A Derivation of Vector and Momentum Matrices

Transcription

1 A Derivation of Vector and Momentum Matrices arxiv:math-ph/040002v 4 Jan 2004 Richard Shurtleff February 23, 2008 Abstract Nonunitary, finite-dimensional matrix representations of the Poincaré algebra are found. These covariant representations may already be known but are less well-known than the unitary, canonical representations or the continuous representations with momentum as a gradiant operator. Given standard angular momentum (spin matrices satisfying the Lorentz algebra of rotations and boosts, the calculation here finds corresponding vector matrices and the momentum matrices that generate translations. The standard spin matrices and momentum matrices together satisfy the commutation rules of the Poincaré algebra. Irreducible Lorentz spin matrices fail to produce vector matrices. General, nontrivial vector matrices can be deduced from those found here for reducible spin matrices with spin (A,B (C,D, where A C = B D = /2. Useful vector matrix formulas that are hard to find elsewhere are displayed here in detail. A problem set is included. Introduction A vector matrix is both a vector and a tensor. Under a Lorentz transformation Λ ν µ the vector matrix V µij transforms like a vector: V µij Λ ν µ V νij. Also, given an n-dimensional representation D ij (Λ of the proper homogeneous Lorentz group, the vector matrix transforms as a second rank tensor: V µij D ij (ΛV µjk D kl (Λ. Equating the results, we get D ij (ΛV µjk D kl (Λ = Λ µ ν V νil, ( affiliation and mailing address: Department of Applied Mathematics and Sciences, Wentworth Institute of Technology, 550 Huntington Avenue, Boston, MA, USA, ZIP 025, telephone number: ( , fax number: ( , address: shurtleffr@wit.edu

2 INTRODUCTION 2 where summation over repeated indices is assumed, indices µ, ν {, 2, 3, 4} with indices {, 2, 3} ( = {x, y, z} labeling a right-handed rectilinear spatial reference frame and 4 ( = t labeling the time component, and i, j {, 2,..., n}. A well known example of a set of vector matrices is the set of the gamma matrices found in the relativistic wave equations for spin /2 particles such as the electron. Equation ( is then a special case of Pauli s Fundamental Theorem. [] Commutation rules follow directly from (. For example, let Λ = R be a rotation of the x, y-plane about the origin through an infinitesimal angle δ. Then cosδ and sin δ δ and Λ µ ν V ν {V x δv y, V y + δv x, V z, V t }, where indices ij can be inferred from the context and are dropped. Now let J z be the z-angular momentum matrix which generates R so that D ± ij(r I ij ± iδj zij, where I ij is the n n unit matrix. Substituting these in ( gives [J z, V x ] ik = J zij V xjk V xij J zjk = iv yik, (2 [J z, V y ] ik = J zij V yjk V yij J zjk = iv xik. Matrix multiplication and summation over repeated indices are understood. These two commutation rules and the commutation rules like it can be used to calculate the vector matrices V µij of any given n n matrix representation of the homogeneous Lorentz group. Since there are three angular momentum components, J i with i {, 2, 3}, and four vector indices, V µ with µ {, 2, 3, 4}, there are twelve commutation rules like (2. Another twelve come from the boost generators K i, for a total of 24 commutation rules to be satisfied by vector matrices. There are not always any nonzero solutions. Irreducible Lorentz representations can be characterized by the spin pair (A, B, with 2A and 2B integers. We show that nontrival vector matrices do not exist for irreducible Lorentz representations. However, there are nonzero vector matrices for reducible representations with spin (A, B (C, D when A C = /2 and B C = /2. The results for more general spin types can be deduced from the results obtained here. Explicit formulas for vector matrix components are collected in an Appendix which are to be used with the standard set of angular momentum and boost generator matrices displayed in Sec. 3 Momentum generates translations and translations, rotations and boosts together make up the inhomogeneous Lorentz group of spacetime symmetries known as the Poincaré group. Translations commute, so the matrix representation of momentum P µij is a set of four vector matrices that commute. Once the expressions for vector matrices are found it is but a short step to apply the rule [P µ, P ν ] = 0 and obtain an n n matrix representation of the Poincaré algebra. The representations of the Poincaré algebra found here are nonunitary, being extensions of nonunitary n n matrix representations of the homogeneous Lorentz group and should

