As I understand, the Wilson line is the operator $W(x) = P\exp(i\int_{xi}^{xf} A.dx)$, where $P$ is path ordering. The Polyakov loop $P(x)$ on the other hand is the trace of the Wilson loop $W(x)$ along the time axis, i.e. = $Tr[T\exp(i\int_0^t A.dt)]$, where $T$ is time ordering.

Now, my question is: is there any relation between the Polyakov loop correlator $\langle P(0)P(d)\rangle$ and the Wilson loop around the closed rectangular curve of length $d$ along the spatial direction, and length $t$ along time axis?

  • $\begingroup$ FWIW, a Wilson loop is a color-traced closed Wilson line. $\endgroup$ – Qmechanic Feb 20 '19 at 15:01
  • $\begingroup$ @Qmechanic : Thanks, but any idea if it is related to the polyakov loop correlator? The two operators look very similar. The only difference seems to be in the trace. The Wilson loop has a overall trace, while the Polyakov correlator has two traces, one for P(0) and the other for P(d). Maybe there is some limiting case, where the trace in the Wilson loop becomes equal to the product of the traces in the polyakov correlator? $\endgroup$ – Angela Feb 21 '19 at 2:59

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