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S-matrix in String Theory

  1. Sep 23, 2013 #1
    Hi there!

    S-matrix is Path Integral with Vertex Operators inserted. I know how to compute Shapiro-Virasoro amplitude. So I don't have problems with calculations but with understanding.

    In this calculations formalism of 2-dimensional CFT is used. But there is no S-matrix in CFT, only correlators (N-point functions).

    I can treat embedding of world sheet into Minkowski space-time like scalar conformal fields with color indices. In this sense it is pure CFT where again no S-matrix is available. In QFT we have assymptoticaly free particles, but due to conformal invariance we can't build such states in CFT.

    What I actually compute when I compute Polyakov's path integral with vertex operators?

    -Anatoly Korybut
     
  2. jcsd
  3. Sep 23, 2013 #2

    fzero

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    In a CFT, we have what is called the state-operator correspondence. What this means is time-dependent states in the CFT are in 1-1 correspondence with operator insertions acting on the vacuum state, ##|\mathcal{O}\rangle = \lim_{z\rightarrow 0} \mathcal{O}(z)|0\rangle##. The standard way to argue this is via the conformal mapping from the complex plane to the cylinder, which can be found, for example, http://web.physics.ucsb.edu/~phys230A/lectures/wk5_virasorouir.pdf [Broken]. This carries over directly to the S-matrix, where we can write a matrix element for given in and out states in terms of a vacuum expectation value of the corresponding correlator.
     
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  4. Sep 23, 2013 #3
    This point is not clear.

    Vertex operators generate single particle/string states without interaction, so this states are obviously free. And these states are our string |in> and |out> states. How to proceed to S-matrix from this point?

    If I understand you right. If we look from state point of view this is pure correlator, from the operator this is string S-matrix
     
    Last edited: Sep 23, 2013
  5. Sep 23, 2013 #4

    fzero

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    The CFT state-operator correspondence is not quite the same as that familiar construction from ordinary QFT. The correspondence here is between local operators and the momentum eigenstates that are generated by the creation operators. From the ordinary QFT point of view, the creation operators are certain Fourier transforms of the local fields, which is a very nonlocal object. In CFT, the correspondence is different.

    We can illustrate this better for a free scalar field, where we have the primary operator ##\partial X(z)##. In the path integral formalism, creating a state (in the corresponding module) on the circle ##|z|=1## corresponds to

    $$ | \partial^m X\rangle \sim \int^{X_f(r)} \mathcal{D} X e^{-S[X]} \partial^m X(z=0).$$

    The integral is over the field configurations with appropriate boundary conditions on the circle. The annihilation operators are

    $$ \alpha_n = \oint \frac{dw}{2\pi i} z^n \partial X(w).$$

    If we act on ##| \partial^m X\rangle## with one of these and apply the appropriate OPE, we find that

    $$ \alpha_n| \partial^m X\rangle \sim \delta_{nm},$$

    so that we can identify

    $$ | \partial^m X\rangle \sim \alpha_{-m}|0\rangle.$$

    This is the operator-state correspondence. It's clear that it doesn't have anything to do with Fourier transforms.

    This correspondence also has nothing to do with ignoring interactions. I chose the free scalar because it made things simpler, but the formalism is just as well suited to interacting theories, where the appropriate OPEs carry the information behind the computation.
     
  6. Sep 24, 2013 #5
    I think I got the idea.
    Due to locality of Vertex Operators I can generate not just a state of a certain momenta but can actually put this state at a certain place.

    My problem was with understanding of constracting asymptotically free states which are needed in QFT as initial and final. In CFT this is done by Vertex Operator and I don't need anything else, don't need to take particles to infinity because operator that generates the state is local.

    Thank you very much for making this tricky(for me) issue clear.
     
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