3 2 THE POINCARÉ COMMUTATION RULES 3 not be confused with the unitary representations of the Poincaré algebra familiar since the work by Wigner. [2] [5] Sections 2 and 3 establish notation. In Sec. 2 the well-known commutation rules of the Poincaré algebra are presented, [6] [0] and the rules for vector matrices are identified. In Sec. 3 formulas for the matrices of the irreducible matrix representations (A, B of the (homogeneous Lorentz group are recalled from the literature. [] The details of the derivation of the vector matrices take up most of the remainder of the article. In Sec. 4 we find that there are no vector matrices with any of the irredicible matrix representations of the Lorentz group, so, in Sec. 5 we consider reducible Lorentz representations with spin (A, B (C, D. In Sections 4 and 5 we take advantage of the diagonal nature of matrices J z and K z. The step-up and step-down matrices J ± give recursion relations in Sec. 6. Then the recursion relations are solved in Sec. 7 for matrices V ± which are combinations of V x and V y. The results determine V z and V t in Sec. 8. Having found expressions for vector matrices, momentum matrices are obtained in Sec. 9. An Appendix collects expressions for the vector matrix components in terms of the spins A, B, C, D and two free parameters AB and t2 CD. The derivation illustrates standard techniques by first applying commutation rules with diagonal matrices and then dealing with step-up and step-down matrices. Also potentially interesting to students is the appearance of the matrices for spin (A, B (C, D. A complication commonly found in the literature [9],[] is found here: we use double indices that reflect the spin structure; matrix indices ij in ( become ab; cd in Sec. 3. The formulas for the components of vector matrices serve as reference for more advanced students and practitioners. A problem set is attached to help assure understanding, to provide the reward of investigation, and to offer an exploration of topics extending beyond the text. 2 The Poincaré Commutation Rules There are ten Poincaré generators, the four components of momentum generate translations, the three components of angular momentum generate rotations and the three components of boost generators generate boosts. The commutator of any two different Poincaré generators is a homogeneous linear combination of generators. The commutation rules of the Poincaré algebra are well known. [6] [0] The fifteen rules that do not involve the momentum form the commutation rules of the homogeneous Lorentz algebra, [J i, J j ] = iǫ ijk J k [J i, K j ] = iǫ ijk K k [K i, K j ] = iǫ ijk J k (3 where J k is the k component of the angular momentum, K k is the k component of the boost generator and ǫ ijk determines the sign: ǫ ijk is antisymmetric in ijk with ǫ xyz = and

4 3 IRREDUCIBLE REPRESENTATIONS OF THE LORENTZ GROUP 4 i, j, k {x, y, z}. Twenty-four rules are linear and homogeneous in momentum components. They can be deduced from ( by appropriate choices for Λ. A vector matrix is a set of four square matrices that obeys the 24 rules linear in momentum. Writing the four components of a vector as V µ = {V i, V t }, µ {x, y, z, t}, the rules are [J i, V j ] = iǫ ijk V k [K i, V j ] = iv t δ ij (4 [J i, V t ] = 0 [K i, V t ] = iv i δ ij (5 References [4] and [5] display rules using the same spacetime metric as here, the diagonal matrix with diagonal {,,, }, but with contravariant vector matrices {V i, V t } = {V i, V t }. This accounts for minus signs in the rules involving V t. The final six rules are quadratic in momentum components and state that momentum components P µ = {P i, P 4 } commute. Thus the P µ obey the rules (4 and (5 for vectors as well as the following six rules that say that the components commute, [P µ, P ν ] = 0 (6 where µ, ν {x, y, z, t}. Eqns. (3 - (6 (with P µ for V µ in (4 and (5 are the commutation rules of the Poincaré algebra. 3 Irreducible Representations of the Lorentz Group Matrices representing the components of angular momentum and boosts of the homogeneous Lorentz group and satisfying (3 are well known and given by standard expressions. [] The irreducible representations of the (homogeneous Lorentz group are catalogued by the pair (A, B, where 2A and 2B are non-negative integers. Following standard procedures [],[2], one begins by writing a 2A + dimensional representation of the rotation group: M +(A σσ = (M (A x + im y (A σσ = r σ (A δ σ,σ + (7 where M (A σσ = (M (A x im y (A σσ = s (A σ δ σ,σ (8 (M (A z σσ = σ δ σσ (9 σ, σ { A, A +,..., A, A} (0

5 4 IRREDUCIBLE LG REPRESENTATIONS - NO SOLUTIONS 5 and with r (A σ s (A σ r σ (A = = r (A σ = (A σ (A + σ + ( (A + σ (A σ + (2 and s (A σ zero outside the allowed range of σ in (0. The matrices {M x (A, M y (A, z } satisfy the commutation rules of the Lie algebra of the rotation group: [M i, M j ] = M (A iǫ ijk M k. Next define the (2A + (2B + dimensional matrices A k and B k, (A k ab,a b (M (A k a,a δ b,b (B k ab,a b (M (B k b,b δ a,a (3 Then the (A, B representation of the Lorentz group can be taken to be the following (2A + (2B + dimensional matrices J (A,B k and K (A,B k By (7-(4, we find J (A,B k = A k + B k K (A,B k = i(a k B k (4 (J +(A,B ab,a b = r (A a δ a,a +δ b,b + r (B b δ a,a δ b,b + (5 (J (A,B ab,a b = s (A a δ a,a δ b,b + s (B b δ a,a δ b,b (6 (K +(A,B ab,a b = i(r (A a δ a,a +δ b,b r (B b δ a,a δ b,b + (7 (K (A,B ab,a b = i(s (A a δ a,a δ b,b s (B b δ a,a δ b,b (8 (J (A,B z ab,a b = (a + bδ a,a δ b,b (9 (K (A,B z ab,a b = i(a bδ a,a δ b,b (20 Note that J z and K z are diagonal (superscript (A, B is suppressed whenever possible; their nonzero components have the double index ab equal to the double index a b. 4 Irreducible LG Representations - No Solutions Suppose we choose as angular momentum and boost matrices the J i and K i of the (A, B irreducible representation of the Lorentz group. One can check that the commutation rules (3 are satisfied.

