- #1

- 199

- 7

1 we add in the integral a delta(F) -meaning that we are going to integrate only in one representative of each class of equivalent configuration-

2 We take some factors out because they are constant and we don´t need them in the Path Integral Approach

3 We arrive to the same action than before but with a Gauge Fixing term "1/(2*epsilon)* (∂

_{μ}A

^{μ})

^{2}.

So the lagrangian given by 3 is the same QED lagrangian up to a multiplicative factor that should not change anything. Nevertheless, if we derive the equations of motion of this new lagrangian, I think that we get a different set of equations.

So, how can it be possible that we only get rid of some constant factor but we changed the path of the field configurations that makes the variation of the lagrangian equal to 0. What am I missing?

As always, thanks in advance for your valuable help!!!!