4
$\begingroup$

I am reading Witten's "Anti-de Sitter Space, Thermal Phase Transitions, And Confinement In Gauge Theories" (see here), in which he connects the confinement-deconfinement transition in $\mathcal{N}=4$ Super Yang-Mills theory with gauge group $SU(N)$ with the Hawking-Page transition via AdS/CFT.

My question is a purely field theoretical one, it is how I should understand the following statement:
"Deconfinement at high temperatures can be usefully described, in a certain sense, in terms of spontaneous breaking of the center of the gauge group" (p.5 of the above mentioned paper).

The center of $SU(N)$ is $\mathbb{Z}_N$ and is composed by the matrices $e^{2\pi i/n}\,\mathbb{I}$, where $\mathbb{I}$ is the unit matrix. As I understand, at high temperatures the vacuum state does not preserve this $\mathbb{Z}_N$-symmetry. How can one see this? What is the significance of the center of the gauge group in general? Why is it important that it is broken?

$\endgroup$

1 Answer 1

5
$\begingroup$

In order to understand the significance of center symmetry, consider Yang-Mills theory with dynamical gluons and static test quarks. Gluons are neutral with respect to the charge corresponding to center symmetry transformations, while test quarks are charged. One can now characterize the phase structure of the theory as follows:

In the deconfined phase, single test quarks can exist, while in the confined phase, there are only colour singlets, i.e. there are no free test quarks. This statement can be reformulated using free energy: in the deconfined phase, the free energy of a single static test quark is finite, while in the confined phase, it is infinite. The free energy $F$ of such a test quark can be described in terms of a Polyakov loop $\Phi$ (a Wilson loop around the Euclidean time direction). Its thermal expectation value is given by $\langle\Phi\rangle=\exp(-F/T),$ where $T$ corresponds to temperature. Finite free energy implies $\langle\Phi\rangle\neq0$, while infinite free energy corresponds to $\langle\Phi\rangle=0$. The action of the theory is center-symmetric at all temperatures, while the expectation value of the Polyakov loop is only invariant under center symmetry transformations if it is zero. Hence, the center symmetry is spontaneously broken in the deconfined phase, where $\langle\Phi\rangle$ acts as an order parameter for the phase transition.

Note that the behaviour of full QCD (including dynamical quarks) is much more complicated and the arguments above cannot be used in the same way.

$\endgroup$
2
  • $\begingroup$ Thank you! How can I see that the Polyakov loop is not center symmetric when its expectation value is not zero? I suppose that the gauge transformations of the center gauge group locally exchange quarks of different colours. Is it because the Polyakov loop is non-local? $\endgroup$
    – physicus
    Oct 27, 2014 at 9:49
  • $\begingroup$ @physicus: Simply write down the Polyakov loop and apply the transformation, which leads to a multiplication by a factor; hence it is not invariant. $\endgroup$ Oct 27, 2014 at 12:12

Your Answer

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

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