PHY 396 K. Solutions for problem set #6. 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 problem set #6. Problem 1(a): 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 1(b): Let F = i 2 Θ αβs αβ, thus M = e F and M 1 = e F. We shall use the multiplecommutator formula for the e F γ µ e +F, so we begin by evaluating the single commutator γ µ, F ] = i 2 Θ αβ γ µ, S αβ] = 1 2 Θ ( αβ g µα γ β g µβ γ α) = Θ αβ g µα γ β = Θ µ β γβ. (S.2) The multiple commutators follow immediately from 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.3)

2 Therefore, by the multiple commutator formula, M 1 γ µ M = 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 + (S.4) = L µ νγ ν. Q.E.D. Problem 1(c): Straightforward algebra gives us { γ ρ, γ λ γ µ γ ν} = { γ ρ, γ λ} γ µ γ ν γ λ{ γ ρ, γ µ} γ ν + γ λ γ µ{ γ ρ, γ ν} = 2g ρλ γ µ γ ν 2g ρµ γ λ γ ν + 2g ρν γ λ γ µ, (S.5) 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 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.6) S ρσ, γ λ γ µ γ ν] = S ρσ, γ λ] γ µ γ ν + γ λ S ρσ, γ µ] γ ν + γ λ γ µ S ρσ, 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.7) Problem 1(d): 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.8) 2

3 Problem 1(e): The covariant derivatives D µ do not commute with each other. Instead, D µ, D ν ] Ψ(x) = iqf µν (x)ψ(x). (S.9) Consequently, squaring the D = γ µ D µ operator gives D 2 = γ µ γ ν D µ D ν = 1 4 {γµ, γ ν } {D µ, D ν } γµ, γ ν ] D µ, D ν ] = 1 2 gµν D µ, D ν ] is µν iqf µν = D 2 + qf µν S µν, (S.10) and hence ( i D m)(+i D m) = D 2 + m 2 = D 2 + qf µν S µν + m 2. (S.11) Therefore, gauge-covariant Dirac equation (i D m)ψ = 0 implies ( D 2 + qf µν S µν + m 2) Ψ = ( i D m)(+i D m)ψ = 0. (S.12) Q.E.D. Problem 2(a): 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 0 γ 1 γ 2 γ 3 and any µ = 0, 1, 2, 3, n = 3 and 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 muν = S µν γ 5. Problem 2(b): First, ( γ 5 iγ 0 γ 1 γ 2 γ 3) = i(γ 3 ) (γ 2 ) (γ 1 ) (γ 0 ) = +iγ 3 γ 2 γ 1 γ 0 = +i((γ 3 γ 2 )γ 1 )γ 0 = ( 1) 3 i γ 0 ((γ 3 γ 2 )γ 1 ) = ( 1) 3+2 i γ 0 (γ 1 (γ 3 γ 2 )) = ( 1) i γ 0 (γ 1 (γ 2 γ 3 )) (S.13) = +iγ 0 γ 1 γ 2 γ 3 +γ 5. 3

4 Second, (γ 5 ) 2 = γ 5 (γ 5 ) = (iγ 0 γ 1 γ 2 γ 3 )(iγ 3 γ 2 γ 1 γ 0 ) = γ 0 γ 1 γ 2 (γ 3 γ 3 )γ 2 γ 1 γ 0 = +γ 0 γ 1 (γ 2 γ 2 )γ 1 γ 0 = γ 0 (γ 1 γ 1 )γ 0 = +γ 0 γ 0 = +1. (S.14) Problem 2(c): Any four distinct γ κ, γ λ, γ µ, γ ν are γ 0, γ 1, γ 2, γ 3 in some order. They all anticommute with each other, hence γ κ γ λ γ µ γ ν = ɛ κλµν γ 0 γ 1 γ 2 γ 3 iɛ κλµν γ 5. The other identity follows from ɛ κλνν ɛ κλνν = 24. Problem 2(d): iɛ κλµν γ κ γ 5 = 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]) = 1 4 (S.15) = 1 4 ( )γλ γ µ γ ν] = γ λ γ µ γ ν]. Problem 2(e): Proof by inspection: In the Weyl basis, the 16 matrices are σ i 1 =, γ 0 =, γ i = σ i, 0 σ k iγ i γ j] = ɛ ijk 0 iσ i 0 σ k, iγ 0 γ i] 0 = 0 +iσ i, (S.16) σ i 1 0 γ 5 γ 0 =, γ 5 γ 1 = +1 0 σ i, γ 5 =, 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.16). Q.E.D. 4

