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Consider the CFT that corresponds to a gauge-fixed closed bosonic string.

Ground level string states are described by vertex operators such as $$V(p) = :\exp(i p_{\mu} X^{\mu}(z, \bar{z})):$$ which are conformal primaries with weight $$ h = \bar{h} = \frac{\alpha'}{4} p^2. $$

The physical states of the strings must have $h = \bar{h} = 1$, therefore, the physical ground state is the tachyon.

Consider the 3-point function of three ground state operators:

$$ G_{p_1, p_2, p_3}(z_1, z_2, z_3; \bar{z}_1, \bar{z}_2, \bar{z}_3) = \left< V(p_1)(z_1, \bar{z}_1) V(p_2)(z_2, \bar{z}_2) V(p_3)(z_3, \bar{z}_3) \right>. $$

Because we're dealing with a free quantum field theory, it isn't hard to calculate this function exactly. Afaik this is called the "Coulomb gas" representation, and the expression is

$$ G_{p_1, p_2, p_3}(z_1, z_2, z_3; \bar{z}_1, \bar{z}_2, \bar{z}_3) = |z_{12}|^{\alpha' p_1 \cdot p_2} \cdot |z_{13}|^{\alpha' p_1 \cdot p_3} \cdot |z_{23}|^{\alpha' p_2 \cdot p_3}, $$

where $z_{ij} = z_i - z_j = -z_{ji}$, and $|z_{ij}| = (z_{ij} \cdot \bar{z}_{ij})^{1/2} = |z_{ji}|$.

However, I expect a general 3-point function to be completely fixed by global conformal symmetries – the Mobius group $SL(2, \mathbb{C})$. The general form for three primary fields with weights $h_i, \bar{h}_i$ is:

$$ \left< \phi_1(z_1, \bar{z}_1) \phi_2(z_2, \bar{z}_2) \phi_3(z_3, \bar{z}_3) \right> = C_{123} z_{12}^{h_3-h_1-h_2} z_{13}^{h_2-h_1-h_3} z_{23}^{h_1-h_2-h_3} \bar{z}_{12}^{\bar{h}_3-\bar{h}_1-\bar{h}_2} \bar{z}_{13}^{\bar{h}_2-\bar{h}_1-\bar{h}_3} \bar{z}_{23}^{\bar{h}_1-\bar{h}_2-\bar{h}_3}, $$

where $C_{123}$ is the 3-point structure constant of the CFT.

For any physical string state, $h = \bar{h} = 1$, therefore Mobius invariance requires that

$$ \left< \phi_1(z_1, \bar{z}_1) \phi_2(z_2, \bar{z}_2) \phi_3(z_3, \bar{z}_3) \right> = C_{123} z_{12}^{-1} z_{13}^{-1} z_{23}^{-1} \bar{z}_{12}^{-1} \bar{z}_{13}^{-1} \bar{z}_{23}^{-1} = C_{123} |z_{12}|^{-2} |z_{13}|^{-2} |z_{23}|^{-2}. $$

This seems to suggest that for the physical ground states participating in scattering (tachyons),

$$\forall i, j, i \neq j: \quad \alpha' p_i \cdot p_j = -2. $$

This restriction seems very odd to me, I've never seen anything like it before.

My questions are:

  1. Have I missed something crucial?
  2. If not, what is the origin of the constraint on the values of tachyon momenta?
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    $\begingroup$ hint: momentum conservation .. $\endgroup$ Commented May 24, 2019 at 14:18
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    $\begingroup$ @Wakabaloola of course, that’s what I was missing! Three spacelike vectors of the same length that sum to zero can only be arranged such that the angle between any two is $2 \pi /3$. That totally makes sense, thank you. Now how about writing this up as an actual answer so that I can upvote and accept it? :) $\endgroup$ Commented May 24, 2019 at 14:33

1 Answer 1

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Using momentum conservation and onshell conditions, $$ p_1+p_2+p_3=0, \qquad p_1^2=p_2^2=p_3^2=\frac{4}{\alpha'}, $$ it follows that for any $i\neq j\neq k$, \begin{equation} \begin{aligned} 2p_i\cdot p_j &= (p_i+p_j)^2-p_i^2-p_j^2 \\ &=p_k^2-p_i^2-p_j^2\\ &=-\frac{4}{\alpha'}, \end{aligned} \end{equation} so this relation is really just kinematics and onshell conditions.

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