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
5,082 questions , 2,232 unanswered
5,353 answers , 22,786 comments
1,470 users with positive rep
820 active unimported users
More ...

  What is the exact relationship between on-shell amplitudes and off-shell correlators in AdS/CFT?

+ 7 like - 0 dislike
1325 views

In this answer to a question, it is mentioned that in the AdS/CFT correspondence, on-shell amplitudes on the AdS side are related to off-shell correlators on the CFT side.

Can somebody explain this to me in some more (technical) details, maybe by an explanatory example?

asked May 6, 2013 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]

2 Answers

+ 3 like - 0 dislike

First, a reference article, by Witten, http://arxiv.org/abs/hep-th/9802150v2.pdf

I'll try to expose the basic idea, with a flat space-time. Suppose you have a relativistic scalar field theory, on a flat space-time domain, with boundary. The equation of the field is :

$$\square \Phi(x) = 0$$ (fields on-shell)

Now, define the partition function

$$Z = e^{−S(\Phi)}$$, where $$S(\Phi) = \int d^nx \,\partial_i \Phi(x)\,\partial^i \Phi(x)$$ is the action for the field $\Phi$

After this, you make a integration by parts (using the above fied equation) , and Stokes theorem, and you get:

$$S(\Phi) = \int d^nx \, \partial_i \Phi(x)\,\partial^i \Phi(x) = \int d^nx \,\partial_i(\Phi(x)\,\partial^i \Phi(x))$$ $$= \int_{Boundary} d \sigma_i \,(\Phi(x)\,\partial^i \Phi(x)) $$

Now, suppose that the field $\Phi(x)$ has the value $\Phi_0(x)$ on the boundary. Then, you can see that $S$ and $Z$ could be considered as functionals of $\Phi_0$, so we could write $Z(\Phi_0)$:

$$ Z(\Phi_0) = e^{ \,( -\int_{Boundary} d \sigma_i \,(\Phi(x)\,\partial^i \Phi(x)))}$$

Now, the true calculus is not with flat space-time, but with Ads or euclidean Ads,so in your calculus, you must involve the correct metrics, but the idea is the same.

The last step is to say that there is a relation between, the Generating function of correlation functions of CFT operators $O(x)$ living on the boundary, and the partition function $Z$

$$<e^{\int_{Boundary} \Phi_0(x) O(x)}>_{CFT} = Z(\Phi_0)$$

The RHS and LHS terms of this equation should be seen as functionals of $\Phi_0$ You can make a development of these terms in powers of $\Phi_0$, and so you got all the correlations functions for the CFT operators $O(x)$

$$<O(x_1)O(x_2)...O(x_n)> \sim \frac{\partial^n Z}{\partial \Phi_0(X_1)\partial \Phi_0(X_2)...\partial \Phi_0(X_n)}$$

So, ADS side, we are using on-shell partitions functions (because field equations for $\Phi$ are satisfied)

Now, CFT/QFT side, the correlations functions $<O(x_1)O(x_2)...>$ are, by definition, off-shell correlation functions (by Fourier transforms, there is no constraint about momentum). To get scattering amplitudes, we simply need to put the external legs on-shell.

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user Trimok
answered May 9, 2013 by Trimok (955 points) [ no revision ]
Comment to the answer (v1): In the future, please link to abstract page rather than pdf file if possible, e.g. arxiv.org/abs/hep-th/9802150

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user Qmechanic
+ 1 like - 0 dislike

The statement can be understood in terms of the GKPW-formula (named after Gubser, Klebanov, Polyakov and Witten), which does exactly that: it relates correlation functions on the CFT side (boundary) to string amplitudes on the AdS side (bulk). Assume that $\phi(\vec{x},z)$ is some field in the bulk, where $z$ is the so-called "holographic coordinate", which reaches its CFT-value $\phi_0$ at the boundary at $z=0$. For some Operator $\mathcal{O}(\vec{x})$, the GKPW-formula is given by $$\langle e^{\int d^4x\,\phi_0(\vec{x})\mathcal{O}(\vec{x})}\rangle_{CFT}=\mathcal{Z}(\phi(\vec{x},z)|_{z=0}=\phi_0(\vec{x})).$$

On the left hand side, you have the correlation function for the CFT and on the right hand side, you find the partition function of the corresponding string theory, evaluated for the boundary value of the field.

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user Frederic Brünner
answered May 6, 2013 by Frederic Brünner (1,130 points) [ no revision ]
Brunner Is there a typo in your 2nd RHS? Shouldn't that $\phi_0 (\vec{x})$ be in the subscript giving the boundary condition for the bulk Z?

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user user6818
It is intended as it is, but one could also write it your way.

This post imported from StackExchange Physics at 2014-03-09 16:21 (UCT), posted by SE-user Frederic Brünner

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$\hbar$ysicsOverfl$\varnothing$w
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
...