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Is there a reversed GKPW for AdS/CFT?

  1. Jun 8, 2015 #1

    Demystifier

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    In Ads/CFT, the famous GKPW equation gives a recipe how to calculate correlators in the boundary theory by using the bulk theory. But is there a reverse? What if I want to calculate the correlators in the bulk theory by using the boundary theory?
     
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  3. Jun 8, 2015 #2

    fzero

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    You use the same formula. However, unlike in a QFT generating functional, where we can describe general correlation functions of off-shell objects, here we can only calculate on-shell bulk amplitudes. On the boundary, we have correlators of off-shell operators, but the sources for the bulk fields are the on-shell boundary values of the fields. Explicitly, we can say something like for an ##n##-point function in the bulk

    $$ \int d\mathbf{x}'_1 dr_1 \cdots d\mathbf{x}_n dr_n A_{1\cdots n}(\mathbf{x}'_1,r_1;\ldots \mathbf{x}'_n,r_n) K(\mathbf{x}'_1,r_1; \mathbf{x}_1) \cdots K(\mathbf{x}'_n,r_n; \mathbf{x}_n) =\langle \mathcal{O}_1(\mathbf{x}_1) \cdots \mathcal{O}_n(\mathbf{x}_n)\rangle,$$

    where ##\mathbf{x}_I## are the boundary coordinates, ##r_I## is the radial variable and ##K(\mathbf{x}'_I,r_I; \mathbf{x}_I) ## are the boundary to bulk propagators. It is not expected that one can manipulate these expressions to obtain the bare ##A_{1\cdots n}##, but the amplitudes for all combinations of physical states are in principle determined from the boundary correlators.
     
  4. Jun 8, 2015 #3

    atyy

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  5. Jun 9, 2015 #4

    Demystifier

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    Fzero, thank you for your illuminating answer, in view of which I can reduce my question to the following one:
    How do I compute them? I mean, can I compute them by using only the boundary theory?
     
  6. Jun 9, 2015 #5

    fzero

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    As usual the propagator is obtained as the Green function for the appropriate equation of motion for the field in AdS. For example, with the upper-half space metric, the bulk-boundary function for AdS##_{d+1}## for a scalar field of mass ##m## is

    $$ K_\Delta(\mathbf{x}',z;\mathbf{x}) = \frac{z^\Delta}{(z^2+|\mathbf{x}'-\mathbf{x}|^2)^\Delta},$$

    where

    $$ \Delta = \frac{1}{2} ( d + \sqrt{d^2+4m^2 })$$

    is also the dimension of the dual operator in the boundary CFT.
     
  7. Jun 9, 2015 #6

    Demystifier

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    By the field, you mean the free field, right? But what about the interacting fields in AdS?

    EDIT: This problem seems to be solved perturbatively in the first paper linked by atty.
     
    Last edited: Jun 9, 2015
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