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.


  • $\begingroup$ 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? $\endgroup$ – user4552 Sep 23 '19 at 2:32
  • $\begingroup$ 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. $\endgroup$ – christianwos Sep 23 '19 at 13:50

Here's an almost solution, based on symmetry relations between the CGs.

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 $\langle 1,1;L 0\vert L, 1\rangle$.

  • $\begingroup$ 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. $\endgroup$ – christianwos Sep 23 '19 at 13:51

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