Formula in Theory of Complex Spectra. II of Giulio Racah

Question

In this article Theory of Complex Spectra II Giulio Racah defines $f(m_{1} m_{2} ; jm)$ by \begin{multline} \left(m_{1} m_{2} \mid j m\right) =(-1)^{j_{1}-m_{1}} f\left(m_{1} m_{2} ; j m\right)\left[\left(j_{1}+m_{1}\right) !\left(j_{2}+m_{2}\right) !(j+m) !\right]^{\frac{1}{2}} /\left[\left(j_{1}-m_{1}\right) !\left(j_{2}-m_{2}\right) !(j-m) !\right]^{\frac{1}{2}} \end{multline} where $\left(m_{1} m_{2} \mid j m\right)$ are the Clebsch-Gordan coefficients.Then, he shows that
\begin{multline} f\left(m_{1} m_{2} ; j m-1\right)=\\ \left(j_{2}+m_{2}+1\right)\left(j_{2}-m_{2}\right) f\left(m_{1} m_{2}+1 ; j m\right)-\\ \left(j_{1}+m_{1}+1\right)\left(j_{1}-m_{1}\right) f\left(m_{1}+1 m_{2} ; j m\right) \tag{1} \end{multline} Now he claims that \begin{multline} f\left(m_{1} m_{2} ; j m\right)=f\left(m_{1} m_{2} ; j j -u\right)=\\ A_{j} \sum_{t}(-1)^{t}\left(\begin{array}{l} u \\ t \end{array}\right) \frac{\left(j_{1}+m_{1}+t\right) !\left(j_{1}-m_{1}\right) !\left(j_{2}+m_{2}+u-t\right) !\left(j_{2}-m_{2}\right) !}{\left(j_{1}+m_{1}\right) !\left(j_{1}-m_{1}-t\right) !\left(j_{2}+m_{2}\right) !\left(j_{2}-m_{2}-u+t\right) !} \tag{2}\end{multline} The summation parameter takes on all integral values consistent with the factorial notation, the factorial of a negative number being meaningless.To demonstrate (2) he says that it suffices to verify that it satisfies (1). This what I tried.We have that \begin{multline} \left(j_{1}+m_{1}+1\right)\left(j_{1}-m_{1}\right) f\left(m_{1}+1 m_{2} ; j m\right)=\\ A_{j} \sum_{t}(-1)^{t}\left(\begin{array}{l} u \\ t \end{array}\right) \left(j_{1}-m_{1}-t\right)\left(j_{1}+m_{1}+1+t\right) \times \\ \frac{\left(j_{1}+m_{1}+t\right) !\left(j_{1}-m_{1}\right) !\left(j_{2}+m_{2}+u-t\right) !\left(j_{2}-m_{2}\right) !}{\left(j_{1}+m_{1}\right) !\left(j_{1}-m_{1}-t\right) !\left(j_{2}+m_{2}\right) !\left(j_{2}-m_{2}-u+t\right) !} \end{multline}

\begin{multline} \left(j_{2}+m_{2}+1\right)\left(j_{2}-m_{2}\right) f\left(m_{1} m_{2} +1 ; j m\right)=\\ A_{j} \sum_{t}(-1)^{t}\left(\begin{array}{l} u \\ t \end{array}\right) \left(j_{2}-m_{2}-u+t\right)\left(j_{2}+m_{2}+1+u -t \right) \times \\ \frac{\left(j_{1}+m_{1}+t\right) !\left(j_{1}-m_{1}\right) !\left(j_{2}+m_{2}+u-t\right) !\left(j_{2}-m_{2}\right) !}{\left(j_{1}+m_{1}\right) !\left(j_{1}-m_{1}-t\right) !\left(j_{2}+m_{2}\right) !\left(j_{2}-m_{2}-u+t\right) !} \end{multline}

