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  Gauge Field Tensor from Wilson Loop

+ 4 like - 0 dislike

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 well-defined 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 well-defined 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 $$\text{exp}\left\{ig\oint_\mathcal{C}dx^\mu\, A_\mu \right\} \leadsto -\frac{1}{4}\left(F_{\mu\nu}\right)^4,$$ 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?

This post imported from StackExchange Physics at 2014-08-15 09:33 (UCT), posted by SE-user freddieknets
asked Aug 15, 2014 in Theoretical Physics by freddieknets (20 points) [ no revision ]

look I have found some quadratic derivative terms that are invariant, now let me fiddle a bit and put its square in L? 

I think that's it, I have not heard of any deeper insight than just that. 

OP's unease comes from

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?

But, calculating from a small loop isn't really more complicated than using Stoke's theorem, remember Stokes theorem is proved exactly by this kind of small loop calculation.

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