Appendix G v3

Appendix G: Rotation Matrices and Related Theorems 1 G.1 Generators and finite rotation matrices 1 G.2 About the general rotation matrix 3 G.3 The Baker-Campbell-Hausdorff and Sandwich Formulas 6 G.4 Two more theorems for the rotation matrix toolbox 9 G.5 Generalizations of the Rotation Group 11 This has been installed, do not edit here. Appendix G: Rotation Matrices and Related Theorems Here we present still more information on rotation matrices and demonstrate some typical manipulations done with such matrices. G.1 Generators and finite rotation matrices The general active rotation matrix for a rotation by angle θ about some n unit vector axis is given by, R(φ) = exp(-i θ n J) , (G.1.1) where the Jk are 3x3 matrices known as the rotation generator matrices, J1 = J2 = J3 = . (G.1.2) The numbers in these three matrices can be summarized in this single statement, (Ja)bc = - i abc // for example, (J1)23 = - i 123 = -i or (iJa)bc = abc (G.1.3) where the εabc permutation tensor is described in (1.5.3). Defining the commutator of two square matrices X,Y as [X, Y] = XY-YX, one can show that the above three matrices satisfy this "commutation relation", [Ji, Jj] = iεijkJk . // [J1,J2] = iJ3 and cyclic (G.1.4) Proof: These easily demonstrated facts will be used in the proof (implied sum on repeated indices) , εabcεABc = δaAδbB - δaBδbA δab ≡ δa,b εabc = εbca = εcab (G.1.5) LHSac = (Ji)ab(Jj)bc - (Jj)ab(Ji)bc = (-i)2[εiabεjbc - εjabεibc ] = - [εiabεcjb - εjabεcib] = - [(δicδaj - δijδac) - (δjcδai - δjiδac)] = δjcδai - δicδaj . RHSac = iεijk(Jk)ac = iεijk[-iεkac] = εijkεkac = εijkεack = δiaδjc - δjaδic . QED The expression shown in (G.1.1) involves the exponentiation of a square matrix to produce a new square matrix of the same dimension. This notion of exponentiating a matrix is straightforward as we demonstrate with a simple example: Rx(θ) = exp(-i θ J) = exp(-i θJ1) ≡ Σn=0∞ (-iθ)n (J1)n /n! = 1 + Σ2,4,6.. (-iθ)n (J1)n /n! + Σ1,3,5.. (-iθ)n (J1)n /n! It is easy to show that (J1)n = J1 for odd n, while (J1)n = for even n > 0. Thus, Rx(φ) = + Σ2,4,6.. (-iθ)n/n! + Σ1,3,5.. (-iθ)n/n! . But Σ2,4,6.. (-iθ)n/n! = - θ2/2! + θ4/4! + ... = (cosθ - 1) Σ1,3,5.. (-iφ)n/n! = (-iθ) + (-iθ)3/3! + ... = (-i) [ θ - θ3/3! + ..] = -i sinθ (G.1.6) and therefore Rx(φ) = + (cosθ - 1) + (-isinθ) = + = . (G.1.7) The three axis-aligned rotations are found in this manner to be Rx(θ) = Ry(θ) = Rz(θ) = . (G.1.8) The general rotation (G.1.1) applied to a vector v produces a forward right-hand-rule rotation of vector v by an angle θ about an arbitrary axis to create v' = Rv. We call this an active rotation. For example, if one applies Rz(θ) shown in (G.1.8) to the unit vector = (1,0,0) one gets (cosθ,sinθ,0) which for small θ is a vector in the first quadrant of the x,y plane. Some authors define rotation matrices which rotate vectors backwards according to the right hand rule, with the connection to our matrices then being θ ↔ -θ. The motivation for doing this is the fact that, whereas v' = Rv in the active view where the vector moves and the axes stay put, one can instead do e'n = R–1en and have the vector stay put and the axes are back-rotated, which is the passive viewpoint. The definition of the rotation matrices is just a convention and (G.1.8) shows our definitions. One says that the matrices Ji "generate" the finite rotations when