by antisymmetry of ɛ s

Transkript

1 PHY 396 K. Solutions for homework set #6. Problem 2a: Ĵi, Ĵj] 1 4 ɛikl ɛ jmn Ĵ kl, Ĵmn] = by eq. 1 = 1 4 ɛikl ɛ jmn ig km Ĵ ln + ig kn Ĵ lm + ig lm Ĵ kn g ln Ĵ km by antisymmetry of ɛ s = ɛ ikl ɛ jmn ig km Ĵ ln = iĵln g km ɛ ikl ɛ jmn = +δ km ɛ ikl ɛ jmn = δ ij δ ln δ in δ lj = 0 iĵji = +iĵij +iɛ ijk Ĵ k. S.1 Ĵi, ˆK j] 1 2 ɛikl Ĵ kl, Ĵ0j] = by eq. 1 = 1 2 ɛikl ig k0 Ĵ lj + ig kj Ĵ l0 + ig l0 Ĵ kj ig lj Ĵ k0 = 2 1ɛikl 0 iδ kj Ĵ l iδ lj Ĵ k0 1 2 ɛikl +iδ kj ˆKl iδ lj ˆKk = 1 2 ɛijl ˆKl 1 2 ɛikj ˆKj = ɛ ijk ˆKk, S.2 ˆKi, ˆK j] Ĵ0i, Ĵ0j] = by eq. 1 = ig 00 Ĵ ij + ig 0j Ĵ i0 + ig i0 Ĵ 0j ig ij Ĵ 00 = iĵij , iɛ ijk Ĵ k. S.3 1

2 Problem 2b: ˆV i, Ĵj] 1 2 ɛjkl ˆV i, Ĵkl] = 1 2 ɛjkl ig ik ˆV l ig il ˆV k = 1 2 ɛjkl iδ ik ˆV l + δ il ˆV k = i 2 ɛjil ˆV l + i 2 ɛjki ˆV k = iɛ ijk ˆV k, S.4 ˆV 0, Ĵj] = 1 2 ɛjkl ˆV 0, Ĵkl] = 1 2 ɛjkl i\g 0k ˆV l i\g 0l ˆV k = 0, S.5 ˆV i, ˆK j] = ˆV i, Ĵ0j] = i\g i0 ˆV j ig ij ˆV 0 = +iδ ij ˆV 0, S.6 ˆV 0, ˆK j] = ˆV 0, Ĵ0j] = ig 00 ˆV j i\g 0j ˆV 0 = +i ˆV j. S.7 Note that the Hamiltonian of a relativistic theory is a member of a 4-vector multiplet ˆP µ = Ĥ, ˆP where ˆP is the net momentum operator. Applying the above equations to the ˆP µ vectors, we obtain ˆP i, Ĵj] = iɛ ijk ˆP k, Ĥ, Ĵ j] = 0, ˆP i, ˆK j] = +iδ ij Ĥ, S.8 Ĥ, ˆKj ] = +i ˆP j. In particular, the Hamiltonian Ĥ commutes with the three angular momenta Ĵj but it does not commute with the three generators ˆK k of the Lorentz boosts. Problem 2c: In the ordinary quantum mechanics, it is often said that generators of continuous symmetries must commute with the Hamiltonian operator. However, this is true only for the symmetries that act in a time independent manner for example, rotating the 3D space by the same angle at all times t. But when the transformation rules of a symmetry depend on time, the Hamiltonian must change to account for this time dependence. 2

3 In a Lorentz boost, the transform x x obviously depends on time, which changes the way the transformed quantum fields such as ˆΦ x, t depend on t. Consequently, the Hamiltonian Ĥ of the theory must change so that the new Heisenberg equations would match the new time dependence. In terms of the generators, this means that the boost generators ˆK i should not commute with the Hamiltonian. Note that this non-commutativity is not caused by Lorentz boosts affecting the time itself, t = L 0 µx µ t. Even in non-relativistic theories where the time is absolute generators of symmetries that affect the other variables in a time-dependent matter do not commute with Ĥ. Indeed, consider a Galilean transform from one non-relativistic moving frame into another, x = x+vt but t = t. This is a good symmetry of non-relativistic particles interacting with each other but not subject to any external potential, Ĥ = a 1 2M ˆp2 a V ˆx a ˆx b. a b S.9 A unitary operator operators as Ĝ realizing a Galilean symmetry acts on coordinate and momentum Ĝˆx a Ĝ = ˆx a + vt, Ĝˆp a Ĝ = ˆp a + Mv, S.10 and it also transforms the Hamiltonian into ĜĤĜ = Ĥ + v ˆP tot M totv 2. S.11 In terms of the Galilean boost generators ˆK G, Ĝ = exp iv ˆK G, ˆx i a, ˆK j G ] = iδ ij t, ˆp i a, ˆK ] j G In particular, the ˆK G i do not commute with the Hamiltonian. = imδ ij, ] Ĥ, ˆKj G = i ˆP i tot. S.12 3

4 Problem 2d: Consider a linear combination 1 2 N µνĵµν of Lorentz generators with some generic coefficients N µν = N νµ. The Lorentz symmetries LN, φ = exp iφ 2 N µνĵµν generated by this combination preserve the momentum p µ of the particle state p if and only if exp iφ 2 N µνĵµν ˆP α exp iφ 2 N µνĵµν p = ˆP µ p. S.13 For an infinitesimal φ, this condition becomes iφ 2 N µνĵµν, ˆP α] p = 0. S.14 Applying eqs. 3 to the left hand side of this formula gives us iφ 2 N µνĵµν, ˆP α] p = iφ 2 N µν ig αµ ˆP ν + g αν ˆP µ p = φn αν ˆPν p S.15 = φn αν p ν p, so the condition for the generator 1 2 N µνĵµν to preserve particle s momentum p µ is simply N αν p ν = 0. S.16 In 3D terms, N ij = ɛ ijk a k and N 0k = N k0 = b k for some 3-vectors a and b, the generator in question is 1 2 N µν Ĵ µν = a Ĵ + b ˆK, S.17 and the condition S.16 becomes a p be = 0 and b p = 0. S.18 Actually, the second condition here is redundant, so the general solution is In terms of eq. S.17, these solutions mean 1 2 N µν Ĵ µν = a Ĵ + a p ˆK E any a, b = a p E. S.19 = a p Ĵ + E ˆK for any a. S.20 In other words, the Lorentz symmetries preserving the momentum p µ have 3 generators, 4

5 namely the components of the 3-vector Ĵ + p E ˆK S.21 For a particle moving in z direction at speed β = p z /E, these components are Ĵ x β ˆK y, Ĵ y + β ˆK x, and Ĵ z. S.22 Naturally, the generators of any symmetry are defined up to linear combinations. Thus, any 3 linearly-independent combination of the operators S.22 will generate the same little group Gp of Lorentz symmetries preserving the momentum p µ. In particular, in eq. 4 I have multiplied 2 of the generators S.22 by the Lorentz slowdown factor γ while leaving Ĵ z as it is. The purpose of this rescaling is to make all 3 generators normalized as angular momenta, with the standard commutation relations with each other. Indeed: Ĵz, J x] Ĵz Ĵz = γ, Ĵx] βγ, ˆK y] = γ iĵy βγ i ˆK x = i J y, Ĵz, J y] Ĵz Ĵz = γ, Ĵy] + βγ, ˆK x] = γ iĵx βγ +i ˆK y S.23 = i J x, Jx, J y] = γ 2 Ĵ x, Ĵy] βγ 2 ˆKy, Ĵy] + βγ 2 Ĵ x, ˆK x, ] = γ 2 iĵz β 2 γ 2 iĵz = iĵz γ 2 1 β 2 = 1 = iĵz. β 2 γ 2 ˆKy, ˆK x] Problem 2e: Eq. S.22 for the three combinations of Lorentz generators preserving some particle s momentum does not care if the particle is massive or massless. The only difference is in the overall normalization factors for these generators. 5

