# Spin in QFT

1. Sep 23, 2014

### TrickyDicky

Do the corrections after renormalization to the gyromagnetic ratio of the electrón, more precisely its g-factor, that slightly deviate from the relativistic Dirac equation prediction of g=2 as corresponds to a spin 1/2 fermion, mean that the fermions spin symmetry is only aproximate? i.e. that it should be something like 1/1.999...?

2. Sep 23, 2014

### vanhees71

The spin of an electron is precisely 1/2. The gyrofactor of 2 you get from the naive Dirac equation is due to relativistic covariance and the principle of minimal coupling of the electromagnetic field (as a gauge field) to the electron.

The deviation of this value in full QED is due to radiative corrections, i.e., Feynman diagrams with loops. Instead of the bare point vertex, symbolized by a Feynman diagram with 1 photon line and two electron-positron lines meeting at a point, which is leading order in perturbation theory and reflecting the classical limit of the electromagnetic interaction of an electron, you have to add more diagrams with the same external lines but with also more internal lines. The most simple of these corrections is a diagram with one loop and three bare vertices. This diagram describes the corrections to the electromagnetic coupling due to the quantum fluctuations of the electromagnetic and the electron-positron field.

The diagram is not so easy to calculate as it seems. First of all the integral over the loop-four-momentum diverges logarithmically. The infinity can be lumped into the bare coupling constant already appearing in the Lagrangian. So the charge value (fine structure constant) gets renormalized and thus dependent on the energy-momentum scale where you subtract the diagram ("running coupling"). Another finite part of this diagram adds a contribution to the magnetic moment of the electron, and this leads to a (small) deviation of its gyro factor from 2.

One can show that all infinitities of the electron-positron-photon vertex can be lumped into the fine-structure constant (coupling constant) and that all contributions to the gyrofactor are finite. So you do not need to introduce an additional magnetic moment into the bare Lagrangian, which is very good, because such a term in the Lagrangian would not be renormalizable in Dyson's sense. QED as it stands is renormalizable, i.e., you can subtract all the infinities by counter terms which look the same as the terms in the bare QED Lagrangian.

3. Sep 23, 2014

### TrickyDicky

Sure, but my point is that what is actually observed is an approximation to 1/2 spin, I know the "bare electron" has precisely 1/2 spin but that is a pure
idealization of the free theory Right?

4. Sep 23, 2014

### The_Duck

The dressed electron also has spin precisely 1/2. Rotational invariance is an exact symmetry, obeyed by the interacting theory and not just the free theory. So all massive particles have to have spin exactly an integer or half-integer because these are the irreducible representations of the rotation group. The radiative corrections change the magnetic moment part of the gyromagnetic ratio, not the spin part.

5. Sep 23, 2014

### TrickyDicky

The magnetic moment part in the electrón is determined by spin S, the g-factor wich is a dimensionles number and constants h and the Bohr magneton. I guess what you are saying is that S is exactly 1/2 and the correction must be in g like it is routinely done, but it is obvious that it is just a proportionality constant without physical meaning that is used to adjust what is observed to the rest of the constants in the magnetic moment. Isn't it more useful to take the spin symmetry as approximate, as we know most of the symmetries in physics are?

Concerning interacting QFT, let's recall that Haag's theorem claims they don't exist.

6. Sep 23, 2014

### dextercioby

Spin symmetry can't be approximate, because the Poincaré covariance would be lost.

7. Sep 23, 2014

### DarMM

As dextercioby has said spin symmetry not being exact would imply Poincaré symmetry isn't exact, which would break relativity.

Secondly $g$ is not without physical meaning. Just look through a good few introductory books on quantum field theory, such as Peskin and Schroeder. An electron can generate a magnetic field in two ways, from its orbital angular momentum and from its spin angular momentum. $g$ measures the coupling between spin angular momentum and the magnetic field. It's value means that spin angular momentum couples slightly more than twice as strongly to the magnetic field than orbital angular momentum does. Just look through an analysis of the three point function in QED to see this.

Finally Haag's theorem does not say interacting theories don't exist (they have been proven to exist). It says that the interaction picture (which you can think of as the "trick" of using the free field in place of the interacting field in the Hamiltonian) does not exist (i.e. this trick does not work) provided the theory is translationally invariant.

8. Sep 23, 2014

### dextercioby

Off topic a little: That g=2 for an electron is one of the remarkable results of non-specially relativistic physics.

9. Sep 23, 2014

### TrickyDicky

Your explanation of Haag's theorem is much better phrased than my one liner, but it implies that the physical meaning for g is clearly an example of the trick the theorem says doesn't exist(minimal coupling, bare lagrangian...).

10. Sep 23, 2014

### DarMM

I don't think so. Haag's theorem just says that using the free field to do calculations in an interacting theory is mathematically dubious. However it is fine at the perturbative level, i.e. the free field can be used to calculate the pertubative expansion of interacting quantities. Constantinescu, Ann. Phys. 108, 37-48, contains a proof of this for $\phi^{4}_{3}$. The papers of Epstein and Glaser contain a proof that this works in general, i.e. the free field computes perturbative quantities correctly.

Besides the physical interpretation of $g$ doesn't come from an analysis with the free field (that's is just used as a practical tool in its computation), it comes from a nonperutrbative analysis of the three-point function of the interaction theory, no bare Lagrangians.

Also Haag's theorem says nothing about minimal couplings.