Dealing with Primes on the Rotation Matrices

Dealing with Primes on the Rotation Matrices PhL 3.2.17 Rerun the starting clip of the movie: en(Φ) = R(Φ)e'n (den(Φ))S' ≡ en(Φ+dΦ) - en(Φ) = R(Φ+dΦ) e'n - R(Φ) e'n // note that (de'n(φ))S' = 0 = [ R(Φ+dΦ) - R(Φ)] e'n = dR e'n dR ≡ R(Φ+dΦ) - R(Φ) At this point in the film, the R's are really operators [ok] , not matrices, so write this last as | (den(Φ))S'> = (R(Φ+dΦ) - R(Φ) ) | e'n> Use time instead of angle to write this as | (den(Φ(t)))S'> = (R(Φ(t+dt)) - R(Φ(t)) ) | e'n> Now take components in Frame S' to get <e'i | (den(Φ(t)))S'> = <e'i | R(Φ(t+dt)) - R(Φ(t)) | e'n> = <e'i | R(Φ(t+dt)) | e'n> - <e'i | R(Φ(t)) | e'n> = [R(Φ(t+dt))]'in - [R(Φ(t))]'in [ok] Now consider <e'i | R(Φ(t+dt)) | e'n> = <e'i |ej><ej| R(Φ(t+dt)) |ek><ek| e'n> = R(Φ(t))ij R(Φ(t+dt))jk R-1(Φ(t))kn = [ R(Φ(t)) R(Φ(t+dt)) R-1(Φ(t)]in Correct but not very useful! In writing 1 = |ej><ej| I am inserting completeness at time t which reads 1 = |en(Φ)><en(Φ)| and then I claim [R(Φ(t+dt))]jk = <ej(Φ)| R(Φ(t+dt)) |ek(Φ)>. The operator and the states do not exist at the same time, so this is not really the matrix element of interest. The matrix of interest is rather [R(Φ(t+dt))]jk = <ej(Φ+dΦ)| R(Φ(t+dt)) |ek(Φ+dΦ)> . I fail to show the dependence on Φ or time. The correct completeness to use here is 1 = | en(Φ+dΦ) > < en(Φ+dΦ) | saying the en are complete at time t+dt. Then So this shows this matrix results: [R(Φ(t+dt))]'= R(Φ(t)) R(Φ(t+dt)) R-1(Φ(t)) [R(Φ(t))]'= R(Φ(t)) R(Φ(t)) R-1(Φ(t)) = R(Φ(t)) So you CANNOT remove the prime on matrix [R(Φ(t+dt))]' . That is the main point so far. You could write R(Φ(t+dt)) ≈ R(Φ(t)) - i J Φ(t) Then you get [R(Φ(t+dt))]'= R(Φ(t)) R(Φ(t+dt)) R-1(Φ(t)) = R(Φ(t)) [ R(Φ(t)) - i J Φ(t)] R-1(Φ(t)) = R(Φ(t)) - i Φ(t) [R(Φ(t))JR-1(Φ(t))] // all matrices So OK then the opening film sequence is this, | (den(Φ))S'> = (R(Φ+dΦ) - R(Φ) ) | e'n> <e'i | (den(Φ(t)))S'> = [R(Φ(t+dt))]'in - [R(Φ(t))]'in = { [R(Φ(t+dt))]' - [R(Φ(t))]' }in = { [R(Φ(t+dt))]' - [R(Φ(t))] }in = { R(Φ(t)) - i Φ(t) [R(Φ(t))JR-1(Φ(t))] - [R(Φ(t))] }in = { - i Φ(t) [R(Φ(t))JR-1(Φ(t))] }in = - i Φ(t) [R(Φ(t))JR-1(Φ(t))]in Here we are looking at the Frame S' evaluation of the opening scene equation. Suppose now that R(Φ) = Rz(ψ)Rx(θ)Rz(φ) // after doing our sign change in G.7 Then you would say <e'i | (den(Φ(t)))S'> = - i Φ(t) [Rz(ψ)Rx(θ)Rz(φ)JRz(-φ)Rx(-θ)Rz(-ψ)]in What is Φ(t) in this case? This path is a little clumsy because Φ(t) is some very messy expression. The good part of this path is that you have Frame S standard matrices appearing. In Section G.7 I take a different path. I say R(Φ(t)) = Rz(ψ(t))Rx(θ(t))Rz(φ(t)) R(Φ(t+dt)) = Rz(ψ(t+dt))Rx(θ(t+dt))Rz(φ(t+dt)) I still have to regard these as operators at this point and not matrices. The movie open goes then, en(Φ) = R(Φ)e'n (den(Φ))S' ≡ en(Φ+dΦ) - en(Φ) = R(Φ+dΦ) e'n - R(Φ) e'n // note that (de'n(φ))S' = 0 = [ R(Φ+dΦ) - R(Φ)] e'n = [ Rz(ψ(t+dt))Rx(θ(t+dt))Rz(φ(t+dt)) - Rz(ψ(t))Rx(θ(t))Rz(φ(t))] e'n = dR e'n dR ≡ R(Φ+dΦ) - R(Φ) and still all operators. I then show that Rz(ψ(t+dt))Rx(θ(t+dt))Rz(φ(t+dt)) - Rz(ψ(t))Rx(θ(t))Rz(φ(t)) = [-idψ Rz(ψ)J3Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ] = dR // operator where things are still all operators. I would then get (computing Frame S' components). <e'i | (den(Φ(t)))S'> = <e'i | dR |e'n> = <e'i| [-idψ Rz(ψ)J3Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ]|e'n> = { [-idψ Rz(ψ)J3Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ] }'in = {dR}'in and take note of the prime there at the right end. This is a matrix of some sort, but I have no idea how to compute the matrix elements of this matrix. Using Dirac, I think I could show that {dR}' = R(Φ) dR R-1(Φ) = Rz(ψ(t))Rx(θ(t))Rz(φ(t)) * [-idψ Rz(ψ)J3Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ] * Rz(-φ(t))Rx(-θ(t))Rz(-ψ(t)) and THEN I think everything is written in standard matrices, nothing has a prime. Now consider the next scene in the film which is this (den)S' = [(dR' )R-1(Φ) ] en = (dA') en dA' ≡ dR' R-1(Φ) This then produces this horrible mess dA' = = Rz(ψ(t))Rx(θ(t))Rz(φ(t)) * [-idψ Rz(ψ)J3Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ] * Rz(-φ(t))Rx(-θ(t))Rz(-ψ(t)) Rz(-φ(t))Rx(-θ(t))Rz(-ψ(t)) where everything is now written in standard rotation matrices. Had I carried out Appendix G.7 correctly, THIS is what I would have for the Frame S' evaluation of the opening scene equation, when written in terms of standard matrices which I regard as Frame S matrices. Now lets go back and instead take the Frame S components of things <ei | (den(Φ(t)))S'> = <ei | dR |e'n> = <ei| [-idψ Rz(ψ)J3Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ]|e'n> = <ei| [-idψ Rz(ψ)J3Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ] |ek><ek|e'n> = { [-idψ Rz(ψ)J3Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ] }ik<ek|e'n> = (dR)ik<ek|e'n> Now <ek|e'n> = R(Φ(t) )nk so we then have <ei | (den(Φ(t)))S'> = (dR)ik R(Φ(t) )nk = [ (dR) R-1(Φ(t) )]in Now consider the next scene in the film which is this (den)S' = [(dR )R-1(Φ) ] en = (dA) en dA ≡ dR R-1(Φ) This then produces, dA = [-idψ Rz(ψ)J3Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ] R-1(Φ) = [-idψ Rz(ψ)J3Rx(θ)Rz(φ) - idθ Rz(ψ) Rx(θ)J1Rz(φ) - idφ Rz(ψ) Rx(θ)Rz(φ)J3 ]Rz(-φ) Rx(-θ) Rz(-ψ) Now THIS is what I have used in Appendix G.7 !!! So I must be evaluating components in Frame S and not in Frame S' as I thought! But THIS in turn leads to the "wrong answer" because it produces an expression for ω in Frame S, but that result then agrees with Goldstein's expression for ω in Frame S' ! So we are back to The Fundamental Paradox