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