they are exponentiated. An alternative approach: First, one can show that Ri(dθ) ≈ 1 - idθJi describes a 3x3 matrix which, when applied to a vector, causes that vector to rotate by the small amount dθ about the i axis (an "infinitesimal rotation"). Setting dθ = θ/n, one shows that limn→∞ (1-i[θ/n]Ji)n = exp(-iθJi). This last limit is analogous to limn→∞(1+x/n)n = ex for a scalar value x. G.2 About the general rotation matrix An explicit expression for the general rotation matrix We present this subsection as a Reader Exercise with a set of steps. The result is stated in (G.2.8). We are interested in the following general rotation matrix: Rn(θ) = exp(-i θ n J) ≡ Σn=0∞ (-iθ)n (n J)n/n! . n = = a unit vector (G.2.1) (a) using (G.1.3) and (G.1.5), show that (nJ)2 = T where Tab = (δab - nanb) . (G.2.2) (b) show that (nJ)n = (nJ) T for n = 3,5,7,.... (nJ)n = T for n = 2,4,6... (G.2.3) (c) show that exp(-iθnJ) = cos(θnJ) - i sin(θnJ) where each term is defined by its series (d) show that cos(θnJ) = 1 + T(cosθ - 1) sin(θnJ) = (θnJ) + (nJ)T (sinθ - θ) (G.2.4) (e) show therefore that exp(-iθnJ) = 1 + T (cosθ - 1) -i (nJ){ θ + T (sinθ - θ) } (G.2.5) (f) show that the terms linear in θ all cancel leaving this simpler result exp(-iθnJ) = 1 + (cosθ - 1) T + sinθ[ -i(nJ)] (G.2.6) Hint: (nJ)abnbnc = ni(-iεiab)nbnc = -inc [ εiabninb] = -inc [0] = 0 since [antisym x sym]. (g) show that T = -i(nJ) = (G.2.7) giving this final result, Rn(θ) = exp(-iθnJ) = + (cosθ - 1) + sinθ . (G.2.8) Note that the first two terms are symmetric matrices, while the third is antisymmetric. (h) verify the special cases shown in (G.1.8) above. These cases have n = (1,0,0), (0,1,0, (0,0,1). Finding n and θ Problem: One is handed a 9-element rotation matrix R and one wants to find n,θ so R = exp(-iθnJ) . Solution: We are unaware of a single magic formula that solves this problem, so we use brute force. One can break R into the sum of two pieces, one symmetric and one antisymmetric by constructing Sab ≡ (Rab+Rba)/2 Aab ≡ (Rab-Rba)/2 Rab = Sab + Aab R = S + A . (G.2.9) Looking at (G.2.8) one then has, S = = + (cosθ - 1) A = = sinθ (G.2.10) which can be written as these nine equations A12 = -n3sinθ S11 = 1 + (cosθ-1)(1-n12) S12 = (cosθ-1)(-n1n2) A13 = n2sinθ S22 = 1 + (cosθ-1)(1-n22) S13 = (cosθ-1)(-n1n3) A23 = -n1sinθ S33 = 1 + (cosθ-1)(1-n32) S23 = (cosθ-1)(-n2n3) . (G.2.11) If θ = 0, then the original matrix must be R = 1 and then we have no work to do. If θ = π, the entire A matrix vanishes according to (G.2.10) leaving only S. The equations above are then S11 = 1 + (-2)(1-n12) S12 = 2n1n2 S22 = 1 + (-2)(1-n22) S13 = 2n1n3 S33 = 1 + (-2)(1-n32) S23 = 2n2n3 . Since n12 + n22 + n32 = 1, not all three ni can vanish. The left column of equations says ni2 = 1 - (1-Sii)/2 i = 1,2,3 . Inspect the three ni2 and select one nr2 which is non-zero. Then select the plus sign to get, nr = +. To be specific, assume nr = n1. Then from the right column, n2 = S12/(2n1) and n2 = S13/(2n1) and we are done. So at least one of the following solutions must be viable: θ = π n1 = n2 = S12/(2n1) n3 = S13/(2n1) θ = π n2 = n1 = S12/(2n2) n3 = S23/(2n2) θ = π n3 = n1 = S13/(2n3) n2 = S23/(2n3) (G.2.12) If θ ≠ 0 or π, then note from (G.2.10) that A122 + A132 + A232 = sin2θ S11+ S22 + S33 = 1 + 2 cosθ . (G.2.13) Using the sign ambiguity in the direction of n, we can assume that 0 < θ < π so sinθ > 0. Then from (G.2.13) and the left column of (G.2.11) we have this solution, sinθ = cosθ = [ S11+ S22 + S33 - 1]/2 n1 = -A23/sinθ n2 = A13/sinθ n3 = -A12/sinθ (G.2.14) and thus a viable θ and n have been found such that R = exp(-iθnJ) . Example: Let R = so then S = and A = . sinθ = = = sinα cosθ = [ S11+ S22 + S33 - 1]/2 = [cosα + cosα + 1 - 1]/2 = cosα θ = α n1 = -A23/sinθ = 0 n2 = A13/sinθ = 0 n3 = -A12/sinθ = sinα/sinθ = 1 n = (0,0,1) Therefore R = exp(-iαJ3) = Rz(α) , as verified in (G.1.8). G.3 The Baker-Campbell-Hausdorff and Sandwich Formulas Statement and proof of the Baker-Campbell-Hausdorff (BCH) formula We now quote a fascinating fact known as the Baker-Campbell-Hausdorff formula involving two square matrices A and B, where commutator [X,Y] ≡ XY-YX , e-ABeA = B + [B,A]/1! + [[B,A],A]/2! + [[[B,A],A],A]/3! + ..... (G.3.1) Outline of BCH proof: LHS = RHS (a) Show that LHS = Σn=0∞Σm=0∞ (-A)nBAm / (n!m!) . (b) Set k = n+m to rewrite as LHS = Σk=0∞Σn=0k (-A)nBAk-n / (n![k-n]!) . (c) Rewrite again as LHS = Σk=0∞Tk/k! where Tk ≡ Σn=0k (-A)nBAk-n . (d) Define C0 = B, C1 = [B,A], C2 = [[B,A],A] , etc. so that RHS = Σk=0Ck/k! . Note that [Ck,A] = Ck+1. The proof LHS = RHS is complete if one can show that Tk = Ck. (e) Show Tk= Ck by induction: show T0 = C0 and Tk= Ck Tk+1= Ck+1 . QED Statement and proof of the Sandwich Formula Now using this BCH formula, along with the commutation relation (G.1.4) that [Ji, Jj] = iεijkJk, one can show that exp(- iθnJ) J exp(+ iθnJ) = cosθ J + sinθ J x n + (1 - cosθ) n(nJ) . (G.3.2) This "vector of matrices" notation is just a shorthand for the following equations for k = 1,2,3 : exp(- iθnJ) Jk exp(+ iθnJ) = cosθJk + sinθ[J x n ]k + (1 - cosθ) nk (nJ) = cosθ Jk + sinθ εkmsns Jm + (1 - cosθ ) nk (nJ) . (G.3.3) This is the "sandwich formula" since Jk on the left is sandwiched between two rotations. Outline of Sandwich proof: LHS = RHS (a) We will use the BCH formula with A = iθ nJ and B = Jk. First, define Ck as in (d) above. (b) show that C0 = Jk, C1 = -θniεkijJj, C2 = θ2 (nkniJi - Jk), C3 = -θ2C1, C4 = -θ2C2 (c) deduce (or use induction to show) that in general, Cn = - (-1)n/2 θn ( nk ni Ji - Jk ) n = 2,4,6.... Cn = - (-1)(n-1)/2 θn ni εkij Jj n = 1,3,,5... At this point we have from the BCH formula, exp(- iθnJ) Jk exp(+ iθnJ) = Σn=0∞ Cn/n! with Cn as in (c) above (d) Show that Σn=0∞ Cn/n! = Jk - ni εkij 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 εkij Jj [θ - θ3/3! + ...] - ( nk ni Ji - Jk ) [ -θ2/2! + θ4/4! + .... ] = Jk - ni εkij Jj sinθ - ( nk ni Ji - Jk ) (cosθ - 1) = Jkcosθ + εkji Jj ni sinθ + nk (nJ)(1-cosθ) . QED Comments A vector v (rank-1 tensor) transforms ("rotates") according to v' = Rv (Active View) . For a matrix M (rank-2 tensor) the corresponding transformation is M' = RMR-1, and this is what one sees on the left side of (G.3.3) where M = Jk and R = exp(-iθ J). In the expression RJkR-1 the rotation generator Jk is "sandwiched" between the two rotations. The above sandwich formulas play a major role in Magnetic Resonance Imaging. The connection is that protons in your body have magnetic moments (spins) which can be lined up by a strong magnetic field. When the proton spins are slammed with a certain radio frequency pulse, they do conical rotation (precession) about the magnetic field axis at the so-called Larmor frequency. After the pulse this