I am not sure if I understand spin operators correctly. Given a two spin system in state $|++\rangle$ and an operator $S = S^{(1)} + S^{(2)}$

Then I have

$$ S_z |++\rangle = (S^{(1)}_z + S^{(2)}_z) (|\tfrac{1}{2}, \tfrac{1}{2}\rangle \otimes |\tfrac{1}{2}, \tfrac{1}{2}\rangle) = (S_z|\tfrac{1}{2}, \tfrac{1}{2}\rangle \otimes S_z|\tfrac{1}{2}, \tfrac{1}{2}\rangle) = (\tfrac{\hbar}{2} |\tfrac{1}{2}, \tfrac{1}{2}\rangle \otimes \tfrac{\hbar}{2} |\tfrac{1}{2}, \tfrac{1}{2}\rangle) = \tfrac{\hbar}{2} (|\tfrac{1}{2}, \tfrac{1}{2}\rangle \otimes |\tfrac{1}{2},\tfrac{1}{2}\rangle) = \tfrac{\hbar}{2} |++\rangle $$

But everywhere I read, I see $$ S_z |++\rangle = \hbar |++\rangle $$

What did I do wrong?

  • 1
    $\begingroup$ Related and might help you understand the notation a bit better: physics.stackexchange.com/a/60414/19976 $\endgroup$ – joshphysics Jan 21 '14 at 18:05
  • $\begingroup$ Also, when typesetting fractions in bras and kets, I find it useful to use the "tfrac" command instead of "frac" because it yields shorter, more readable fractions when they're being used as labels. I'd encourage giving it a try! $\endgroup$ – joshphysics Jan 21 '14 at 19:10
  • $\begingroup$ Yep, replaced them. $\endgroup$ – iblue Jan 21 '14 at 19:19
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    $\begingroup$ In a nutshell: changing $+$ for $\otimes$. The latter is a product, not a sum. $\endgroup$ – Emilio Pisanty Jan 21 '14 at 19:23

Clearly, calling $\mathbb{I}^{(j)}$ the identity matrix acting on subspace $j$ of the tensor space,

$$S_z|++\rangle= \left(S^{(1)}_z\otimes \mathbb{I}^{(2)} + \mathbb{I}^{(1)} \otimes S^{(2)}_z\right) |\frac{1}{2},\frac{1}{2}\rangle \otimes |\frac{1}{2},\frac{1}{2}\rangle = \left( S_z^{(1)}|\frac{1}{2},\frac{1}{2}\rangle\otimes \mathbb{I}^{(2)} |\frac{1}{2},\frac{1}{2}\rangle \right) + \left( \mathbb{I}^{(1)} |\frac{1}{2},\frac{1}{2}\rangle\otimes S_z^{(2)} |\frac{1}{2},\frac{1}{2}\rangle \right) = \left( \frac{\hbar}{2}|\frac{1}{2},\frac{1}{2}\rangle\otimes |\frac{1}{2},\frac{1}{2}\rangle \right) + \left( |\frac{1}{2},\frac{1}{2}\rangle\otimes \frac{\hbar}{2} |\frac{1}{2},\frac{1}{2}\rangle \right) = 2 \left( \frac{\hbar}{2}|\frac{1}{2},\frac{1}{2}\rangle\otimes |\frac{1}{2},\frac{1}{2}\rangle \right) = \hbar |++\rangle$$

  • $\begingroup$ Why is the last equality sign true? $\endgroup$ – iblue Jan 21 '14 at 11:28
  • $\begingroup$ Does the $\otimes$ behave like a product? So $\alpha |a\rangle \otimes \beta |b\rangle = \alpha\beta (|a\rangle \otimes |b\rangle)$? $\endgroup$ – iblue Jan 21 '14 at 11:38
  • $\begingroup$ I guess you factor the kets and use 1/2+1/2=1 $\endgroup$ – Your Majesty Jan 21 '14 at 11:41
  • $\begingroup$ The last equality is $\hbar/2 |\mathrm{two spins}\rangle + \hbar/2 |\mathrm{two spins}\rangle = \hbar |\mathrm{two spins}\rangle$. The $\otimes$ is a tensor product between objects dwelling in two different spaces (you can think of $x$ and $y$ if you like). $\hat{S}$ should be written as $\hat{S}_z^{(1)}\otimes \mathbb{I}^{(2)} + \mathbb{I}^{(1)}\otimes\hat{S}_z^{(2)}$ ($\mathbb{I}$ is the identity matrix) where each operator acts on a different parts of the tensor space. $\endgroup$ – perplexity Jan 21 '14 at 13:35
  • $\begingroup$ I have edited the answer and hopefully now is clearer for you. $\endgroup$ – perplexity Jan 21 '14 at 13:42

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