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The Clebsch–Gordan coefficients had a recursion relation: Sakurai Eq 3.8.45

$$J_\pm|j_1j_2; jm\rangle =(j_{1\pm}+j_{2\pm}) \sum_{m_1} \sum_{m_2} |j_1j_2;m_1m_2\rangle\langle j_1j_2;m_1m_2| j_1j_2;jm\rangle$$

Thus if one started say $|j_1=1j_2=1|J=2M=2\rangle$ one could obtain the representation of $|J=2$, $M=2,1,0,-1,-2\rangle$.

However, why there's not a recursion relationship for $J$? i.e. although one could use orthogonal relationship to calculate for $|J=1M=1\rangle$, why there's not a recursion relationship like that for $M$?

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    $\begingroup$ $J_{\pm}$, in the Lie algebra of the rotation group, move you within a rep labelled by J, so do not hop to another rep. To hop to another rep, different J, you need operators outside the algebra, such as Runge-Lenz operators. This is exactly how Pauli solved the Hydrogen atom before Schroedinger and his equation! $\endgroup$
    – Cosmas Zachos
    Commented May 5, 2020 at 21:48
  • $\begingroup$ @CosmasZachos Did you mean $I=(L+N)/2$ and $K=(L-N)/2$ operator (Sakurai chapter 4.1 Symmetries from page 266 to Eq 4.1.37 $E=-\frac{mZ^2 e^4}{2\hbar^2} \frac{1}{(2k+1)^2}$)? But the ket space was orthogonal. The orthogonal should be able to be expressed in some sort of relationship. $\endgroup$
    – ShoutOutAndCalculate
    Commented May 5, 2020 at 22:09
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    $\begingroup$ I'm not familiar in detail with his treatment, but that sounds right; to the extent that N does not commute with $L^2$, it connects different Js in your language. I have seen this hopping in the spectrum-generating chapter of B Wybourne's Classical Groups for physicists book. $\endgroup$
    – Cosmas Zachos
    Commented May 5, 2020 at 23:20

1 Answer 1

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They do. Obviously they are not generated by the action of $J_\pm$ since this action cannot change the $J$ quantum number but if my typesetting is right two of them are \begin{align} &\sqrt{\frac{2c(-a+b+c)(a+b+c+1)}{2c+1}}C^{c\gamma}_{a\alpha b\beta}\\ &=\sqrt{(b-\beta)(c-\gamma)}C^{c-1/2,\gamma-1/2}_{a\alpha,b\beta-1/2} +\sqrt{(b+\beta)(c+\gamma)}C^{c-1/2,\gamma-1/2}_{a\alpha-1/2,b\beta} \end{align} and \begin{align} &\sqrt{\frac{(-a+b+c)(a-b+c)(a+b-c+1)(a+b+c+1)(2c-1)}{2c+1}} C^{c\gamma}_{a\alpha,b\beta}\nonumber \\ &\quad =\sqrt{(b+\beta)(b-\beta+1)(c+\gamma)(c+\gamma-1)}C^{c-1,\gamma-1}_{a\alpha,b\beta-1}-2\beta\sqrt{c^2-\gamma^2}C^{c-1,\gamma}_{a\alpha,b\beta}\\ &\qquad -\sqrt{(b-\beta)(b+\beta+1)(c-\gamma)(c-\gamma-1)}C^{c-1,\gamma+1}_{a\alpha,b\beta+1} \end{align} and more are given in

Varshalovich, D. A., Moskalev, A. N., & Khersonskii, V. K. M. (1988). Quantum theory of angular momentum.

I believe are consequences of recursion relations of Regge symbols. See also along those lines

Bincer, A. M. (1970). Interpretation of the Symmetry of the Clebsch‐Gordan Coefficients Discovered by Regge. Journal of Mathematical Physics, 11(6), 1835-1844.

Another paper

Smorodinskiĭ, Y. A., & Shelepin, L. A. (1972). Clebsch-Gordan coefficients, viewed from different sides. Soviet Physics Uspekhi, 15(1), 1.

gives a completely cool approach to obtaining some unexpected relations which can be transformed into recursion relations.

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