# Regarding Clebsch-Gordan Coefficients using Recursive method

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 $$J_1z$$?

• Why are you assuming that $m_1=m-\frac12$ isn’t an integer? What do you think the allowed values of $m$ are? Nov 26, 2023 at 5:36
• 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 $m_1=m-\frac{1}{2}$ Nov 26, 2023 at 13:59
• I think there is still an issue with my answer. Nov 26, 2023 at 14:31
• 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 $m_1=m-\frac12$. Nov 26, 2023 at 18:04
• 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? Nov 26, 2023 at 18:32

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?