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If we consider two angular momentum operators $\hat{J}_{1}$ and $\hat{J}_{2}$ and where $J := J_{1} \otimes 1 + 1 \otimes J_{2}$ where respectively we have common eigenstates $|j_1j_2;m_1 m_2 \rangle$ of $\hat{J}_{1}$ and $\hat{J}_{2}$ and $|j_1,j_2; jm \rangle$ is a common eigenstate of $\hat{J}^2$ and $\hat{J}_{z}$.

Does it follows generally that $-j \leq m_{1} + m_{2} \leq j$? Or is this a special case where we consider the Clebsch-Gordan coefficients defined as $\langle j_1j_2; m_1 m_2 | j_1 j_2; jm \rangle$, which vanish unless $m = m_1 + m_2$ and would therefore imply that $m = m_{1} + m_2 \leq j$?

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  • $\begingroup$ It holds true in all cases. $\endgroup$ – Abhijeet Melkani May 14 '17 at 14:53
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It sort of depends on how you are construing that $m_1+m_2$. If you're thinking it can range over all the possible values of $m_1$ and $m_2$, independently of whether $m_1+m_2$ has anything to do with the total magnetic quantum number $m$, then no, it's not the case. There are combinations that do not conform to that bound, and they are only discarded by the fact that the corresponding Clebsch-Gordan amplitudes vanish.

As a simple example, consider two spin-1/2 particles, for which one of the two resulting representations is the singlet state at $j=0$. Here you still have the state $|j_1j_2;m_1m_2{=}↑↑\rangle$, for which $m_1+m_2>j$, so obviously the bound is false in general. For this state, as you note, the Clebsch-Gordan coefficient is zero (because $m_1+m_2\neq m$ for all the possible $m$s in the representation) and that state does not contribute to the singlet representation.

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  • $\begingroup$ Thanks for your answer. I understand I think. In your example I assume you are considering no orbital angular momentum of the spin 1/2 particles? So in summary the basic idea is that writting $|j_{1}j_{2}; j m \rangle$ in terms of eigenstates $| j_{1}j_{2}; m_{1}m_{2} \rangle$ requires that $m_{1} + m_{2} = m$ but it is not true generally (and once we assume $m_{1} + m_{2} = m$ we have $m \leq j$)? $\endgroup$ – user100411 May 15 '17 at 12:08
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    $\begingroup$ Yeah, pretty much. Note that this is all abstract (you're just adding together one $j$ with another) so it doesn't need to be instantiated as 'the spin of a particle where we're ignoring the orbital angular momentum' - we can just take it on its face and do the relevant calculations. $\endgroup$ – Emilio Pisanty May 15 '17 at 12:29
  • $\begingroup$ In the text it state that we can take the Clebsch-Gordan Coefficients to be real matrix elements. Are we allowed this freedom because the important quantity is $|\langle \cdot | \cdot \rangle|^2$ rather than the matrix element $|langle \cdot| \cdot \rangle$ itself? $\endgroup$ – user100411 May 15 '17 at 14:09
  • $\begingroup$ @JohnDoe No, that's not enough, because the relative phase between the $|jm\rangle$s is fixed by the action of the $L_\pm$, and ditto for the individual particles. The fact that you can make the CGCs real tells you additional information that you didn't know before. (Not fully sure what, though.) $\endgroup$ – Emilio Pisanty May 15 '17 at 14:33
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This is generally the case. You have created a new system that can be fully described by $|j,m>$. Hence, the usual rules apply, so that $ -j \leq m \leq j$. The Clebsch-Gordan coefficients are derived from quantum mechanics and, therefore, reflect that $ -j \leq m_1+m_2 \leq j$.

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