0
$\begingroup$

I read that in the limit of large $N$, the CFT on the boundary becomes classical. My question is if in such limit the physics in the bulk also becomes classical, or if we can still have a quantum field theory in the bulk (so that entanglement is possible in the bulk but not in the boundary).

$\endgroup$
1
  • 1
    $\begingroup$ It sounds like you read imprecise sources. AdS / CFT is concerned with situations where a classical bulk calculation gives the same answer as some (quantum) CFT at large N. $\endgroup$ Commented Sep 1, 2023 at 14:12

1 Answer 1

2
+50
$\begingroup$

The precise statement is that in AdS/CFT we have a relation between the parameters of the bulk and boundary theories. For example, in the first example of the correspondence Type IIB superstring theory on AdS$_5\times S^5$ with $N$ units of five-form flux has been related to ${\cal N}=4$ SYM theory with gauge group ${\rm SU}(N)$ subject to the relation between parameters: $$g_{YM}^2=2\pi g_s,\quad 2g_{YM}^2N=\frac{R^4}{\ell_s^4},$$

where $g_s$ is the string coupling, $g_{YM}$ is the Yang-Mills coupling, $R$ is the common radius of AdS$_5$ and $S^5$ and $\ell_s$ is the string length. It is also common to introduce $\lambda = g_{YM}^2N$ the t'Hooft coupling.

One can then study the large $N$ limit with either $\lambda$ fixed or taken to be large as well. In the first case $g_s\to 0$ and $\ell_s^2/R^2\neq 0$, so that one gets classical string theory in the bulk. In the second case, $g_s\to 0$ and $\ell_s^2/R^2\to 0$, so that we get classical supergravity, since the string length shrinks to zero compared to the AdS radius.

So what really happens is that the large $N$ limit of the CFT is connected to classical string theory or supergravity in the bulk, but nevertheless, there is no classical limit involved in the CFT side. We just have a quantum CFT at large $N$.

For more details I suggest the book Gauge/Gravity Duality by Martin Ammon and Johanna Erdmenger.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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