6 For a massive particle, multiplying the two transverse combinations of Ĵ and ˆK by the γ factor makes the three generators normalized as three components of an angular momentum J. But for a massless particle γ =, so we do not have that option. Instead, we simply leave the generators exactly as in eq. S.22 for β = 1, hence eq. 5. Without the γ factors, the commutation relations Ĵz, Îx ] = iîy and Ĵz, Îy ] = iîx work exactly as in the first two eqs. S.23, but in the third equation we get Îx, Îy] = iĵz 1 β 2 without the γ 2 factor = 0. S.24 Problem 2f: The quantum state p, λ has definite momentum p µ, thus ˆP α p, λ = p α p, λ and likewise ɛ αβγδ Ĵ βγ ˆP δ p, λ = ɛ αβγδ p δ Ĵ βγ p, λ. S.25 For simplicity, let s assume the particle moves in the z direction and spell out the operator ˆQ α = ɛ αβγδ p δ Ĵ βγ in components: ˆQ 0 = ɛ 0ij3 p 3 Ĵij = 2p 3 Ĵ12 = 2p z J z, ˆQ 1 = ɛ 1023 p 3 Ĵ02 + ɛ 1203 p 3 Ĵ20 + ɛ 1jk0 p 0 Ĵjk = 2p 3 Ĵ02 2p 0 Ĵ23 = 2p z ˆK y 2p 0 Ĵx, ˆQ 2 = ɛ 2013 p 3 Ĵ01 + ɛ 2103 p 3 Ĵ10 + ɛ 2jk0 p 0 Ĵjk S.26 = +2p 3 Ĵ01 2p 0 Ĵ31 = +2p z ˆK x 2p 0 Ĵy, ˆQ 3 = ɛ 3ij0 p 0 Ĵij = 2p 0 Ĵ12 = 2p 0 Ĵ z. For a massless particle with p z = p 0 = E, these components simplify to ˆQ 0 = +2E Ĵz, ˆQx = 2E Îx, ˆQy = 2E Îy, ˆQz = 2E Ĵz. S.27 Besides definite momentum, the state p, λ has definite helicity λ, Ĵ z p, λ = λ p, λ. Moreover, it is annihilated by other two generators of the little group of its momentum, 6

7 Î x p, λ = Îy p, λ = 0. Therefore, when we apply the operators S.27 to this state, we obtain ˆQ 0 p, λ = +2Eλ p, λ, ˆQ x p, λ = 0, ˆQ y p, λ = 0, S.28 ˆQ z p, λ = 2Eλ p, λ, Comparing the right hand sides here to the particle momentum with a lower index p α = +E, 0, 0, E we see that ˆQ α p, λ = 2λp α p, λ. S.29 In other words, ɛ αβγδ Ĵ βγ ˆP δ p, λ = 2λ ˆP α p, λ. 8 become What if a massless particle moves in another direction? In 3-vector notations, eqs. S.26 ˆQ 0 = 2E β Ĵ, ˆQ = 2E Ĵ + β ˆK. S.30 Up the overall factor 2E, the components of ˆQ are the generators S.22 of the little group of the momentum p µ. The definite-helicity state p, λ is annihilated by the two generators p and is an eigenstate of the third generator p. Indeed, for a massless particle β Ĵ is the helicity operator, hence ˆQ 0 p, λ = 2Eλ p, λ, β ˆQ p, λ = 2E β Ĵ p, λ = 2Eλ p, λ, S.31 while β ˆQ p, λ = 0. S.32 Consequently, ˆQ p, λ = β β ˆQ p, λ β β ˆQ p, λ = β 2Eλ p, λ = 2pλ p, λ, S.33 7

8 and combining this formula with ˆQ 0 p, λ = 2Eλ p, λ we obtain ˆQ µ p, λ = 2λp µ p, λ. S.34 This proves eq. 8 for a massless particle moving in any direction. Problem 2g: Consider a continuous Lorentz transform x µ x µ = L µ νx ν acting on the p, λ state of a massless particle. The operators on both sides of both sides of eq. 8 transform as Lorentz vectors, ˆDL ˆP α ˆD L = L µ α ˆP µ, ˆDL ɛ αβγδ Ĵ βγ ˆP δ ˆD L = L µ α ɛ µβγδ Ĵ βγ ˆP δ. S.35 Consequently, the transformed state ˆD p, λ = Lp,?? S.36 satisfies the same eq. 8 as the original state p, λ. Indeed, L α µ ɛ µβγδ Ĵ βγ δ ˆP Lp,?? = ˆDL ɛ αβγδ Ĵ βγ δ ˆP ˆD L ˆD p, λ ɛ αβγδ Ĵ βγ δ ˆP p, λ = ˆDL by eq. 8 = ˆDL 2λ ˆP α p, λ S.37 and hence = 2λ ˆDL ˆP α ˆD L ˆD p, λ = 2λ L µ α ˆP µ Lp,??, ɛ µβγδ Ĵ βγ ˆP δ Lp,?? = 2λ ˆP µ Lp,??. S.38 Note that this equation for the transformed state Lp,?? has exactly the same helicity eigenvalue λ as the original eq. 8. In other words, the transformed state has the same 8

