Homework 4 Solutions Weyl or Chiral representation for γ-matrices. Phys624 Dirac Equation Homework 4

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1 Homework 4 Solutions Weyl or Chiral representation for γ-matrices 4.1.1: Anti-commutation relations We can write out the γ µ matrices as where ( ) 0 σ γ µ µ = σ µ 0 σ µ = (1, σ), σ µ = (1 2, σ) The anticommutator is ( )( ) ( )( ) 0 σ {γ µ, γ ν µ 0 σ ν 0 σ ν 0 σ µ } = σ µ 0 σ ν + 0 σ ν 0 σ µ 0 ( ) ( ) σµ σ = ν 0 σν σ 0 σ µ σ ν + µ 0 0 σ ν σ µ ( ) σµ σ = ν + σ ν σ µ 0 0 σ µ σ ν + σ ν σ µ Consider the upper-left component, σ µ σ ν + σ ν σ µ. For µ = ν = 0, σ 0 σ 0 + σ 0 σ 0 = For µ = 0 and ν 0, σ 0 σ i + σ i σ 0 = 0 For µ 0 and ν 0, we get σ i σ j + σ j σ i = σ i σ j σ j σ i = { σ i, σ j} = 2δ ij Putting all of these together, we get In exactly the same way, σ µ σ ν + σ ν σ µ = 2g µν 1 2 so σ µ σ ν + σ ν σ µ = 2g µν 1 2 {γ µ, γ ν } = 2g µν 1 4 1

2 4.1.2: Boost and rotation generators We can write S µν = i 4 [γµ, γ ν ] = i 4 ({γµ, γ ν } 2γ ν γ µ ) = i 2 (gµν γ ν γ µ ) (1) And, ( )( ) 0 σ γ ν γ µ ν 0 σ µ = σ ν 0 σ µ 0 ( ) σν σ = µ 0 0 σ ν σ µ Then, S µν = i 2 ( g µν σ ν σ µ 0 ) 0 g µν σ ν σ µ K i = S 0i = i 2 ( ) σ i 0 0 σ i (2) (3) J k = 1 2 ɛijk S ij (4) = 1 i ( ) g ij σ j σ i 0 2 ɛijk 2 0 g ij σ j σ i (5) = i ( ) σ j σ i 0 4 ɛijk 0 σ j σ i (6) = i ( ) [σ j, σ i ] 0 8 ɛijk 0 [σ j, σ i (7) ] = i ( ) 2iɛ jimσ m 0 8 ɛijk 0 2iɛ jim σ m (8) = 1 ( ) σ k σ k (9) 2

