# Spin of the particle and degrees of freedom

Wigner showed that irreducible representations of the Poincare group can be listed, depending of the mass being zero or larger then zero, as $$2J+1$$ dimensional representations where $$J$$ is half-integer (for mass non-zero, positive) and just 2 dimensional for mass zero.

Now, if we have four-vector or more generally, tensor, representation, we can decompose 4d vector representation into 1d spin 0 and 3d spin 1 representation.

Now, when talking about particles, we define fields which determine specific properties of particles. In order to describe particles we try to introduce spin representation together with vector representation. So for spin 1 we define a condition which defines so called polarization vectors and we end up with three of them. Now, under Poincare transformations, am I to assume that these 3 basic vectors transform among themselves under the Poincare group in vector representation?

So, to conclude, can we define transformations of spin 1 field as vector ones? And are we to define transformations of spin 1/2 particles as not vector because with spin 1/2 the components do not transform as vectors? And, for spin 1, these 3 polarization vectors, do they define or constrain, directions of the field?

• Could anyone please say something? – Žarko Tomičić Nov 15 '19 at 17:41

## 1 Answer

So basically, due to Wigner, determining the unitary, infinite dimensional irreps of Poincare group in a sense can be reduced to specifying the unitary irreps of its little groups, which in the massive case is so(3), and hence for the massive case the irreps are all finite-dimensional and given by the spin.

Using Wigner's Little group one can classify particles using irreps of the little group, which in the massive case is so(3). So given an arbitrary mass, a particle is specified by its spin(in the rest frame) that specifies which irrep of so(3) it is in.

But fields transform under irreducible representations of Lorentz group(in fact really of Poincare group, but the translation part is represented trivially). In general, an irrep of Lorentz group might be a reducible representation of the little group. In our massive case(so(3)), this is the case except for 2 cases: the trivial scalars, or spin-1/2, either (1/2,0) or (0,1/2). This mimatch is the origin of the extra degrees of freedom issue as explained below.

For relativistic systems, we require both fields and states to be irreps of Poincare group. Thus to have a field of a massive spin-1 particle, we need a field(which is an irrep of Lorentz group) that when decomposed into irreps of so(3), contains a spin-1 component. The natural choices are (1,0) or (1/2,1/2); exactly which one to represent your "vector" is a physical choice(e.g. (1,0) and (0,1) describe (anti-)self-dual particles, but we do not need them as fundamental particles in Standard Model, although not that they are not important even in gauge theories).

Now how to impose the requirement that we only want the relevant part that transform under the spin-1 representation of so(3)? We impose polarization conditions, which is the content of "get rid of the redundant degrees of freedom".

p.s I have always been very much confused by this part of QFT too myself, so my explanation might quit well be incomplete or mistaken.