Problem 1(a): Starting with eq. (3) proved in class and applying the Leibniz rule, we obtain

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1 PHY 396 K. Solutions for homework set #5. Problem 1a: Starting with eq. 3 proved in class and applying the Leibniz rule, we obtain [ γ κ γ λ, S µν] γ κ [ γ λ, S µν] + [ γ κ, S µν] γ λ γ κ ig λµ γ ν ig λν γ µ + ig κµ γ ν ig κν γ µ γ λ ig λµ γ κ γ ν ig κν γ µ γ λ ig λν γ κ γ µ + ig κµ γ ν γ λ ig λµ γ κ γ ν g κν ig κν γ µ γ λ g λµ ig λν γ κ γ µ g κµ + ig κµ γ ν γ λ g λν S.1 2g λµ S κν 2g κν S µλ 2g λν S κµ + 2g κµ S νλ, where the last line follows from γ κ γ ν g κν 2iS κν proved in class. And since g κλ are c-numbers which commute with everything, [ S κλ, S µν] 2 i [ γ κ γ λ, S µν] ig λµ S κν ig κν S µλ ig λν S κµ + ig κµ S νλ. 4 Q.E.D. Problem 1b: By the multiple commutator formula AKA Hadamard Lemma, for any matrix F e F γ µ e +F γ µ + [ γ µ, F ] [[ γ µ, F ], F ] [[[ γ µ, F ], F ], F ] +. S.2 Let F 2 i Θ αβs αβ so that M D e F and MD 1 e F ; then MD 1 γµ M D follows via eq. S.2 from the multiple commutators of F with the γ µ. It turns out that all such commutators are linear combinations of the γ ν. Indeed, the single commutator here is [ γ µ, F ] i 2 Θ αβ [ γ µ, S αβ] 2 1Θ αβ g µα γ β g µβ γ α Θ αβ g µα γ β Θ µ β γβ, S.3 while the multiple commutators follow by iterating this formula: [[ γ µ, F ], F ] Θ µ λ Θλ νγ ν, [[[ γ µ, F ], F ], F ] Θ µ λ Θλ ρθ ρ νγ ν,.... S.4 1

2 Combining all these commutators as in eq. S.2, we obtain M 1 D γµ M D e F γ µ e +F γ µ + [ γ µ, F ] [[ γ µ, F ], F ] [[[ γ µ, F ], F ], F ] + γ µ + Θ µ νγ ν Θµ λ Θλ νγ ν Θµ λ Θλ ρθ ρ νγ ν + δ ν µ + Θ µ ν + 2 1Θµ λ Θλ ν + 6 1Θµ λ Θλ ρθ ρ ν + γ ν S.5 L µ νγ ν. Q.E.D. Problem 1c: This is an exercise in Leibniz rules for commutators and anticommutators, [A, BC] [A, B]C + B[A, C] {A, B}C B{A, C}, {A, BC} [A, B]C + B{A, C} {A, B}C B[A, C]. S.6 Applying these formulae to Dirac matrices we immediately obtain [γ ρ, γ µ γ ν ] {γ ρ, γ µ } γ ν {γ ρ, γ ν } γ µ 2g ρµ γ ν 2g ρν γ µ S.7 and hence { γ ρ, γ λ γ µ γ ν} { γ ρ, γ λ} γ µ γ ν γ λ [γ ρ, γ µ γ ν ] 2g ρλ γ µ γ ν γ λ 2g ρµ γ ν 2g ρν γ µ 2g ρλ γ µ γ ν 2g ρµ γ λ γ ν + 2g ρν γ λ γ µ, S.8 [γ ρ, γ κ γ λ γ µ γ ν] {γ ρ, γ κ } γ λ γ µ γ ν γ κ { γ ρ, γ λ γ µ γ ν} 2g ρκ γ λ γ µ γ ν γ κ 2g ρλ γ µ γ ν 2g ρµ γ λ γ ν + 2g ρν γ λ γ µ 2g ρκ γ λ γ µ γ ν 2g ρλ γ κ γ µ γ ν + 2g ρµ γ κ γ λ γ ν 2g ρν γ κ γ λ γ µ, S.9 etc., etc. 2

