# I How do we know the spin of a field?

1. Jun 19, 2016

### carllacan

If we use the Weyl representation the solutions to the Dirac equation turn out to be eigenfunctions of the S3 operator with eigenvalues 1/2 and -1/2, so we say that the field has spin 1/2.

But what about other fields? Why do we say the scalar real and complex field have spin 0? I tried following the same approach and see if their solutions are eigenfunctions of that operator, but I don't know how to do it.

And side question (should I open another thread?): why is the spin related to their statistics? Everything I've read so far just shows that using commutation relations in the Dirac field quantization gives rise to a non positive-definite hamiltonian, whereas using anticommutation relations doesn't. Then they show that commutators imply Bose-Einstein statistics and anticommutators imply Dirac-Fermi statistics.

In light of this it seems that if the field has integer spin we must use commutators, and if it has half-integer spin we must use anticommutators, but nobody goes on to explain why this is.

2. Jun 20, 2016

### vanhees71

That's representation theory of the Poincare group. You first have to get the angular-momentum operator and then diagonalize it in the zero-momentum subspace of single-particle states since relativistically the spin is defined by the representation theory of the rotation group on this zero-momentum subspace.

The spin-statistics theorem is proven by the demand that energy should be bounded from below, i.e., that a stable ground state exists. It turns out that this is achievable with commutator relations for integer-spin and with anticommutator relations for half-integer-spin only. For a detailed explanation, see Weinberg QT of fields, vol. I. A less general explanation can be found in my qft manuscript:

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf