Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I could understand the derivation of the "bulk-to-boundary" propagators ($K$) for scalar fields in $AdS$ but the iterative definition of the "bulk-to-bulk" propagators is not clear to me.

On is using the notation that $K^{\Delta_i}(z,x;x')$ is the bulk-to-boundary propagator i.e it solves $(\Box -m^2)K^{\Delta_i}(z,x;x') = \delta (x-x')$ and it decays as $cz^{-\Delta _i}$ (for some constant $c$) for $z \rightarrow 0$. Specifically one has the expression, $K^{\Delta_i}(z,x;x') = c \frac {z^{\Delta _i}}{(z^2 + (x-x')^2)^{\Delta_i}}$

  • Given that this $K$ is integrated with boundary fields at $x'$ to get a bulk field at $(z,x)$, I don't understand why this is called a bulk-to-boundary propagator. I would have thought that this is the "boundary-to-bulk" propagator! I would be glad if someone can explain this terminology.

  • Though the following equation is very intuitive, I am unable to find a derivation for this and I want to know the derivation for this more generalized expression which is written as,

$\phi_i(z,x) = \int d^Dx'K^{\Delta_i}(z,x;x')\phi^0_i(x') + b\int d^Dx' dz' \sqrt{-g}G^{\Delta_i}(z,x;z',x') \times$ $\int d^D x_1 \int d^D x_2 K^{\Delta_j}(z,x;x_1)K^{\Delta_k}(z,x;x_2)\phi^0_j(x_1) \phi^)_k(x_2) + ...$

where the "b" is as defined below in the action $S_{bulk}$, the fields with superscript of $^0$ are possibly the values of the fields at the boundary and $G^{\Delta_i}(z,x;z',x')$ - the "bulk-to-bulk" propagator is defined as the function such that,

$(\Box - m_i^2)G^{\Delta_i}(z,x;z',x') = \frac{1}{\sqrt{-g}} \delta(z-z')\delta^D(x-x')$

  • Here what is the limiting value of this $G^{\Delta_i}(z,x;z',x')$ that justifies the subscript of $\Delta_i$.

Also in this context one redefined $K(z,x;x')$ as,

$K(z,x;x') = lim _ {z' \rightarrow 0} \frac{1}{\sqrt{\gamma}} \vec{n}.\partial G(z,x;z',x')$ where $\gamma$ is the metric $g$ restricted to the boundary.

  • How does one show that this definition of $K$ and the one given before are the same? (..though its very intuitive..)

  • I would also like to know if the above generalized expression is somehow tied to the following specific form of the Lagrangian,

$S_{bulk} = \frac{1}{2} \int d^{D+1}x \sqrt{-g} \left [ \sum _{i=1}^3 \left\{ (\partial \phi)^2 + m^2 \phi_i^2 \right\} + b \phi_1\phi_2 \phi_3 \right ]$

Is it necessary that for the above expression to be true one needs multiple fields/species? Isn't the equation below the italicized question a general expression for any scalar field theory in any space-time?

  • Is there a general way to derive such propagator equations for lagrangians of fields which keep track of the behaviour at the boundary?
share|improve this question
    
I've re-edited backslash-box as backslash-Box (capitalized). Note that backslash-square is also possible. Also, there was a mistake in the big equation. The braces are not just braces - which are given a special role in TeX. They are written as backslash-braces (both left and right). –  Luboš Motl Jan 7 '12 at 8:23
    
@Lubos Thanks for the edits! Can you also answer the question? :) –  user6818 Jan 7 '12 at 21:01
    
This is taken from McGreevy's review, he actually explains that he solves the classical equations for a particular Lagrangian –  John Aug 30 '13 at 17:05

1 Answer 1

Your first question is just semantics; I agree that boundary-to-bulk is more intuitive.

The equation below your italicized question is the iterative solution to the field equations of the $\phi^3$ Lagrangian, to first order in the coupling constant. It would in general depend on the specific bulk theory. For example, if you had a $\lambda \phi^4$ interaction, you would find \begin{align*} \phi=\int dx K\phi^0+\lambda\int dz dx G\int dx_1dx_2dx_3 K_1K_2K_3\phi^0_1\phi_2^0\phi_3^0+\ldots, \end{align*} and so on.

The reason that finding the classical solution to the field equations is useful is that in the semiclassical limit, the bulk path integral with the source $\phi_0$ turned on can be written as \begin{align*} Z[\phi_0]=\int d\phi|_{\phi(z=\epsilon)=\epsilon^{d-\Delta}\phi_0} \exp(-S[\phi])\sim \exp(-S[\phi_{\text{cl}}]), \end{align*} where $\phi_{\text{cl}}$ is the extension of $\phi_0$ to a solution of the bulk field equations which is regular at the horizon. Differentiating both sides with respect to $\phi_0$ then evaluates the tree level Feynman diagrams of the boundary theory.

I think that the notation $G^{\Delta_i}$ is meant to show that $G$ depends on the masses of the fields (and therefore the dual dimensions of the bulk operators). Its limiting behavior can be seen from the next equation that you wrote, which in turn follows from Stokes' theorem.

share|improve this answer
    
I can see the iterative nature of the solutions but that is only intuitive. Is there is a clean derivation for the expansion in the spirit of deriving Feynman diagrams? In Feynman diagrams the situation is much cleaner since there one is taking of an expansion for the propagators (and not the fields like here!) and the propagators are defines as derivatives of the partition function which has a natural power-series form. I can't see this structure in AdS/CFT. May be I am missing something! –  user6818 Jan 8 '12 at 21:02
    
I clarified my answer a little, hope this helps. –  Matthew Dodelson Jan 8 '12 at 21:34
    
You start with writing the equations of motion, say $(\square +m^2)\phi= g \phi^2$ for the $g\phi^3/3$ potential. Then you formally expand the solution in coupling $g$, $\phi=\phi_0+g\phi_1+..$ solve the linearized equations $(\square +m^2)\phi_0=0$ by $\phi_0 =K \bar{\phi}$, $\bar{\phi}$ is the boundary data. Then to the order $g$ you find $(\square +m^2)\phi_1= \phi_0^2$, hence $\phi_1=G \phi_0\phi_0=G K\bar{\phi} K \bar{\phi}$, so we recover the formula you wrote in the question –  John Aug 30 '13 at 17:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.