# Problem understanding Wigner's classification

1. May 10, 2013

### jinawee

Wigner classified particles in function of the eigenvalues of $P_\mu P^\mu$ and $W_\mu W^\mu$. Then, it can be proved that for massless particles spin values can be only $\pm s_{max}$. But for a particle with mass could have intermediate spin values.

1. If we think that a massless particle is the limit where $m\rightarrow0$ (very small mass), how can we have this sudden change of the spin values (I think this is just an intuition error)?

2. What is the difference between polarization and spin (or helicity)?

3. Is it reasonable to say that massless particles have no spin but just helicity (I've read that this is because that don't have a center of mass and also because spin could point in any direction)?

4. Should we consider photons of different helicity different particles?

5. Is there any nice demonstration of why $W^2$ eigenvalues are $-m^2s(s+1)$, most books just refer to Wigner (1939).

Last edited: May 10, 2013
2. May 10, 2013

### dextercioby

4. No, it's one particle but with 2 different quantum states. It's like an electron in the H atom for n=1 and n=2, it's still the same electron, but in two different energy eigenstates.

3. May 10, 2013

### vanhees71

4. May 10, 2013

### DrDu

Hendrik, that is a nice presentation of the Wigner classifications. I have one question:
The group ISO(C) is non-compact. I thought that non-compact groups only allow for infinite dimensional unitary representations. But the representation you construct is one-dimensional, characterized by the only parameter λ. How can that be?

5. May 10, 2013

### vanhees71

This is the point! Of course, you can always construct more general unitary (ray) representations of ISO($\mathbb{C}$) (which is of course equivalent to the symmetry group of the two-dimensional Euclidean plane).

The most general one would lead to representations like in first-quantized quantum mechanics for two-dimensional systems. The construction would be very similar to the representation theory of the Poincare group itself. However, these representations imply continuous representations for the "momentum operators" (i.e., the operators generating translations in the plane). The physical meaning of this little group for massless particles, however is to provide something similar as "spin" for massive particles. Using representations with continuous eigenvalues of observables would imply that you had particles with a continous spin-like quantum number. I don't know, whether ever somebody has investigated this possibility further, but it seems that nature has not made use of this possibility and that's why it's not treated. So, you assume that the "translations" (i.e., the null rotations contained in the little group for massless particles) are represented trivially, and only the rotations around the three-momentum direction of the massless particle may be represented non-trivially. That's an O(2) rotation (or U(1)). On the first glance you can have any helicity you like, but in fact you want lift this little-group representation to a full representation of the Poincare group including three-dimensional rotations. Thus the helicity is bound to be a half-integer number, and thus any massless particle has then helicity states with $h=\pm s$ with $s \in \{0,1/2,1,\ldots \}$.

6. May 10, 2013

### jinawee

I understand your point, but in some places they say that because in this case helicity (here quirality) is a Lorentz invariant, you could consider them different.

When you change a photon's quirality are you destroying the photon?

7. May 10, 2013

### dextercioby

I don't know what you mean by <destroying> the photon. At the fundamental level there's no Feynman diagram with 1 photon "in" and no photon "out".

8. May 10, 2013

### DrDu

Last edited: May 10, 2013
9. May 10, 2013

### dextercioby

Which article ? The PDF you linked to is a report from the German Ministry of Eduation & Research. :)

10. May 10, 2013

### DrDu

11. May 10, 2013

### jinawee

I've heard that in string theory it can be proved that spin has to be discrete, but I haven't found any information.

I was meaning that you have you destroy the photon with helicity +1 and immediately you create one of helicity -1 (now I see that this would mean that the photon hasn't been destroyed).

I still think that since you can't change the photon's helicity you can consider them like different particles. It would be like inversing the charge of a particle.

I have two more questions, photons are said to be their own anti-particles, but since parity changes helicity, haven't you changed their helicity?
To get an anti-particle do you apply CP or CPT?

12. May 10, 2013

### DrDu

I am confused. Doesn't the electromagnetic field couple to e.g. electrons via a term pA, with p being a bilinear expression of electron creation and destruction operators and A a superposition of photon generators and anihilators?
That a photon cannot be absorbed by a free electron is more a question of energy and momentum conservation.
In scattering events, like Bhabha scattering a photon is first created and then destroyed.

13. May 10, 2013

### strangerep

The problem with the approach of Kim et al is that they then try to interpret the translation-like generators in E(2) as gauge transformations. But they never show how these same transformations can act on the electron field as position-dependent phase transformations. Thus, Kim puts himself at odds with the mainstream, imho.

As Hendrik briefly alluded to, the mainstream approach (cf. Maggiore or Weinberg) is to simply assume that the E(2) translation-like generators act trivially on physical fields, since there are apparently no elementary fields in nature with continuous spin.

The Poincare group serves up some dishes that nature didn't order.

14. May 11, 2013

### vanhees71

However, if you want to describe a massless field with spin $\geq 1$ as a local quantum field, i.e., as a field operator that transforms under the Poincare group as the corresponding classical field. E.g., for the photon field you want to have
$$A'^{\mu}(x')={\Lambda^{\mu}}_{\nu} A^{\nu}(x)={\Lambda^{\mu}}_{\nu} A^{\nu}[\Lambda^{-1}(x'+a)],$$
where I assumed the Poincare transformation taking the form
$$x'=\Lambda x-a$$
with an arbitrary constant four-vector $a$ and $\Lambda \in \mathrm{SO}(1,3)^{\uparrow},$
you have reduntant degrees of freedom, since the physical vector field has only two polarization states, but $A^{\mu}$ has four field degrees of freedom.

The way out is to define the vector field as a gauge field, and the null rotations, i.e., the continuous part of E(2) (or SL(2,C) for that matter) are represented by pure gauge transformations, which doesn't lead to a new physical field configuration and thus is represented trivially. The point is that one has to introduce the idea of gauge invariance to represent the null rotations trivially.

15. May 12, 2013

### jinawee

@vanhees71, I've found an article where it's explained how String Theory doesn't consider the possiblity of continous spin, but I haven't got any knowledge of String Theory in order to understand it (I should start reading Zweibach one of these days). The paper is: http://arxiv.org/abs/1302.4771