Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Related and might help you understand the notation a bit better: – joshphysics Jan 21 '14 at 18:05
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! – joshphysics Jan 21 '14 at 19:10
Yep, replaced them. – iblue Jan 21 '14 at 19:19
In a nutshell: changing $+$ for $\otimes$. The latter is a product, not a sum. – Emilio Pisanty Jan 21 '14 at 19:23

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

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

share|cite|improve this answer
Why is the last equality sign true? – iblue Jan 21 '14 at 11:28
Does the $\otimes$ behave like a product? So $\alpha |a\rangle \otimes \beta |b\rangle = \alpha\beta (|a\rangle \otimes |b\rangle)$? – iblue Jan 21 '14 at 11:38
I guess you factor the kets and use 1/2+1/2=1 – Faq Jan 21 '14 at 11:41
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. – perplexity Jan 21 '14 at 13:35
I have edited the answer and hopefully now is clearer for you. – perplexity Jan 21 '14 at 13:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.