and \begin{multline} f\left(m_{1} m_{2}; j m-1\right)=\\ A_{j} \sum_{t}(-1)^{t}\left(\begin{array}{l} u \\ t \end{array}\right) \times \\ \frac{u+1}{u+1-t}\left(j_{2}+m_{2}+u+1-t\right)\left(j_{2}-m_{2}-u+t\right) \times \\ \frac{\left(j_{1}+m_{1}+t\right) !\left(j_{1}-m_{1}\right) !\left(j_{2}+m_{2}+u-t\right) !\left(j_{2}-m_{2}\right) !}{\left(j_{1}+m_{1}\right) !\left(j_{1}-m_{1}-t\right) !\left(j_{2}+m_{2}\right) !\left(j_{2}-m_{2}-u+t\right) !} \end{multline} and so we arrive at the following \begin{multline} \frac{(j_{1}+m_{1}+u+1)!(j_{1}-m_{1})!}{(j_{1}-m_{1}-u-1)!(j_{1}+m_{1})!}A_{j}(-1)^{u+1}+\\ A_{j} \sum_{t}(-1)^{t}\left(\begin{array}{l} u \\ t \end{array}\right) \frac{\left(j_{1}+m_{1}+t\right) !\left(j_{1}-m_{1}\right) !\left(j_{2}+m_{2}+u-t\right) !\left(j_{2}-m_{2}\right) !}{\left(j_{1}+m_{1}\right) !\left(j_{1}-m_{1}-t\right) !\left(j_{2}+m_{2}\right) !\left(j_{2}-m_{2}-u+t\right) !}\times \\ \Bigg[ \frac{u+1}{u+1-t}\left(j_{2}+m_{2}+u+1-t\right)\left(j_{2}-m_{2}-u+t\right)+\left(j_{1}-m_{1}-t\right)\left(j_{1}+m_{1}+1+t\right) \\-\left(j_{2}-m_{2}-u+t\right)\left(j_{2}+m_{2}+1+u -t \right) \Bigg] =0 /tag{3} \end{multline}

where $m=m_1+m_2$ , $u=j-m$ and $j=j_1+j_2$

Now $j_1-m_1=j_1-(m-m_2)=j_1-(j-u-m_2)=-j_2+m_2+u$ and so $-\left(j_{1}-m_{1}-t\right)=(j_2 -m_2-u +t)$

After some algebra last equation becomes \begin{multline} \frac{(j_{1}+m_{1}+u+1)!(j_{1}-m_{1})!}{(j_{1}-m_{1}-u-1)!(j_{1}+m_{1})!}A_{j}(-1)^{u+1}+ \\ A_{j} \sum_{t}(-1)^{t}\left(\begin{array}{l} u \\ t \end{array}\right) \frac{\left(j_{1}+m_{1}+t\right) !\left(j_{1}-m_{1}\right) !\left(j_{2}+m_{2}+u-t\right) !\left(j_{2}-m_{2}\right) !}{\left(j_{1}+m_{1}\right) !\left(j_{1}-m_{1}-t\right) !\left(j_{2}+m_{2}\right) !\left(j_{2}-m_{2}-u+t\right) !}\times \\ (j_2 -m_2-u +t)\Bigg[ \frac{u+1}{u+1-t}\left(j_{2}+m_{2}\right) -j-m-1 \Bigg] =0 \end{multline}

Can anyone tell why this last expression is zero?

Answer 1

Staring from the given definition of $ f(m_{1}, m_{2} ; j ,m)$: \begin{multline} \tag{1} f(m_{1}, m_{2} ; j ,m)=f(m_{1}, m_{2} ; j, j -u)=\\ A_{j} \sum_{t}(-1)^{t} \left( \begin{array}{l} u \\ t \end{array} \right) \frac{(j_{1}+m_{1}+t) ! (j_{1}-m_{1}) ! (j_{2}+m_{2}+u-t) ! (j_{2}-m_{2}) !}{(j_{1}+m_{1}) ! (j_{1}-m_{1}-t) ! (j_{2}+m_{2}) ! (j_{2}-m_{2}-u+t) !} \end{multline}

We are aiming at proving the following equation: \begin{multline}\tag{2} f(m_{1}, m_{2} ; j, m-1)=\\ (j_{2}+m_{2}+1) (j_{2}-m_{2}) f(m_{1} ,m_{2}+1 ; j ,m)-\\ (j_{1}+m_{1}+1) (j_{1}-m_{1}) f(m_{1}+1, m_{2} ; j ,m) \end{multline}