6 5 REDUCIBLE LG REPRESENTATIONS 6 Next, we seek vector matrices V µ that solve the 24 commutation rules (4 and (5. Let us begin with the rules (4 and (5 that involve the diagonal matrices J z and K z. By (4, we have [J z, V x ] = iv y and [J z, V y ] = iv x which imply that V x = [J z, [J z, V x ]]. Hence V x = [J (A,B z, [J (A,B z, V x ]] = (J (A,B z We also have [K z, V x ] = 0, which implies [K (A,B z 2 V x 2J z (A,B V x J z (A,B + V x (J z (A,B 2 (2, V x ] = K z (A,B V x V x K z (A,B = 0 (22 With (9 and (20 for J (A,B z and K (A,B z, we get from (2 (V x ab,a b = [(a + b (a + b ] 2 (V x ab,a b (23 and from (22 we get [(a b (a b ](V x ab,a b = 0. (24 One can quickly show that (23 and (24 imply that either a a = b b = ± 2 or (V x ab,a b = 0 (25 But, by (0, the indices a and a must differ by an integer n aa since a and a range over the set { A, A +,...,A, A}. Since the indices cannot differ by ±/2, we get the trivial result, a a = b b = n aa ± (V x ab,a b 2 = 0. (26 Thus V x vanishes and that implies that all four V µ vanish because the others can be written as commutators of V x with J z +(A,B, J y +(A,B, and K x +(A,B by (4. We have V µ = 0. (Irreducible J (A,B, K (A,B (27 There are no non-zero vector matrices V µ, and therefore also no non-zero momentum matrices P µ, satisfying the Poincaré commutation rules with the angular momentum and boost matrices of an irreducible representation of the homogeneous Lorentz group. 5 Reducible LG Representations Since irreducible representations fail to produce interesting results, we next try reducible representations of the Lorentz group. The simplest of these have angular momentum and

7 5 REDUCIBLE LG REPRESENTATIONS 7 boost matrices in block diagonal form with two irreducible representations along the diagonal. Consider the (2A+(2B + + (2C +(2D+ dimensional representation of the Lorentz group which combines an (A, B and a (C, D representation. In block matrix notation, we have ( J 0 J = = 0 J 22 ( J (A,B 0 0 J (C,D ( K 0 K = = 0 K 22 ( K (A,B 0 0 K (C,D with J (A,B and K (A,B as in Section 3 and similar expressions for J (C,D and K (C,D. Putting the unknown vector matrices V µ in block matrix form, we get ( Vµ V V µ = µ2 V µ2 V µ22 The commutator [J, V ] in block matrix form is ( J.V [J, V ] = V.J J.V 2 V 2.J 22 J 22.V 2 V 2.J J 22.V 22 V 22.J 22 The commutation rules (4 and (5, by inspection of the diagonal blocks of (30, do not mix (A, B and (C, D representations. Thus the results of Section 4 for irreducible Lorentz representations apply to V and V 22 and we infer that V = 0 and V 22 = 0. However, the 2 and 2 blocks of [J, V ] do mix the (A,B and the (C,D representations and so the results of Section 4 do not apply to the off-diagonal blocks of V. Thus we have, by Section 4 and without loss of generality, V µ = ( 0 Vµ2 V µ2 0 (28 (29 (30. (3 The unknowns are now the components of the off-diagonal blocks V µ2 and V µ2 of the four vector matrices V µ. As in Section 4, we consider the equations in (4 and (5 that involve the diagonal matrices J z and K z. Here, as there, we have V x = [J z, [J z, V x ]] from [J z, V x ] = iv y and [J z, V y ] = iv x. For the 2 block, by (28, (30, and (3, these commutation rules give V x2 = [J z, [J z, V x ]] 2 = (J (A,B z And we have [K z, V x ] = 0, which gives By (9 and (20 for J (A,B z 2 V x2 2J z (A,B V x2 J z (C,D + V x2 (J z (C,D 2 (32 [K z, V x ] 2 = K z (A,B.V x2 V x2.k z (C,D = 0. (33, J (C,D z, K z (A,B, and K z (C,D in (32 and (33 we get (V x2 ab,cd = [(a + b (c + d] 2 (V x2 ab,cd (34

