Physics Asked on July 28, 2021
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 mright)
=(-1)^{j_{1}-m_{1}} fleft(m_{1} m_{2} ; j mright)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 $left(m_{1} m_{2} mid j mright)$ are the Clebsch-Gordan coefficients.Then, he shows that
begin{multline}
fleft(m_{1} m_{2} ; j m-1right)= left(j_{2}+m_{2}+1right)left(j_{2}-m_{2}right) fleft(m_{1} m_{2}+1 ; j mright)- left(j_{1}+m_{1}+1right)left(j_{1}-m_{1}right) fleft(m_{1}+1 m_{2} ; j mright) tag{1}
end{multline}
Now he claims that
begin{multline}
fleft(m_{1} m_{2} ; j mright)=fleft(m_{1} m_{2} ; j j -uright)= A_{j} sum_{t}(-1)^{t}left(begin{array}{l}
u
t
end{array}right) frac{left(j_{1}+m_{1}+tright) !left(j_{1}-m_{1}right) !left(j_{2}+m_{2}+u-tright) !left(j_{2}-m_{2}right) !}{left(j_{1}+m_{1}right) !left(j_{1}-m_{1}-tright) !left(j_{2}+m_{2}right) !left(j_{2}-m_{2}-u+tright) !}
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}+1right)left(j_{1}-m_{1}right) fleft(m_{1}+1 m_{2} ; j mright)= A_{j} sum_{t}(-1)^{t}left(begin{array}{l}
u
t
end{array}right) left(j_{1}-m_{1}-tright)left(j_{1}+m_{1}+1+tright) times frac{left(j_{1}+m_{1}+tright) !left(j_{1}-m_{1}right) !left(j_{2}+m_{2}+u-tright) !left(j_{2}-m_{2}right) !}{left(j_{1}+m_{1}right) !left(j_{1}-m_{1}-tright) !left(j_{2}+m_{2}right) !left(j_{2}-m_{2}-u+tright) !}
end{multline}
begin{multline}
left(j_{2}+m_{2}+1right)left(j_{2}-m_{2}right) fleft(m_{1} m_{2} +1 ; j mright)= A_{j} sum_{t}(-1)^{t}left(begin{array}{l}
u
t
end{array}right) left(j_{2}-m_{2}-u+tright)left(j_{2}+m_{2}+1+u -t right) times frac{left(j_{1}+m_{1}+tright) !left(j_{1}-m_{1}right) !left(j_{2}+m_{2}+u-tright) !left(j_{2}-m_{2}right) !}{left(j_{1}+m_{1}right) !left(j_{1}-m_{1}-tright) !left(j_{2}+m_{2}right) !left(j_{2}-m_{2}-u+tright) !}
end{multline}
and
begin{multline}
fleft(m_{1} m_{2}; j m-1right)= 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-tright)left(j_{2}-m_{2}-u+tright) times frac{left(j_{1}+m_{1}+tright) !left(j_{1}-m_{1}right) !left(j_{2}+m_{2}+u-tright) !left(j_{2}-m_{2}right) !}{left(j_{1}+m_{1}right) !left(j_{1}-m_{1}-tright) !left(j_{2}+m_{2}right) !left(j_{2}-m_{2}-u+tright) !}
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}+tright) !left(j_{1}-m_{1}right) !left(j_{2}+m_{2}+u-tright) !left(j_{2}-m_{2}right) !}{left(j_{1}+m_{1}right) !left(j_{1}-m_{1}-tright) !left(j_{2}+m_{2}right) !left(j_{2}-m_{2}-u+tright) !}times
Bigg[ frac{u+1}{u+1-t}left(j_{2}+m_{2}+u+1-tright)left(j_{2}-m_{2}-u+tright)+left(j_{1}-m_{1}-tright)left(j_{1}+m_{1}+1+tright) -left(j_{2}-m_{2}-u+tright)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
$-left(j_{1}-m_{1}-tright)=(j_2 -m_2-u +t)$
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}+tright) !left(j_{1}-m_{1}right) !left(j_{2}+m_{2}+u-tright) !left(j_{2}-m_{2}right) !}{left(j_{1}+m_{1}right) !left(j_{1}-m_{1}-tright) !left(j_{2}+m_{2}right) !left(j_{2}-m_{2}-u+tright) !}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?
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 + tright) = 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).
Correct answer by ytlu on July 28, 2021
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