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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    $\begingroup$ 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). $\endgroup$ Commented May 1, 2012 at 18:47
  • $\begingroup$ Aw, come on Qmechanic, how isn't this stuff fun? You even have it written in your name! $\endgroup$ Commented May 1, 2012 at 18:56
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    $\begingroup$ And thanks @Luboš Motl, I think that about answers it. Sorry for being dense. Do you want to post it as an answer? $\endgroup$ Commented May 1, 2012 at 18:57
  • $\begingroup$ It is indeed fun, but unfortunately only 5 tags are allowed, cf. physics.stackexchange.com/posts/24698/revisions $\endgroup$
    – Qmechanic
    Commented May 1, 2012 at 19:09
  • $\begingroup$ Yeah, I'm aware of that. It was a joke. $\endgroup$ Commented May 1, 2012 at 19:15

1 Answer 1


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.

  • $\begingroup$ This is a placeholder for an answer until Lubos decides to make his comment into an answer--- it answers the question, so it shouldn't be a comment. But it looks like Sakurai just screwed up, not that he meant something else. $\endgroup$
    – Ron Maimon
    Commented May 2, 2012 at 2:26
  • $\begingroup$ I spent 10mins last night scrolling through Sakurai chapter 3 online looking for the promised explanation :( $\endgroup$
    – twistor59
    Commented May 2, 2012 at 7:33
  • $\begingroup$ @twistor59: Don't bother, people make stupid mistakes. Sakurai understands this completely, but sometimes you get confused by the chalk dust when teaching and say silly things. There is no reason he should say this. $\endgroup$
    – Ron Maimon
    Commented May 2, 2012 at 15:52
  • $\begingroup$ I was waiting for him to turn his comment into an answer, but since it doesn't seem to be happening, have some bonus points. $\endgroup$ Commented May 23, 2012 at 5:07

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