Computing specific Clebsch-Gordan coefficients

I am trying to compute some Clebsch-Gordan coefficients with some specific values. My coefficients look like

$$\left< L,0;1,\lambda\middle|J,M\right>$$

I know that $$M=0+\lambda=\lambda$$ and that $$J$$ ranges from $$|L-1|$$ to $$L+1$$, so that I have three cases: $$J=L-1$$, $$J=L$$, and $$J=L+1$$. For instance, if I take the second case, I have

$$\left< L,0;1,\lambda\middle|L,\lambda\right>$$

I know that the answer should be

$$-\frac{\lambda}{\sqrt{2}},~\lambda=-1,0,+1$$

However, I cannot get the expression myself. I was thinking of using the recursion relation given for Clebsch-Gordan, for instance in Sakurai, which is

$$\sqrt{(J\mp M)(J\pm M+1)}\left< L,M_L;1,\lambda\middle|J,M\pm1\right> = \sqrt{(L\mp M_L)(L\pm M_L+1)}\left< L,M_L\mp1;1,\lambda\middle|J,M\right>+\sqrt{(1\mp \lambda)(1\pm \lambda+1)}\left< L,M_L;1,\lambda\mp1\middle|J,M\right>$$

For $$J=L$$, $$M=\lambda$$, this becomes

$$\sqrt{(J\mp \lambda)(J\pm \lambda+1)}\left< J,0;1,\lambda\middle|J,\lambda\pm1\right> = \sqrt{J(J+1)}\left< J,\mp1;1,\lambda\middle|J,\lambda\right>+\sqrt{(1\mp \lambda)(2\pm \lambda)}\left< J,0;1,\lambda\mp1\middle|J,\lambda\right>$$

but now I am at loss, since none of the coefficients looks like my original coefficient. What am I doing wrong? Is there another method to compute these functions?

Any help will be greatly appreciated.

Thanks.

• There's various open-source software for this, e.g., docs.sympy.org/latest/modules/physics/quantum/cg.html . Is that all you want, the ability to calculate specific CG coefficients numerically? – Ben Crowell Sep 23 '19 at 2:32
• Hi, thanks for your answer. I am looking for a way to compute them using known recursion relations or identities, without relying on software. I would like to understand how to approach the problem generally. – christianwos Sep 23 '19 at 13:50

One has quite genrally \begin{align} \langle \ell_1m_1;\ell_2 m_2\vert LM\rangle = \sqrt{\frac{2L+1}{2\ell_1+1}}(-1)^{\ell_2+m_2} \langle L,-M ;\ell_2 m_2\vert \ell_1 -m_1\rangle \end{align} so set $$\ell_2=1$$ and $$m_2=0$$. Now, if $$m_1=m_2=M=0$$, then one obtains, with $$\ell_1=L$$: \begin{align} \langle L 0;10\vert L 0\rangle = (-1)\langle L 0;10\vert L 0\rangle \end{align} so that this one is $$0$$. Next, use this again with $$m_2=M\ne 0$$ but with $$m_1=0$$ to find \begin{align} \langle L 0;1M\vert LM\rangle = (-1)^{1-M}\langle L,-M ;1M\vert L0\rangle\, . \end{align} The CGs must satisfy \begin{align} \sum_{m_2(M)} \vert \langle L,-M;1M\vert L0\rangle\vert^2=1 \end{align} and using $$\langle L,-M;1 M\vert L0\rangle = (-1)\langle L M;1,-M\vert L0\rangle$$ you can see that:
1. For $$M=0$$ the CG is $$0$$ as discussed before,
2. For $$M=1$$ the CG is in absolute value the same as for $$M=-1$$.
As they must all sum to $$1$$ you can conclude that one is $$+1/\sqrt{2}$$ while the other is $$-1/\sqrt{2}$$.
I'm not quite sure how to get the phase: I thought of starting from the Condon-Shortley convention $$\langle 1,1;L,L-1\vert L,L\rangle >0$$ but I don't quite see how to get it down to $$\langle 1,1;L 0\vert L, 1\rangle$$.
• Thank you very much for your answer. This is probably what I need. I will try to apply it to the cases $J=L-1,~L+1$ as well, and see what I get. – christianwos Sep 23 '19 at 13:51