Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Derivation of the basic equation for Witten diagrams

+ 6 like - 0 dislike
754 views

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?
This post has been migrated from (A51.SE)
asked Jan 6, 2012 in Theoretical Physics by user6818 (960 points) [ no revision ]
retagged Apr 19, 2014 by dimension10
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).

This post has been migrated from (A51.SE)
@Lubos Thanks for the edits! Can you also answer the question? :)

This post has been migrated from (A51.SE)

1 Answer

+ 2 like - 0 dislike

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.

This post has been migrated from (A51.SE)
answered Jan 7, 2012 by Matthew Dodelson (20 points) [ no revision ]
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!

This post has been migrated from (A51.SE)
I clarified my answer a little, hope this helps.

This post has been migrated from (A51.SE)

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOv$\varnothing$rflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...