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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:

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However, I don't quite get how this implies that $j$ must have half-integer values. What is the reasoning behind this?

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3 Answers 3

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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 $-(-1/3)$.

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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.

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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 $1/2$.

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