PROOF

First write down the left-hand-side of the equation (2), defining $u_1 \equiv j -m +1$.: \begin{multline}\tag{3} f(m_{1}, m_{2} ; j, m-1)=\\ A_{j} \sum_{t}(-1)^{t} \left( \begin{array}{l} u_1 \\ t \end{array} \right) \frac{(j_{1}+m_{1}+t) ! (j_{1}-m_{1}) ! (j_{2}+m_{2}+u_1-t) ! (j_{2}-m_{2}) !}{(j_{1}+m_{1}) ! (j_{1}-m_{1}-t) ! (j_{2}+m_{2}) ! (j_{2}-m_{2}-u_1+t) !}, \end{multline}

Then, examine the two terms in the right-ahnd-side, using $u_2 \equiv j -m = u_1 - 1$. The frist term in the RHS: \begin{multline}\tag{4} (j_{2}+m_{2}+1) (j_{2}-m_{2}) f(m_{1} ,m_{2}+1 ; j ,m) = (j_{2}+m_{2}+1) (j_{2}-m_{2}) \times\\ A_{j} \sum_{t}(-1)^{t} \left( \begin{array}{l} u_2 \\ t \end{array} \right) \frac{(j_{1}+m_{1}+t) ! (j_{1}-m_{1}) ! (j_{2}+m_{2}+1+u_2-t) ! (j_{2}-m_{2}-1) !}{(j_{1}+m_{1}) ! (j_{1}-m_{1}-t) ! (j_{2}+m_{2}+1) ! (j_{2}-m_{2}-1-u_2+t) !}\\ = A_{j} \sum_{t}(-1)^{t} \left( \begin{array}{l} u_2 \\ t \end{array} \right) \frac{(j_{1}+m_{1}+t) ! (j_{1}-m_{1}) ! (j_{2}+m_{2}+1+u_2-t) ! (j_{2}-m_{2}) !}{(j_{1}+m_{1}) ! (j_{1}-m_{1}-t) ! (j_{2}+m_{2}) ! (j_{2}-m_{2}-1-u_2+t) !} \end{multline} The two multipliers are combined into the factorial functions in Eq.(4).

The second term in the RHS: \begin{multline}\tag{5} (j_{1}+m_{1}+1) (j_{1}-m_{1}) f(m_{1}+1 ,m_{2} ; j ,m) = (j_1+m_1+1) (j_1-m_1) \times\\ A_{j} \sum_{t}(-1)^{t} \left( \begin{array}{l} u_2 \\ t \end{array} \right) \frac{(j_{1}+m_{1}+1+t) ! (j_{1}-m_{1}-1) ! (j_{2}+m_{2}+u_2-t) ! (j_{2}-m_{2}) !}{(j_{1}+m_{1}+1) ! (j_{1}-m_{1}-1-t) ! (j_{2}+m_{2}) ! (j_{2}-m_{2}-u_2+t) !}\\ = A_{j} \sum_{t}(-1)^{t} \left( \begin{array}{l} u_2 \\ t \end{array} \right) \frac{(j_{1}+m_{1}+1+t) ! (j_{1}-m_{1}) ! (j_{2}+m_{2}+u_2-t) ! (j_{2}-m_{2}) !}{(j_{1}+m_{1}) ! (j_{1}-m_{1}-1-t) ! (j_{2}+m_{2}) ! (j_{2}-m_{2}-u_2+t) !} \end{multline} The two multipliers are also combined into the factorial functions in Eq.(5).

Now we will proceed to show that Eq.(4)-Eq.(5) = Eq.(3). 