9 helicity as the original state, and since the momentum and the helicity completely determine the quantum state of a particle up to an overall phase, it follows that ˆDL p, λ = Lp, same λ e i phase. 9 For massless particles, the continuous Lorentz transforms preserve helicity! Problem 3a: In light of eqs. 2, Ĵi ±, Ĵj ±] = 1 4 Ĵi, Ĵj] ± 4 i ˆKi, Ĵj] ± 4 i Ĵi, ˆK j] 4 1 ˆKi, ˆK j] = 1 4 iɛijk Ĵ k ± i 4 iɛijk ˆKk ± i 4 iɛijk ˆKk 1 4 iɛijk Ĵ k Ĵi ±, Ĵj ] = ɛ ijk i 2 Ĵ k 1 2 ˆK k = iɛ ijk Ĵ k ±, = 1 4 Ĵi, Ĵj] ± 4 i ˆKi, Ĵj] 4 i Ĵi, ˆK j] ˆKi, ˆK j] S.39 = 1 4 iɛijk Ĵ k ± i 4 iɛijk ˆKk i 4 iɛijk ˆKk iɛijk Ĵ k = 0. Problem 3b: All three Pauli matrices σ i are hermitian; specifically, the σ 1 and the σ 3 are real and symmetric while the σ 2 is imaginary and antisymmetric. Consequently, all three matrices are related to their complex conjugates as σ 2 σ i σ 2 = σ i. S.40 Indeed, by inspection σ 2 σ 1σ 2 = +σ 2 σ 1 σ 2 = σ 2 σ 2 σ 1 = σ 1, S.41 σ 2 σ 2σ 2 = σ 2 σ 2 σ 2 = σ 2 σ 2 σ 2 = σ 2, S.42 σ 2 σ 3σ 2 = +σ 2 σ 3 σ 2 = σ 2 σ 2 σ 3 = σ 3. S.43 9

10 Now, let a and b be two real 3-vectors parametrizing a Lorentz symmetry exp ia Ĵ ib ˆK. Let A be the exponent in eq. 11, A = 1 2 ia + b σ = ML = expa. S.44 Then in light of eq. S.40 σ 2 A σ 2 = 1 2 ia + b σ 2 σ σ 2 = 1 2 +ia + b σ = 1 2 ia b σ, S.45 where the RHS is the exponent in eq. 12, Thus, ML = exp σ 2 A σ 2 = σ2 expa σ2 = σ 2 M σ 2 S.46 where the second equality follows from expanding the exponential into the power series. Indeed, since σ 2 σ 2 = 1, we have σ2 A σ 2 2 = σ2 A σ 2 σ 2 A σ 2 = σ 2 A A σ 2 = = σ 2 A 2 σ 2 = σ 2 A 2 σ 2 S.47 and likewise for any integer power n, σ2 A σ 2 n = σ2 A n σ 2, S.48 hence exp σ 2 A σ 2 = n 1 σ2 A n 1 σ 2 = σ2 n! n! An 2 σ 2 = σ 2 expa σ2. n S.49 This completes the proof of M = σ 2 M σ 2 10

11 The other equality in eq. 13 is even easier. By hermiticity of the Pauli matrices, A = 1 2 ia b σ, S.50 which is the exponent in eq. 12. Thus, M = exp A = M 1 = exp+a = expa = M, S.51 and therefore M = M 1. S.52 Q.E.D. Problem 3c: Let s start with reality. The matrix V = V µ σ µ is hermitian if and only if the 4-vector V µ is real. For any matrix M SL2, C, the transform V V = MV M S.53 preserves hermiticity: if V is hermitian, then so is V ; indeed V = MV M = M V M = MV M = V. S.54 In terms of the 4-vectors, this means that if V µ is real than V µ = L µ νv ν is also real. In other words, the 4 4 matrix L µ νm is real. Next, let s prove that L µ νm O3, 1 it preserves the Lorentz metric g αβ, or equivalently, for any V µ, g αβ V αv β = gαβ V α V β. In 2 2 matrix terms, the Lorenz square of 11

