# Significance of massive states in string theory

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A free superstring has an infinite tower of states with increasing mass. The massless states correspond to the fields of the corresponding SUGRA. In "Quantum Fields and Strings: A Course for Mathematicians", vol. II p. 899 we find that the massive states do not contribute anything new to the possible string backgrounds. Terms in the string action corresponding to coupling to a massive background field are nonrenormalizable and therefore disappear when we RG-flow to the IR fixed point, which is the CFT we actually use in quantum string theory. Actually it is explained for the bosonic string but I don't think the difference is essential

What is the physical meaning of this result?

Does it mean massive string states are solitons of the massless fields? If so, do these solitons exist in classical SUGRA?

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asked Jan 21, 2012

## 2 Answers

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Massive string modes have masses of order the string mass $M_s$, independent of the string coupling $g_{str}$, whereas solitons have masses of order $1/g_{str}$ or $1/g_{str}^2$, depending on whether they are open or closed string solitons. So that putative matching does not work (the exponential degeneracy of massive string states would be another obstacle).

I believe the statement you are referring to does not have the wide ranging implications you draw from it, it has to do specifically with mechanics of computing S-matrix elements via string perturbation theory. In such computations in the background of massless modes, the contribution of the massive strings is already accounted by the usual procedure of summing over Riemann surfaces. This is explained nicely by this classic paper by Dine and Seiberg, Microscopic Knowledge From Macroscopic Physics In String Theory.

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answered Jan 26, 2012 by (2,405 points)
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Thx, do you know whether this article can be obtained freely online?

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I looked at the article but I am still confused, probably due to my own stupidity. I find it hard to reconcile the following 3 statements (maybe one of them is wrong):

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1. The moduli space of (appropriate) CFTs is the space of solutions of the "classical string equations of motion"

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2. String theory "regarded as a field theory" has an infinite number of fields

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3. The choice of a CFT has the same number of degrees of freedom as the choice of solution for the equations of motion of the massless fields only, of which there is a finite number

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I think the distinction you are making is between linearized perturbations and exact solutions. Even in ordinary nonlinear field theory, like GR, the phase space of the theory is spanned, to leading order, by the linear perturbation (say plane wave solutions of the linearized eom), but the exact phase space may be completely different (e.g. Smaller). In ST the linear perturbations are massive string modes, and exact CFTs are solutions to the full eom.

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OK but the space of linearized perturbations is the tangent space of the phase space so it must have the same dimension (of course dimension is infinite either way but heuristically the number of degrees of freedoms must be preserved). Of course this has to take gauge symmetry into account.

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The interactions of a massive particle fall off exponentially with distance (massless particles have long-range interactions), the exponent determined by the mass. Mathematically, this dependence is governed by the quadratic term of the field in the action.

Now let's lump all fields together into a multi-index field. Then the vacuum state(s) corresponds to the minimum energy configuration(s), and the nearby shape of the landscape around that minimum (those minima) is determined by the massive fields. Adding more of these fields won't change the space of minima or the long-range behavior of interactions.

Or is that too nontechnical and hand-wavy?

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answered Jan 26, 2012 by (385 points)

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