It was said here that Veneziano derived:

open-string tachyon scattering amplitude from principles of Regge theory and S-matrix theory

and used the Euler beta-function to make all the critical conditions work. I know that this was very hard and I am curious to learn more about string theory and especially the maths behind it.

In simple terms and not using really complex or obscure mathematical notation where possible, what is meant by open-string tachyon scattering amplitude?

  • 1
    $\begingroup$ The bosonic open string has a tachyonic ground state with $m^2 = - \frac{1}{\alpha'} < 0$. The open-string tachyon that Veneziano derived is a $2\to2$ scattering amplitude of 4 such tachyonic states. $\endgroup$
    – Prahar
    Jan 24, 2016 at 15:27

1 Answer 1


Veneziano amplitude is a 4 tachyon amplitude in bosonic open string theory. Two tachyons are ingoing and two are outgoing. From the stringy point of view, tachyons are present in both closed and open bosonic theory and are the lowest particle in the spectrum, in particular they have negative mass squared: $m_\textrm{open}^2=\frac{-1}{\alpha'}$.

In general a 4-point open string amplitude in bosonic string theory is something like:

\begin{equation} \mathcal{A}_{\phi^4 }=g_s^{-\chi+n_c+\frac{1}{2}n_o} \int \frac{d^{4}z}{V_{CKV}} \langle V_{\phi}(z_1) V_{\phi}(z_2) V_{\phi}(z_3) V_{\phi}(z_4) \rangle \end{equation}

First of all you need to choose the surface for the string worldsheet on which you want to perform the calculation. Surface with high genus number correspond to high order in perturbative $g_s$ expansion, where $g_s$ is the string coupling constant (actually is a field, the dilaton). In the formula, $\chi=2-2g-b-c$ is the Euler Number, with $g$ the genus of the surface, $b$ the boundaries, $c$ the crosscaps. Veneziano amplitude is the analog of a tree level amplitude in field theory (no loops), so you have to take the surface with lowest genus and with 1 boundary (open strings lives on the boundary of the worldsheet, while closed strings are inserted inside the worldsheet), so we need the disk ($g=0, b=1, c=0)$. For the closed strings, tree level calculations are instead on the sphere. Notice that the disk is conformally equivalent to the upper half of the complex plane. $n_c$ and $n_o$ are the number of closed and open strings, so in this case $n_c=0$, $n_o=4$.

Then you insert a vertex operator $V_{\phi}$ for every particle. The vertex operator of tachyons, is very simple, it doesn't involve polarizations or other stuff:

$ \begin{equation} V_{\phi}(z_1)= e^{ik_1X} \end{equation}$

where $k_1$ is the tachyon momentum and $X$ the string coordinates. The open vertex depends on a real variable $z_1$. The amplitude is the integral over all the possible position of the vertex operators on the worldsheet, so $\mathrm d^4z=\mathrm dz_1 \mathrm dz_2 \mathrm dz_3 \mathrm dz_4\;.$ For open strings the order in which they are putted on the boundary of the worldsheet is relevant, so the final answer is the sum of all the possible inequivalent permutations of the positions.

The subtle part is the $V_{CKV}$, that is the conformal killing volume. To make it simple, you have a $SL(2,R)$ freedom to place your vertex on the disk. Basically you can fix 3 vertex positions, in an arbitrary way (for instance $z_1=-2$, $z_2=0$, $z_3=4$). Doing this, there is only one integration left. The integral left can be recast in terms of Beta functions or Euler functions and gives you the well known formula of the Veneziano Amplitude. Usually one use Mandelstam variables $s,t,u$ to express the string momenta, so you will find something like:

\begin{equation} \mathcal{A}_{\phi^4 } \propto \frac{\Gamma(-1-\alpha' s)\Gamma(-1-\alpha' t)}{\Gamma(-2-\alpha' (s+t))} + (t,u)+(u,s) \end{equation}

Magically, due to the property of Euler function that has infinite poles on negative integers, the poles of this amplitudes are exactly a tower of infinite massive particles. The interpretation is the that massive excitations are exchanged during the scattering process, in the same way as virtual particles are exchanged in QFT scattering.

The same amplitude in closed string theory is, modulo some subtlety, very similar. In some sense it's the same calculation repeated twice (left and right modes of the string).


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