12 a 4-vector becomes the determinant: g αβ V α V β = det V = V µ σ µ. S.55 Indeed, from the explicit form of the 4 matrices σ 0 = , σ 1 = , σ 2 = 0 i +i 0, σ 3 = S.56 we have V0 + V 3 V 1 iv 2 V = V µ σ µ = V 1 + iv 2 V 0 V 3 and hence detv = V 0 + V 3 V 0 V 3 V 1 iv 2 V 1 + iv 2 = V 2 0 V 2 3 V 2 1 V 2 2 = g αβ V α V β. S.57 The determinant of a matrix product is the product of the individual matrices determinants. Hence, for the transform S.53, detv = detm detv detm = detv detm 2. S.58 The M matrices of interest to us belong to the SL2, C group they are complex matrices with units determinants. There are no other restrictions, but detm = 1 is enough to assure detv = detv, cf. eq. S.58. Thanks to the relation S.55, this means g αβ V αv β = detv = detv = g αβ V α V β S.59 which proves that the matrix L µ νm is indeed Lorentzian. To prove that the Lorentz transform L µ νm is orthochronous, we need to show that for any V µ in the forward light cone V 2 > 0 and V 0 > 0 the V µ is also in the forward light cone. In matrix terms, V 2 > 0 and V 0 > 0 mean detv > 0 and trv > 0; together, 12

13 these two conditions means that the 2 2 hermitian matrix V is positive-definite. The transform S.53 preserves positive definiteness: if for any complex 2-vector ξ 0 we have ξ V ξ > 0, then ξ V ξ = ξ MV M ξ = M ξ V M ξ > 0. S.60 Note that M ξ 0 for any ξ 0 because detm 0. Thus, for any M SL2, C the Lorentz transform V µ V µ preserves the forward light cone in other words, the L µ νm is orthochronous, L µ νm O + 3, 1. Problem 3c : The simplest proof the L ν µ M is proper as well as orthochronous involves the group law part d of this problem and the explicit examples of a pure rotation and a pure boost in parts e and f, both of which are manifestly proper. For any SL2, C matrix M we may decompose M = HU where H = MM is hermitian and U = H 1 M is unitary. Proof: UU = H 1 MM H 1 = H 1 H 2 H 1 = 1. Furthermore, both H and U are unimodular deth = detu = 1, or in other words H, U SL2, C, which allows us to define two separate Lorentz transforms LH and LU. According to the group law, together these two transform accomplish the LM transform, LM = LH LU. S.61 Now, H is hermitian, unimodular, and positive definite, hence it has a well-defined logarithm which is hermitian and traceless, trlog H = logdeth = 0. For the 2 2 matrices, this means log H = 2 1rσ for some real 3 vector r, or equivalently H = exp 2 1rnσ. As we shall see in part f below, this means that LH is a pure Lorentz boost of rapidity r in the direction n. This boost manifestly does not invert space or time, thus LU is proper. Likewise, U is unitary and unimodular, thus U SU2 and defines a pure rotation of space. Indeed, any U SU2 can be written as U = exp 2 i θn σ for some angle θ and some axis n, and we shall see in part e that for such U, LU is indeed a rotation of the 3D space by angle θ around axis n. Again, this rotation is proper it does not invert space or time. Thus, LH and LU are both proper Lorentz transforms, hence their product LM must also be proper. Proof: detlm = detlh detlu = +1. Q.E.D. 13

14 And by the way, since any proper, orthochronous Lorentz transform L SO + 1, 3 can be realized as LM for some M SL2, C, it follows that any such transform is a product of a pure space rotation LH followed by a pure Lorentz boost LU. Problem 3d: Let s plug L ν µ MV ν = V µ into eqs. 20: σ µ L ν µ MV ν = σ µ V µ = Mσ ν V ν M = Mσ ν M V ν. S.62 Since this equation holds true for any 4-vector V ν, we must have σ µ L ν µ M = Mσ ν M. S.63 This formula defines the Lorentz transform LM in a V µ independent way, which is more convenient for verifying the group law. Indeed, for any two SL2, C matrices M 1 and M 2, it gives us σ µ Lµ ν M 2 M 1 = M 2 M 1 σ ν M 2 M 1 = M 2 M 1 σ ν M 1 M 2 = M 2 M 1 σ ν M 1 M 2 = M 2 σ ρ Lρ ν M 1 M 2 = M 2 σ ρ M 2 L ρ ν M 1 = σ µ Lµ ρ M 2 Lρ ν M 1 = σ µ Lµ ρ M 2 Lρ ν M 1 S.64 and hence L ν µ M 2 M 1 = L ρ µ M 2 L ν ρ M 1. S.65 In short, LM 2 M 1 = LM 2 LM 1. Q.E.D. 14

