sandwich rules page

Rules Page PhL 3.26.05 1. Rules for a single J. The general rule for R() J R(-) is this: ( see separate "evaluate n dot J rotation of J.doc") exp(- i J) Jk exp(+ i J) = Jk cos + kmsJmns sin + nk ( J) (1 - cos) (1) Let's specialize to =, a unit vector in the j direction where j = x, y or z. Then we know that nk = kj and we get exp(- i Jj) Jk exp(+ i Jj) = Jk cos + kmjJm sin + kj Jj (1 - cos) (2) The non-trivial cases are those with i j for which we have exp(- i Jj) Jk exp(+ i Jj) = Jk cos + jkm Jm sin i j (3) Now let's just for the moment further assume that j and k follow each other in forward cyclic order, so that the ordered set of indices is j,k,l . In this case we get exp(- i Jj) Jk exp(+ i Jj) = Jk cos + Jl sin j,k,l cyclic Now suppose j,k are in reverse follow order. In this case we would say that j,k,l was in anti-cyclic order and our second term changes sign due to kjm = - jkm. Then we have exp(- i Jj) Jk exp(+ i Jj) = Jk cos Jl sin j,k,l anti-cyclic So let's group the above next to each other: exp(- i Jj) Jk exp(+ i Jj) = Jk cos + Jl sin j,k,l cyclic exp(- i Jj) Jk exp(+ i Jj) = Jk cos Jl sin j,k,l anti-cyclic (4) We can now write down various special cases using either (4) or (5): exp(- i Jx) Jy exp(+ i Jx) = Jy cos + Jz sin exp(- i Jx) Jz exp(+ i Jx) = Jz cos Jy sin So the rule is this: if you keep the same bread but change the filling, the sine term changes sign, and you set the J's to the obvious values on the right. Going back to (4) with j = y we get exp(- i Jy) Jz exp(+ i Jy) = Jz cos + Jx sin exp(- i Jy) Jx exp(+ i Jy ) = Jx cos Jz sin And finally we have exp(- i Jz) Jx exp(+ i Jz) = Jx cos + Jy sin exp(- i Jz) Jy exp(+ i Jz ) = Jy cos Jx sin So let's put all these results together: special cases: /2 -/2 exp(- i Jx) Jy exp(+ i Jx) = Jy cos + Jz sin Jz Jy Jz exp(- i Jx) Jz exp(+ i Jx) = Jz cos Jy sin Jy Jz Jy exp(- i Jy) Jz exp(+ i Jy) = Jz cos + Jx sin Jx Jz Jx exp(- i Jy) Jx exp(+ i Jy ) = Jx cos Jz sin Jz Jx Jz exp(- i Jz) Jx exp(+ i Jz) = Jx cos + Jy sin Jy Jx Jy exp(- i Jz) Jy exp(+ i Jz ) = Jy cos Jx sin Jx Jy Jx (5) 3. Two subspaces J and j. Now suppose, in this last case, we has = jn where jn is an angular momentum operator in some subspace different from the Jn. Since [J,j] = 0, the results above all apply, except then each trig function is an operator mess, such as cos(jn ) = k=even (-1)k/2 (1/k!) jnk k sin(jn ) = k=odd (-1)(k-1)/2 (1/k!) jnk k 3A. The case of jn being spin-1/2. For jn being spin-1/2, however, we know (see "rotation group generators.doc") that (jn)k = (1/2)k 1 k = even (jn)k = (1/2)k-1 jn k = odd // = (1/2)k (1/2)-1 jn = (1/2)k 2jn so we can then write cos(jn ) = k=even (-1)k/2 (1/k!) (1/2)k k = cos(/2) * 1 sin(jn ) = k=odd (-1)(k-1)/2 (1/k!) (1/2)k k * 2 jn = sin(/2) * 2 jn We can then go back to our general formula exp(- i Jj) Jk exp(+ i Jj) = Jk cos + jkm Jm sin jk [ (3) ] and replace = jn to get exp(- i jn Jj) Jk exp(+ i jn Jj) = Jk cos(/2) + jkm 2jnJm sin(/2) jk but it now seems best to replace 2 and we get exp(- i [2 jn Jj]) Jk exp(+ i [2 jn Jj]) = Jk cos() + jkm [2jnJm ] sin() jk where jn goes along for the ride. Here are special cases: exp(- i [2jnJj]) Jk exp(+ i [2jnJj]) = Jk cos + [2jnJl ] sin() j,k,l cyclic j = 1/2 exp(- i [2jnJj]) Jk exp(+ i [2jnJj]) = Jk cos [2jnJl ] sin() j,k,l anti-cyclic where we have jn = n/2 and n are the Pauli matrices, so the generators are jx = (1/2) jy = (1/2) jz = (1/2) (6) In Levitt's work on secular AX J-coupling, we use the above with n = z and j = z so we have: exp(- i [2jzJz]) Jx exp(+ i [2jzJz]) = Jx cos + [2jzJy ] sin() exp(- i [2jzJz]) Jy exp(+ i [2jzJz]) = Jy cos [2jzJx ] sin() where, in his application, j = I1 and J = I2 and vice versa. Write this out as exp(- i [2 I1zI2z]) I2x exp(+ i [2 I1zI2z]) = I2x cos + [2 I1zI2y ] sin() exp(- i [2 I1zI2z]) I2y exp(+ i [2 I1zI2z]) = I2y cos [2 I1zI2x ] sin() where usually = J . 