3 Finally, applying the Leibniz rule to the left side of the bracket, we obtain [ γ σ γ ρ, γ λ γ µ γ ν] γ σ { γ ρ, γ λ γ µ γ ν} { γ σ, γ λ γ µ γ ν} γ ρ γ σ 2g ρλ γ µ γ ν 2g ρµ γ λ γ ν + 2g ρν γ λ γ µ 2g σλ γ µ γ ν 2g σµ γ λ γ ν + 2g σν γ λ γ µ γ ρ S.1 and hence [ S ρσ, γ λ γ µ γ ν] ig σλ γ ρ γ µ γ ν + ig σµ γ λ γ ρ γ ν + ig σν γ λ γ µ γ ρ ig ρλ γ σ γ µ γ ν ig ρµ γ λ γ σ γ ν ig ρν γ λ γ µ γ σ. S.11 Problem 1d: γ α γ α 1 2 {γα, γ β }g αβ g αβ g αβ 4; γ α γ ν γ α 2g αν γ α γ ν γ α γ α 2γ ν γ ν 4 2γ ν ; γ α γ µ γ ν γ α 2g αµ γ ν γ α γ µ γ α γ ν γ α 2γ ν γ µ γ µ 2γ ν 2{γ ν, γ µ } 4g µν ; γ α γ λ γ µ γ ν γ α 2g αλ γ µ γ ν γ α γ λ γ α γ µ γ ν γ α 2γ µ γ ν γ λ γ λ 4g µν 2 γ µ γ ν 2g µν γ λ 2γ ν γ µ γ λ. S.12 Problem 2a: γ µ γ ν ±γ ν γ µ where the sign is + for µ ν and otherwise. Hence for any product Γ of the γ matrices, γ µ Γ 1 n Γγ µ, where n is the number of γ ν µ factors of Γ. For Γ γ 5 iγ γ 1 γ 2 γ 3, n 3 for any µ, 1, 2, 3, hence γ µ γ 5 γ 5 γ µ. As to the spin matrices, γ 5 γ µ γ ν γ µ γ 5 γ ν +γ µ γ ν γ 5 and therefore γ 5 S µν +S µν γ 5. 3

4 Problem 2b: First, γ 5 iγ γ 1 γ 2 γ 3 iγ 3 γ 2 γ 1 γ +iγ 3 γ 2 γ 1 γ +iγ 3 γ 2 γ 1 γ 1 3 i γ γ 3 γ 2 γ i γ γ 1 γ 3 γ i γ γ 1 γ 2 γ 3 S.13 +iγ γ 1 γ 2 γ 3 +γ 5. Second, γ 5 2 γ 5 γ 5 iγ γ 1 γ 2 γ 3 iγ 3 γ 2 γ 1 γ γ γ 1 γ 2 γ 3 γ 3 γ 2 γ 1 γ +γ γ 1 γ 2 γ 2 γ 1 γ γ γ 1 γ 1 γ +γ γ +1. S.14 Problem 2c: Since the four Dirac matrices γ, γ 1, γ 2, γ 3 all anticommute with each other, ɛ κλµν γ κ γ λ γ µ γ ν γ [ γ 1 γ 2 γ 3] 24γ γ 1 γ 2 γ 3 24iγ 5. S.15 To prove the other identity, we note that a totally antisymmetric product γ [κ γ λ γ µ γ ν] vanishes unless the Lorentz indices κ, λ, µ, ν are all distinct which makes them, 1, 2, 3 in some order. For such indices, the anticommutativity of Dirac matrices implies γ κ γ λ γ µ γ ν ɛ κλµν γ γ 1 γ 2 γ 3 and hence γ [κ γ λ γ µ γ ν] 24 ɛ κλµν γ γ 1 γ 2 γ 3 24 iɛ κλµν γ 5. S.16 Problem 2d: 6iɛ κλµν γ κ γ γ κ γ [κ γ λ γ µ γ ν] 4 1 γ κ γ κ γ [λ γ µ γ ν] γ [λ γ κ γ µ γ ν] + γ [λ γ µ γ κ γ ν] γ [λ γ µ γ ν] γ κ 4γ [λ γ µ γ ν] + 2γ [λ γ µ γ ν] + 4g [λµ γ ν] + 2γ [ν γ µ γ λ] γ[λ γ µ γ ν] γ [λ γ µ γ ν]. S.17 4

