# Racah's Clebsch Gordan J dependence in Theory of Complex spectra ii

I have a question concerning Giulio Racah's derivation of the closed algebraic form of the Clebsch-Gordan coefficients in his Theory of complex spectra ii paper (Phys. Rev. v 62, pp 438 1942).

I have followed his proof until equation (10), and am lost at the transition from this equation to (11).

The gist seems to be that an equation for a general CG coefficient is used to calculate the case where m -> j and j-> j+1, but the resulting equation appears to have been much simplified, and I do not understand how.

Eq 10:

\begin{align}(m_1m_2|jm)=\delta (m1+m2,m)A_j \left[ \frac{(j_1-m_1)!(j_2-m_2)!(j-m)!(j+m)!}{(j_1+m_1)!(j_2+m_2)!}\right]^{0.5}\sum_{t}(-1)^{j_1-m_1+t}\times \frac{(j_1+m_1+t)!(j+j_2-m_1-t)!}{t!(j-m-t)!(j_1-m_1-t)(j_2-j+m_1+t)!} \end{align}

$$A_j$$ is a constant to be determined at this point, and not of any real concern. I am not interested in re-iterating how this equation was reached, but I will leave a link to the original paper if you are curious (https://journals.aps.org/pr/abstract/10.1103/PhysRev.62.438).

What confuses me is the next bit. And I quote:

"In order to obtain from (a previous equation,4) the dependence of $$A_j$$ on j, we calculate at first from (1) the expression of $$(m_1m_2|j+1j)$$: owing to the $$\delta$$ factor and to the expression of $$(j\vdots J_1\vdots j)$$ (TAS $$10^3 2a$$), we have

(11)

\begin{align} (m_1m_2|j+1 j)=\delta (m1+m2,j)(-1)^{j_1-m_1}A_{j+1} \left[ \frac{(j_1+m_1)!(j_2+m_2)!(2j+1)!}{(j_1-m_1)!(j_2-m_2)!}\right]^{0.5} \times 2(j+1)[m_1-j(j\vdots J_1\vdots j)] \end{align}"

The reference he makes to TAS $$10^3 2a$$ is an equation provided in Condon and Shortley's book Theory of atomic spectra, equation 2a in chapter 3, subchapter 10. It reads verbatim:

\begin{align} (\gamma j_1j_2j\vdots J_1\vdots \gamma j_1j_2j)=\frac{j_1(j_1+1)-j_2(j_2+1)+j(j+1)}{2j(j+1)}\hbar \end{align}

We can ignore $$\hbar$$ because we are in units of $$\hbar=1$$. Likewise the $$\gamma$$ is a placeholder for any other observable quantum numbers in that state.

The procedure appears to be taking j to j+1, and m to j. The alternating sum then includes a term in the denominator of the form

$$(j-m-t)! \rightarrow (j+1-j-t)!$$

For which only t = 0,1 yield meaningful results. Hence we compute these two terms and perform the multiplication.

My results for the sum look like: $$\frac{(j_1+m_1)! (j+j_2-m_1)!}{(j_1-m_1) (j-m)! (-j+j_2+m_1)!}-\frac{(j_1+m_1+1)! (j+j_2-m_1-1)!}{(j_1-m_1-1) (j-m-1)! (-j+j_2+m_1+1)!}$$

Which upon making the substitutions $$j \rightarrow j+1$$ and $$m \rightarrow j$$ yields:

$$\frac{(j_1+m_1)!(j+j_2-m_1+1)!}{(j_1-m_1)(-j+j_2+m_1-1)!}-\frac{(j_1+m_1+1)! (j+j_2-m_1)!}{(j_1-m_1-1)(-j+j_2+m_1)!}$$

The square root in (10) likewise becomes:

$$\sqrt{\frac{(2 j+1)!(j_1-m_1)!(j_2-m_2)!}{(j_1+m_1)! (j_2+m_2)!}}$$

And hence the product (full-simplified in Mathematica) is :

$$\left(\frac{j+j_2-m_1+1}{j_1-m_1}-\frac{j_1+m_1+1}{(j_1-m_1-1) (-j+j_2+m_1)}\right)\times\sqrt{\frac{(2 j+2)!(j_1-m_1)! (j_2-m_2)!}{(j_1+m_1)! (j_2+m_2)!}}\times \frac{(j_1+m_1+1)! (j+j_2-m_1+1)!}{(-j+j_2+m_1)!}$$

In short, it's a big mess. I have no idea how this hodge-podge is supposedly equivalent to

$$\left[ \frac{(j_1+m_1)!(j_2+m_2)!(2j+1)!}{(j_1-m_1)!(j_2-m_2)!}\right]^{0.5} \times 2(j+1)[m_1-j\frac{j_1(j_1+1)-j_2(j_2+1)+j(j+1)}{2j(j+1)}]$$

Which, omitting the Kronecker function, $$A_{j+1}$$, and the (-1) term, is what we should have.

Regardless of whether my simplifications in Mathematica are correct (which I have a mind to think they aren't), Racah is asserting an equality of this kind, which I have not been able to show. If anyone has a familiarity with this source material, or a mastery of combinatorics and a leniency with notation, I would much appreciate the assistance.

The point of this particular exercise (calculating $$$$ in this manner), is to determine the relationship between $$A_j$$ and $$A_{j+1}$$ such that we can fully determine the form of this equation and get the closed form of the CG . It is in fact, the only real boundary to my understanding the rest of his proof.

I am aware through my other posts and my readings that there are other proofs for the CG coefficients, but I would appreciate an answer to this question, and not a referral to others if at all possible.

All help is greatly appreciated!

Edit1: As @LonelyProf pointed out, we can take the terms $$(j_1-m_1-1)$$ and $$(j_1-m_1)$$ in the denominator of Eq 10 to be misprinted in that they are not factorials. If this is true, then we get the following:

$$\sqrt{\frac{(2 j+1)!(j_1-m_1)!(j_2-m_2)!}{(j_1+m_1)! (j_2+m_2)!}}\times\frac{(j_1+m_1)!(j_2+m_2)!}{(j_1-m_1)! (j_2-m_2)!} = \sqrt{\frac{(2 j+1)!(j_1+m_1)!(j_2+m_2)!}{(j_1-m_1)! (j_2-m_2)!}}$$

• It seems quite likely to me that the term $(j_1-m_1-t)$ in the denominator of eqn (10) of Racah's paper is a misprint, and should be $(j_1-m_1-t)!$. You seem to be dutifully carrying this term through your derivation, in the terms $(j_1-m_1)$, $(j_1-m_1-1)$, without the factorial sign. Have you tried working through it, but with these terms written as factorials? I think you are probably better set up to check this than me (otherwise I would have offered an answer!).
– user197851
Dec 24, 2018 at 14:14
• I did think that at some point. By my reckoning, if we take that as a misprint, we can expand the $(j_1-m_1-1)!$ term to $(j_1-m_1-1)(j_1-m_1)!$ and factor out all the factorials. This gives us the necessary terms to flip the arguments in the square root as Racah does, by $\sqrt{\frac{(2 j+1)!(j_1-m_1)!(j_2-m_2)!}{(j_1+m_1)! (j_2+m_2)!}}\times\frac{(j_1+m_1)!(j_2+m_2)!}{(j_1-m_1)! (j_2-m_2)!} = \sqrt{\frac{(2 j+1)!(j_1+m_1)!(j_2+m_2)!}{(j_1-m_1)! (j_2-m_2)!}}$ Dec 24, 2018 at 15:11
• However, it leaves us with a term of the form $\sqrt{\frac{(2 j+1)!(j_1+m_1)!(j_2+m_2)!}{(j_1-m_1)! (j_2-m_2)!}}(1-\frac{(j_1+m_1+1)}{(j_2+m_2)(j_1-m_1-1)(j_2-m_2+1)})$ And now the form implies that $(1-\frac{(j_1+m_1+1)}{(j_2+m_2)(j_1-m_1-1)(j_2-m_2+1)})$ is equivalent to $2(j+1)[m_1-j\frac{j_1(j_1+1)-j_2(j_2+1)+j(j+1)}{2j(j+1)}]$, which does not appear to be true either. Dec 24, 2018 at 15:13
• @LonelyProf, I will edit the question to reflect this part. Thank you! Dec 24, 2018 at 15:14