8 5 REDUCIBLE LG REPRESENTATIONS 8 By (34 and (35, we find that either [(a b (c d](v x2 ab,cd = 0 (35 a c = b d = ± 2 or (V x2 ab,cd = 0 (36 Compare (36 with Eqn. (25. In (25 a and a could not differ by a half, so there are no nontrivial solutions to (25. But a and c can differ by a half, so there may be nontrivial solutions to (36. Since spins A, B, C, D must be integers or half-integers and by (0 A a, B b, C c, D d are all integers, we find from (36 that when V x2 0, A = C + n + /2 B = D + m + /2, (37 where n and m are integers. (Proof: In (36, let a = A n and c = C n 2, so that A C = n 2 n ±/2. Thus either A or C is an integer while the other is half an odd integer and likewise for B and D. Using deltas to enforce (36, we have can write the ab, cd component of the block V x2 as (V x2 ab,cd = abδ a,c+/2 δ b,d+/2 + u 2 abδ a,c /2 δ b,d /2, (38 where ab and u2 ab are as-yet-undetermined coefficients. From J z and V x we can get V y ; by (4, we have iv y = [J z, V x ]. By (9, (28, (29, (30 and (38, we get the 2 block of iv y, i(v y2 ab,cd = ab δ a,c+/2δ b,d+/2 u 2 ab δ a,c /2δ b,d /2 (39 Comparing (38 and (39 suggest defining simpler matrices V + and V by (V + 2 ab,cd 2 (V x2 + iv y2 ab,cd = abδ a,c+/2 δ b,d+/2 (40 (V 2 ab,cd 2 (V x2 iv y2 ab,cd = u 2 ab δ a,c /2δ b,d /2 (4 To determine the 2-block of V + and V, exchange the (A, B and (C, D blocks in the matrices J k and K k, (28. The effect is to exchange a c, b d, 2 in (40 and (4. We get (V + 2 cd,ab 2 (V x2 + iv y2 cd,ab = t 2 cd δ c,a+/2δ d,b+/2 (42 (V 2 cd,ab 2 (V x2 iv y2 cd,ab = u 2 cdδ c,a /2 δ d,b /2 (43

9 6 STEP-UP AND DOWN; RECURSIONS 9 Thus we have and V + = V = ( 0 V + ( 2 V 2 + = 0 ( 0 V ( 2 V2 = 0 0 ab δ a,c+/2δ b,d+/2 cd δ c,a+/2δ d,b+/2 0 t 2 u 2 0 u 2 cd δ c,a /2δ d,b /2 0 abδ a,c /2 δ b,d /2 Note carefully that there may not be a variable ab for every allowed value of a and b because the factors δ a,c+/2 and δ b,d+/2 in (40. These factors imply that (V 2 + ab,cd depends on ab only when c = a /2 and d = b /2. Thus the indices of t2 ab are restricted also by the ranges of c and d, forcing the indices a and b of ab to be in the following ranges (44 (45 t (2 ab : max[ A, C + 2 ] a min[a, C + 2 ] (46 t (2 ab : max[ B, D + 2 ] b min[b, D + 2 ] (47 Similar restrictions apply to the ranges of the variables t 2 cd, u 2 ab, and u 2 cd. In this section we have considered the commutation rules involving the vector matrices V x and V y and the diagonal matrices J z and K z. The resulting simplification is evident in Eqns. (44 and (45. Next we consider the commutation rules of V x and V y with the given matrices J x and J y. 6 Step-Up and Down; Recursions Define the step-up and step-down matrices J + and J, J + = J x + ij y and J = J x ij y, (48 so-named by their effect on eigenvectors of J z. The commutation rules (4 applied to the commutators of J + and J with the matrices V + and V yield [J ±, V ± ] = 0 and [J ±, V ] = ±V z, (49 which are simpler than the equivalent equations with J x and J y. By (5, (6, and (28, we find the needed formulas for J ±, J + = ( r (A a δ a,a +δ b,b + r (B b δ a,a δ b,b r c (C δ c,c +δ d,d + r (D d δ c,c δ d,d + (50

