# Why helicity of photon is 1 but not 3?

1. Feb 14, 2016

### fxdung

Why helicity of phon is 1 but not 3 or higher?Is there any quantity relation between the circular polarization of light and spin of photon?Why spin of graviton is 2?Is there any relation with vector and tensor charater of electromagnetic and gravitation fields and of P symmetry?Why do the elementary Fermi particles have smallest spin 1/2 but not higher?

2. Feb 14, 2016

### vanhees71

That's a good question. It's just an empirical finding. There's no deeper principle (like symmetries) that "explains" the properties of the known elementary particles, described by the Standard Model.

3. Feb 14, 2016

### A. Neumaier

The spin of the photon is 1; its helicity is +1 or -1 (if circularly polarized) or a superposition of these (otherwise).
Because the metric is a symmetric tensor of order 2.
Yes. A field in a vector representation has spin 1, a field in a symmetric tensor representation has spin 2.
Spin 3 would correspond to a completely symmetric tensor of order 3; but there are no known interacting local quantum field theories describing such fields.
Because by current agreement, ''elementary'' means ''described by a local quantum field theory''. This essentially forces fermions to have spin 1/2. (But there are theoretical models for interacting local quantum field theories with spin 3/2 involving supersymmetry.)

4. Feb 14, 2016

### fxdung

Which book says about this topic? What about the book ''Quantum Field Theory Vol 3'' of Weinberg?

5. Feb 14, 2016

### fxdung

In representation 3-vector field,so there are 3 components states,then spin is 1.But because Lorentz symmetry,so we must consider 4-vector,then why the spin still is equal 1?

6. Feb 14, 2016

### vanhees71

The problem with fields at higher spin is that they contain redundant field-degrees of freedom. E.g., using a four-vector field $A^{\mu}$ to represent a massless spin-1 field you have four field-degrees of freedom, of which only two are physical. This leads to the gauge principle: If you want the massless four-vector field to represent particles with discrete spin-like degrees of freedom, you must treat it necessarily as a gauge field. This comes out of the analysis of the unitary representations of the Poincare group.

7. Feb 14, 2016

### A. Neumaier

It is about supersymmetry. But begin with volume 1 - it has more than enough to digest, and explains spin 1/2 and spin 1.

8. Feb 14, 2016

### vanhees71

...and also fields of any spin. In fact it's the only book I'm aware of which treats the fields of arbitrary spin in full generality. Of course Weinberg has written original papers on the subject much earlier than the marvelous textbooks:

S. Weinberg. Feynman Rules for Any Spin. Phys. Rev., 133:B1318–B1332, 1964.
http://dx.doi.org/10.1103/PhysRev.133.B1318

S. Weinberg. Feynman Rules for Any Spin. II. Massless Particles. Phys. Rev., 134:B882–B896, 1964.
http://dx.doi.org/10.1103/PhysRev.134.B882

S. Weinberg. Feynman Rules for Any Spin. III. Phys. Rev., 181:1893–1899, 1969.
http://dx.doi.org/10.1103/PhysRev.181.1893

Last edited: Feb 14, 2016
9. Feb 14, 2016

### A. Neumaier

In the free case only.

10. Feb 14, 2016

### fxdung

Why local field theory forces spin of Fermion particle equal 1/2 but not higher?

11. Feb 14, 2016

### A. Neumaier

For spin >2, the field equations from local Lagrangians do not define irreducible representations of the Poincare algebra. And spin 3/2 is problematic for the same reason, except in case of supersymmetry. This leaves spin 0, 1/2, 1, 2, which are the ones observed in Nature for the fundamental fields (aka particles).

12. Feb 14, 2016

### fxdung

Which book says about this?It seem to me the books of Weinberg does not say about thing that Prof.Neumaier point out above.

13. Feb 14, 2016

### A. Neumaier

Weinberg and other books gives Lagrangians only for spin $\le 1$. Books usually ignore discussing no-go theorems and cover instead the constructions that were found useful. But there is an extended literature about (failed) attempts to create a consistent local field theory for other interacting fields. Only gravity results, due to special circumstances (diffeomorphism invariance and coupling to the energy-momentum-tensor of other fields). I don't remember appropriate references.

14. Feb 14, 2016

### vanhees71

I've quoted literature in #8, and it's simply not true that there are no field theories for spin $\geq 3/2$. There are non Dyson-renormalizable ones, and indeed one of the problems is that the local field theories do not provide irreducible representations of the proper orthocrhonous Poincare group and that you have to deal with unphysical degrees of freedom. As an exampe for spin 3/2 (modelling the $\Delta$ resonance in effective hadronic theories), see e.g.,

http://arxiv.org/abs/0712.3919

15. Feb 14, 2016

### A. Neumaier

I didn't know any examples of such theories; thanks for the pointer. I'd like to understand the paper just cited. How does one see the spin 3/2 nature of the particle? (I don't see any spin indices.) Where is the free part of the $\Delta$ field Lagrangian? I only saw the interactions (p.4). How are the unphysical degrees of freedom handled in the perturbative treatment? Surely this is all very noncanonical and not treated in textbooks. But is is also not treated in the paper itself, it seems. So where can one see how to handle all this in a consistent way?

