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Gauge Field Tensor from Wilson Loop 
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#1
Aug1514, 06:35 AM

P: 4

It is possible to introduce the gauge field in a QFT purely on geometric arguments. For simplicity, consider QED, only starting with fermions, and seeing how the gauge field naturally emerges. The observation is that the derivative of the Dirac field doesn't have a welldefined transformation, because:
$$n^\mu \partial_\mu \,\psi = \lim\limits_{\epsilon\rightarrow 0}\big[\psi(x+\epsilon n)\psi(x)\big],$$ i.e. the derivative combines two fields at different spacetime points (having different transformation rules). We need to introduce a parallel transporter ##U(y,x)## that transforms as $$U(y,x) \rightarrow e^{ig\alpha(y)}U(y,x) e^{ig \alpha(x)},$$ such that we can adapt the defintion of the derivative into a covariant derivative, that transforms in a welldefined way: $$n^\mu D_\mu \,\psi = \lim\limits_{\epsilon\rightarrow 0}\big[\psi(x+\epsilon n) U(x+\epsilon n,x)\psi(x)\big].$$ From geometric arguments, it is straightforward to show that the parallel transporter is a Wilson line: $$U(y,x) = \mathcal{P}\,e^{ig\int\limits_x^y dz^\mu\, A_\mu(z)},$$ which introduces a new field, namely the gauge field ##A_\mu##. See e.g. Peskin & Schroeder Chapter 15 for more detail. However.. Where the interaction term ##\bar{\psi} A_\mu \psi## emerged in a natural way, I totally don't see how the kinetic terms emerge. The standard way to proceed, is to consider a Wilson loop (a Wilson line on a closed path), and use Stokes' theorem: $$\text{exp}\left\{ig\oint_\mathcal{C}dx^\mu\, A_\mu \right\} = \text{exp}\left\{ig\int_\Sigma dx^\mu \wedge dx^\nu\,\left(\partial_\mu A_\nu  \partial_\nu A_\mu \right)\right\},$$ where of course ##\partial_\mu A_\nu  \partial_\nu A_\mu \equiv F_{\mu\nu}##. In Peskin & Schroeder, they then consider a small rectangular loop, and see that in the limit ##\epsilon\rightarrow 0##, ##F_{\mu\nu}## is invariant. But what's the point? I mean, the transformation law for ##A_\mu## is easily calculated from the definition of the Wilson loop: $$A_\mu\rightarrow A_\mu+\partial_\mu \alpha,$$ making ##F_{\mu\nu}## invariant by definition: $$F_{\mu\nu}\rightarrow \partial_\mu A_\nu  \partial_\nu A_\mu +\square \alpha\square \alpha. $$ I would have liked to see a calculation, starting from a particular loop parameterisation, that naturally leads to the correct kinetic terms in the Lagrangian, as was the case for the interaction term. In other words $$\text{exp}\left\{ig\oint_\mathcal{C}dx^\mu\, A_\mu \right\} \leadsto \frac{1}{4}\left(F_{\mu\nu}\right)^2,$$ but I have no idea how to do it. Or is the idea simply 'look I have found some quadratic derivative terms that are invariant, now let me fiddle a bit and put its square in ##\mathcal{L}##'? If yes, then why do Peskin & Schroeder bother calculating a loop parameterisation (p484), if using Stokes' theorem would have been enough to find ##F_{\mu\nu}## somewhere? 


#2
Aug1814, 03:50 PM

P: 883




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