10 6 STEP-UP AND DOWN; RECURSIONS 0 ( s J (A a = δ a,a δ b,b + s (B b δ a,a δ b,b 0 0 s (C c δ c,c δ d,d + s (D (5 d δ c,c δ d,d Consider the 2-block [J +, V + ] 2 = 0, and, with V + from (44 and J + from (50, we have (r a t (A 2 a,b r(c c ab δ a,c+3/2δ b,d+/2 = 0 (52 (r (B b t2 a,b r (D d abδ a,c+/2 δ b,d+3/2 = 0 (53 From [J, V ] 2 = 0, with V from (45 and J from (5, we get (s (A a+u 2 a+,b s (C c u 2 abδ c,a+3/2 δ d,b+/2 = 0 (54 (s (B b+ u2 a,b+ s (D d u 2 abδ c,a+/2 δ d,b+3/2 = 0 (55 Note that these are recursion relations in the indices a and b. From the right-hand equation in (49, we have [J +, V ] 2 = [J, V + ] 2. With V ± and J ± from (44, (45, (50 and (5, we have (r a u (A 2 a,b r(c c u 2 ab + s(b b+ t2 a,b+ s(d d ab δ a,c+/2δ b,d /2 = 0 (56 (r (B b u2 a,b r(d d u 2 ab + s(a a+ a+,b s(c c ab δ c,a+/2δ d,b /2 = 0 (57 Without loss of generality we can assume that A > C, because we can interchange the AB and CD blocks in the J and K matrices whenever C > A. Thus, by interchanging the AB and CD blocks if necessary, we can assume that n is positive or zero in (37. We have A = C + n + 2 (n 0. (58 The factor δ a,c+3/2 in Eqn. (52 implies that the recursion relation is trivial unless c = a 3/2 = /2+a = C +n A+a, where we use /2 = C +n A from (58. By definition (, r c (C vanishes for c C 0, i.e. for c C = n A + a 0. Thus r c (C vanishes for a A n + and the recursion relation (52 implies that (r (A a a,b r (C c abδ b,d+/2 = r (A a a,bδ b,d+/2 = 0 (a A n +. (59 By (46 and (59 and since r (A a is not zero for A a A +, we have a,b = 0 for min[a, C + /2] a max[a n +, A +, C + /2]. If n, then min[a, C + /2] = A and max[a n +, A +, C + /2] A, implying a = A is in the range and therefore A,b = 0. The recursion formula (52 makes all the t2 s vanish. We have A+2,b = = t2 A,b = 0 (A C + 3/2 (60

11 7 APPLYING THE RECURSION RELATIONS Substituting ab = 0 into Eqn. (56, now gives an equation similar to (52, but with us intead of ts. The same reasoning that gave (60 gives all u 2 ab = 0. And similar reasoning applied to the 2-block gives t 2 cd = u 2 cd = 0, and V + = V = 0, i.e. V x = V y = 0. By (4, all four vector matrices vanish and we have, for n, V µ = 0 (A C + 3/2. (6 Since n is positive or zero by assumption in (58, we conclude that V µ = 0 unless n = 0. When n = 0 in (58, there are no recursion relations that reduce to a single term as in (59. Note that the requirement a A n + would require a A + and that exceeds the range of index a, by (0. Thus there may be nonzero vector matrices for A = C + /2. Dropping the restriction that A be larger than C we see that C = A + /2 could give nonzero matrices V µ. By investigations of the recursion relations for the B and D indices we find that the vector matrices V µ are zero unless A = C ± 2 and B = D ± 2 (62 Thus the requirement that there be nonzero vector matrices severely limits the choice of spins (A, B (C, D. 7 Applying the Recursion Relations The problem of finding vector matrices V µ has reduced to finding the quantities ab, t2 cd, u2 ab, and u 2 cd in V + and V, (44 and (45. In this section we find that the recursion relations (52-(57 allow the ts and us to be written as functions of A, B, C, D with two of the ts, AB and t 2 CD, remaining as free parameters. We need only consider the two of the possibilities in (62 with A > C, since we can interchange the (A, B and (C, D blocks of the J and K matrices if C > A. The two cases are A = C + /2; B = D ± /2. We take B > D as Case and D > B as Case 2. Case : A = C + /2; B = D + /2. Consider the recursion relation (52 and start with a = A. Then the delta function factor δ a,c+3/2 in (52 determines c = a 3/2 = A 3/2 = C. For any allowed index b, by ( and (52, we have A,b = r(c C r (A A A,b = [C (C ][C + (C + ] [A (A ][A + (A + ] t2 A,b = 2A 2A A,b. (63 Substituting successive values of a, a = A,..., A + 2, into the recursion relation, we find that A + a a,b = A,b (64 2A

12 7 APPLYING THE RECURSION RELATIONS 2 The recursion (53 gives similar results for the b indices. We get a,b = A + a 2A B + b 2B A,B. (Case (65 for all values of the indices, A + a A and B + b B allowed by Eqn. (46. Recursions (54 and (55 give a formula for the us. We get A a B b u 2 a,b = u 2 A, B. (Case (66 2A 2B for all allowed values of the indices, A a A and B b B. Substituting ab and u 2 ab from (65 and (66 into Eqn.(56, we get A + a B b ( A,B + 2A 2B u2 A, B = 0. (67 For a A and b B in (67, we have u 2 A, B = A,B (Case (68 Eqn. (57 produces the same result. Equations (65, (66 and (68 determine the quantities ab and u 2 ab for Case, A = C + /2 and B = D + /2. Case 2: A = C + /2; B = D /2. For Case 2, start the recursion relation (53 with b = B. The delta function factor δ b,d+3/2 in (53 determines d = b 3/2 = B 3/2 = D 2. Then for any allowed index a, by ( and (53 we have a,b = r(d D 2 r (B B a,b = [D (D 2][D + (D 2 + ] [B (B ][B + (B + ] t2 a,b = 2 a,b (69 Substituting successive values of b, b = B,..., B +, into the recursion relation, we find that a,b = B b + a,b. (70 Applying the recursion (64 for the a-index, we get a,b = A + a 2A B b + t 2 A,B. (Case 2 (7 for all allowed indices, A + a A and B b B in Eqn. (46.

