I have what is hopefully a quick question. In Racah's 1942 paper Theory of Complex Spectra II, the author utilizes the action of $J_+$ on the Clebsch Gordan expression:

$$ \begin{align*} &|j m\rangle = \sum_{m_1,m_2} \langle j_1 m_1, j_2 m_2| j m \rangle |jm\rangle\\ &J_+ |j m\rangle = (J_{1+} + J_{2+}) \sum_{m_1,m_2} \langle j_1 m_1, j_2 m_2| j m \rangle |jm\rangle\\ \end{align*} $$

The subsequent operation on the either side and expansion allows us to equate like coefficients of $|j_1 m_1,j_2 m_2\rangle$, giving us the expression that Racah labels (5):

$$ \begin{align*} &[(j-m)(j+m+1)]^{\frac{1}{2}}\langle m_1m_2|jm+1\rangle = [(j_1-m_1+1)(j_1+m_1)]^{\frac{1}{2}}\langle m_1-1m_2|jm\rangle + [(j_2-m_2+1)(j_2+m_2)]^{\frac{1}{2}}\langle m_1m_2-1|jm\rangle \\ \end{align*} $$

This makes sense, however, I do not understand the next step, where Racah states:

Quote: In order to avoid the irrational factors we put:

$$ \begin{align*} &\langle m_1 m_2 | jm \rangle = (-1)^{j_1-m_1} f(m_1m_2;jm)\frac{[(j_1+m_1)!(j_2+m_2)!(j+m)!]^{\frac{1}{2}}}{[(j_1-m_1)!(j_2-m_2)!(j-m)!]^{\frac{1}{2}}}\\ \end{align*} $$

First is that I do not see how irrational factors would arise. Although there are square roots in equation (5) above, the only differences which could give rise to a value such as $\sqrt{-z}$ are between $j_i$ and $m_i$ values. The minimum value of $m$_i is $-j_i$, such that at worst we would get zeros in these brackets, nothing like $\sqrt{-z}$.

So in the first place I do not understand the need to protect against irrational factors. In the second, I have no idea how this substitution would do so.

The last point of confusion is the appearance of the term: $f(m_1m_2;jm)$. I have no idea what this is, and it is not defined in the paper before this point.

Any help on this one step would be appreciated. I know that I am missing something obvious, but I simply don't know what it is.

  • 1
    $\begingroup$ BTW: sorry to be picky but it’s “Gordan” with an “a” and not “Gordon” with an “o”. Paul Gordan was advisor to Emmy Noether: en.m.wikipedia.org/wiki/Paul_Gordan $\endgroup$ Sep 24, 2018 at 1:15

1 Answer 1


The quote states all there is to know. With 100% foresight, irrational factors are "taken out" by factoring the complicated $m$-dependent square root on the right of your last expression, meaning the remaining factor $f$ is basically a rational function and "easier" to determine ("easier" here is quite a relative notion). In general CG's are square roots since matrix elements of raising and lowering operators are square roots.

Thus, for instance, you can guess how making this substitution in the expression \begin{align} \sqrt{(j-m)(j+m+1)}\langle m_1m_2\vert j m+1\rangle &= \sqrt{(j-m)(j+m+1)}(-1)^{j_1-m_1}f(m_1,m_2;jm+1) \\ &\qquad \times \sqrt{\frac{(j_1+m_1)!(j_2+m_2)!(j+m+1)!}{(j_1-m_1)!(j_2-m_2)!(j-m-1)!}} \\ &=(j+m+1)(j+m) (-1)^{j_1-m_1}f(m_1,m_2;jm+1) \\ &\qquad \times \sqrt{\frac{(j_1+m_1)!(j_2+m_2)!(j+m)!}{(j_1-m_1)!(j_2-m_2)!(j-m)!}} \tag{1} \\ \end{align} is likely to lead to a common factor that includes the messy factorial in (1) as a common factor, leaving you with a more "manageable" form for $f$.

  • $\begingroup$ So to make sure I understand you correctly...we are factoring the expression $\langle m_1 m_2 | jm \rangle$ into $(-1)^{j_1-m_1} f(m_1m_2;jm)\frac{[(j_1+m_1)!(j_2+m_2)!(j+m)!]^{\frac{1}{2}}}{[(j_1-m_1)!(j_2-m_2)!(j-m)!]^{\frac{1}{2}}}$ by definition? We can do this because we've isolated a "rational" factor $f(m_1m_2;jm)$, to be discovered later, and an irrational part, $(-1)^{j_1-m_1}\frac{[(j_1+m_1)!(j_2+m_2)!(j+m)!]^{\frac{1}{2}}}{[(j_1-m_1)!(j_2-m_2)!(j-m)!]^{\frac{1}{2}}}$? If this is what you mean, what allows us to do this? Where does the factorial factor come from? $\endgroup$ Sep 24, 2018 at 0:06
  • $\begingroup$ Also I am not sure that I understand your phrase: "..likely to lead to a common factor that includes the messy factorial in (1) as a common factor.." Are you saying that we factor into an equation which itself then has another common factor? $\endgroup$ Sep 24, 2018 at 0:08
  • $\begingroup$ @Glycoholic Racah made a very educated guess that, by writing $\langle m_1m_2\vert j m\rangle$ in this way, things would work out so yes we are factoring this expression by definition so that $f$ has a simpler form. By using "likely" I'm attempting to justify the guess. Racah was a master at combinatorics so I can't say by what process he had insight that this factorization would work but it did. I'd like to think that Racah was aware there out to be some sort of symmetry between terms in $j_1m_1$, $j_2m_2$ and $jm$ but that's supposition on my part. No doubt experience was a guide. $\endgroup$ Sep 24, 2018 at 0:23
  • $\begingroup$ @Glycoholic I doubt very many people, besides Racah and possibly Wigner or some old masters, would have guessed this factorization. I can safely say I would not have. See aapt.scitation.org/doi/abs/10.1119/1.1935073 for an alternate derivation which also requires dexterity with combinatorial identities. $\endgroup$ Sep 24, 2018 at 0:26
  • $\begingroup$ If the factorization was an (extremely) educated guess, then by what metric do we say that it "would (did) work"? Are you implying that the verification of the resulting formula is empirical? If not, then what derivation substantiates the fact that the factorization "works"? $\endgroup$ Sep 24, 2018 at 1:13

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