proton precession decays away (time T1) and bulk-decoheres (time T2) producing a certain return RF signal which can be analyzed. These return signals are sensitive to the local environment of the protons. The location of a particular response is determined by giving the magnetic field a spatial gradient which affects the Larmor frequency. In this manner, an image can be formed. The sandwich formulas are not applied directly to individual spin angular momenta J, but to the average spin (polarization) density in the object being scanned. The analysis is quite complicated since it must take into account thermal and statistical effects which are managed with the use of the density matrix formalism. See the excellent text of Levitt for all the details. Special cases of the sandwich formula We shall have our own purposes for the sandwich formulas in Appendix H. For rotations about the i axis we set ns = δs.i n J = Ji and kmsJmns = kmiJm = εikmJm . (G.3.4) Then from the sandwich formula (G.3.3), exp(- iθnJ) Jk exp(+ iθnJ) = cosθ Jk + sinθ εkmsnsJm + (1 - cosθ) nk(n J), (G.3.3) we find that (there is no implied sum on i in the rightmost term and εkmi = εikm ) exp(- iθJi) Jk exp(+ iθJi) = cosθ Jk + sinθεikmJm + (1 - cosθ) δk,i Ji so then Ri(θ) Jk Ri(-θ) = cosθ Jk + sinθ εikm Jm + (1 - cosθ) δk,i Ji . (G.3.5) In the case i = k we know that the left side is just Jk since everything then commutes. This is verified on the right since εiim = 0 and the other two terms then add up to Jk. So although obvious, we state: Rk(θ) Jk Rk(-θ) = Jk . i = k (G.3.6) In the case i ≠ k the third term in (G.3.5) does not contribute and we have Ri(θ) Jk Ri(-θ) = cosθ Jk + εikmsinθ Jm i ≠ k We shall now construct a table of all the cases for which i ≠ k : (G.3.7) R1(θ) J2R1(-θ) = cosθ J2 + ε12m sinθ Jm = cosθ J2 + sinθ J3 R1(θ) J3R1(-θ) = cosθ J3 + ε13m sinθ Jm = cosθ J3 - sinθ J2 R2(θ) J1R2(-θ) = cosθ J1 + ε21m sinθ Jm = cosθ J1 - sinθ J3 R2(θ) J3R2(-θ) = cosθ J3 + ε23m sinθ Jm = cosθ J3 + sinθ J1 R3(θ) J1R3(-θ) = cosθ J1 + ε31m sinθ Jm = cosθJ1 + sinθ J2 R3(θ) J2R3(-θ) = cosθ J2 + ε32m sinθ Jm = cosθJ2 - sinθ J1 . (G.3.8) G.4 Two more theorems for the rotation matrix toolbox Theorem 1: RJR-1 = R-1J (G.4.1) This is another vector/matrix notation theorem which makes a claim about rotating a vector of matrices. The above ambiguous notation is a shorthand for the following, RJiR-1 = [R-1J]i = R–1ijJj i = 1,2,3 . (G.4.2) The object on the left is a product of three 3x3 matrices, while the right side is a linear combination of 3x3 matrices, so at least the theorem's claim is dimensionally reasonable. More generally Ji might be an abstract "operator" and R a rotation which acts on that operator, see Section G.5. Proof: Start with the general rotation form given in (G.2.6), R = exp(-iθnJ) = 1 + (cosθ - 1) T + sinθ[ -i(nJ)] where Tab = (δab - nanb) . (G.4.3) Sandwich this rotation around Ji and then use the sandwich formula (G.3.3), RJiR-1 = exp(- iθn J) Ji exp(+ iθn J) = cosθ Ji + sinθ εijknk Jj + (1 - cosθ ) ni (n J) = cosθ δijJj + sinθ εijknk Jj + (1 - cosθ ) ninjJj = [ cosθ δij + sinθ εijknk + (1 - cosθ ) ninj ] Jj . (G.4.4) Our theorem is proved if, comparing (G.4.2) and (G.4.4), we can show that R–1ij = cosθ δij + sinθ εijknk + (1 - cosθ ) ninj . (G.4.5) Looking back at (G.4.3), one has R-1 = exp(+iθnJ) = 1 + (cosθ - 1) T + sinθ[ +i(nJ)] so R-1ij = δij + (cosθ - 1) Tij + sinθ[+nk (iJk)ij] = δij + (cosθ - 1) (δij - ninj) + sinθ nk εkij // (G.1.3) and def of Tij = δij cosθ + (1-cosθ) ninj + sinθ εijknk (G.4.6) But this is the same as (G.4.5) so the theorem is proved. Theorem 2: R exp(-iθnJ) R-1 = exp(-iθn'J) where n' = Rn (G.4.7) or R Rn(θ) R-1 = Rn'(θ) where n' = Rn The matrix R is an arbitrary rotation. On the left we have a product of three 3x3 matrices, while the right side is a 3x3 matrix. One proof of this theorem might be to say "what else could it be? ". We shall provide a more substantial proof below. Proof: Start by inserting the general rotation form (G.2.6) into the left side of (G.4.7), R exp(-iθnJ) R-1 = R [ 1 + (cosθ - 1) T - i sinθ (nJ) ] R-1 = 1 + (cosθ - 1)RTR-1 - i sinθ nk RJkR-1 . (G.4.8) Now consider [RTR-1]ad = RabTbcR-1cd = Rab(δbc - nbnc)RTcd = δad - Rab nbncRdc = δad - ( Rab nb)(Rdcnc) = δad - n'an'd n' = Rn ≡ T'ad . // in other words, RTR-1 = T' (G.4.9) From Theorem 1 (G.4.2) we know that RJkR-1 = R–1kjJj . (G.4.2) Inserting this last item and (G.4.9) into (G.4.8) gives R exp(-iθnJ) R-1 = 1 + (cosθ - 1)T' - i sinθ nkR–1kjJj = 1 + (cosθ - 1)T' - i sinθ Rjknk Jj = 1 + (cosθ - 1)T' - i sinθ [Rn]j Jj = 1 + (cosθ - 1)T' - i sinθ n'jJj = 1 + (cosθ - 1)T' - i sinθ (n' J) = exp(-iθn'J) // using (G.2.6) with n → n' (G.4.10) Thus the theorem is proved. The fact (G.4.9) that RTR-1 = T' is in fact just the standard rule for the transformation of a rank-2 tensor under rotations, see (1.1.21) [ also (J.24) and (J.25)]. That is to say, if an Observer in Frame S sees T, an Observer in rotated Frame S' will see T' = RTR-1. This same statement applies at the higher level of our theorem R Rn(θ) R-1 = Rn'(θ) n' = Rn // Theorem 2 restated (G.4.11) The rotation Rn(θ) seen by an Observer in Frame S appears as Rn'(θ) in Frame S' where e'n = R-1en as in (1.1.30). Like T, the rotation Rn(θ) transforms as a rank-2 tensor under rotations. Applying (G.4.11) to a vector v, one finds, R Rn(θ) R-1v = Rn'(θ)v where n' = Rn (G.4.12) Instead of rotating a vector v by amount θ about axis ', one can first back-rotate the vector v by R-1, then rotate by amount θ about axis , the forward rotate the result by R. Example: Rz(π/2)Rx(θ)Rz(-π/2) = ?? Here n = and n' = Rz(π/2) = so we conclude that Rz(π/2)Rx(θ)Rz(-π/2) = Ry(θ) . (G.4.13) Here is a graphical interpretation of (G.4.13) applied to a particular vector v : (G.4.14) G.5 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.5.1) 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.5.1). 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.5.2) 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.5.1) 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.5.1) 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.5.3) 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.5.4) 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.5.5) 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.5.6) = 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 (sideways) 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.5.7) 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.5.3) and (G.5.6) 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.5.7) 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.5.8) [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.5.8), 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