In Eq.(5), we change the summation dummy index $t+1 = t'$. \begin{multline}\tag{6} A_{j} \sum_{t'=1}^{t'=u_2+1} (-1)^{t'-1} \left( \begin{array}{l} u_2 \\ t'-1 \end{array} \right) \frac{(j_{1}+m_{1}+t') ! (j_{1}-m_{1}) ! (j_{2}+m_{2}+u_2-t'+1) ! (j_{2}-m_{2}) !}{(j_{1}+m_{1}) ! (j_{1}-m_{1}-t') ! (j_{2}+m_{2}) ! (j_{2}-m_{2}-u_2+t'-1) !} \\ = A_{j} \sum_{t'=0}^{t'=u_1} (-1)^{t'-1} \left( \begin{array}{l} u_2 \\ t'-1 \end{array} \right) \frac{(j_{1}+m_{1}+t') ! (j_{1}-m_{1}) ! (j_{2}+m_{2}+u_1-t') ! (j_{2}-m_{2}) !}{(j_{1}+m_{1}) ! (j_{1}-m_{1}-t') ! (j_{2}+m_{2}) ! (j_{2}-m_{2}-u_1+t') !} \end{multline} In the last expression, we use $u_2 = j-m = u_1 - 1$, and extend the summation lower limit from $1$ to $0$, since the term $t'=0$ vanishes having $(-1)!$ in the denominator of the combinatory function.

Also in Eq.(4), we replace $u_2 +1 = u_1$, and extended the summation upper limit from $u_2$ to $u_1 = u_2 +1$, the extra term has a $-1 !$ in the denominator of the combinatory function, thus vanishes. \begin{multline}\tag{7} A_{j} \sum_{t=0}^{t=u_1} (-1)^{t} \left( \begin{array}{l} u_2 \\ t \end{array} \right) \frac{(j_{1}+m_{1}+t) ! (j_{1}-m_{1}) ! (j_{2}+m_{2}+u_1-t) ! (j_{2}-m_{2}) !}{(j_{1}+m_{1}) ! (j_{1}-m_{1}-t) ! (j_{2}+m_{2}) ! (j_{2}-m_{2} -u_1+t) !} \end{multline}

Finally, the Eq.(4) - Eq.(5) now becomes Eq.(7) - Eq.(6):

\begin{multline}\tag{8} \text{Eq.(7) } - \text{ Eq.(6) } = A_{j} \sum_{t=0}^{t=u_1} (-1)^{t} \left\{ \left( \begin{array}{l} u_2 \\ t \end{array} \right) - (-1) \left( \begin{array}{l} u_2 \\ t-1 \end{array} \right) \right\} \times \\ \frac{(j_{1}+m_{1}+t) ! (j_{1}-m_{1}) ! (j_{2}+m_{2}+u_1-t) ! (j_{2}-m_{2}) !}{(j_{1}+m_{1}) ! (j_{1}-m_{1}-t) ! (j_{2}+m_{2}) ! (j_{2}-m_{2} -u_1+t) !} \end{multline}

Evaluate the combinatroy functions in the curry braket: \begin{multline} \tag{9} \left( \begin{array}{l} u_2 \\ t \end{array} \right) + \left( \begin{array}{l} u_2 \\ t-1 \end{array} \right) = \frac{u_2 !}{t! (u_2-t)!} + \frac{u_2 !}{(t-1)! (u_2 - t + 1)!} \\ = \frac{u_2 !}{t! (u_2-t+1)!} \left( u_2 - t + 1 + t\right) = \frac{(u_2+1) !}{t! (u_2-t+1)!} = \frac{(u_1) !}{t! (u_1-t)!} = \left( \begin{array}{l} u_1 \\ t \end{array} \right) \end{multline}

Using the result of Eq.(9), Eq.(8) becomes: \begin{multline}\tag{10} A_{j} \sum_{t=0}^{t=u_1} (-1)^{t} \left( \begin{array}{l} u_1 \\ t \end{array} \right) \frac{(j_{1}+m_{1}+t) ! (j_{1}-m_{1}) ! (j_{2}+m_{2}+u_1-t) ! (j_{2}-m_{2}) !}{(j_{1}+m_{1}) ! (j_{1}-m_{1}-t) ! (j_{2}+m_{2}) ! (j_{2}-m_{2} -u_1+t) !} \end{multline}

Eq. (10) is exactly the same as Eq.(3), the left hand side of Eq.(2). Therfore Eq.(4) - Eq.(5) = Eq.(3), conclude the proof of the equality in Eq.(2).