# Representation theory of $SU(2)$ and half-integer eigenvalues

I'm reading "Symmetries and Standard Model" by Matthew Robinson. On pages 88-91, the author shows the representation of $$SU(2)$$ will in general have $$2j+1$$ states: However, I don't quite get how this implies that $$j$$ must have half-integer values. What is the reasoning behind this?

If you start at $$-j$$ and add 1 $$N$$ times and end up at $$+j$$ then $$-j+N=j$$, therefore $$2j=N$$, so j has to be a multiple of $$1/2$$. Just as an example if you were to start at $$-1/3$$ and add one then you end up at $$2/3$$ which is obviously not $$-(-1/3)$$.

What constrains $$j$$ to be a half-integer is the fact that in each representation, we have $$2 j + 1$$ states. Since the number of states is an integer, then $$j$$ is a half-integer.

One way to prove this is by using the so-called highest weight method. The outline of the steps is this:

1. Since we care about finite-dimensional irreducible representations of $$SU(2)$$, then there exists a state $$| j \rangle$$ with highest $$S_z$$ eigenvalue $$j$$. Otherwise, we could keep acting with the raising operator $$S_+ \propto S_x + i S_y$$ on any eigenstate of $$S_z$$ to get an infinite number of orthogonal states, contradicting the finite dimension premise.
2. By using the commutation relations of the $$SU(2)$$ Lie algebra, we can conclude that $$J_{\pm} |m \rangle = \sqrt{j(j+1) - m(m \pm 1)} |m \pm 1 \rangle,$$ where $$|m \rangle$$ is an eigenstate of $$S_z$$ with eigenvalue $$m$$.
3. From this expression, we see that $$J_- |-j \rangle = 0$$. Hence, we have $$2 j + 1$$ states in (the vector space of) the irreducible representation.

PS: It is correct that just saying the set of eigenvalues of an operator is $$\{ -j, -j+1, \ldots, j \}$$ does not imply j is a half-integer.

If $$2j+1$$ is integer because the number of states is (obviously) an integer, then $$2j$$ is an integer and just $$j$$ can be (but need not be) a half integer, and must be an integer multiple of $$1/2$$.