# Why aren't scalar fields called spin-infinity fields?

A spin-N field goes back to its original state after a rotation of $$360^\circ/N$$.

So a spin-1 field takes $$360^\circ$$, a spin-1/2 field takes $$720^\circ$$ and a spin-2 field takes $$180^\circ$$.

But this breaks down for scalar fields (called spin-0 fields). As it seems to suggest you must rotate it infinity times to get back to its original state. Whereas in fact you can rotate it an arbitarily small amount.

Hence shouldn't scalar fields be called "spin-infinity fields"? And a true spin-0 field should not look the same under any amount of rotation.

Why is this logic wrong?

• So a spin-2 field has twofold rotational symmetry about every possible axis? – J. Murray Nov 12 '20 at 2:36
• @Murray Perhaps not. – zooby Nov 12 '20 at 2:50
• Because that's not the definition of spin? – Javier Nov 12 '20 at 3:00

That is, whenever we say "spin-$$j$$", we mean that the field transforms under the irreducible representation of $$SU(2)$$ (or $$SO(3)$$ for integer spin, the Lie algebras are isomorphic) indexed by the number $$j$$. The finite irreducible representations of this Lie algebra are all classified by the highest eigenvalue of the $$J_z$$ matrix, which is $$j$$. The dimension of the spin-$$j$$ representation is $$2j+1$$. In particular, the $$j=0$$ representation is the trivial one, meaning that the field does not actually transform.
All of this only applies to finite dimensional representations. If you were to take $$j$$ to infinity, then you would be dealing with an infinite dimensional representation of $$SU(2)$$, and the rules for these objects change, so your analysis of the periodicity actually breaks down in precisely the limit you're asking about.