exp nJ calculation

This is the Title PhL 3.26.05 Note that page numbering is turned on in this template. (Ji)n = Ji odd powers = -i εijk (Ji)n = (Ji)2 even powers = δjk( 1 - δij) (Ji)n+2 = (Ji)n all powers (Kin) = (-1)(n-1)/2 Ki odd powers = (-1)(n-1)/2 i ( giαg0β – g0αgiβ) (Kin) = (-1)n/2+1 (Ki)2 even powers = (-1)n/2+1 (δiαgiβ – δ0αg0β) (Kin+2) = – (Kin) all powers! exp(-i[rJ + bK]) = 1 - i[rJ + bK] + (-i)2/2 [rJ + bK]2 + ... [rJ + bK]2 = ( riJi + biKi)2 = ( riJi + biKi) ( raJa + baKa) = riraJiJa + biraKiJa + ribaJiKa + bibaKiKa ???? rJ + b K = Note: On 11.19.16 I found a sign error in the second term of my final result below. Everything is now corrected. I hope I did not use the wrong result in any of my docs. I searched all my PDFs for exp(- and did not find anything. ***************************************************** Calculation of exp(-iθnJ) where n is a unit vector. 1. Powers of nJ (nJ)2ab = niJinjJj = ninj (-iεiac)(-iεjcb) = - ninjεciaεcbj= ninjεciaεcjb = ninj( δijδab - δibδaj) = (δab - nanb) ≡ Tab => (nJ)2 = T Notice that ( on this line I use ncnc = 1, so I am assuming n is a unit vector) TacTcb = (T2)ab = (δac - nanc) (δcb - ncnb) = δab - nanb - nanb + nanb = (δab - nanb) = Tab which says T2 = T. Now, moving right along (nJ)3ab = (nJ)ac (nJ)2cb = (nJ)ac Tcb => (nJ)3 = (nJ) T = T (nJ) (nJ)4 = (nJ)2 (nJ)2 = T T = T2 = T (nJ)5 = (nJ)T ok The conclusion is then that (nJ)n = (nJ) T for n = 3,5,7,.... (nJ)n = T for n = 2,4,6... Now write exp(-iθnJ) = cos(θnJ) - i sin(θnJ) // how is this justified? Maybe as all series The first term is this: cos(θnJ) = 1 - (θnJ)2/2! + (θnJ)4/4! - ... = 1 - T ( θ2/2! - θ4/4! + ...) = 1 +T ( - θ2/2! + θ4/4! + ...) = 1 +T (1 - θ2/2! + θ4/4! + ... - 1) = 1 + T(cosθ - 1) The second term sin is this sin(θnJ) = (θnJ) - (θnJ)3/3! + (θnJ)5/5! - = (θnJ) + (nJ)T (- θ3/3! + θ5/5! - ...) = (θnJ) + (nJ)T (θ - θ3/3! + θ5/5! - ... - θ) = (θnJ) + (nJ)T (sin(θ) - θ) Thus, our result is exp(-iθnJ) = cos(θnJ) - i sin(θnJ) = = 1 + T (cos(θ) - 1) -i { (θnJ) + (nJ)T (sin(θ) - θ)} = 1 + T (cos(θ) - 1) -i (nJ){ θ + T (sin(θ) - θ)} We can write this out in specific components as exp(-iθnJ)ab = δab + (Cθ - 1) Tab - i (nJ)ac { θ δcb + (Sθ- θ) Tcb} where (nJ)ac = ni(Ji)ac = ni(-iεiac) Tab = (δab - nanb) So we can write the whole thing out in fine detail as: exp(-iθnJ)ab = δab + ( (Cθ - 1) (δab - nanb) - i ni(-iεiac) { θ δcb + (Sθ- θ) (δcb - ncnb)} = δab + (Cθ - 1) (δab - nanb) - niεiac { θ δcb + (Sθ- θ) (δcb - ncnb)} The terms linear in θ seem unusual. These terms are: - niεiac { θ δcb - θ (δcb - ncnb)} = -θ ni εiac [δcb - δcb + ncnb ] = -θ ni εiac ncnb = -θ εiac ni nc nb = 0 by symmetry! This leaves us with exp(-iθnJ)ab = δab + (Cθ - 1) (δab - nanb) - niεiac { Sθ (δcb - ncnb)} where we have set the linear θ terms to zero. The very last term also vanishes by the same symmetry argument, so we can simplify further to get = δab + (Cθ - 1) (δab - nanb) - ini(-iεiab) Sθ = δab + (Cθ - 1) (δab - nanb) + Sθ [ -i(nJ)ab] which says exp(-iθnJ) = 1 + (Cθ - 1) T + Sθ[ -i(nJ)] It is now a simple matter to just construct the matrices where(δab - nanb) = Tab and (nJ)ac = ni(-iεiac) T = which of course is symmetric and then use (nJ)ac = ni(-iεiac) = - i niεiac = - i niεaci to get (nJ) = i => -i(nJ) = so our final result becomes exp(-iθnJ) = 1 + (Cθ - 1) T + Sθ[ -i(nJ)] = + (Cθ - 1) + Sθ which is a pretty good closed form I think! I see no benefit to combining all into one matrix. Now, special cases: n = (1,0,0) = + (Cθ - 1) + Sθ = which is correct for Rx(θ) For each axis case, the T matrix has a very simple form, and the rightmost matrix is the generator matrix with the i's set to 1. n = (0,1,0) = + (Cθ - 1) + Sθ = which is correct for Ry(θ) Conclusion: We have computed the general 3x3 rotation matrix in a symmetrical closed form: Rn(θ) = exp(-iθnJ) = + (Cθ - 1) + Sθ