evaluate n dot J rotation of J

Evaluate n dot J rotation of J PhL maybe 1.10.08 Problem 1: Is there some way to evaluate this thing: Q = exp(- i J) J exp(+ i J) We can always write this: exp(- i J) = exp( -i [ xJx + yJy + zJz ]) where i = ni but we cannot factor this into the product of three rotations because the operators don't commute! Comment: I give two evaluations here, the second uses Campbell-Hausdorff and is more substantial. The first arrives at the answer by computing it in a special frame, then rotating out of that frame to a general frame. I think both methods are valid. Method #1: I think we can do it by using some coordinate system gibberish. As we write this, we already have in mind that the components of J and of n are x,z,y components and is some arbitrary direction. Suppose we define a new set of coordinate axis x', y' and z' such that points in the z' direction, say. I think we have to be very specific about now we define these new coordinates. But for now, let's assume we have found a rotation such that R = R is a 3x3 matrix Now the equation Q = exp(- i J) J exp(+ i J) says vector = vector, and we can look at this in any coordinate system we want. If we have a vector V in the unprimed system, then we know that V' = RV For example, if V = , we find that V' = R = . So, lets start with Q = exp(- i J) J exp(+ i J) and apply R to both sides. This R ix just a 3x3 matrix, not an operator, so I think it can pass through the exponential operators which are 3x3-space rotational scalars [huh??]. If so, we would have Q' = exp(- i J) J' exp(+ i J) and we can write out the scalars in the primed system as well Q' = exp(- i J') J' exp(+ i J') = exp(- i J'3 ) J' exp(+ i J'3 ) To evaluate this, use component k as follows: Q'k = exp(- i J'3 ) J'k exp(+ i J'3 ) = J'k cos + 3kJ' sin according to Levitt top page 144. It agrees if we use k = 1 because then things are in standard A,B,C order and we have 3k = 31 = 2 so yes, a plus sign as Levitt shows. With k = 2 we will get a minus sign as expected. BUT: Levitt's result assumes that k 3. If we want to make this result be true also for k=3 as well, we can add a little correction term, = J'k cos + 3kJ' sin + k3 (1 - cos) Jk' // no summation on k in last term! Then when k = 3, the we have 33 = 0 and then the cosines cancel and we get the right answer J3. However, in order to cast the above into vector notation , we have to cast the last term into something that is in fact a vector with component k and we don't have that as written above. So we write the factor of interest as, k3 Jk' = k3 J3' = k ( J') // since k = k3 and ( J') = J3 Then we can write this in vector notation like so: Q' = J' cos - ( x J' ) sin + ( J ') (1 - cos) ( Detail: the kth component of x J' is km m J' but m = m3 so get k3J' . ) Now, this appears to be a 3-space vector equation. We should be able to write this in the unprimed system as follows: Q = J cos - x J sin + ( J ) (1 - cos) and this then appears to be our "evaluation". It is much simpler than I thought it would be. So here then is our final result. [ see written in more ways at the end of this document ] Q = exp(- i J) J exp(+ i J) = J cos + J x sin + ( J) (1 - cos) Note: the vector covariant form for the correction term was not obvious until I found it in the C-H method below. Method #2: Let's try a separate proof using Campbell-Hausdorff. We have Q = exp(- i J) J exp(+ i J) so define A = i J = i ni Ji and B = Jk We now need to compute a series of commutators. C1 = [B,A] = [ Jk , i J ] = i [ Jk, J] But we know that [ Jk, J ] = i J x where is the unit vector along the k axis, k = 1,2,3 so we get [B,A] = [ Jk , i J ] = i [ Jk, J] = i ( J x ) i But the cyclic rule says that ( J x ) = ( x J ) so we have so far that: C1 = [B,A] = i ( x J ) i and with some interest we see an ( x J ) factor appearing on the right. Now, let's do the next commutator: C2= [ [B,A] , A ] = [ i ( x J ), i J ] We really cannot maintain vector notation here so need now to resort to summations, so lets go back to the start: C0 = B = Jk C1 = [B,A] = [Jk, i J ] = i ni [Jk, Ji ] = i ni kij Jj i = - ni kij Jj Then the next one is: C2 = [ [B,A] , A ] = [ - ni kij Jj , i nn Jn] = -i ()2 ni kij nn [Jj , Jn] = ()2 ni kij nn jnsJs = ()2 ni nnkij jns Js = ()2 ni nn jki jns Js = ()2 ni nn Js [ kn is - ksin] = ()2 ni nn Js kn is - ()2 ni nn Js ksin = ()2 ni nk Ji - ()2 ni ni Jk = ()2 ni nk Ji - ()2 Jk = ()2 ( nk ni Ji - Jk ) Then the next one is C3 = [ ()2 ( nk ni Ji - Jk ) , i nn Jn ] = i ()3 [ ( nk ni Ji - Jk ) , nn Jn ] = i ()3{ nk ni nn i insJs - nn i knsJs } = ()3 nn knsJs = ()3 ni kijJj = -()2 C1 Here the first term vanishes because is contracted on two indices with a symmetric tensor n n. Let's do just one more to get the pattern; // ni nnkij jns Js C4 = [ ()3 nn knsJs, i nr Jr ] = i ()4 nn kns nr i srtJt = - ()4 ni nn kij jnsJs = - ()2 C2 Here is what we have found so far: C0 = Jk C1 = - ni kij Jj C2 = ()2 ( nk ni Ji - Jk ) C3 = -()2 C1 = θ3ni kij Jj C4 = - ()2 C2 = - θ4 ( nk ni Ji - Jk ) __________________________________________________________ Noted added 2.14.17: The conjecture is then that for n = 2,4,6... Cn = (-1)n/2-1 θn ( nk ni Ji - Jk ) = - (-1)n/2 θn ( nk ni Ji - Jk ) And for n = 1,3,,5... Cn = (-1)(n+1)/2 θn ni kij Jj = - (-1)(n-1)/2 θn ni kij Jj Then e-ABeA = B + [B,A]/1! + [[B,A],A]/2! + [[[B,A],A],A]/3! + ..... (G.3.1) = B + C1/1! + C2/2! + C/3! + .... In our case we have exp(- i J) Jk exp(+ i J) = Jk + C1/1! + C2/2! + C/3! + ... = Jk - Σn=1,3,5.. (-1)(n-1)/2 θn ni kij Jj/n! - Σn=2,4,6.. (-1)n/2 θn ( nk ni Ji - Jk )/n! = Jk - ni kij Jj Σn=1,3,5.. (-1)(n-1)/2 θn/n! - ( nk ni Ji - Jk ) Σn=2,4,6.. (-1)n/2 θn/n! = Jk - ni kij Jj [θ - θ3/3! + ...] - ( nk ni Ji - Jk ) [ -θ2/2! + θ4/4! + .... ] = Jk - ni kij Jj sinθ - ( nk ni Ji - Jk ) (cosθ - 1) = Jk - ni kij Jj sinθ - nk ni Ji(cosθ - 1) + Jk(cosθ - 1) = Jkcosθ - ni kij Jj sinθ - nk ni Ji(cosθ - 1) = Jkcosθ - ni εkij Jj sinθ - nk (nJ)(cosθ - 1) = Jkcosθ - εkij Jj ni sinθ - nk (nJ)(cosθ - 1) = Jkcosθ + εkji Jj ni sinθ + nk (nJ)(1-cosθ) and this is the correct result comparing to the box below. __________________________________________________________ So the result is going to be this: answer = C0 + C1/1! + C2/2! + C3/3! + C4/4! + ... = Jk +()2 ( nk ni Ji - Jk ) /2! - ()4 (nk ni Ji - Jk )/4! + .... + - ni kij Jj /1! + ()3 ni kij Jj/3! + ... The second series of the odd terms is this: - ni kij Jj [ - 3/3! + ... ] = - ni kij Jj sin = - kij ni Jj sin = - [ n x J ]k sin and this agrees exactly with my expected second term in the Q expression above. What about the first series of even terms? Jk +()2 ( nk ni Ji - Jk ) /2! - ()4 nk ni Ji - Jk )/4! + .... Let's add and subtract ( nk ni Ji) to get = nk ni Ji + ( Jk - nk ni Ji) +()2 ( nk ni Ji - Jk ) /2! - ()4 (nk ni Ji - Jk )/4! + .... = nk ni Ji + ( Jk - nk ni Ji) [ 1 - ()2/2! + 4 + ....] = nk ni Ji + ( Jk - nk ni Ji) cos = nk nJ + ( Jk - nk nJ) cos This makes the final result seem to be this: Qk = exp(- i J) Jk exp(+ i J) = nk J + ( Jk - nk J) cos - [ x J ]k sin = Jk cos - [ x J ]k sin + nk J (1 - cos) So this agrees exactly with the boxed result found by Method #1 ! Let's test this result with n in the z direction: Jk cos - [ x J ]k sin + k3 J3 (1-cos) and [ x J ]k = kij i Jj = kij i3 Jj = k3jJj giving the result as Jk cos - k3jJj sin + k3 J3 (1-cos) Now, suppose k = 1. We then get J1 cos - 13jJj sin = J1 cos + J2 sin and this is the correct result Levitt p 144. Now let's try k = 3 to activate the last term. Jk cos - k3jJj sin + k3 J3 (1-cos) = J3 cos - 33jJj sin + J3 (1 - cos) = J3 cos + J3 (1 - cos) = J3 which again is the correct answer! Now we see the need for that extra term. So I will now write this out in a few different ways: // remember that is a unit vector (1) Q = exp(- i J) J exp(+ i J) = J cos + J x sin + ( J) (1 - cos) (2) Qk = exp(- i J) Jk exp(+ i J) = Jk cos + [J x ]k sin + k ( J) (1 - cos) (3) Qk = exp(- i J) Jk exp(+ i J) = Jk cos + kmsJmns sin + nk ( J) (1 - cos) To get the third version we used these facts: [J x ]k = kmsJmns and k = nk in our notation Addendum 1: After writing the above, I realized that you can write the above in this way (rightmost entry) Q = RJR-1 = J cos + J x sin + ( J) (1 - cos) = R-1 J where R = R() exp(-iJ) and R = R() exp(-iJ) = 3x3 Cartesian rotation matrix. We therefore can "read off" what this matrix R must be. Use the component form to say Jk cos + kmsJmns sin + nk ( J) (1 - cos) = [R-1 J]k = [R-1]kl Jl Rewrite the LHS like so: [ cos kl + kls ns sin + nk nl (1 - cos) ] Jl Therefore we have [R(-)]kl = cos kl + kls ns sin + nk nl (1 - cos) or [R()]kl = cos kl + ksl ns sin + nk nl (1 - cos) I verified that this gives the right matrix for = . Addendum 2. Proof of theorem: [R()]ij = (1/2) tr [ iR()jR†()] From work above we know that , since is a vector operator, R()jR†() = RjR-1= [R-1]jl l Therefore: (1/2) tr [ iR()jR†()] = (1/2) tr [ i[R-1]jl l ] = [R-1]jl { (1/2) tr [ il ]} = [R-1]jl il = [R-1]ji = [R†]ji = Rij QED. So restate the theorem in general terms: // this appeared in my 547 notes. Rij = (1/2) tr [ iRjR†]