1. Jul 30, 2015

### Paul Draw

hi~I am a beginner of Gauge theories,and I have some questions~
Why we need loacal phase invariance?
Because i cannot understand why we assume phase depend on the position.Is there any theory leading to this asssumption?

2. Jul 30, 2015

### the_pulp

There are several ideas that may be related to the reason why the gauge principle is needed. I will write down some of them that, as I said, do not answer your question but may be hints related to it:

1) There are two famous theorems of Emily Noether. The most famous is his first theorem in which she discovers the relation between global symmetries and charges (that is to say, "things" that remains constant over time). Nevertheless, her second theorem talks about local symmetries and she finds that any action with local symmetries has related a Charge that it is, in a way, "more strong" than in the first case (in a theory with global symmetry, the existence of a charge is a consequence of the equations of motion that arise extremizing the action. In a theory with local symmetry, the charge exist not only if the matter field and the electromagnetic field satisfy the equations of motion but also if one of them does not satisfy it -the other fields adapts in order to "give birth to the charge"). Im not so sure if it is related to your problem but it sounds to me that when we go to the quantum behaviour of, for example, the electromagnetic interactions and we evaluate not only the path related to the equation of motion but also other paths (having in mind the path integral approach), the paths where there exists a charge "appear" more often than what they would do in a non Gauge theory.
2) There is a paper about it of Carlo Rovelli that studies the need of working with redundant variables (like in Gauge theories). In a way, he says that if we are working with, let´s say, just one field, there is no need to work with redundant variables, but when we work with, lets say, two fields, AND THEIR INTERACTION, then having redundant degrees of freedom is useful in order to make them interact.

There are probably more hints. Nevertheless I think that the best answer still is "It is not known".

PS: Im not a physicist, this is just a hobby for my. Please anybody correct myself if you see anything wrong (Im probably not very rigorous because of this and because I dont want to make sentences too long).

3. Jul 30, 2015

### fzero

Here's one argument for why gauge invariance requires local phase invariance:

1. Classical electromagnetism is invariant under both gauge and Lorentz symmetries. The gauge invariance is a local symmetry. The EM potential is shifted by a 4-gradient, but the source 4-current does not transform with the change of gauge.
2. If we want to quantize the EM field in a Lorentz covariant way, the quantum theory must also have local gauge invariance.
3. If we want to consistently couple a charged matter field to the EM field, the gauge symmetry on the EM potential must be enlarged to include a local phase transformation of the matter field, or else the Lagrangian will not be invariant.

4. Jul 30, 2015

### the_pulp

Sorry fzero, but I dont think that your answer is what it is being asked. I think that you start from classical electromagnetism and you deduce how QED should be. But I think that the opener is looking for a reason why QED is like it is, not depending on classical ED (CED), given that, in reality, CED should be a consequence of QED and not the other way around.
I mean, there is not an obvious-everyday classical cromodinamics or classical electroweak dinamics. So your argument I think that does not apply in general in order to understand why every theory seems to be coherent with the gauge principle.

5. Jul 30, 2015

### fzero

You are correct that one might prefer to think that way. Then we can just start with (2) and leave (1) as a statement to be deduced in the classical limit. It is still the case that the Lorentz invariant quantum dynamics of a massless vector field requires local gauge invariance.

6. Jul 30, 2015

### the_pulp

Sorry, i cant see why lorentz covariance implies local symmetry. Arent there theories that are lorentz covariant with just global symmetries? Why not? Where can I find a proof?

Ive been asking this for a lot of time and Im suprised that there could be such a short answer!

Thanks

7. Jul 30, 2015

### fzero

Any modern QFT text that covers gauge fields will explain this. For the photon, we can look at
$$- \frac{1}{4} F_{\mu\nu}F^{\mu \nu } = \frac{1}{2} \left( - (\partial_\mu A_0)^2 + (\partial_\mu A_1)^2 + (\partial_\mu A_2)^2 + (\partial_\mu A_3)^2 \right) + \cdots.$$
I could be precise and write the rest of the terms down, but the main thing I want to point out is that the kinetic term for $A_0$ (called the time-like polarization) is opposite from the space-like components. However, the sign of these terms relates directly to the sign of probability amplitudes, so if we do our quantum mechanics so that the space-like components lead to positive probabilities, then the time-like component will lead to negative probabilities.