15 Problem 3e: Let M = exp 2 i θ nσ = cos 2 θ i sin 2 θ nσ and hence M = M 1 = cos 2 θ + i sin 2 θ nσ. Since σ 0 = 1 while the Pauli matrices σ 1,2,3 are traceless, for any unitary M Mσ 0 M = MM = 1 = σ 0 S.66 while MσM = MσM 1 = tr Mσ i M = tr Mσ i M 1 = tr σ i = 0 = Mσ i M = R ij σ j for some R ij. S.67 In terms of the Lorentz transform 20, this means that V 0 = V 0 while V depends only on the space components of the V. In other words, this Lorentz transform is a rotation of space that does not affect the time. Specifically, σ V = MVσM = cos 2 θ + i sin 2 θ nσ Vσ cos 2 θ i sin 2 θ nσ = cos 2 2 θ Vσ i sin 2 θ cos 2 nσ, θ Vσ] = 2in V σ + sin 2 2 θ nσvσnσ = 2nvnσ Vσ S.68 = cos θ Vσ + sin θn Vσ + 1 cos θnvnσ, hence V = cos θ V nnv + sin θ n V + nnv. S.69 This is indeed a rotation through angle θ around axis n. 15

16 Problem 3f: Now consider the Lorentz transforms LM for a hermitian matrix M = M. Specifically, plugging M = exp 2 r nσ = cosh 2 r sinh 2 r nσ into eq. 20, we obtain σ µ V µ = V 0 σ V = MV 0 σ VM = cosh r 2 sinh r 2 nσ V 0 Vσ cosh r 2 sinh r 2 nσ = cosh 2 2 r V 0 Vσ {nσ, sinh 2 r cosh 2 r V0 Vσ } = 2V 0 nσ 2nV + sinh 2 2 r nσv 0 Vσnσ = V 0 2nVnσ + Vσ = cosh r V 0 + sinh r nv σn sinh r V 0 + cosh r nv In other words, σ V nnv. S.70 V 0 = cosh r V 0 + sinh r nv, V = n sinh r V 0 + cosh r nv + V nnv, S.71 which is precisely the Lorentz boost of rapidity r in the direction n. The rapidity r is related to the usual parameters of a Lorentz boost according to β = tanh r, γ = cosh r, γβ = sinh r. For several boosts in the same directions, the rapidities add up, r tot = r 1 +r 2 +. Problem 3g: Q.E.D. For any Lie algebra equivalent to an angular momentum or its analytic continuation, the product of two doublets comprises a triplet and a singlet, 2 2 = 3 1, or in j notations, = 1 0. Furthermore, the triplet 3 = 1 is symmetric with respect to permutations of the two doublets while the singlet 1 = 0 is antisymmetric. For two separate and independent types of angular momenta J + and J we combine the j + quantum numbers independently from the j and the j quantum numbers independently from the j +. For two bi-spinors, this gives us 2 1, , 1 2 = 1, 1 1, 0 0, 1 0, 0. S.72 Furthermore, the symmetric part of this product should be either symmetric with respect to 16

17 both the j + and the j indices or antisymmetric with respect to both indices, thus 1 2, , 1 2 ] sym = 1, 1 0, 0. S.73 Likewise, the antisymmetric part is either symmetric with respect to the j + but antisymmetric with respect to the j or the other way around, thus 1 2, , 1 2 ] antisym = 1, 0 0, 1. S.74 From the SO + 3, 1 point of view, the bi-spinor 1 2, 1 2 is the Lorentz vector. A general 2-index Lorentz tensor transforms like a product of two such vectors, so from the SL2, C point of view it s a product of two bi-spinors, which decomposes to irreducible multiplets according to eq. S.72. The Lorentz symmetry respects splitting of a general 2-index tensor into a symmetric tensor T µν = +T νµ and an asymmetric tensor F µν = F νµ. The symmetric tensor corresponds to a symmetrized square of a bi-spinor, which decomposes into irreducible multiplets according to eq. S.73. The singlet 0, 0 component is the Lorentz-invariant trace T µ µ while the 1, 1 irreducible multiplet is the traceless part of the symmetric tensor. Likewise, the antisymmetric Lorentz tensor F µν eq. S.74. = F νµ decomposes according to Here, the irreducible components 1, 0 and 0, 1 are complex but conjugate to each other; individually, they describe antisymmetric tensors subject to complex duality conditions 1 2 ɛκλµν F µν = ±if κλ, or in 3D terms, E = ±ib. 17