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why no double counting of s- and t-channels in string theory?

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In string theory for the four particle tree diagram exchange, why is there some mysterious crossing duality between the s- and t- and u-channels? Why isn't there a double counting in the Feynman diagrams? How is this consistent with a quantum field theoretic description?

This post has been migrated from (A51.SE)
asked Nov 3, 2011 in Theoretical Physics by York (20 points) [ no revision ]

1 Answer

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Let's consider the scattering of four (two to two) open strings, for the sake of concreteness.

Using Feynman's approach to quantum mechanics in terms of the sum over histories, string theory commands us to compute the tree-level diagram as the sum over all histories – world sheets – where two initial open strings become two other open strings.

By conformal transformations and the Wick rotation, each such history may be transformed to a disk diagram where the 4 external strings may be mapped to 4 points on the boundary of the disk.

The field theory limit is reproduced if these 4 insertions are pairwise close to each other. If 1-2 are close to each other as well as 3-4, we have an $s$-channel. Similarly for $t$-channel and $u$-channel. However, the string diagram unifies all these channels into a single integral; all these contributions are continuously connected to each other within string theory.

If one strictly rewrites string theory as a field theory with infinitely many fields and if one includes all the fields (infinitely many of them) that may occur as intermediate particles in the $s$-channels, then one may reproduce the Veneziano amplitude in this way. However, including all the infinitely many intermediate particles means that we are allowed to go away from the limit where 1-2 are close to each other, much like 3-4. The summing over all the heavy intermediate particles reproduces the whole stringy diagram with arbitrary positions of the 4 insertions.

That's why the $s$-channel, even in the field-theoretical language with infinitely many fields, already contains all the tree-level contributions and including the $t$-channel and $u$-channel diagrams separately would be a double- (or triple-) counting. This may be counterintuitive because in field theory, we're used to the idea that all three channels have to be included separately and again for unitarity.

However, string theory, because of its infinite number of fields and their new relationships, isn't quite equivalent to a field theory in this sense and one may check that the right unitarity is reproduced if the stringy diagram is only calculated once.

This post has been migrated from (A51.SE)
answered Nov 3, 2011 by Luboš Motl (10,178 points) [ no revision ]

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