Section G_9 generalizations INSTALLED

This has been installed into Appendix G, do not edit here. G.9 Generalizations of the Rotation Group Here we consider some generalizations of the ideas presented above in Section G.1 N dimensional representations of the rotation group The generators Ji shown in (G.1.2) are a special case of a more general idea which starts with the commutation relation [Ji, Jj] = iεijkJk . // for example, [J1,J2] = iJ3 (G.1.4) (G.9.1) One first thinks of the Ji as abstract "operators" in some abstract "operator space". One can show that it is possible to find a set of three NxN matrices of any integer dimension N which satisfy (G.1.4). The matrices are not unique, so (G.1.2) for the Ji in three dimensions is not unique, but it is a standard form. The three NxN generator matrices Ji are said to form an N-dimensional "irreducible representation" of the abstract generators Ji. One can always create new viable generator matrices by taking a "direct sum" of existing viable generator matrices, such as in this block-diagonal-form picture (G.9.2) This generator matrix is "reducible" into a direct sum of S, T and R. An "irreducible" representation is one that cannot be reduced in this manner. The same comment applies to rotation matrices. The commutation relation (G.9.1) is an example of a Lie Algebra. Our particular Lie Algebra is called so(3), so we can represent the algebra elements Ji of this algebra by three NxN matrices Ji. It is possible to write down a formula analogous to (G.1.3) (iJa)bc = abc which works for any N, but (G.1.3) itself only applies to N = 3. This is so because εabc has no meaning for N ≠ 3. But it always has meaning in (G.9.1) because there are only three generators regardless of the value of N. For general N, the object Ri(θ) = exp(-iθJi) is an NxN matrix which represents the action of a rotation of an N-vector in a Euclidean space EN. The set of such rotation matrices forms an "irreducible representation" of the rotation group SO(3) in N dimensions. The integer N is usually written N = 2j+1 where j = 0, 1/2, 1, 3/2 ... and this j then serves as a label for a given matrix representation. The value of j in N = 2j+1 is associated with "angular momentum" or "spin". In the case N=2 (having j=1/2) the generator matrices are the 2x2 "Pauli matrices". In this case the 2x2 matrices exp(-iθJi) describe the rotations of spin-1/2 particles such as electrons or protons. Here are the details for N=2 : Jx = (1/2) Jy = (1/2) Jz = (1/2) Rx = Ry = Rz = (G.9.3) The "vectors" for spin-1/2 particles have two components . The special case is called "spin up" and is "spin down". For the Lie Algebra so(3) one can show that J2 ≡ J12 + J22 + J32 and any particular Ji commute with each other, so [J2,Ji] = 0. J2 is called a Casimir operator of this Algebra, and fancier Lie Algebras can have several such Casimirs. A Differential operator representation of the rotation group It is also possible to "represent" the three rotation generators Ji by three differential operators in spherical coordinates θ and φ. These operators satisfy [Ji, Jj] = iεijkJk and in this context they are usually called Li but we shall stick with Ji. In this case, one can compute the differential operator J2 and one can ponder differential equations which take the form J2 fjm(θ,φ) = j(j+1) fjm(θ,φ) and Jz fjm(θ,φ) = m fjm(θ,φ). [The facts that the eigenvalue of J2 is j(j+1) and not j2, and that m runs from -j to j, derive from the structure of the Lie Algebra.] The solutions fjm(θ,φ) are able to have well-defined eigenvalues j(j+1) and m because [J2,J3] = 0. If we instead had [J2,J3] ≠ 0, then [J2,J3] fjm(θ,φ) = j(j+1)m - mj(j+1) = 0 is a contradiction and the two eigenfunction equations could not exist. The solutions fjm(θ,φ) of these equations are called the spherical harmonics and are usually written Yjm(θ,φ). Just for the record, here is what the differential operators look like, where C = cos and S = sin (for example, Sφ = sinφ and ∂φ = ∂/∂φ) : J1 = i [ Sφ ∂θ + cotθ Cφ ∂φ] J = eiφ [ ∂θ + i cotθ ∂φ ] J2 = i [ Cφ ∂θ cotθ Sφ ∂φ] J2 = [ ∂θ2 + cotθ ∂θ + (1/Sθ)2 ∂φ2] J3 = -i ∂φ J2 = [ (1/S) ∂θ [ Sθ ∂θ ] + (1/Sθ)2 ∂φ2 ] (G.9.4) Reader Exercise: Verify that the three operators on the left satisfy the Lie Algebra [Ji, Jj] = iεijkJk. For the hydrogen atom with a spinless electron, there are three mutually commuting quantities H, J2 and J3 where H is the Hamiltonian. This means that the solution eigenfunctions can have well defined E, j and m values and these eigenfunctions are those painful "orbitals" appearing in chemistry books. When the Hamiltonian commutes with some other operator like J2, that operator is called a "symmetry". Solution functions then bear a label for each such symmetry, such as j for J2. The Lie Algebra so(3) is isomorphic (one-to-one related) to another Lie Algebra called su(2). The Lie Group SO(3) is isomorphic to another Lie Group called SU(2). In the above we discuss only the Lie Group SO(3) with its three