the_pulp said:
Ok, thanks for all your kind answers. I've read Peskin, Srednicky and other books of QFT but surely not in a rigorous way. As a consequence, I still have some doubts:
1) -diff(A0)^2 has a negative sign, so what? how do you arrive that it develops negative probabilities? How something like: <Φ|exp(i*-(∂A0)^2)|ψ> may give a negative number? Sorry, it seems something extremely trivial but I can't see it (there should be something extremely trivial that I can't see and it is bothering me).*
The easiest way to see it is actually not use the action (though we could). Let's just suppose that we take the field ##A_\mu## and introduce creation and annihilation operators ##a_\mu^\dagger(\vec{k}),a_\mu(\vec{k})## for modes with momentum ##\vec{k}##. Then the one particle states are ##|\vec{k},\mu\rangle = a_\mu^\dagger(\vec{k}) | 0\rangle##, where ##|0\rangle## is the zero particle ground state. The scalar product of one-particle states must be determined from Lorentz covariance to be
$$ \langle \vec{k} , \mu|\vec{k} , \nu \rangle = c \eta_{\mu\nu} $$
for some constant ##c##. Now the metric has indefinite signature, so if we choose the space-like polarizations to have positive norm, then the time-like polarization will have negative-norm. A negative norm is very bad. If we went back to position space we would find that the probability to find the time-like photon at a position ##\vec{x}## would be a negative number.
2) Supposing I get to follow your point 1), then, I can't see how the gauge principle solves this problem. At least, it would be good to know some reference or a short explanation. (Most probably this is my fault, it sounds to me that you are talking about things like BRST symmetry and stuff that I did not study in a very profound way)
Well BRST is a nice way to do it, but the more important point is that the gauge symmetry can be used to remove all of the negative norm states. It's not that hard to show, but it is a bit lengthy, so I would suggest you read pgs 133-135 of
http://www.damtp.cam.ac.uk/user/tong/qft/six.pdf I am happy to try to answer questions about it, but it just saves some time to let you read a well-organized explanation first.
3) Nevertheless I think that this argument works the other way around. I mean, I have negative probabilities, Gauge principle is a way in which this can be solved. However, what the opener is looking is something like (or, at least, what I've been looking for is) "Gauge principle is the only way in which this can be solved", but I guess that this is not the case, am I right?.
I think we are going in some circles. The QFT treatment of the photon does require gauge symmetry. The original question seemed to be why this also requires local phase symmetry, which is relevant only when we introduce matter. In any case, I do not know of any other way to solve the problem.
*While I was writting this I started to think that if <Φ|exp(i*(∂A1)^2)|ψ> is positive, then, it should be natural that <Φ|exp(i*-(∂A0)^2)|ψ> will be negative. I can't see the total link here but most probably your point is in this direction. Can you help me to complete / formalize the idea?
I'm not sure that I follow. ##A_\mu## should be treated as an operator when acting on states, and as we saw above, we'd actually have to specify the states we're using in the matrix element in order to comment on the positivity.