5 Problem 2e: Proof by inspection: In the Weyl basis and 2 2 block form, the 16 matrices are 1 1 +σ i 1 4 4, γ, γ i 1 1 σ i, 1 σ k 2 γ[i γ j] iɛ ijk 1 σ i σ k, 2 γ [ γ i] +σ i, S.18 1 σ i 1 γ 5 γ, γ 5 γ 1 +1 σ i, γ 5, +1 and their linear independence is self-evident. Since there are only 16 independent 4 4 matrices altogether, any such matrix Γ is a linear combination of the matrices S.18. Q.E.D. Algebraic Proof: Without making any assumption about the matrix form of the γ µ operators, let us consider the Clifford algebra γ µ γ ν + γ ν γ µ 2g µν. Because of these anticommutation relations, one may re-order any product of the γ s as ±γ γ γ 1 γ 1 γ 2 γ 2 γ 3 γ 3 and then further simplify it to ±γ or 1 γ 1 or 1 γ 2 or 1 γ 3 or 1. The net result is up to a sign or ±i factor one of the 16 operators 1, γ µ, 1 2 γ[µ γ ν], 1 6 γ[λ γ µ γ ν] iɛ λµνρ γ 5 γ ρ cf. d or i 24 γ[κ γ λ γ µ γ ν] ɛ κλµν γ 5 cf. c. Consequently, any operator Γ algebraically constructed of the γ µ s is a linear combination of these 16 operators. Incidentally, this proof explains why the Dirac matrices are 4 4 in d 4 spacetime dimensions: the 16 linearly-independent products of Dirac matrices require matrix size to be Technically, we may also use matrices of size 4n 4n, but then we would have γ µ γ µ n n, and ditto for all their products. Physically, this means combining the Dirac spinor index with some other index i 1,..., n which has nothing to do with Lorentz symmetry. Nobody wants such an index confusion, so physicists always stick to 4 4 Dirac matrices in 4 spacetime dimensions. Problems 3f and 3g: As in part a of this problem, γ µ Γ 1 n Γγ µ where n is the number of γ ν µ factors of Γ. For Γ γ γ 1 γ d 1 moduli the overall sign or ±i factor, each γ µ appear once, with 5

6 the remaining d 1 factors being γ ν µ. Thus, n d 1 and γ µ Γ 1 d 1 Γγ µ : for even spacetime dimensions d, Γ anticommutes with all d γ µ matrices, but for odd d, Γ commutes with all the γ µ instead of anticommuting. Now consider the dimension of the Clifford algebra made out of Dirac matrix products and their linear combinations. As in part e, any product can be re-arranged as ±γ or 1 γ 1 or 1 γ d 1 or 1, and the number of distinct products of this type is clearly 2 d. Alternatively, we count a single unit matrix, d 1 d of γ µ matrices, d 2 of independent 1 2 γ [µ γ ν] products, d 3 of 1 6 γ [λ γ µ γ ν] products, etc., etc., all the way up to d d 1 d of γ [β γ ω] ɛ αβ ω γ α Γ, and d d 1 of γ [α γ ω] ɛ αβ ω Γ; the net number of all such matrices is d d k k 2 d. For even dimensions d, all the 2 d matrix products are distinct, and we need matrices of size 2 d/2 2 d/2 to accommodate them. But for odd d, the Γ matrix commutes with the entire Clifford algebra since it commutes with all the γ µ, so to avoid redundancy we should set Γ 1 or Γ 1. Consequently, any product of k distinct Dirac matrices becomes identical up to a sign or ±i with the product of the other d k matrices, for example γ γ k 1 i?? γ k γ d 1. This halves the net dimension of the Clifford algebra from 2 d to 2 d 1 and calls for matrix size 2 d 1/2 2 d 1/2. Thus, the Dirac matrices in 2 or 3 space time dimensions are 2 2, in 4 or 5 dimensions they are 4 4, in 6 or 7 dimensions 8 8, in 8 or 9 dimensions 16 16, in 1 or 11 dimensions 32 32, etc., etc. Problem 3: Eq. S.18 gives explicit form of various products of Dirac γ matrices in the Weyl convention. In particular, the second like of S.18 spells out 1 2 γ[µ γ ν] 1 2 [γµ, γ ν ] 2iS µν. In terms of the S µν matrices themselves, S ij ɛijk 2 σ k σ k, S i S i 1 2 iσ i +iσ i. S.19 Note that all these matrices are block-diagonal, hence all the ML exp i 2 Θ µνs µν ma- 6

7 trices are also block-diagonal as in eq. 6. Specifically, let Θ ij ɛ ijk a k, Θ i Θ i b i S.2 for some real 3-vectors a and b, then Θ µν S µν a + ib σ 1 2 a ib σ S.21 and hence M D exp i 2 Θ µνs µν exp 12 ia + b σ exp 1 2 ia b σ. S.22 Moreover, comparing the diagonal blocks here to eqs. 8 and 9 from the last homework, we immediately see that M D ML M R where M L and M R denoted M and M in the last homework represent the same Lorentz symmetry L expθ in 2 and 2 irreducible multiplets of the SL2, C Spin3, 1. 6 In particular, a pure rotation through angle ϕ around the z axis 1 L µ cos ϕ sin ϕ ν sin ϕ cos ϕ exp +ϕ ϕ 1 S.23 corresponds to a,, ϕ, b,,, hence M L M R exp i 2 ϕσz S.24 More generally, for a pure rotation around any axis n we have a ϕn, b, hence M L M R exp i ϕ 2 n σ. 7 7

