matrix theorems with expo of matrix

Matrix theorems with expo of matrix PhL 11.20.16 I guess I don't have this stuff written down anywhere, so do it here today. 1. Theorem 1: eAeB = eBeA = eA+B provided A and B commute. Proof: If A and B commute, then clearly An and Bm commute, so we have eAeB = (ΣnAn/n!) (ΣmBm/m!) = (ΣmBm/m!)(ΣnAn/n!) = eBeA Next, consider this simpler problem of showing for non-matrix x and y exey = ex+y We have to show then that (Σnxn/n!) (Σmym/m!) = Σn(x+y)n/n! . First write the left side as LHS = (Σnxn/n!) (Σmym/m!) = ΣnΣm xnym / (n!m!) On the right side we know that (x+y)n = Σk=0n (n,k) xn-kyk = Σk=0n xn-kyk (**) So then the right side is this RHS = Σn(x+y)n/n! = Σn Σk=0n xn-kyk /n! = Σn=0∞ Σk=0n xn-kyk So now we want to show that LHS = RHS Σn=0∞Σm=0∞ xnym / (n!m!) = Σn=0∞ Σk=0n xn-kyk (*) Now forget (*) for a moment and recall this general summation reordering Theorem 3 from integrals and series/ double sum... doc: Σm=0∞ Σn=0∞ fn,m = Σk=0∞ Σm=0k fk-m,m m +n = k Theorem 3 (1) Now swap order on the left, then rename dummies on the right so that k→n and m→k Σn=0∞ Σm=0∞ fn,m = Σn=0∞ Σk=0n fn-k,k m +n = k Theorem 3 (2) Now let fn,m = xnym / (n!m!) fn-k,k = xn-kyk / ((n-k)!k!) The theorem in form (2) then says Σn=0∞ Σm=0∞ xnym / (n!m!) = Σn=0∞ Σk=0n xn-kyk / ((n-k)!k!) But this is exactly equation (*) so we have thus proven that (*) is correct and therefore that exey = ex+y Now how is this different if x → A and y→ B are matrices? Look at step (**) (x+y)n = Σk=0n (n,k) xn-kyk With matrices this would be for example (A+B)2 =(A+B)(A+B) = A2 + BA + AB + B2 only if A and B commute can you write this as A2 + 2AB + B2 = Σk=02 (2,k) An-kBk So to make the above proof work for matrices, they must commute. Then we have eAeB = eA+B iff A and B commute QED Corollary: exp(i [ n + dn]J) ≠ exp(i dnJ) exp(i nJ) Ji = generators Write A = in J = iniJi B = i dnJ = i dniJi Then [A,B] = [ iniJi,i dnjJj] = - nidnj [Ji,Jj] = - nidnj i εijkJk ≠ 0 [Ji, Vj ] = iijkVk