# Addition of angular momenta and Clebsch-Coefficients

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 $m = m_{1} + m_2 \leq j$?

• It holds true in all cases. May 14, 2017 at 14:53

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.

• 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 $m \leq j$)?
– user100411
May 15, 2017 at 12:08
• 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. May 15, 2017 at 12:29
• 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?
– user100411
May 15, 2017 at 14:09
• @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.) May 15, 2017 at 14:33

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 $-j \leq m_1+m_2 \leq j$.