# How could $\textbf{S}^2$ not be a multiple of the identity?

I'm self-studying quantum mechanics with Sakurai's book (Modern Quantum Mechanics, 2nd edition) and came across the following in reference to the operator $\textbf{S}^2$:

As will be shown in Chapter 3, for spins higher than $\frac{1}{2}$, $\textbf{S}^2$ is no longer a multiple of the identity operator; however, $[\textbf{S}^2, S_i] = 0$ still holds (for $i = x, y, z$). (page 28)

The square of the total spin commuting with the components, I'm comfortable with. But the first part just confuses me: for a system with spin $s$, is it not true that

$$\textbf{S}^2|\cdot\rangle = \hbar^2s(s+1)|\cdot\rangle$$

whether or not $s = \frac{1}{2}$? Or do I have a fundamental misunderstanding of the situation? (I have read through Chapter 3, but apparently I kept missing the part where the book addresses this.)

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I suppose that Sakurai wants to say that the Hilbert space may be reducible i.e. a direct sum of $SU(2)$ representations with different values of $s$, so $s(s+1)$ is different for the subspaces, so the matrix of $S^2$ isn't a multiple of the unit matrix (with a universal factor). – Luboš Motl May 1 '12 at 18:47
Aw, come on Qmechanic, how isn't this stuff fun? You even have it written in your name! – Vandermonde May 1 '12 at 18:56
And thanks @Luboš Motl, I think that about answers it. Sorry for being dense. Do you want to post it as an answer? – Vandermonde May 1 '12 at 18:57
It is indeed fun, but unfortunately only 5 tags are allowed, cf. physics.stackexchange.com/posts/24698/revisions – Qmechanic May 1 '12 at 19:09
Yeah, I'm aware of that. It was a joke. – Vandermonde May 1 '12 at 19:15

It is a multiple of the identity, assuming you have a fixed spin, which is implied by context. So you are absolutely right, and Sakurai just made a typo or a blunder--- he might have meant that $S_z^2$ is not a multiple of the identity, or he might have had a non-irreducible representation in mind. In any case, it is confusing at best, and most likely just wrong.