16. Feb 14, 2016

### vanhees71

Perhaps this helps

http://arxiv.org/abs/hep-ph/0008026

The Rarita-Schwinger field $\psi^{\mu}$ carries only a Lorentz-vector index but these components are Dirac-spinor valued.

17. Feb 14, 2016

### A. Neumaier

Yes, this looks like a good technical reference; I'll study it.

@fxdung: This means that my reasoning is limited to renormalizable fields. Renormalizability is usually assumed to distinguish elementary particles from effective ones. So this still answers your question - in the renormalizable case, simple power counting rules out high spin fields.

18. Feb 14, 2016

### fxdung

How to demonstrate that the non-renormalizable local quantum field theories(e,g hadrond fields) do not provide irreducible presentations of proper Poincare algebra?

19. Feb 15, 2016

### A. Neumaier

Just count the number of components of the fields in the Lagrangian representation and of the fields in Weinberg's form of the iirrep.

20. Feb 15, 2016

### vanhees71

Yes, part of the trouble with fields of higher spin $s \geq 1$ is that you have redundant degrees of freedom. E.g., a spin-1 field has 4 field components $V^{\mu}$, but spin 1 means that in fact you have only 3 spin degrees of freedom (for a massive boson). That's why you have to make sure that your equations of motion get rid of the unphysical degrees of freedom.

One realization of a massive spin-1 particle is the "naive" one, called the Proca Lagrangian,
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{m^2}{2} V_{\mu} V^{\mu}$$
with
$$F_{\mu \nu}=\partial_{\mu} V_{nu} - \partial_{\nu} V_{\mu}.$$
From Hamilton's principle of least action you get the Proca equation
$$\partial_{\mu} F^{\mu \nu} + m^2 V^{\nu}=0.$$
Taking the divergences by contracting with $\partial_{\mu}$ yields
$$m^2 \partial_{\mu} V^{\mu}=0,$$
i.e., as long as $m \neq 0$ the field equations imply an appropriate constraint to project out the unphysical scalar field degree.

The case $m=0$ is different, and you get an Abelian gauge theory.

Since gauge theories are not only aesthetically nice but have better renormalizability features, Stückelberg came up with an alternative formulation also for the massive field, which in the Abelian case leads to a gauge theory of massive vector fields without an additional Higgs boson.

21. Feb 15, 2016

### Demystifier

I find it counterintuitive that one has 3 degrees of freedom for arbitrarily small mass, but 2 degrees of freedom for mass exactly zero. My intuition suggests me that, for very small mass, the effect of the third degree should be negligible, so that the transition from small to zero mass is effectively continuous. Can someone confirm this idea by a more concrete result in the literature?

For example entropy is, in general, proportional to the number (2 or 3 for vector particles) of degrees of freedom per point in space. But consider entropy of a gas made of massive vector particles in a thermodynamic equilibrium at temperature much larger than the mass. One expects that the mass effect should be negligible at high temperature. Does it mean that the contribution of the third degree of freedom to the entropy is somehow suppressed in this high-temperature regime? From the thermodynamic point of view that sounds very strange.

Last edited: Feb 15, 2016
22. Feb 15, 2016

### A. Neumaier

In the Foldy representation one can see how the longitudinal degree of freedom disappears as the mass tends to zero. What happens (instructive exercise!) is that the representation becomes reducible and splits into two irreps, one with helicity 0, and one with helicity $\pm 1$.

23. Feb 15, 2016

### Demystifier

Hm, I still don't see how exactly the helicity-0 component disappears. The splitting by itself is not a disappearance. Any hint?

24. Feb 15, 2016

### A. Neumaier

It doesn't disappear - it is no longer part of the model. In QED you never have the longitudinal modes.

In massive QED you have them, but if the mass is tiny and one calculates cross sections where the input is transversal, the output will (most likely, I didn't do the calculations) be essentially transversal with very high probability, tending to 1 as the mass goes to zero. At mass exactly zero, the longitudinal states are therefore completely unobservable since they never come into being (if they aren't present initially).

25. Feb 15, 2016

### A. Neumaier

One has precisely the same situation in a well-understood case, namely collinear molecules made of three atoms. here one of the rotation degrees of freedom disappears due to the appearance of a rotational symmetry. As a consequence, the entropy contribution decreases by one. $CO_2$ is a good example.

The tiniest deviation from collinearity should make the entropy jump! But of course, calculating the entropy in this simple-minded way is an approximation; so in reality there is no jump but a smooth transition! I am sure that (almost likely, again I didn't do the calculations) this can be seen if you calculate the entropy from corresponding simulations in which the bond angle is varied continuously.

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