5 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 0 γ 0 γ 1 γ 1 γ 2 γ 2 γ 3 γ 3 and then further simplify it to ±(γ 0 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, iγ µ γ ν], iγ λ γ µ γ ν] = ɛ λµνρ γ 5 γ ρ (cf. (d)) or iγ 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 16 = 4. 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 Dirac spinor index with some other index i = 1,..., n which have nothing to do with Lorentz symmetry. This is technically possible, but rather silly, and nobody does this. Problems 2(f) and 2(g): 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 0 γ 1 γ d 1 (moduli the overall sign or ±i factor), each γ µ appear once, with 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 Dirac matrix products and their linear combinations. As in part (e), any product can be re-arranged as ±(γ 0 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 γ µ γ ν] products, ( d 3) of 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 γω] ɛ αβ ω Γ; the net number of all such matrices is d ( d k=0 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 sign 5

6 or ±i) with the product of the other d k matrices, for example γ 0 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. Problem 3(a): In even spacetime dimensions such as d = 3 + 1, parity reflects the space coordinates but not the time. Consequently, it changes signs of the space derivatives but not of the time derivatives, = but 0 = + 0. Hence, Ψ(x) = (γ γ )Ψ(x, t) transforms into (γ γ ) Ψ (x ) = (γ 0 0 γ ) ±γ 0 Ψ(x) ] = ±γ 0 (γ γ )Ψ(x). (S.17) In other words, P : (i m)ψ(x) (i m) Ψ (x ) = ±γ 0 (i m)ψ(x)] (S.18) and the Dirac equation is covariant under parity. Also, parity transforms Ψ(x) into Ψ (x ) = ±Ψ(x)γ 0. Consequently, the Dirac Lagrangian L = Ψ(i m)ψ at point x = (t, x) becomes a similar expression at point x = (t, x), and the Dirac action d 4 x L remains invariant. Q.E.D. Problem 3(b): For parity acting according to eq. (8) which works in any spacetime dimension the Dirac field Ψ(x) should transform as P : Ψ(x) Ψ (x ) = ±P ψ(x) (S.19) where P is some matrix which anticommutes with the γ 1 but commutes with all the other γ µ 1. Given such a matrix, we would have (i m) Ψ (x ) = ±P (i m)ψ(x) (S.20) similarly to part (a), and hence covariant Dirac equation and invariant Dirac action. 6

7 In even spacetime dimensions we may use P = γ 1 Γ = M(R)γ 0, where R is the space rotation which turns ( x 1, +x 2,...+x d 1 ) into ( x 1, x 2,..., x d 1 ), cf. the difference between eqs. (8) and (6). But in odd dimensions, there are no matrices which anticommute with the γ 1 but commutes with all the other γ µ 1. The best we can do instead is to use P = γ 1 which does exactly the opposite. Thus, we let P : Ψ(x) Ψ (x ) = ±γ 1 ψ(x), (S.21) which gives us Ψ (x ) = γ 1 Ψ(x) (S.22) and hence (i m) Ψ (x ) = γ 1 (i + m)ψ(x). (S.23) For a massive field, this is not covariant note the wrong sign of m in the parentheses on the right hand side. But for m = 0, the Dirac equation transforms similar to the field except for the overall sign, which is as good as covariant. As to the Dirac action, the Dirac conjugate of eq. (S.21) gives us Ψ (x ) = ±Ψ(x)γ 1 (note γ 1 = +γ 1 ) and consequently L = Ψ(i m)ψ L = Ψ (i m) Ψ x = Ψγ 1 γ 1 (i + m)ψ. = +Ψ(i + m)ψ. x x x (S.24) For a massless Dirac fermion, L = iψ Ψ is invariant (modulo x x ), and the action is invariant. But for a massive fermion, the mass term mψψ changes sign, and this breaks the parity symmetry. However, even the massive Dirac fermions can have a parity symmetry in an odd spacetime dimensions, but this requires two Dirac fields Ψ 1 (x) and Ψ 2 (x) with identical charges (if any) and exactly opposite masses, m 1 = m 2. The parity works by swapping the two fields 7

8 according to Ψ 1(x ) = ±γ 1 Ψ 2 (x), Ψ 2(x ) = γ 1 Ψ 1 (x), (S.25) which makes the Lagrangian L = Ψ 1 (i m)ψ 1 + Ψ 2 (i + m)ψ 2 \eq invariant. But for a single Dirac field in odd dimension, one may have mass or parity, but never both. Problem 3(c): The free part L free = iψ Ψ ( µφ) M 2 Φ 2 (S.26) of the Lagrangian (9) is invariant under parity, provided the fermion field Ψ transforms as in part (a) for even d or as in part (b) for odd d; for the odd dimensions, it s important that the fermion is massless. Now consider the interaction term gφψψ. We saw in part (b) that P : ΨΨ Ψ Ψ x x = ( 1) d ΨΨ : x (S.27) the sign is + in even dimensions but in odd dimensions. Consequently, to to make the interaction term invariant (modulo x x ), we need P : Φ(x) Φ (x ) = ( 1) d Φ(x). (S.28) In other words, Φ should be a true scalar for even d but a pseudoscalar for odd d. 8