the full J couple expo

Evaluation of the full J-coupling exponential PhL 1.8.08 Evaluate: exp(- I1 I2) I1 exp(+ I1 I2). To simplify, define L and S: I1 = L I2 = S so evaluate exp(- L S) L exp(+ L S) Let's try a brute force application of Campbell-Hausdorff: A = L S = Li Si = D D = L S B = Lk C0 = B = Lk C1 = [B,A] = [Lk , Li Si ] = [ Lk ,Li ] Si = i kis Ls Si // [Li, Lj] = i ijkLk = [Lk , D ] = [Lk , D] = - iksi Ls Si = -i [ L x S]k I have written out C1 about every way I can think of. Now move on to the next one: C2 = [ C1 , A] = [ - iksi Ls Si , D] = - i2 ksi [ Ls Si , D] At this point we digress to obtain a little theorem. First, quote this known theorem: [ab,cd] = a[b,c]d + ac[b,d] + [a,c]db + c[a,d]b Then apply as follows: [ don't be confused by reuse of symbols ] [ La Sb, Lc Sd ] = 0 + LaLc [ Sb, Sd ] + [ La, Lc] SdSb + 0 = LaLc ibdeSe + iacfLf SdSb Now suppose we have c = d and a summation on the resulting index. We then get: [ La Sb, Lc Sc ] = LaLc ibceSe + iacf Lf ScSb = [LaSb , D ] Now apply this to our case above to get [ Ls Si, Lc Sc ] = LsLc iiceSe + iscf Lf ScSi = [ Ls Si, D ] Then we have C2 = i2 kis [ Ls Si, D] C2 = i2 kis [ Ls Si, Lj Sj ] = i2 kis { Ls Lj iije Se + isjf Lf Sj Si } But we then have kis ije = isk ije = sjke - sejk kissjf = ski sjf = kjif - kfij Therefore C2 =(i)2 { ( sjke - sejk) Ls Lj Se + (kjif - kfij) Lf Sj Si } = (i)2 { 0 - Le Lk Se + Li Sk Si - 0 } = - (i)2 { Le Lk Se - Li Sk Si } = - (i)2 { Le Se Lk - Li Si Sk } = - (i)2 { D Lk - D Sk } = - (i)2 D { Lk - Sk } // D = Li Si Summary to this point: C0 = B = Lk C1 = [Lk , D] = - iksi Ls Si = -i [ L x S]k C2 = - i2 ksi [ Ls Si , D] = - (i)2 D { Lk - Sk } Next we have C3 = [ C2 , D] = [ - (i)2 D { Lk - Sk } , D ] = - (i)2 [ D { Lk - Sk }, D] = - (i) D [ { Lk - Sk }, D] where I used the rule [ab,a] = a[b,a] to pull out the D to the left. Now we know that [Lk , D] = - i ksi Ls Si => [Sk , D] = - i ksi Li Ss where to get the second item we just switch L S. So we can say that C3 = (i) D [ { Lk - Sk }, D] = - (i)2 D ksi { Ls Si - Li Ss} = - (i)2 2D ksi Ls Si = (i) 2D{ -iksi Ls Si } = (i) 2D C1 We are starting to see the recursion form. Let's go another step: C4 = [ C3 , D] = [ (i) 2D C1 , D] = (i) 2 [D C1 , D] But we again use our rule [ab,a] = a[b,a] to pull D out to the left to get C4 = (i) 2D [C1 , D] = (i) 2D C2 and we have yet another recursion. Lets try another term: C5 = [C4, D] = [(i) 2D C2, D] = (i) 2 [ D C2, D] and we again pull D out to the left using the same rule and we have C5 = (i) 2D [ C2, D] = (i) 2D C3 I think I am now convinced of this fact: Cn+2 = (i) 2D Cn starting with n = 1 So let's try to build up the result: C1 = - iksi Ls Si C3 = (i) 2D C1 = (2iD) C1 C5 = (i) 2D C3 = (2iD)2 C1 C7 = (i) 2D C5 = (2iD)3 C1 So the sum of all the odd terms is going to be (include the factorials now) odd terms = C1 + C3/3! + C5/5! + ... = [ 1 + x/ 3! + x2/5! + x3/7! + ...... ] C1 x = (2iD) Now the even look like this: C0 = B = Lk C2 = - (i)2 D { Lk - Sk } C4 = (i) 2D C2 = (2iD) C2 C6 = (2iD) C4 = (2iD)2 C2 So ignoring the first term, we get even terms = C2/2! + C4/4! + C6/6! + ... = [ 1/2! + x/4! + x2/6! + ....] C2 So we can at this point state our answer as follows: exp(- LS) Lk exp(+ LS) = exp(- D) Lk exp(+ D) = Lk + [ 1 + x/ 3! + x2/5! + x3/7! + ...... ] { - iksi Ls Si } + [ 1/2! + x/4! + x2/6! + ....] ({ - (i)2 D { Lk - Sk }) where x = (2iD) = 2i LS. Now let = i or -i = so we have what we really want, exp(- i LS) Lk exp(+ i LS) = exp(- i D) Lk exp(+ i D) = Lk + [ 1 + x/ 3! + x2/5! + x3/7! + ...... ] { ksi Ls Si } + [ 1/2! + x/4! + x2/6! + ....] ({ - 2 D { Lk - Sk }) where x = (-2D) = - 2 LS. Now define y = 2D = -x so we get exp(- i LS) Lk exp(+ i LS) = exp(- i D) Lk exp(+ i D) = Lk + [ 1 - y/ 3! + y2/5! - y3/7! + ...... ] { ksi Ls Si } + [ 1/2! - y/4! + y2/6! + ....] ({ - 2 D { Lk - Sk }) Now consider the first bracket: [ 1 - y/ 3! + y2/5! - y3/7! + ...... ] // y = 2D Now define y = z2 so we get = [ 1 - z2/ 3! + z4/5! - z6/7! + ...... ] = z-1 [ z - z3 /3! + z5/5! - z7/7! ] = z-1 sin(z) z = Now consider the second bracket: [ 1/2! - y/4! + y2/6! + ....] = [ 1/2! - z2/4! + z4/6! + ....] = z-2 [ z2/2! - z4/4! + z6/6! + ... ] = z-2 [ 1 -1 + z2/2! - z4/4! + z6/6! + ... ] = z-2 [ 1- cos(z) ] So we arrive at our final result which is exp(- i LS) Lk exp(+ i LS) = exp(- i D) Lk exp(+ i D) = Lk + z-1 sin(z) { ksi Ls Si } + z-2 [ 1- cos(z) ] ({ - 2 D { Lk - Sk }) = Lk + z-1 sin(z) ksi Ls Si - 2 z-2 D [ 1- cos(z) ] ( Lk - Sk ) where z = . In vector notation we get exp(- i D) L exp(+ i D) = = L + z-1 sin(z) L x S - 2 z-2 D [ 1- cos(z) ] ( L - S ) where D = LS and z = . This is certainly an exceedingly obscure result. Notice that D and z are operators. Let's restate this in Malcolm language exp(- i I1 I2) I1 exp(+ i I1 I2) = = I1 + z-1 sin(z) I1 x I2 - 2 z-2 D [ 1- cos(z) ] ( I1 - I2 ) where D = I1 I2 and z = . So this is now a non-secular J-coupling Hamiltonian would transform one of the spins in Levitt. If we were in the L-S coupling scheme where L2 and S2 and J2 are good, we could say for matrix elements in |JLS; ml,ms> states that D = LS = (1/2)[ J(J+1) - L(L+1) - S(S+1)] = a number and then things like sin(z) are more reasonable.