3B. The case of jn being spin-1. We repeat the above for the case j = 1. As before, we start with cos(jn ) = k=even (-1)k/2 (1/k!) jnk k sin(jn ) = k=odd (-1)(k-1)/2 (1/k!) jnk k For jn being spin-1, however, we know (see "rotation group generators.doc") that jn = jn (jn)2 = (jn)2 (jn)3 = jn (jn)4 = (jn)2 .... (jn)k = (jn)2 k = even (jn)k = jn k = odd so we then get: cos(jn ) = k=even (-1)k/2 (1/k!) (jn)2 k = (jn)2 cos() sin(jn ) = k=odd (-1)(k-1)/2 (1/k!) jn k = jn sin() We can then go back to our general formula exp(- i Jj) Jk exp(+ i Jj) = Jk cos + jkm Jm sin jk [ (3) ] and replace = jn to get exp(- i jn Jj) Jk exp(+ i jn Jj) = (jn)2 Jk cos() + jkm jn Jm sin() jk which we can write in the two cases: ( for j = spin-1, J = any spin) exp(- i jn Jj) Jk exp(+ i jn Jj) = (jn)2 Jk cos() + jn Jl sin() j,k,l cyclic j = 1 exp(- i jn Jj) Jk exp(+ i jn Jj) = (jn)2 Jk cos() jn Jl sin() j,k,l anti- cyclic where the generators are as follows: Jx = (1/) Jy = (1/i) Jz = (7) 3C. The case of jn being spin-j for j>1 . For j > 1, I don't think there are any simple closed forms like this, see "rotation group generators.doc". 4. Rules applied to raise and lower operators. Recall that [J, Jz] = ∓ J from our ang mom formulas sheet. Consider then this particular sandwich, exp(- i Jz) J exp(+ i Jz) We could use our existing rules, or just run a fresh C-H on this, so let's do that: A = iJz B = J : C0 = J C1= [J, iJz] = ∓iJ C2 = [∓iJ, iJz] = (i)2J and so on with the ∓ appearing only in the odd terms. This as have = J [ 1 ∓i + (i)2/2! ∓ (i)3/3! + .... ] = J [ 1 + (∓i) + (∓i)2/2! + (∓i)3/3! + ...] = J exp(∓i) So we get an interesting and very simple result which we add to our list of formulas: exp(- i Jz) J exp(+ i Jz) = exp(∓i) J This explains, for example, why a "+ coherence" propagates with a negative phase. We could derive other rules involving J but let's wait until they are needed. // Well, they soon became "needed"! exp(- i Jx) J exp(+ i Jx) = exp(- i Jx) [ Jx iJy] exp(+ i Jx) = Jx i exp(- i Jx) Jy exp(+ i Jx) = Jx i [ Jy cos + Jz sin] Special case of interest is = in which case get Jx ∓ iJy = J∓ Next: exp(- i Jy) J exp(+ i Jy) = exp(- i Jy) [ Jx iJy] exp(+ i Jy) = exp(- i Jy) Jx exp(+ i Jy) i Jy = i Jy + [ Jx cos Jz sin] Again, if = we get = i Jy - Jx = - Jx i Jy = [ Jx ∓ i Jy ] = J∓ Now summarize all these last results: exp(- i Jx) J exp(+ i Jx) = Jx i [ Jy cos + Jz sin] // = J∓ if = exp(- i Jy) J exp(+ i Jy) = i Jy + [ Jx cos Jz sin] // = J∓ if = exp(- i Jz) J exp(+ i Jz) = exp(∓i) J // = J if = Here are the special cases for = /2: exp(- i Jx) J exp(+ i Jx) = Jx i [ Jy cos + Jz sin] = Jx i Jz exp(- i Jy) J exp(+ i Jy) = i Jy + [ Jx cos Jz sin] = i Jy Jz exp(- i Jz) J exp(+ i Jz) = exp(∓i) J = exp(∓i/2 ) = ∓i J Rx(/2) J Rx(-/2) = Jx i Jz Ry(/2) J Ry(-/2) = i Jy Jz Rz(/2) J Rz(-/2) = ∓i J Rx(-/2) J Rx(/2) = Jx ∓ i Jz Ry(-/2) J Ry(/2) = i Jy + Jz Rz(-/2) J Rz(/2) = i J Rx() J Rx(-) = J∓ Ry() J Ry(-) = J∓ Rz() J Rz(-) = J