13 8 COMPLETING THE SET OF VECTOR MATRICES 3 Similar work with Eqn. (54 and Eqn. (55 gives a formula for the us. We get u 2 a,b = A a 2A B + b + u 2 A, B (Case 2 (72 for all allowed indices, A a A and B b B. Substituting ab and u2 ab from (7 and (72 in Eqns.(56 and (57, we get Equations (7, (72 and (73 determine the quantities ab and B = D /2. To determine the 2-block quantities t 2 cd and u 2 u 2 A, B = t2 A,B (Case 2. (73 and u2 ab for Case 2, A = C + /2 cd, interchange the (A, B and (C, D blocks in the J k and K k matrices (28. The effect is to exchange a c, b d, 2, Case Case 4, and Case 2 Case 3. Thus the equations for t 2 cd and u 2 cd can be transcribed from (65, (66 and (68 for Case and from (7, (72 and (73 for Case 2. At this point we could write expressions for the vector matrices V + and V. However, we wait until we can also write V z and V t. 8 Completing the Set of Vector Matrices Expressions for vector matrices V z and V t in terms of V + and V follow from the commutation rules. By (49, V z = [J +, V ] and, with J + and V from (45 and (50, we have V z2 = 2 [(ra a u 2 a,b r C c u 2 a,bδ a,c+/2 δ b,d /2 + (r B b u 2 a,b r D d u 2 a,bδ a,c /2 δ b,d+/2 ]. (74 The formula expresses the components of the matrix V z in terms of known functions. An expression for V t can be found by (4, [ik +, V ] = i[k x + ik y, V x iv y ]/2 = V t. By (7 and (28, the needed matrix ik + is found to be ik + = ( r (A a δ a,a +δ b,b r (B b δ a,a δ b,b r c (C δ c,c +δ d,d r (D d δ c,c δ d,d + This together with V from (45 gives V t2 = 2 [(ra a u2 a,b rc c u2 a,b δ a,c+/2δ b,d /2 (r B b u2 a,b rd d u2 a,b δ a,c /2δ b,d+/2 ]. (76 To obtain the 2-blocks of V z and V t make the following exchanges in the 2-blocks: 2, {A, a} {C, c}, and {B, b} {D, d} (75

14 9 MOMENTUM MATRICES 4 Note that the terms in brackets in these expressions for V z and V t differ only in sign, so it is preferable to display the combinations V z ± V t since the many cancellations simplify the displayed expressions. The explicit formulas for the matrix components of V µ are lengthy so they have been placed in an appendix. The collection of formulas in Appendix A is the main result of this paper. 9 Momentum Matrices Momentum matrices P µ satisfy the commutation rules (4 and (5 for vector matrices as well as rule (6 which says that momentum matrices commute, [P µ, P ν ] = 0. For Case, with A = C + /2; B = D + /2 and by (82 - (85, the -block of [P +, P ] gives [P +, P Ab + ab ] ;a,a,b,b = ABt 2 CDδ a,a δ b,b = 0, (A = C + /2; B = D + /2 (77 AB where P ± = (P x ± ip y /2. Choosing a = A and b = B, we get [P +, P ] ;A,a,B,b = 2 AB AB t2 CD δ A,a δ B,b = 0. (78 Since neither A nor B can be zero in Case and the delta functions are nonzero for a = A and b = B, we have AB = 0 or t 2 CD = 0. (79 Inspection of (82 - (97 shows that AB is a common factor in all components of P µ 2 and t 2 CD is a common factor in all components of P 2. µ Thus, by (79, the momentum matrix has one of the following two block matrix forms, either ( 0 0 P µ = Vµ 2 0 or P µ = ( 0 V 2 µ. ( One can show that the same result holds for all cases, A = C ± /2; B = D ± /2: the momentum matrices P µ are the vector matrices V µ, (82-(97, with either the 2-block or the 2-block set equal to zero. That P 2 should be zero relates back to the definition of momentum as the generator of translations. Let T(a be a translation through displacement a. Then P 2 = 0 implies that T(a = exp ( iap = iap and allows T(bT(a = T(a+b to have a matrix representation.

