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So, I am reading the section on Recursion relation for Clebsch-Gordan Coefficents from Sakurai. He gives an example where he adds $J_1=l$, $l$ being an integer, and $J_2=\frac{1}{2}$. I am following the notations given in the text book. For the case $J=l+ \frac{1}{2}$ he writes $m_1=m-\frac{1}{2}$ and $m_2=\frac{1}{2}$. I don't understand why $m_1$ is taking that value in equation $3.7.55$. Should not it be an integer value because $m_1$'s are associated with the eigen values of $J_1z$?

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    $\begingroup$ Why are you assuming that $m_1=m-\frac12$ isn’t an integer? What do you think the allowed values of $m$ are? $\endgroup$
    – Ghoster
    Nov 26, 2023 at 5:36
  • $\begingroup$ Thanks for pointing that out. I finally get it. So we are using the relation $m_1+m_2=m-1$ Taking $m_2$ to the other side, I get $m_1=m-\frac{1}{2}$ $\endgroup$
    – zerozero
    Nov 26, 2023 at 13:59
  • $\begingroup$ I think there is still an issue with my answer. $\endgroup$
    – zerozero
    Nov 26, 2023 at 14:31
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    $\begingroup$ I don’t know how you got $m_1+m_2=m+1$. From (3.7.55) you can see that $m_1+m_2=m$. This is always true when you combine angular momenta. Furthermore, if what you wrote was true then when you take $m_2$ (which is 1/2) to the other side you would get $m_1=m-\frac32$, not $m_1=m-\frac12$. $\endgroup$
    – Ghoster
    Nov 26, 2023 at 18:04
  • $\begingroup$ Isnt this the equation that we are using? $\sqrt{(j+m)(j-m+1)} \bra{j_1j_2;m_1,m_2}\ket{j_1 j_2,m-1}=\sqrt{(j_1-m_1. )(j_1+m_1+1)}\bra{j_1,j_2;m_1+1, m_2}\ket{j_1 j_2;j m}+ \sqrt{(j_2+m_2+1)(j_2-m_2)}\bra{j_1 j_2;m_1,m_2+1}\ket{j_1 j_2;jm}$ . When i put $m_2=\frac{1}{2}$, the last term becomes 0 and we must have $m_1+1+m_2=m \implies m_1=m-\frac{3}{2}$? why is bra/ket notation not working here? $\endgroup$
    – zerozero
    Nov 26, 2023 at 18:32

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You are told $$ J=\ell +1/2, \\ m= m_1+m_2= m_1+1/2, \\ \implies ~~ m-1/2= m_1, $$ obviously an integer, no?

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