I Does the electron have charge?

[radium]

I should have clarified that the symmetry remains in the Lagrangian but is broken in the ground state.

 

[ftr]

(Dirac) Lagrangian is invariant under the continuous group U(1)

As an internal symmetry the U(1) still indicates an interaction, Otherwise U(1) is global and it only indicates conserved probability, correct?

 

[bhobba]

As an internal symmetry the U(1) still indicates an interaction, Otherwise U(1) is global and it only indicates conserved probability, correct?

I wouldn't put it that way - but local U(1) gauge symmetry is the basis of EM.

The details are not suitable for a thread - you need to study it. I like (see chapter 7):
https://www.amazon.com/dp/3319192000/?tag=pfamazon01-20

Basically symmetry determines the Lagrangian's for spin 0, spin 1/2 and spin one particles (the fields are complex numbers). However you notice, for say, spin 0 particles, it has a global U(1) symmetry. But when you check local U(1) symmetry a term appears added to the Lagrangian - you need something to cancel the term if you want it locally U(1) invariant. Then you look at spin 1 particles and you notice it has a local symmetry by adding the derivative of an arbitrary function to it. You compare the two and notice easily how to get this term you add to it to cancel the term that appears in the spin 0 Lagrangian so you write down a combined Lagrangian that is locally U(1) invariant. This is the EM Lagrangian from which Maxwell's equations and the Lorentz force law easily follows. The way its done in the above text can be simplified quite a bit - its a an interesting exercise doing it.

Now of relevance here is there is this number q that appears multiplying the interaction term in the Lagrangian and gives the strength of the coupling of the two fields - it called the coupling constant. This pretty obviously is related to charge in EM. But it turns out, and this is the twist in the answer to your question, its value depends on the energy scale you are probing:
https://en.wikipedia.org/wiki/Coupling_constant

Not taking that things like mass and charge in fact depend on that, which mathematically means you are introducing a cutoff in your theory, is what lead to the infinities that plagues QED. Once you introduce a cutoff, then get rid of the cutoff terms by replacing them with actually measured values (called the re-normalized values) by the trick in my paper on simple re-normalization you can get finite answers in your calculations.

Thanks
Bill

 

[strangerep]

Charge-conjugation is not the correct operation to consider here, because [...]

Thanks Sam. I had a definite suspicion I was on the wrong track.
It's always beneficial to receive another,... er,... suppository of wisdom.

Hopefully all these sorts of things will be explained in your (eventual) book?

 

[bhobba]

Hopefully all these sorts of things will be explained in your (eventual) book?

I think I will want a copy to.

I just wish Samalkhaiat would post more - his posts are always GOOD.

Thanks
Bill

 

[vanhees71]

This is true of any quantum theory--multiplying everything by a constant phase changes nothing. So if that's what's meant by global U(1) symmetry, I don't see how the concept is of any use. I also don't see how you get a conserved charge from it.

Of course, this global U(1) symmetry defines, as any continuous global symmetry, a conserved charge, called the Noether charge of the corresponding symmetry. It's the starting point to define interactions via gauge theory, i.e., you make this global symmetry local, which means to introduce a vector gauge boson (which is massless in the most simple realization of the gauge principle). The result is, roughly speeking, QED. Of course, the global symmetry is still a symmetry, and thus the Noether charge is still conserved, and indeed now that we coupled the photon to the Dirac field, it's interpreted as the electric charge, and its conservation is necessary for local gauge invariance, as is also known from classical electrodynamics, where the Maxwell equations alone imply necessarily charge conservation as an integrability condition, i.e., it follows without using the equations of motion for the charges.

In non-relativistic QT, the U(1) symmetry also leads to a conserved charge. For the Schrödinger field the charge density is $|\psi|^2$. Thus in the case of non-relativistic QT the global U(1) symmetry ensures that a once normalized single-particle state stays normalized via time evolution.

 

[samalkhaiat]

Hopefully all these sorts of things will be explained in your (eventual) book?

Yeah, to keep everyone happy and to avoid harsh criticism from my colleagues, the trade dinosaurs , I started writing long chapter on group theory.

 

[samalkhaiat]

I think I will want a copy to.

I send you a signed copy when it is ready.

I just wish Samalkhaiat would post more

I wish I can.

 

[bhobba]

I send you a signed copy when it is ready.

Of course many thanks.

But I must admit to feeling strongly about getting textbooks for 'free'. You guys aren't exactly millionaires and you do need compensation for writing it so I do like paying for such things.

Thanks
Bill

 

[vanhees71]

@samalkhaiat Yep, just tell us when your book is published. Then I'm pretty sure, I'll buy a copy. The author should benefit from writing textbooks!

 

[ftr]

Not taking that things like mass and charge in fact depend on that, which mathematically means you are introducing a cutoff in your theory, is what lead to the infinities that plagues QED. Once you introduce a cutoff, then get rid of the cutoff terms by replacing them with actually measured values (called the re-normalized values) by the trick in my paper on simple re-normalization you can get finite answers in your calculations.

Thanks for adding the twist. I always loved the song by King Crimson "Confusion will be my epitaph". Earlier, I was reading about why charge has the same value in different frames of reference and then I was about to ask about your twist the energy dependence. It is not clear to me if the values(for both mass and charge) are "cut off" dependent i.e. technical or "true" energy dependent. Although it is clear how mass is treated by making it invariant by doing away with initial Einstein's relativistic mass. But the charge, well, it seems to be less of an intrinsic (interaction) as relative to mass.

 

[vanhees71]

In modern physics for good reasons we do not use the idea of a "relativistic mass" anymore. Since 1908, with Minkowski's analysis of the mathematical structure of special relativistic spacetime, it is clear that we should formulate our theories in a covariant way, i.e., in terms of tensors on four-dimensional Minkowski spacetime. That's why mass is exclusively defined as the invariant mass. What was called relativistic mass is nowadays understood as energy (times $c^2$, where $c$ is a fundamental conversion parameter that enters the theory only, because we measure space and time intervals in different units, which is convenient for practical purposes but artificial from a fundamental physics point of view).

Contrary to mass, electric charge has always been described as a scalar. So there's no unnecessary confusion for electric charge.

In renormalized QFT the physical mass of a particle is defined by the pole of the propagator of the corresponding quantum field. This pole is independent of the chosen renormalization scheme, i.e., the renormalized parameters of the theory are fit to experimental data on the particles.

 

[bhobba]

That's why mass is exclusively defined as the invariant mass.

John Baez wrote a good article about it:
http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

As well as re-normalisation:
http://math.ucr.edu/home/baez/renormalization.html

Its pretty obvious you can only make re-normalization work using the invariant mass, which as mentioned above, is a pole in this thing called the propagator.

Thanks
Bill