15 A FORMULAS FOR VECTOR MATRICES 5 A Formulas for Vector Matrices We now give formulas for the matrix components of V µ for spin (A, B (C, D. Results for V µ in the general case of spin (A i, B i can be deduced from the formulas here, see Problem 4. The angular momentum matrices J i and the boost matrices K i can be found by (28 and (5 through (20. The matrices V µ are n n square matrices with n = (2A+(2B ++(2C +(2D+, with components put in block matrix form, ( ( (V V = ab,ab (V 2 ab,cd = (V 2 cd,ab (V 22 cd,cd 0 (V 2 ab,cd (V 2 cd,ab 0, (8 since the and 22 blocks are null in all four cases. The indices {ab, cd} and {cd, ab}are pairs of double indices originating from the (A, B (C, D representation of the Lorentz group. Replacing the pairs of double indices with pairs of single indices, e.g. {ab, cd} {i, j}, is discussed in the problem set, see Problem 3. The indices have ranges a { A, A +,...,A}, b { B, B +,...,B}, c { C, C +,..., C}, d { D, D +,..., D}. The formulas for the (V 2 ± ab,cd blocks for Case, A = C + /2; B = D + /2, come from substituting (65, (66 and (68 in (40 and (42. The (V z2 ±V t2 ab,cd blocks for Case come from substituting (66 and (68 in (74 and (76. For Case 2, A = C + /2; B = D /2, use (7, (72 and (73 instead of (65, (66 and (68. Cases and 2 become Cases 4 and 3 when we swap (A, B (C, D (C, D (A, B. Thus the formulas of Case 4 and 3 can be obtained from those of Cases and 2 by the exchanges {A, a} {C, c}, {B, b} {D, d}, 2. Four cases give nontrivial solutions, A = C ± /2; B = D ± /2. Case : A = C + /2; B = D + /2. (V 2 ± ab,cd = A ± a B ± b 2 (V x2 ± iv y2 ab,cd = ± ABδ a,c±/2 δ b,d±/2 (82 2A 2B A ± a B b 2 (V z2 ± V t2 ab,cd = AB δ a,c /2δ b,d±/2 (83 2A 2B (V ± 2 cd,ab = 2 (V x2 ± iv y2 cd,ab = ± A a B b t 2 CD δ c,a±/2δ d,b±/2, (84 2 (V z2 ± V t2 cd,ab = + A a B ± b t 2 CD δ c,a±/2δ d,b /2 (85

16 A FORMULAS FOR VECTOR MATRICES 6 Case 2: A = C + /2; B = D /2. (V ± 2 ab,cd = 2 (V x2 ± iv y2 ab,cd = A ± a 2A D d t 2 AB δ a,c±/2 δ b,d±/2 (86 A ± a 2 (V z2 ± V t2 ab,cd = ± D ± d t 2 AB δ a,c±/2 δ b,d /2 (87 2A (V ± 2 cd,ab = 2 (V x2 ± iv y2 cd,ab = A a D ± d 2D t 2 CDδ c,a±/2 δ d,b±/2 (88 2 (V z2 ± V t2 cd,ab = D d A a t 2 CD δ c,a±/2δ d,b /2 (89 2D Case 3: A = C /2; B = D + /2. (V 2 ± ab,cd = 2 (V x2 ± iv y2 ab,cd = B ± b C c 2B AB δ a,c±/2δ b,d±/2 (90 2 (V z2 ± V t2 ab,cd = B b C c AB δ a,c±/2δ b,d /2 (9 2B (V ± 2 cd,ab = 2 (V x2 ± iv y2 cd,ab = C ± c 2C B b t 2 CD δ c,a±/2 δ d,b±/2 (92 C ± c 2 (V z2 ± V t2 cd,ab = ± B ± b t 2 CD δ c,a±/2 δ d,b /2 (93 2C Case 4: A = C /2; B = D /2. (V ± 2 ab,cd = 2 (V x2 ± iv y2 ab,cd = ± C c D d AB δ a,c±/2δ b,d±/2 (94 2 (V z2 ± V t2 ab,cd = + C c D ± d (V ± 2 cd,ab = 2 (V x2 ± iv y2 cd,ab = ± AB δ a,c±/2δ b,d /2 (95 C ± c D ± d t 2 CD δ c,a±/2δ d,b±/2 (96 2C 2D 2 (V z2 ± V t2 cd,ab = C ± c 2C D d 2D t 2 CDδ c,a±/2 δ d,b /2 (97

