# Normal Order in Free Boson Vertex Operator

In string theory, if we consider the CFT of a free boson $X$ and consider the (momentum) vertex operator:

$$V_k(z_1,z_2)=:e^{ikX(z,\bar{z})}:$$

Then we have the OPE:

$$:e^{ik_1 X(z_1,\bar{z_1})}::e^{ik_2X(z_2,\bar{z_2})}:~=|z_1-z_2|^{\alpha' k_1 k_2}:e^{ik_1X(z_1,\bar{z_1})}e^{ik_2X(z_2,\bar{z_2})}:$$

where $: :$ denotes normal ordering and $\alpha'$ is the Regge slope.

What I don't understand is the computation of VEV's of such vertex operators:

$$\langle0|V_{k_1}(z_1,\bar{z_1})V_{k_2}(z_2,\bar{z_2})|0\rangle=|z_1-z_2|^{\alpha' k_1 k_2}\langle 0|:e^{ik_1X(z_1,\bar{z_1})}e^{ik_2X(z_2,\bar{z_2})}:|0\rangle$$

it seems to me that this should vanish, as any VEV of normal ordered products. Nevertheless my String lecture notes say it doesn't.

Could anyone explain why this normal ordered VEV doesn't vanish?

• Comment to the question (v1): It is not true that any VEV of normal ordered products should vanish. E.g. for the identity operator $\langle0|:\hat{\bf 1}:|0\rangle~=~1$. Aug 4, 2016 at 12:56
• @Qmechanic I see, do you suggest that the non-zero part comes from the zeroth-power in the exponential? Then we would have simply $\langle 0 | 0\rangle$. Aug 4, 2016 at 13:05

$$\langle 0|:e^{ikX_1(z_1,\bar{z}_1)}e^{ik_2X(0,0)}:|0\rangle=\langle 0|\left(:e^{i(k_1+k_2)X(0,0)}:+\,O(z_1,\bar{z}_1)\right)|0\rangle$$
all the $O(z_1,\bar{z}_1)$ terms will contain operator that differ from $1$, so annihilate the $|0\rangle$ or $\langle 0|$ state. The first term does not annihilate the states if $k_1+k_2=0$, actually we get $\langle 1\rangle$. This is what is going to produce the conservation of momentum of the scattering amplitudes.
$$\langle0|V_{k_1}(z_1,\bar{z_1})V_{k_2}(z_2,\bar{z_2})|0\rangle=|z_1-z_2|^{\alpha' k_1 k_2}(2\pi)^{d}\delta^{d}(k_1+k_2)\langle 1\rangle$$