# On the uniqueness of QED

1. Dec 14, 2005

### dextercioby

Using this convention
$$A \overleftrightarrow{\partial }_{\mu} B =:A \overrightarrow{\partial}_{\mu} B - A \overleftarrow{\partial}_{\mu} B$$
one can write the QED Lagrangian density simply as
$$\mathcal{L}_{QED} =\frac{i}{2} \bar{\Psi}_{\alpha} \left(\gamma^{\mu}\right)^{\alpha}{}_{\beta} \overleftrightarrow{\partial }_{\mu} \Psi^{\beta} -m \bar{\Psi}_{\alpha}\Psi^{\alpha} +g \bar{\Psi}_{\alpha} \left(\gamma^{\mu}\right)^{\alpha}{}_{\beta} \Psi^{\beta} A_{\mu} -\frac{1}{4} F_{\mu\nu} F^{\mu\nu} -\frac{1}{2\xi} \left(\partial^{\mu}A_{\mu}\right)^{2} + \left(\partial^{\mu}\bar{\eta}\right) \left(\partial_{\mu} \eta}\right)$$
,where "g" is the coupling constant (plus/minus the electron's charge depending on the convention), $\eta$ is the ghost field associated to the gauge parameter $\epsilon$ and $\bar{\eta}$ is a ghost field from the nonminimal spectrum.
The question doesn't concern the uniqueness of the gauge-fixing term (one can implement various gauges, chosing a gauge-fixing fermion is, up to a point, arbitrary), but the consistent cross-interaction term, Dirac field - gauge abelian one-form field. The question is
Is $g \bar{\Psi}_{\alpha} \left(\gamma^{\mu}\right)^{\alpha}{}_{\beta} \Psi^{\beta} A_{\mu}$ the only consistent cross-interaction between a massive Dirac field and a gauge abelian one-form field ? If so, how would one go about & prove it...?
Daniel.

2. Dec 14, 2005

### vanesch

Staff Emeritus
Naive answer: dimensional power counting. Do you see another way to have a polynomial term that will give rise to a renormalizable interaction ?

3. Dec 14, 2005

### Physics Monkey

I think one could in principle have other terms like $$\bar{\Psi} \sigma^{\mu \nu} \Psi F_{\mu \nu}$$ or indeed higher powers and so on. This term respects Lorentz invariance, gauge invariance, and CP invariance, and there is no reason, from the point of view of symmetries, why it shouldn't be in the Lagrangian. The term I mention here would give, among other things, a contribution to the electron's magnetic moment. Of course, as vanesch said, these terms are all non-renormalizable which is why they were historically excluded, but in the spirit of effective field theory, non-renormalizability isn't such a big deal. Such a term could actually be present and we wouldn't even know it unless we made it to sufficiently high energy or otherwise looked sufficiently closely.

I know this is discussed in the Weinberg, and if I remember right I think he even talks about this particular term.

Last edited: Dec 14, 2005
4. Dec 15, 2005

### dextercioby

Renormalizability, in the context of the question, is not an issue. One can couple gravitons to a large spectrum of fields, yet such theories are not renormalizable.

Well, the term $c \bar{\Psi} \sigma^{\mu\nu} \Psi F_{\mu\nu}$ is both $\delta$ and $\gamma$ closed, so it describes a trivial interaction.

Daniel.

5. Dec 15, 2005

### dextercioby

I have come across very solid reasons that $g \bar{\Psi} \gamma^{\mu} \Psi A_{\mu}$ is the only possible cohomologically non-trivial cross interaction between a gauge abelian one-form field & a massive Dirac field, regardless of "number of derivatives + renormalizability" issues. For now i'm interested if there are any other justifications.

Daniel.

Last edited: Dec 15, 2005
6. Dec 16, 2005

### akhmeteli

I don't know if the following meets your requirements, but in my work http://www.arxiv.org/abs/quant-ph/0509044 I consider the standard Dirac-Maxwell Lagrangian, but vary the action under the constraint that the axial current vanishes, so the resulting equations of motion are different. Alternatively you may impose a weaker constraint, requiring that divergence of the axial current vanishes (in this case initial values may be arbitrary). These two theories seem to differ from the standard one but have the required symmetries.

7. Dec 17, 2005

### huhuayu

Is QED a superior theory to quantum mechanics? Forgive my ignorance, but I think QED itself(the annihilation operator, the Lagrange and so on) is hard to get the scene of electrons jumping between bound state. Is it true?

8. Dec 17, 2005

### reilly

A question always to ask in problems of interaction structures is, "What else could it be?" There are only two possible terms for a local nucleon parity conserving vector current; first, the standard gamma nu, second is the anomolous magnetic moment contribution (q mu) (sigma mu nu) (See any particle physics book on electromagnetic structure of the nucleon, and the Rosenbluth cross section.)

Regards,
Reilly Atkinson

9. Dec 17, 2005

### lonelyphysicist

Could you explain what you meant by the above statements?

10. Dec 19, 2005

### dextercioby

Well, "this particular term" cannot be in the classical lagrangian, since it's nothing but a (first order in $\hbar$) quantum correction to the original classical (nonquantized) lagrangian. U can see that $g\bar{\Psi} \gamma^{\mu} \Psi A_{\mu}$ and $g\bar{\Psi} \sigma^{\mu\nu} \Psi \partial_{[\mu} A_{\nu]}$ are dimensionally not the same. Even if, by absurd they were, "this particular term" would still be found when discussing radiative corrections to the QED vertex $g\bar{\Psi} \gamma^{\mu} \Psi A_{\mu}$.