3 4.2 - General representation of γ-matrices 4.2.1: Lorentz group algebra In order to have a compact notation, let us evaluate the following, [S µν, S ρσ ] (10) We look at this because the Lorentz generators are made out of S µν,and their commutation will follow from the quantity above. We can write S µν = i 4 [γµ, γ ν ] = i 4 ({γµ, γ ν } 2γ ν γ µ ) = i 2 (gµν γ ν γ µ ) (11) Now, since g µν is implicitly multiplied with the identity spinor space, the commutator we are after is [S µν, S ρσ ] = 1 4 [gµν γ ν γ µ, g ρσ γ σ γ ρ ] = 1 4 [γσ γ ρ, γ ν γ µ ] The strategy to evaluate this commutator is roughly as follows. We keep anti-commuting the γ-matrices in the first term, till we get the second term. Each anti-commutation gives us something in the form g µν γ ρ γ σ. We collect all these terms in the end, and rewrite them in terms of S µν. Carrying out the calculation, [S µν, S ρσ ] = 1 4 (γσ γ ρ γ ν γ µ γ ν γ µ γ σ γ ρ ) = 1 4 ((2gσρ γ ρ γ σ )γ ν γ µ γ ν γ µ γ σ γ ρ ) = 1 4 (2gσρ γ ν γ µ γ ρ γ σ γ ν γ µ γ ν γ µ γ σ γ ρ ) = 1 4 (2gσρ γ ν γ µ γ ρ (2g σν γ ν γ σ )γ µ γ ν γ µ γ σ γ ρ ) = 1 4 (2gσρ γ ν γ µ 2g σν γ ρ γ µ + γ ρ γ ν γ σ γ µ γ ν γ µ γ σ γ ρ ) = 1 4 (2gσρ γ ν γ µ 2g σν γ ρ γ µ + γ ρ γ ν (2g σµ γ µ γ σ ) γ ν γ µ γ σ γ ρ ) = 1 4 (2gσρ γ ν γ µ 2g σν γ ρ γ µ + 2g σµ γ ρ γ ν γ ρ γ ν γ µ γ σ γ ν γ µ γ σ γ ρ ) = 1 4 (2gσρ γ ν γ µ 2g σν γ ρ γ µ + 2g σµ γ ρ γ ν (2g ρν γ ν γ ρ )γ µ γ σ γ ν γ µ γ σ γ ρ ) = 1 4 (2gσρ γ ν γ µ 2g σν γ ρ γ µ + 2g σµ γ ρ γ ν 2g ρν γ µ γ σ + γ ν γ ρ γ µ γ σ γ ν γ µ γ σ γ ρ ) = 1 4 (2gσρ γ ν γ µ 2g σν γ ρ γ µ + 2g σµ γ ρ γ ν 2g ρν γ µ γ σ + γ ν (2g ρµ γ µ γ ρ )γ σ γ ν γ µ γ σ γ ρ ) = 1 4 (2gσρ γ ν γ µ 2g σν γ ρ γ µ + 2g σµ γ ρ γ ν 2g ρν γ µ γ σ + 2g ρµ γ ν γ σ γ ν γ µ γ ρ γ σ γ ν γ µ γ σ γ ρ ) = 1 4 (2gσρ γ ν γ µ 2g σν γ ρ γ µ + 2g σµ γ ρ γ ν 2g ρν γ µ γ σ + 2g ρµ γ ν γ σ γ ν γ µ (2g ρσ γ σ γ ρ ) γ ν γ µ γ σ γ ρ ) = 1 4 (2gσρ γ ν γ µ 2g σν γ ρ γ µ + 2g σµ γ ρ γ ν 2g ρν γ µ γ σ + 2g ρµ γ ν γ σ 2g ρσ γ ν γ µ = 1 2 (gνσ γ ρ γ µ g µσ γ ρ γ ν + g νρ γ µ γ σ g µρ γ ν γ σ ) Now we can add g νσ g ρµ g µρ g σν and g νρ g σµ g µσ g νρ : [S µν, S ρσ ] = i[ i 2 (gµρ γ ρ γ µ )g νσ + i 2 (gνρ γ ρ γ ν )g µσ + γ ν γ µ γ σ γ ρ γ ν γ µ γ σ γ ρ ) i 2 (gσµ γ µ γ σ )g νρ + i 2 (gσν γ ν γ σ )g µρ ] 3

4 Using the above and the fact that S µν is antisymmetric, we get [S µν, S ρσ ] = i(g νρ S µσ g µρ S νσ g νσ S µρ + g µσ S νρ ) In principle, we are done already, because one can show that this is the same commutation relation that the J µν matrices (defined in Problem 4.2.2) satisfy, and hence S µν satisfies the same commutation relation as Lorentz transformation generator. However, let us calculate the commutators explicitly in terms of J i, K i etc. [J i, J j ] = 1 4 ɛmni ɛ pqj [S mn, S pq ] (12) = i 4 ɛmni ɛ pqj (g np S mq g mp S nq g nq S mp + g mq S np ) (13) = iɛ mni ɛ pqj g mp S nq (14) = iɛ mni ɛ mqj S nq (15) = i (δ nq δ ij δ jn δ iq ) S nq (16) = i δ jn δ iq S nq (17) = i 2 (δ jnδ iq δ jq δ in )S nq (18) = i 2 ɛijk ɛ nqk S nq (19) = iɛ ijk J k (20) where we have used the ɛ-tensor contraction identity ɛ ijk ɛ imn = δ jm δ kn δ jn δ km (21) and the anti-symmetry of S µν in the above derivation. The other two commutation relations follow from similar manipulations. [K i, K j ] = [S 0i, S 0j ] = is ij = iɛ ijk J k (22) [K i, J j ] = 1 2 ɛmnj [S 0i, S mn ] (23) = i 2 ɛmnj (g im S 0n g in S 0m ) (24) = iɛ inj S 0n (25) = iɛ ijk J k (26) 4

