2

So for spinless particle, we have $$(\square^2 + m^2)\phi = 0$$

$\phi$ is a scalar.

For half spin particle, we have $$(i \gamma^\mu \partial_{\mu} - m)\psi = 0$$

$\psi$ is a spinor with four components.

For 1 spin particle, we have $$\partial_\mu (\partial^\mu B^v - \partial^v B^\mu) + m^2 B^v = 0$$

$B_v$ is a vector

For 2 spin particle, we have $$h_{\mu v} = 0$$

$h$ is a $4\times4$ matrix, symmetric and traceless.

My question is, as soon as i spot a equation, how do i know what is the spin of the particle we are dealing? I mean, is it necessary to evaluate the spin tensor everytime, or can we guess the spin only by the format of the equation?

Suppose i find something like

$$(\gamma_0 \square^2 + \gamma_1 \partial_{\mu} + \gamma_2 ) \alpha = 0$$

Can we guess, only from this, the spin quantity of the particle described by the mathematical thing $\alpha$?

For example, $\gamma_1 = 0$ i could guess the particle is spinless, $\gamma_0=0$ the particle would be half spin (again, guess), but and for general $\gamma$? Can i get information only from this?

LSS
  • 1,018

1 Answers1

5

It depends on the transformation properties of the field under Lorentz transformations. But actually in most cases the study of the transformation properties under rotations is already sufficient.

The knowledge of representation theory is very helpful in this context.

The representations of rotation group SO(3) are characterized by a single weight. The spin depends mostly on the weight of the representation of the field under study.

Actually, to be precise the field has to be decomposed in its irreducible components in order to see which spin component(s) it might have. However, in the majority of the cases the field under consideration just contains one irreducible component.

In most cases given the weight s of a representation of the rotation group SO(3), the number of independent components of an irreducible field is 2s+1.

weight s=0: The representation is the identity operator, so field is a scalar, it has only $2\cdot 0+1$ components and the spin is zero.

weight 1/2: The representation matrix the fields transforms under is a spinor representation and the field is a (Weyl)-spinor, so the spin of the field is 1/2. It has 2 components.

weight s=1: the fields transforms under rotations like a vector and the spin is 1. It has 3=2*1+1 components.

In case of tensors one has to be a bit careful since the most general 2-tensor can be decomposed in its irreducible parts which are an antisymmetric part, a traceless symmetric part and the trace part. All parts transform differently, for instance an anti-symmetric part is equivalent to a vector, so one would assign a spin 1 to it. The trace part transforms like a scalar, so would assign a spin 0 to it. The traceless symmetric part transforms as a real tensor and its representation matrix would have the weight 2, and therefore a particle described by such a field would have spin 2.

A part from a simple spinor one could also have a bi-spinor field. In this case a study of the field under pure rotations would not be sufficient. The transformation behaviour under Lorentz transformations has to be looked at. It turns out that a bispinor consists of 2 spinors which transform differently and each irreducibly under Lorentz-transformations. So the total transformation is reducible and its two irreducible parts transform each like a simple (Weyl)-spinor, so therefore would attribute spin 1/2 to it.

Recap: Look in a given field for that irreducible component whose weight is the highest: That is the spin one would attribute to it. Actually, in field-theoretical descriptions one tendentially tries to get rid of additional field's irreducible components of smaller weight by the means of gauge transformations. However, this is not always done/the case.

Lagrangian
  • 10,439