the secular J couple expo

Evaluation of the secular exponential PhL 1.8.08 Note added: A generalized version of this sandwich rule is given as equation (6) in document "sandwich rules page". There instead of having I1zI2z, we have the more general I1nI2j, and both are spin-1/2 I's. ________________________________________________________________________________ Evaluate: exp(- i I1zI2z) I1 exp(+ i I1zI2z ). To simplify, define L and S: I1 = L I2 = S so evaluate exp(- i LzSz) L exp(+ i LzSz ) In this particular problem, I think we can treat I2z = Sz as a constant (albeit a constant operator) because it commutes with everything else in the expression! We could not do this when we had I1 I2 = L S the exponential because in that case the components of I2 don't commute with each other so we cannot make the simplification. So consider then: exp(- i LzSz) L exp(+ i LzSz ) and look at our general result Qk = exp(- i J) Jk exp(+ i J) = Jk cos + kmsJmns sin + nk ( J) (1 - cos) So, set = and set J = L and set = Sz (which is an operator). Qk = exp(- i Lz) Jk exp(+ i Lz) = Lk cos + kmsLm ns sin + nk Lz (1 - cos) = Lk cos + kms Lm sz sin + kz Lz (1 - cos) = Lk cos + kmz Lm sin + kz Lz (1 - cos) In particular, we have Qx = Lx cos + xyz Ly sin = Lx cos + Ly sin Qy = Ly cos + yxz Lx sin = Ly cos - Lx sin Qz = Lz In fact, we can treat our exponential operator as Rz() and instantly write the above result. Now, there may be one more simplification we can do. Consider that sin = - 3/3! + 5/5! - ... We know that in the spin-1/2 representation we have Sz2 = 1/4. Thus we have = Sz = Sz / 20 3 = ( Sz)3 = 3 Sz (1/4) = 3 Sz / 22 5 = ( Sz)5 = 5 Sz (1/4 )2 = 5 Sz / 24 2n = (/2)n So consider: (1/2) sin = (1/2) [ - 3/3! + 5/5! - ...] = Sz / 21 - 3 Sz / 23 3! + 5 Sz / 25 5! = Sz [ (/2) - (/2)3/3! + (/2)5/5! - ... ] = Sz sin(/2) Similarly, 2 = ( Sz)2 = 2/4 = (/2)2 4 = ( Sz)4 = 4/42 = (/2)4 so cos = [ 1 - 2/2! + 4/4! - ... ] =[ 1 - (/2)2/2! + (/2)4/4! - ...] = cos(/2) So our result now simplifies to become Qk = exp(- i LzSz) Lk exp(+ i LzSz ) = Lk cos + kmz Lm sin + kz Lz (1 - cos) = Lk cos(/2) + kmz Lm 2Sz sin(/2) + kz Lz (1 - cos(/2)) So finally I am seeing where the "factor of two" comes in that Levitt uses everywhere. Suppose we change things so that = /2 . Then we get exp(- i 2LzSz) Lk exp(+ i 2LzSz ) = Lk cos + kmz 2LmSz sin + kz Lz (1 - cos)) Another way to look at this is that we expect to find that [ 2LzSz , Lk ] = izkm 2LmSz or a b c [ 2LzSz , Lx ] = izxm 2LmSz = izxy 2LySz = i 2LySz The question is: does this form a cyclic set of operators with the same relationship? Let's look at the other two cases: [ Lx , 2LySz ] = i 2LzSz ? lhs = 2Sz i xyzLz = i 2 Lz Sz [ 2LySz, 2LzSz ] = i Lx ? lhs = 4 Sz2 i Lx = iLx yes So we see that it is essential to have the factor of 2 if you want to have the operators a,b,c form a cyclic set! This is how we get those Levitt pictures on page 377. So one more time for our final result: exp(- i 2LzSz) Lk exp(+ i 2LzSz ) = Lk cos + kmz 2LmSz sin + kz Lz (1 - cos) How do we write the above as a vector equation? Think of kz = k then write exp(- i 2LzSz) L exp(+ i 2LzSz ) = L cos + 2 L x S sin + ( L )(1 - cos)