17 B PROBLEMS 7 B Problems. Use Appendix A with (A, B = (0, /2 and (C, D = (/2, 0 and an appropriate choice of free parameters AB and t2 CD to obtain a set of Dirac gamma matrices V µ = γ µ. Show that the matrices satisfy V µ V ν + V ν V µ = 2η µν I, where η is the metric and I is the 4x4 unit matrix. (The matrix indices on the V µ and I are suppressed. 2. The Euclidean group in two dimensions, E 2. [3] Rotations R in the x, y plane about the origin can be represented with the z-angular momentum generator J z given by 0 i 0 J z = i 0 0, which acts on vector components {x, y, }. The constant third component indicates the direct product of {x, y} with a scalar which doesn t change under rotations. Let V xij and V yij be vector matrices obeying the commutation rules [J z, V x ] ij = iv yij and [J z, V y ] ij = iv xij. The group of motions in the x, y-plane including the rotations R and x and y translations is called E 2. (a Find V xij and V yij from the commutation rules. (b Find the momentum matrices P xij and P yij. 3. Single index notation. One may reduce the double index (a, b to a single index i. For example, with A = B = /2, we can write (a, b i by {( /2, /2, ( /2, +/2 2, (+/2, /2 3, (+/2, +/2 4}, so that i {, 2, 3, 4}. With this numbering system and the formulas of Case 3 in Appendix A, find the ten 7 7 matrices J k, K k, V µ for spin (A, B (C, D = (/2, /2 (, 0. Verify one commutation rule from each set of rules (3 - (6. 4. What spins A, B, C, D, E, F give non-zero vector matrices V µ when the Js and Ks form the (A, B (C, D (E, F reducible representation of the Lorentz algebra (3? Argue from the linearity of V µ in rules (4 and (5 or, as in Sec. 4, use the diagonal matrices J z and K z to decide which blocks of the vector matrices are possibly non-zero or argue otherwise. 5. Vector Representation. Calculate the momentum matrices P µ for (A, B = (/2, /2 and (C, D = (0, 0 with Pµ 2 0 and Pµ 2 = 0. Find the similarity transformation S that takes these P µ s to the following set of matrices, SPS = P, where δ µ,x P µ = δ µ,y δ µ,z δ µ,t

18 B PROBLEMS 8 Show that and ( + ix µ P µ 2 = ( + i2x µ P µ e ixµp µ = + ixµ P µ. 6. Now show that the new generators in the previous problem (the J s, K s and P s produce matrices D(L, a that represent Poincaré transformations in the form ( L µ D(L, a = ν a µ 0 where L µ,ν is a (4x4 homogeneous Lorentz transformation and a = {a x, a y, a z, a t } is the translation vector. [4] (First, verify that the AB block of J s and K s form the regular representation for which the matrices D(L, 0 are well known. Then use the off-diagonal P s from Problem 6 to finish. Or proceed some other way. 7. Find all spins (A, B (C, D such that the commutators of the vector matrices are proportional to the angular momentum and boost generators expressed in four dimensional notation, i.e. [V µ, V ν ] = kj µν, where k is constant, J µν is antisymmetric in µ, ν, J ij = ǫ ijk J k, and K it = K i. [There is at least one solution, the Dirac gamma matrices of Problem.] 8. Find all spins (A, B (C, D such that the anti-commutator of the vector matrices is proportional to the metric, that is, {V µ, V ν } = V µ V ν + V µ V ν = kη µν I, where k is constant, η µν = diag{,,, +} is the metric and I is the unit matrix. [There is at least one solution, the Dirac gamma matrices of Problem. Can you show that the Dirac gammas are the only vector matrices for spin (A, B (C, D whose commutators and anticommutators produce the angular momentum matrices and the metric as in problems 7 and 8?]

19 REFERENCES 9 References [] J.J.Sakurai, Advanced Quantum Mechanics (Addison-Wesley, Reading, 967, Appendix C (Paulis fundamental theorem. [2] Wigner, E., Annals of Mathematics, 40, 49 (939. [3] Kim, Y. S. and Noz, M. E., Theory and Applications of the Poincaré Group (D. Reidel Publishing Co., Dordrecht, Holland, 986, Chapter III. [4] Tung, W., Group Theory in Physics (World Scientific Publishing Co. Pte. Ltd., Singapore, 985, Sec [5] Weinberg, S., The Quantum Theory of Fields, Vol. I (Cambridge University Press, Cambridge, 995, Sec. 2.5 and references therein. [6] The Resource Letter in Ref. [9] contains many references to articles published prior to 98. References [7] to [0] are only some of the many resources available to those interested in the Poincaré group and its algebra. The same comment applies to the references for other topics listed below. [7] Kim, Y. S. and Noz, M. E., op. cit. [8] Tung, W., op. cit., Chapt. 0; [9] The Resource Letter: Stuewer, R. H., Am. J. Phys., Vol. 49, No. 4, April 98, pp [0] Hamermesh, M., Group Theory and its Application to Physical Problems (Addison- Wesley Publishing Co., Reading, Massachusetts, 964. [] Weinberg, S., op. cit., Sec. 5.6 and references therein. [2] Tung, W., op. cit., Sec [3] Tung, W., op. cit., Sec. 9.; [4] Hamermesh, M., op. cit., Sec. 2.4.