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  Why does tachyon arise in bosonic string theory?

+ 6 like - 0 dislike
  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?

This post imported from StackExchange Physics at 2014-09-07 04:36 (UCT), posted by SE-user user10001
asked Jul 6, 2013 in Theoretical Physics by user10001 (635 points) [ no revision ]
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 $m^2 \sim (26 - D)$

This post imported from StackExchange Physics at 2014-09-07 04:36 (UCT), posted by SE-user Trimok

1 Answer

+ 1 like - 0 dislike

Firstly, it is a misconception that tachyons do not exist in the naive (i.e. non-GSO, inconsistent) RNS superstring theory. It is just that, RNS superstring theory can go under a GSO truncation, leading to the Type IIB and Type IIA string theories, which do not have tachyons.

In the Bosonic String theory, the mass spectrum of closed strings is given by (in naturaol units where $\ell_s=\hbar =c_0=1$:

$$m=\sqrt{N+\tilde N-2}$$

$N$ and $\tilde N$ can only take discrete values as non-negative, either half-integers, or integers. For example, if you set $N=\tilde N = 0$, which is clearly for the ground state $|0\rangle $. :

$$m=\sqrt{-2}=\sqrt2 i$$

I.e. an imaginary mass. Therefore, a tachyon. This is also for the open string sector, whose mass spectrum is $m=\sqrt{N-1}$, then, when $N=0$, at the ground state $|0\rangle $, i.e., it' is a tachyon.

The same problem holds in the RNS string theory.

If you analyse the mass spectrum of the RNS superstring theory, you see that the same problem holds.

$$m=\sqrt{N+\tilde N - A}$$

Where $A=0$ in the RR sector, $A=1$ in the N-SN-S sector, and $A=\frac{1}{2} $ in both the RN-S and N-SR sectors. Clearly, in all sectors; but the RR sector, there is a tachyonic ground state.

This initiates the need for the GSO Truncation/(Projection).

answered Jul 27, 2013 by dimension10 (1,985 points) [ revision history ]
But the theory is inconsistent without GSO projection. The easiest argument is Witten's--- the theory has a gravitino, so it must be space-time supersymmetric, and therefore have no tachyon and come in equal number of bosonic and fermionic particles at each level. The GSO projection is really necessary, it is interpreted as summing over spin structures on the worldsheet, which means certain of the naive modes don't contribute.

This post imported from StackExchange Physics at 2014-09-07 04:36 (UCT), posted by SE-user Ron Maimon
@RonMaimon: Isn't that what I said, actually?; That there needs to be a GSO, or the naive RNS is wronmg.

This post imported from StackExchange Physics at 2014-09-07 04:36 (UCT), posted by SE-user Dimensio1n0
Oh, but then you shouldn't say "it's a misconception that ..." because then people think you are under the impression that RNS strings exist as theories. Ok, no worries, it's just miscommunication.

This post imported from StackExchange Physics at 2014-09-07 04:36 (UCT), posted by SE-user Ron Maimon
With other projections, one obtains the so-called type 0A and type 0B string theories. These have tachyons as their ground state and are non-supersymmetric. This paper (arxiv.org/abs/hep-th/0111212) argues that the end-point of tachyon condensation is the supersymmetric type II string. Tachyonic theories are not necessarily inconsistent -- they are unstable and the end-point of their decays, if one can determine them, are stable. Classic examples are the open-string tachyons discussed by Sen (arxiv.org/abs/hep-th/9805170).

This post imported from StackExchange Physics at 2014-09-07 04:36 (UCT), posted by SE-user suresh

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