8 For a pure Lorentz boost in the z direction γ +βγ 1 +r L µ 1 ν 1 exp +βγ γ +r S.25 where r is the rapidity of the boost; it is related to the β v/c and γ 1/ 1 β 2 parameters as β tanhr and γ coshr. For successive boosts in the same direction, their rapidities add up. Anyhow, the boost S.25 corresponds to a,, while b,, r, where the minus sign comes from Θ i +Θi Θi b i. Similarly, a Lorentz boost in a general direction n has a while b rn, hence M L exp r 2 n σ while M R exp + r 2 n σ. S.26 In terms of β and γ parameters of the boost, M L 2 exp rn σ coshr sinhr n σ S.27 and likewise hence γ βγ n σ M R 2 exp +rn σ coshr + sinhr n σ γ + βγ n σ, S.28 M L γ 1 βn σ, M R γ 1 + βn σ. 8 Equivalently, for a boost L that turns a particle at rest to a moving particle with energymomentum p µ E, p, m ML L E p σ, m MR L E + p σ. S.29 We will need this formula in problem 4 below parts b and c. 8

9 Problem 4a: For p, p +m and p m mγ 1. Hence, up, s satisfy γ 1u, or in the Weyl basis u u ζ S.3 ζ where ζ is an arbitrary two-component spinor. It s normalization follows from u u 2ζ ζ, so if we want u u 2E p 2m for p we need ζ ζ m. Equivalently, we want ζ mξ and hence u as in eq. 1 for a conventionally normalized spinor ξ, ξ ξ. Note that there are two independent choices of ξ, normalized to ξ sξ s δ s,s so that u, su, s 2mδ s,s 2E p δ s,s. They correspond to two spin states of a p electron. In terms of the spin vector, S 2 1ξ sσξ s. Problem 4b: The Dirac equation is Lorentz-covariant, so we may obtain solutions for all p µ +E p, p by simply Lorentz-boosting the solutions 1 from the rest frame where p µ +m,. Thus, up, s M D L up, s ML M R m ξs m ξs m ML ξ s m MR ξ s S.31 where M D, M L, and M R are respectively Dirac-spinor, LH Weyl-spinor, and RH Weyl-spinor representations or the Lorentz boost m, E, p. Eq. S.29 above gives the explicit form of the M L and M R matrices for this boost. Plugging them into eq. S.31 immediately gives us up, s E p σ ξ s. 11 E + p σ ξ s Q.E.D. Problem 4c: The negative-frequency solutions e +ipx vp, s have Dirac spinors v satisfying p + mv. 9

10 For a particle at rest, p µ +m,, this equation becomes mγ + 1v, or in the Weyl basis ηs vp, s vp, s m η s S.32 for a pair of independent two-component spinors η s. As in part a, the m factor translates the normalization v p, svp, s 2E p δ s,s 2mδ s,s for p to η sη s δ s,s. For p we proceed similarly to part b, namely Lorentz-boost the rest-frame solution S.32 to the frame where p µ +e p, p: ML + m ηs V p, s M D L V p, s M R m η s + m ML ξ s m M R ξ s + E p σ η s 11 E + p σ η s. Q.E.D. Problem 4d: Physically, the holes in a Fermi sea have opposite free energies, momenta, and spins from the missing particle states. I shall explain this point in class. A positron is a hole in the Dirac sea of electrons, so it should have opposite p µ E, p from the missing electron states that s why the vp, s spinors accompany the e +ipx e +iet ipx plane waves instead of the e ipx E ietipx factors of the up, s spinors. Likewise, the positron should have opposite spin state from the missing electron, and that s why the η s should have opposite spin from the ξ s. More accurately, the η s should carry an opposite spin vector η ssη s from ξ ssξ s. The solution to this spin relation is η s σ 2 ξs where denotes complex conjugation. Indeed, using the Pauli matrix relation σ 2 σ σ 2 σ from the last homework, we see that η σ 2 ξ η ξ σ 2 η ση ξ σ 2 σσ 2 ξ ξ σ 2 σ σ 2 ξ ξ σξ ξ σξ 1

