# Help on QFT

1. Aug 1, 2004

### marlon

Can anyone give me an explanation in layman-terms of the concept of a Polyakov-loop in QFT? What does it mean fysically. I know that this is often used in lattice QCD, but why ???

I have the same questions for a socalled Ward-identity

regards
marlon

2. Aug 1, 2004

### vanesch

Staff Emeritus
I think you are beyond me concerning lattice QCD and so on (and in QFT in general!). But I can say something about the Ward identity - but maybe you know this very well. The Ward identity (not the Ward-Takahashi identity ; we suppose the external fermion lines on shell) states that a complicated piece of Feynman diagram to which you couple a photon can be written as epsilon_mu(k) M^mu(k), then k_mu M^mu(k) = 0. M^mu is the electromagnetic current that will couple to our photon, and the ward identity guarantees that this current is conserved. So the Ward identity just says that no matter how complicated the interaction with and EM field, the part that acts as a source is a conserved current. This has then the advantage that we NEVER couple to an eventual longitudinal mode in the EM field. At least I think so

cheers,
Patrick.

3. Aug 1, 2004

### marlon

hmm, i don't get it to be honest.

Doesn't a foton always couple to an EM-current ??? Fotons make up the that EM-current. So why tlaking about their coupling that always is there.

Can a foton not couple to an EM-current.

Anybody knows something interesting on Polyakov-loops ??? What are they in QFT ?

regards

marlon

4. Aug 2, 2004

### vanesch

Staff Emeritus
We're talking next to eachother I'm affraid. I give it one more try.

The EM field (A^mu) couples to something, and that's called the EM current. The EM current is of course NOT the EM field (or the photons). It is the charged fermionic (and if you want to, bosonic!) matter that constitutes the current.
Classically, charge is conserved, and the Ward identity is this claim on the quantum level (that means, the part of any set of Feynman graphs that couple to a photon). After fourier transformation, current conservation (d_mu J^mu = 0) is written k_mu J^mu = 0.
The nice thing about the fact that this identity also holds on the quantum level is that _even if there were a longitudinal component in the EM field_ it wouldn't couple to any matter, so you only have to consider the two transversal degrees of freedom of the photon (epsilon^1 and epsilon^2, spacelike and perpendicular to k_photon).
But notice the peculiarity: this transversality of the photon's degrees of freedom doesn't come from the EM field itself, but rather from the fermionic matter fields that couple to it.

cheers,
patrick.

5. Aug 2, 2004

### marlon

nice, very nice. Now i got the point.

Thanks for this cristalclear explanation

any ideas on this Polyakov-loop-thing yet ???

regards
marlon

6. Aug 3, 2004

### vanesch

Staff Emeritus
I should add something. I wrote:
In fact, when you have massless fermions, there is not only a conserved current, but there is also a conserved CHIRAL current (for memory, the current is psi-bar gamma^mu psi and the chiral current is psi-bar gamma-5 gamma^mu psi). At least classically. The funny thing is that this chiral current conservation is NOT conserved on a quantum level, so there is no quantum equivalent of the Ward identity for the chiral current conservation. This observation, where a classical symmetry is gone on the quantum level is called an anomaly, in our case the chiral anomaly.

cheers,
Patrick.