generators Ji. There are many other Lie Groups which have physics applications. Some Other Lie Groups of Interest The group SO(n) is the orthogonal group in n dimensions and it has n(n-1)/2 generators. The group SO(3,1) is the Lorentz Group which has 6 generators Ji and Ki which generate 3 rotations and 3 "boosts" (velocity transformations). The Lie Algebra is this, [ Ji, Jj] = +i εijkJk [ Ji, Kj] = +i εijk Kk [ Ki, Kj] = -i εijkJk (G.9.5) There are two Casimirs : J2 - K2 and JK . In the Lorentz Group "vector representation" known as 1/21/2 the generators are represented as 4x4 matrices. When these are exponentiated, one obtains the finite rotation and boost matrices used in special relativity. For example, with space-time vectors ordered xμ = (ct,x,y,z) one has = exp(-irJ1) where (J1)μν = // rotation Rx(r) (G.9.6) = exp(-ibK1) where (K1)μν = // boost Bx(b) The Poincare Group is basically the Lorentz group SO(3,1) bolted onto the group T(4) of translations in four directions ct,x,y,z. It thus has 10 generators Ji, Ki and Pμ where these last four are momentum and energy. The Poincare algebra has two Casimir operators whose eigenvalues are associated with mass and spin. For example, the mass Casimir is PμPμ . The irreducible representations of the Poincare Group are associated with "elementary particles" which have well defined mass and spin. The group SU(3) has 8 generators and 2 Casimirs. In one key representation the generators are represented by 3x3 matrices which act on 3-vectors. Instead of having up and down states as with spin-1/2 noted above, these vectors have up, down and strange states called u,d.s which are associated with quarks. The full symmetry group for the Standard Model of elementary particles is SU(3) x SU(2) x U(1) in which SU(3) plays its part. Special Groups have Traceless Generators It is desirable that "rotation matrices" have unit determinant because such matrices then do not change the "length" of a vector on which they act. When the representation matrices are restricted to have unit determinant, they are called "special" and the group name is prefixed by the letter S, as in SO(3) for the rotation group. For the 3x3 representation of the rotation group we know that the matrices are real orthogonal which means RRT = 1 which in turn means [det(R)]2 =1 and in SO(3) we select only those R with det(R) = +1. A "rotation" which just negates z (reflection) still satisfies RRT = 1 but has det(R) = -1. An elegant theorem concerning exponentiated square matrices is this (proved below): det(eA) = etr(A) (G.9.7) where det is the determinant and tr(A) ≡ ΣiAii is the "trace" or "spur" of the matrix A -- the sum of the diagonal elements. If we want the matrix exp(-i θ J) to have unit determinant so it is "Special", the exponent must be traceless, and in this case that means that the generators Ji must all be traceless. One can see from examples (G.1.2) and (G.9.3) and (G.9.6) that this is indeed the case. Offhand, the matrix identity det(eA) = etr(A) seems very unlikely and almost too simple. For that reason, we include here a straightforward proof which we feel is "one for the Book". Proof of (G.9.7) There are no implied sums in this proof! Write A = ΣijAij s(ij) where [s(ij)]ab ≡ δiaδjb . To verify, Aab = ΣijAij [s(ij)]ab = ΣijAijδiaδjb = Aab . The matrix s(ij) is all zeros except for a single 1 located in row a and column b. Using ex+y.. = exey ... we can write eA = exp( ΣijAij s(ij)) = Πi,j exp(Aij s(ij)) . Then using det(XY..) = det(X)det(Y).... , det(eA) = det {Πi,j exp(Aij s(ij)) } = Πi,j det [ exp(Aij s(ij)) ] . (G.9.8) [Case i ≠j :] We first note that [s(ij)]2 = 0 : [s(ij)]2ac = Σb[s(ij)]ab [s(ij)]bc = Σbδiaδjb δibδjc = δiaδjiδjc = 0 since i ≠ j Then [s(ij)]n = 0 for n ≥ 2. In this case we have exp(Aij s(ij)) = Σn=0∞ (Aij)n [ s(ij)]n/n! = 1 + Aij [ s(ij)] and so det [exp(Aij s(ij)) ] = det [ 1 + Aij s(ij)] = 1 . This is so because the matrix in question has all 1's on the diagonal and one non-vanishing off-diagonal element at (i,j). In fact, any triangular matrix (one side all zeros) with 1's on the diagonal has det = 1. [Case i=j : ] In this case we have [s(ii)]2 = [s(ii)] : [s(ii)]2ac = Σb[s(ii)]ab [s(ii)]bc = Σbδiaδib δibδic = δiaδiiδic = δiaδic = [s(ii)]ac . Note that [s(ii)]ab = δiaδib = 1 only when a = b = i, so s(ii) is an all-zero matrix with a single 1 at location i on the diagonal. Since [s(ii)]2 = s(ii) it follows that [s(ii)]n = s(ii) for n ≥ 1. Then, exp(Aii s(ii)) = Σn=0∞ (Aii)n [ s(ii)]n / n! = 1 + s(ii) [Σn=1∞(Aii)n / n!] = 1 + s(ii) [ -1 + Σn=0∞(Aii)n/n! ] = 1 + s(ii)[-1 + exp(Aii) ] . This last item is the unit matrix with the ith diagonal 1 replaced by exp(Aii) . Therefore, det [exp(Aii s(ii)) ] = det { 1 + s(ii)[-1 + exp(Aii)] } = exp(Aii) . Now go back to (G.9.8), det(eA) = Πi,j det [exp(Aij s(ij)) ] = (Πi≠j det [exp(Aij s(ij)) ] ) * (Πi=j det [exp(Aij s(ij)) ] ) = (1*1*1*1...... ....*1*1) * ( exp(A11)exp(A22) ..... ) = 1 * exp(A11 + A22 + ...) = exp (tr(A)) = etr(A) . QED