# Why does tachyon arise in bosonic string theory?

1. I am looking for precise mathematical and physical reasons which cause the presence of tachyon in bosonic string theory(specially closed bosonic string theory). Has it to do with the specific form of the conformal field theory of free bosons?

2. In particular suppose one makes use of a conformal field theory of central charge 26 but different from the conformal field theory of 26 free bosons to formulate a string theory without local worldsheet supersymmetry. Will the tachyon problem still persist?

• For 1.1 : The zero mode of the set of harmonic oscillators (the string excitation) gives a negative energy for each dimension. So the ground state has a negative energy (if the "classical" center-of-mass momentum is zero) and a negative squared mass - $m^2 \sim (2 - D)$ - while $D=26$ for the coherence of the theory (there are $D-2$ excited states at first level, so it it a representation of $SO(D-2)$ which must be massless (the mass of the first level is $m^2 \sim (26 - D)$ – Trimok Jul 6 '13 at 19:31

The mass spectrum of closed strings in the bosonic theory is given by $m=\sqrt{N+\tilde N-2}$ (where $N$, $\tilde N$ are non-negative integers or half-integers), which clearly gives you ten imaginary-mass particles, e.g. at the ground state where $N=\tilde N=0$ etc.
The same problem would've held true with the RNS superstring, whose mass spectrum is $m=\sqrt{N+\tilde N - A}$ ($A=0$ in the RR sector, $A=1$ in the N-SN-S sector, and $A=\frac{1}{2}$ in the RN-S, N-SR sectors) -- but it turns out that the theory needs a GSO projection to remain consistent (see Ron Maimon's comments for an explanation), which also eliminates the tachyons.