5 4.2.2: Meaning of µ index on γ µ We have the commutator [γ µ, S ρσ ] = i 2 [γµ, g ρσ γ σ γ ρ ] = i 2 [γµ, γ σ γ ρ ] = i 2 γσ [γ µ, γ ρ ] + [γ µ, γ σ ]γ ρ = i{γ σ (g µρ γ ρ γ µ ) + (g µσ γ σ γ µ )γ ρ } = ig µρ γ σ ig µσ γ ρ + iγ σ γ ρ γ µ + iγ σ γ µ γ ρ = ig µρ γ σ ig µσ γ ρ + 2ig µρ γ σ iγ σ γ µ γ ρ + iγ σ γ µ γ ρ = ig µρ γ σ ig µσ γ ρ We can write the right-hand side down in the same form by substituting the explicit representation of (J ρσ ) µ ν: Thus, (J ρσ ) µ νγ ν = ig µα (δ ρ αδ σ ν δ σ αδ ρ ν)γ ν = ig µα δ ρ αδ σ ν γ ν ig µα δ σ αδ ρ νγ ν = ig µρ γ σ ig µσ γ ρ [γ µ, S ρσ ] = (J ρσ ) µ νγ ν 4.2.3: Chirality projection operator Following the steps in Problem 4.2.1, we can write, S µν = i 2 (gµν γ µ γ ν ) (27) Note that γ 5 anti-commutes with all the γ µ. {γ 5, γ µ } = i(γ 0 γ 1 γ 2 γ 3 γ µ + γ µ γ 0 γ 1 γ 2 γ 3 ) (28) For a given value of µ, we can anti-commute the γ µ in each term all the way to the corresponding γ in γ 5. In each anti-commutation, we pick up a negative sign. There are even anti-commutations in one term, and odd in the other, and thus they always cancel. Let us do the steps for µ = 1. {γ 5, γ 1 } = i(γ 0 γ 1 γ 2 γ 3 γ 1 + γ 1 γ 0 γ 1 γ 2 γ 3 ) (29) = i(( 1) 2 γ 0 γ 1 γ 1 γ 2 γ 3 + ( 1)γ 0 γ 1 γ 1 γ 2 γ 3 ) (30) = 0 (31) 5

6 Therefore, [γ 5, S µν ] = i 2 [γ5, (g µν γ µ γ ν )] (32) = i 2 [γ5, γ µ γ ν ] (33) = i 2 [γ5, γ µ ]γ ν + γ µ [γ 5, γ ν ] (34) = i 2 [γ5, γ µ ]γ ν + γ µ [γ 5, γ ν ] (35) = i(γ 5 γ µ γ ν + γ µ γ 5 γ ν ) (36) = i(γ 5 γ µ γ ν γ 5 γ µ γ ν ) = 0 (37) 6

7 4.3 - Lorentz Transformations 4.3.1: Modified spinor (Ex 4.6 Lahiri and Pal) Please ignore the text above the line in the scan. 7

