AB\rangle$$ so$$\langle:AB:\rangle=\langle AB\rangle-\langle AB\rangle=0$$

since average of single vertex operator is normal ordered it's automatically zero.

in your calculation the $$1$$ in the expansion is also an operator its an identity operator you should normally order it as well $$\langle :1:\rangle=\langle 1\rangle-\langle 1\rangle=0$$

• There are several definitions of normal ordering and this causes some confusion. I think the one you used is not common in CFT. For example, the following formula for vertex operators is often cited $:e^{i\alpha_1\phi(z_1,\bar{z}_1)}::e^{i\alpha_2\phi(z_2,\bar{z}_2)}:=:e^{i\alpha_1\phi(z_1,\bar{z}_1)}e^{i\alpha_2\phi(z_2,\bar{z}_2)}:e^{-\alpha_1\alpha_2\langle\phi(z_1,\bar{z}_1)\phi(z_2,\bar{z}_2)\rangle}$. With your prescription correlator of any number of vertex correlators would be zero, independent of the neutrality condition. Mar 1 '19 at 16:11
• if a and b depends on some coordintaes you have to take limint x-x'=0 so i think there is no problem. Mar 1 '19 at 16:52