11 η Sη 1 2 η ση ξ Sξ. S.33 Also, for η s σ 2 ξ s we have + E p σ η s + E p σ vp, s σ 2 ξs E + p σ η s E + p σ σ 2 ξs +σ2 σ 2 E p σ σ 2 ξs σ 2 σ 2 E + p σ σ 2 ξ s using σ 2 σσ 2 σ σ 2 E p σ σ 2 +σ2 E + p σ ξ s σ 2 E p σ ξ s +σ2 + E p σ ξ s σ 2 + E + p σ ξ s γ 2 u p, s. E ± p σ S.34 Similarly, up, s γ 2 v p, s. Indeed, γ 2 v p, s γ 2 γ 2 u p, s γ 2 γ 2 up, s up, s since γ 2 γ 2 γ 2 γ Problem 4e: The 3D spinors ξ λ of definite helicity λ 1 2 satisfy p σξ p ξ. S.35 Plugging these ξ λ into the positive-energy Dirac spinors 11, we obtain E ± p ξ up, λ 1 2. S.36 E p ξ In the ultra-relativistic limit E p m the square roots here simplify to E + p 2E and E p by comparison with the other root. we have E p and hence 11

12 E + p 2E; by comparison, E p. Consequently, eq. S.36 simplifies to up, L 2E ξl, up, R 2E ξ R. S.37 In other words, positive-energy ultra-relativistic Dirac spinors of definite helicity are chiral dominated by the LH Weyl components for the left helicity or by the RH Weyl components for the right helicity. Now consider the negative-energy Dirac spinors 12. Because the η s spinors have exactly opposite spin from the ξ s, their helicities are also opposite and hence p ση ± p η S.38 note opposite sign from eq. S.35. Therefore, the negative-energy Dirac spinors v of definite helicity are and in the ultra-relativistic limit they become vp, L 2E + E p η vp, λ 1 2, S.39 E ± p η η L, vp, R + 2E ηr. S.4 Again, the ultra-relativistic spinors are chiral, but this time the chirality is opposite from the helicity the left-helicity spinor has dominant RH Weyl components while the right-helicity spinor has dominant LH Weyl components Problem 4f: First, let s check the normalization conditions for the Dirac spinors 11 and 12 for general moments: u p, sup, s ξ s E p σ 2 + E + p σ 2 ξ s ξ s2eξ s 2Eδ s,s, v p, svp, s η s + E p σ 2 + E + p σ 2 η s η s+2eη s 2Eδ s,s. S.41 Note that in the second equation η sη s ξ s σ 2 σ 2 ξ s ξ sξ s δs,s. 12

13 Now consider the Lorentz invariant products ūu and vv. The ū and v are given by ūp, s u p, sγ E p σ ξ s 12 2 E + p σ ξ s ξ s E + p σ, ξ s E p σ, + E p σ vp, s v p, sγ η s 12 2 E + p σ η s S.42 η s E + p σ, +η s E p σ. Consequently, ūp, s up, s ξ s E + p σ E p σ ξ s + ξ s E p σ E + p σ ξ s S.43 2m ξ sξ s 2mδ s,s because E + p σ E p σ E p σ E + p σ E 2 p σ 2 E 2 p 2 m. S.44 Likewise, vp, s vp, s η s E + p σ E p σ η s η s E p σ E + p σ η s 2m η sη s 2mδ s,s. S.45 Problem 4g: In matrix notations column row matrix, we have up, s up, s E pσ ξs ξ E + pσ s E + pσ, ξ s E pσ ξs E pσ ξs ξ s E + pσ ξs ξ s E + pσ E + pσ E pσ ξ s ξ s E + pσ ξ s ξ s E pσ E pσ, S.46 13

14 vp, s vp, s + E pσ ηs E + pσ η s E pσ η s η s η s E + pσ, +η s E pσ E + pσ + E pσ η s η s E pσ S E + pσ η s η s E + pσ E + pσ η s η s E pσ Summing over two spin polarizations replaces ξ s ξ s with s ξ s ξ s η s η s with s η s η s Consequently, and likewise up, s up, s s [ E pσ s ξ s ξ s ] E + pσ E pσ [ E s ξ s ξ s] pσ ] E + pσ E + pσ [ E s ξ s ξ s] pσ [ E + pσ s ξ s ξ s E pσ E + pσ E pσ E pσ E + pσ E + pσ E + pσ E pσ m E pσ E + pσ m m E γ p γ p + m. vp, s vp, s s [ E pσ s η s η s ] E + pσ + E pσ [ E s η s η s] pσ + E + pσ [ ] E s η s η s + pσ E + pσ [ E s η s η s] pσ E pσ E + pσ + E pσ E pσ + E + pσ E + pσ E + pσ E pσ m E pσ E + pσ m m E γ p γ S.48 p m. S.49 Q.E.D. 14