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I'm starting to learn about quantum spin so this might be a trivial question. From a section explaining Clebsch-Gordan Coefficients it states that generally we have $$|sm\rangle = \sum_{m_1 +m_2 = m}C^{s_1s_2s}_{m_1m_2m}|s_1 m_1\rangle| s_2 m_2 \rangle.$$ So for example, we can write $$| 3 0 \rangle = \frac{1}{\sqrt{5}}|2 1 \rangle |1 -1 \rangle + \sqrt{\frac{3}{5}}|2 0 \rangle |1 0 \rangle + \frac{1}{\sqrt{5}}|2 -1\rangle |1 1 \rangle.$$

Would it then be fine to write something like this for some two spin-$\frac{1}{2}$ system:

$|1 1 \rangle|1 0 \rangle = |1 1\rangle \otimes |1 0 \rangle = \frac{1}{\sqrt{2}}(| \uparrow \uparrow \rangle) \otimes ( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle ) = \frac{1}{\sqrt{2}}(| \uparrow \rangle \otimes |\uparrow \rangle) \otimes (| \uparrow \rangle \otimes | \downarrow \rangle + | \downarrow \rangle \otimes | \uparrow \rangle)$

Is everything in my expression valid?

Thanks.

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  • $\begingroup$ A state of four spin 1/2 (doublets) with partial symmetry. What of it? $\endgroup$ – Cosmas Zachos Dec 1 '16 at 20:28
  • $\begingroup$ @CosmasZachos Nothing of it, I'm trying to understand if this type of expression is valid. As I have stated I am new to this and I'm not sure if this type of expression "$\frac{1}{\sqrt{2}}(| \uparrow \rangle \otimes |\uparrow \rangle) \otimes (| \uparrow \rangle \otimes | \downarrow \rangle + | \downarrow \rangle \otimes | \uparrow \rangle)$" is valid or makes sense as a tensor product. It seems like something that would be in a tensor space of the form $(V_i \otimes V_j) \otimes (V_i \otimes V_j)$... $\endgroup$ – user100411 Dec 1 '16 at 20:37
  • $\begingroup$ Are you asking whether you understand the notation taught? I am not sure why you don't just write V1⊗V2⊗V3⊗V4. $\endgroup$ – Cosmas Zachos Dec 1 '16 at 20:46
  • $\begingroup$ @CosmasZachos I asked the question as is in my original question. If the answer is yes everything is fine then I would ask the question that I asked in the comments, that's all. $\endgroup$ – user100411 Dec 1 '16 at 20:57
  • $\begingroup$ Except for the wrong √2 , in the previous eqn, it appears fine; and the parentheses should be distributed away. But the answer is neither here nor there and itching to go awry... It is a very limited state of the 16-dim Kronecker space. $\endgroup$ – Cosmas Zachos Dec 1 '16 at 22:31

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