As I noted in my comment, it seems highly likely that eqn (10) in Racah's paper has a misprint: $$(j_1-m_1-t)$$ should be $$(j_1-m_1-t)!$$. As the OP notes in the edited question, this simplifies the expression, and allows various factorials to be taken inside the square root.

Having done this, I have verified that eqn (10) leads to eqn (11) in Racah's paper. Matching up the Kronecker delta, the $$A_{j+1}$$ coefficient, the sign term, and the square root, what remains is $$-j^2 + (2m_1-1)j -j_1^2 +j_2^2 -j_1 +j_2 +2m_1$$ which is equal to $$2(j+1)[m_1-jJ_1]$$ where $$J_1$$ is the matrix element defined in Condon and Shortley's book.

Following OP request:

Including the factor needed to flip four of the terms inside the square root, and taking out the sign factor, the sum over $$t$$ gives $$\left[\frac{(j_1-m_1)!(j_2-m_2)!}{(j_1+m_1)!(j_2+m_2)!}\right] \left( \frac{(j_1+m_1)!(j+j_2-m_1+1)!}{(j_1-m_1)!(j_2-j-1+m_1)!}- \frac{(j_1+m_1+1)!(j+j_2-m_1)!}{(j_1-m_1-1)!(j_2-j+m_1)!} \right)$$ Rearranging, i.e. pushing factors $$(j_1\pm m_1)$$ from the left-hand term into the right-hand term and pulling factors $$(j_2\pm j\mp m_1)$$ in the reverse direction: $$\left[\frac{(j_2-m_2)!(j+j_2-m_1)!}{(j_2+m_2)!(j_2-j+m_1)!}\right] \bigl( (j+j_2-m_1+1)(j_2-j+m_1) - (j_1-m_1)(j_1+m_1+1) \bigr)$$ The term in square brackets is unity, because of the condition $$m_1+m_2=j$$. Expanding the remaining terms gives the desired result.

• Hey LonelyProf! Could you quickly point out how you got $-j^2 + (2m_1-1)j -j_1^2 +j_2^2 -j_1 +j_2 +2m_1$? I don't seem to arrive at the same expression, and it indeed does simplify as you say. If I can understand how you got to that expression, then that is indeed the answer! And thank you so much for answering my question! Dec 24, 2018 at 16:55
• For clarification, when I proceed with the factorization of the factorials (do $\frac{(a+b)}{(a+b)!}=\frac{1}{(a+b-1)!}$ and the inverse) and the square root edit, I am left with an expression of the kind $(1-\frac{(j_1+m_1+1)(j_1-m_1)}{(j_2+m_2)(j_2-m_2+1)})$ for the remaining term. I do not see how this can be made equal to your $-j^2 + (2m_1-1)j -j_1^2 +j_2^2 -j_1 +j_2 +2m_1$ expression, even if I take $j\rightarrow j_1+j_2$ and group all terms under the same denominator. Explicitly, I am getting $\frac{j_2^2+m_2^2+j_2+m_2-j_1^2-m_1^2-j_1+m_1}{j_2^2-m_2^2+j_2+m_2}$ doing so. Dec 24, 2018 at 17:39
• I have done this. The rearrangement step involves identifying factorials such as $(X+1)!=(X+1)\times X!$ and $(X-1)!=X!/X$. Hopefully that's clear now.
– user197851
Dec 24, 2018 at 17:52
• Thank you so much! For anyone who is reading this exchange later on, my fatal error was making the substitution $j\rightarrow j+1$ and $m\rightarrow j$ in all places Except the sum. This must be done first, and then you get exactly what @LonelyProf indicates. Thank you so much! Dec 24, 2018 at 18:23