We can't make physical sense of negative probabilities, but it turns out that if we include gauge symmetry, we can eliminate the problematic time-like component from the photon field. For additional reasons having to do with Lorentz covariance, gauge symmetry also removes the component of $A_\mu$ corresponding to the direction of the photons momentum (called the longitudinal mode). We are then left with two physical transverse polarizations, which we can represent as right and left-circular polarizations if we want.

Now, you could ask, why didn't we just postulate the two physical polarizations from the start? Well, we could have made a sensible quantum theory that way, but at the price of completely obscuring Lorentz invariance. An arbitrary Lorentz transformation would generate time-like and longitudinal components from the transverse ones. So it is crucial to have gauge symmetry in order to demonstrate Lorentz symmetry.

8. Jul 31, 2015

### the_pulp

Ok, thanks for all your kind answers. Ive 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 cant see it (there should be something extremely trivial that I cant see and it is bothering me).*
2) Supposing I get to follow your point 1), then, I cant 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)
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 Ive 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?.

*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 cant see the total link here but most probably your point is in this direction. Can you help me to complete / formalize the idea?

9. Jul 31, 2015

### fzero

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.

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.

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.

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.

10. Jul 31, 2015

### the_pulp

The only thing I would mention (about point "3") is that what I think that has no explanation is why is it proposed that Strong Interactions and Electroweak interactions mediate through a Lagrangian that follows from the gauge principle. I mean, if QED can be deduced from the Gauge principle, it is natural that the rest of the theories would follow from the Gauge principle, but this line of reasoning is just an extrapolation, not a deduction from first principles. This is what I am asking and what I understood is your answer is "there is no other way to make interacting theories between spin 1 and 1/2 particles that avoids negative probabilities, at least in such an elegant way, at least that I know". Am I right?

PS: During all my posts, from "Gauge Principle" I wanted to mean "Invariance of the Lagrangian after performing a local rotation in the inner space of a spin 1/2 field and adding the corresponding terms in the spin 1 fields".

11. Jul 31, 2015

### fzero

The strong and electroweak interactions also involve massless spin 1 fields analogous to the photon (in the EW case, the photon is actually constructed from a combination of electroweak gauge bosons), so gauge symmetry also deals with the unphysical timelike polarizations there.

It seems that the only consistent Lorentz invariant quantum theory of a massless spin 1 field is a gauge theory, yes.

12. Aug 1, 2015

### atyy

I've seen it argued that this is not necessarily the case. The heuristic idea is that one could use gauge-invariant degrees of freedom, so the gauge invariance is just a convenience for a local action and manifest Lorentz covariance. For example, the discussion on p81-82 of http://isites.harvard.edu/fs/docs/icb.topic521209.files/QFT-Schwartz.pdf is in this spirit.

13. Aug 1, 2015

### fzero

You're right, I should have said mainifest Lorentz covariance, I made the point in post #7 above.

14. Aug 1, 2015

### Gvido_Anselmi

Weinberg gives detailed explanation of why we need gauge invariance to construct QED in his book (vol.I section 5.9).
You may find it nice enough.

15. Aug 3, 2015

### vanhees71

Indeed, Weinberg has the best logic in his arguments, starting with a general discussion of the ray representations of the proper orthochronous Poincare group. If you want to avoid "exotics" like quanta with continuous spin-like degrees of freedom, a massless vector field must necessarily be realized as a gauge field. The idea to realized gauge fields not only with a U(1) gauge group but more general non-Abelian groups was just an ingenious idea by Yang and Mills, which later turned out to be the right concept to describe all (so far discovered) particles.