Daniel.

11. Dec 19, 2005

### dextercioby

The way the problem was presented, phenomenology is not an issue, as well...

Daniel.

12. Dec 19, 2005

### dextercioby

When discussing the idea behind the construction of QED Lagrangian, people usually get exposed to the "good ole Noether procedure" which requires gauging the global matter field's symmetry (for the case of a single Dirac field the global symmetry group is $U(1)$). The interesting fact is that they take it for granted. One never asks him/herself why we do it and whether the result of this (namely the coupling one gets) is unique...Incidentally, why cannot one have a "seagull" term like in $SED$ ...?

Daniel.

Last edited: Dec 19, 2005
13. Dec 19, 2005

### Physics Monkey

dextercioby,

Thanks for the reply, but I don't understand your objection. The two terms have to have the same dimensionality if they are to be added in the Lagrangian. You seem to indicate that I would put the same coupling in front of each, but this is certainly not true. In particular, you can easily see that the coupling to the term I mentioned would have to have a dimension of -1, where as the usual coupling is dimensionless. This dimensional coupling is, of course, a sign of non-renormalizability. My point is that this term is completely consistent with gauge and Lorentz symmetry, and it leads to a non-trivial correction to the electron's magnetic moment.

Also, you need not go the route of gauging the global symmetry. Instead, you can start by trying to deal with the peculiarities of defining the propagator of a massless spin 1 particle, and then observe that you need a conserved current to couple the vector field to. With the minimal coupling assumption you end up at the same place, but this route makes it clear that there is no reason, from the point of view of gauge symmetry, why terms like the one I mentioned shouldn't be included.

I feel like I must be missing part of your argument, so what am I missing?

Last edited: Dec 19, 2005
14. Dec 19, 2005

### Haelfix

Simple answer here. Even if you don't care about renormalizability too many weird cross terms can break partial wave unitarity bounds from experiment. Its a bit of a long detailed calculation from many feynman diagrams to get all the relevant constraints, but it has been done (theres a book that does this)

15. Dec 19, 2005

### Physics Monkey

Haelfix,

Thanks for your reply, but I'm not sure I follow. The unitarity bounds are automatically respected as a consequence of the unitarity of the S-matrix. In non-renormalizable theories, it makes sense to expand the theory in a perturbation series in $$E/M$$ where $$E$$ is the energy scale of interest and $$M$$ is the mass scale of the relevant non-renormalizable interactions. Now, simple extrapolation of the expansion to $$E > M$$ does, in fact, lead to a naive violation of unitarity bounds. The Fermi theory of the weak decay had this problem, but everything is really ok. We know that the naive Fermi vertex develops structure at higher energies i.e. the W bosons appear, and this new physics prevents the violation of unitarity bounds. There is nothing wrong with the Fermi theory when viewed as effective field theory, and it is a useful description for energies $$E << M_W$$. A similar thing is done all the time when developing various effective field theories based on QCD, those effective field theories contain all kinds of horrible interactions. Except in theories where the non-renormalizable effects somehow saturate, we expect that new physics will appear at the mass scale $$M$$. If this is the case, then our non-renormalizable quantum field theory is simply an effective field theory, but it isn't useless or invalid just because it can't be used in every regime.

Am I missing your point?

P.S. I would love to take a look at the book you mentioned, can you remember the title? Thanks.

Last edited: Dec 19, 2005
16. Dec 20, 2005

### dextercioby

I like your arguments. The problem i was trying to address is simply if one knows any way of generating interactions among free (of any self-interactions) fields and by means of this method if one can prove the uniqueness of couplings both in QED (simple part), SED (a bit more difficult) and QCD (not too difficult)...As i said b4, renomalizability, phenomenology and unitarity are not issues to be taken into consideration.

Daniel.

Daniel.

17. Dec 20, 2005

### humanino

This is a very interesting discussion.
I would like to state that I agree very much with Physics Monkey. But then, I like QCD at low energy too

Could you elaborate on this ? One must be careful with such statements, because only at the perturbative level can total derivative terms be ignored.

18. Dec 21, 2005

### dextercioby

Well, that term doesn't contain antifields, so it's obviously $\delta$-closed. Moreover, the exterior differential along the gauge orbits has the properties:

$$\gamma \bar{\Psi} = 0 \ ;\gamma \Psi = 0 \ \mbox{and} \ \gamma A_{\mu} = \partial_{\mu} \eta$$

from where my second assertion follows simply.

Daniel.

19. Dec 21, 2005

### humanino

Dear Daniel,

1) Could you present your "very solid reasons that [...] is the only possible cohomologically non-trivial cross interaction" ? That sounds quite interesting.

2) I was earlier refering to instanton-like phenomena. It seems to me your arguments are classical. Are you sure that quantum-tunneling cannot spoil them ?

20. Dec 21, 2005

### humanino

but then of course, I do expect you to reply that you do not care about that, just as you did not care about renormalizability or phenomenology
Are you interested in non-commutative geometry ?
My point is also that you might have the wrong forum ! The all argument (which, as I already stated, is very interesting by itself) might be too mathematical. (just a suggestion : some more competent people in algebraic geometry can be found either in the "Strings, Branes, & LQG" subforum, or even the "Tensor Analysis & Differential Geometry" where you've been proven to hang around )