8 8

9 4.3.2: Bilinears Part (i) Ex 4.5 Lahiri and Pal 9

10 Part (ii) We have that In infinitesimal form, this reads ψ ψ Λ 1 2 ψ ψ [ 1 + i 2 ω µν(s µν ) ] If we define ψ := ψ γ 0 we have the transformation ψ ψ [ 1 + i 2 ω µν(s µν ) ] γ 0 Since S ij is given by S ij = 1 ( ) σ k 0 2 ɛijk 0 σ k we have (S ij ) = (S ij ). Also, [γ 0, S ij ] = 1 [( )( ) ( )( )] 0 1 σ k 0 σ k ɛijk σ k 0 σ k 1 0 = 1 [( ) ( )] 0 σ k 0 σ k 2 ɛijk σ k 0 σ k 0 = 0 In terms with µ or ν zero, we have ( ) S 0i = 2( i ) σ i 0 0 σ i, ( ) ( ) S 0i i σ i 0 = 2 0 σ i = ( S 0i) Now consider the anti-commutator, { γ 0, S 0i} = 2[( i )( ) ( )( )] 0 1 σ i 0 σ i σ i + 0 σ i 1 0 = i [( ) ( )] 0 σ i 0 σ i 2 σ i + 0 σ i 0 Therefore, = 0 (S µν ) γ 0 = γ 0 (S µν ) 10

11 and the transformation for ψ becomes For finite rotations, this is just ψ γ 0 ψ γ 0[ 1 + i 2 ω µν(s µν ) ] ψ From this relationship, we immediately see that ψλ so ψψ ψλ Λ 1 2 ψ ψψ ψψ that is, ψψ transforms like a Lorentz scalar. 11

12 4.4 - Gordon identity We begin with the following two equations, (/p m)u(p) = 0 (38) ū(p )(/p m) = 0 (39) Multiplying with appropriate spinors and adding these two together, We can write the first term as, ū(p )γ µ (/p m)u(p) + ū(p )(/p m)γ µ u(p) = 0 (40) ū(p )(γ µ γ α p α + γ α γ µ p α)u(p) 2mū(p )γ µ u(p) = 0 (41) ū(p )(γ µ γ α 1 2 ((p α + p α) + (p α p α)) + γ α γ µ 1 2 ((p α + p α) (p α p α)))u(p) (42) = 1 2 (p α + p α)ū(p )({γ µ, γ α })u(p) (p α p α)ū(p )[γ µ, γ α ]u(p) (43) = (p µ + p µ)ū(p )u(p) iσ µα (p α p α)ū(p )u(p) (44) Therefore, Gordon identity, ū(p ) [ (p µ + p µ) iσ µα (p α p α) ] u(p) 2mū(p )γ µ u(p) = 0 (45) Since the equation for the v spinor looks exactly like the u spinor case except for a relative negative sign on the mass parameter m, it is simple to write down the Gordon identity for v. ū(p ) [ (p µ + p µ) iσ µα (p α p α) ] u(p) + 2mū(p )γ µ u(p) = 0 (46) Completeness of spinors Part (i) Ex 4.8 Lahiri and Pal This is a special case of the Gordon identity. For completeness sake, we do the calculation fully. We are given the following equations, Proceeding as suggested in the problem, (/p m)u(p) = 0 (47) ū(p)(/p m) = 0 (48) ū(p)γ µ (/p m)u(p) + ū(p)(/p m)γ µ u(p) = 0 (49) p α ū(p)(γ µ γ α + γ α γ µ )u(p) 2mū(p)γ µ u(p) = 0 (50) p µ ū(p)u(p) m ū(p)γ µ u(p) = 0 (51) The same calculation with the v spinors yields, p µ v(p)v(p) + m v(p)γ µ v(p) = 0 (52) 12

13 If we put µ = 0 in the u equation, we get, p 0 ū s (p)u r (p) m ū s (p)γ 0 u r (p) = 0 (53) Eū s (p)u r (p) m u s(p)γ 0 γ 0 u r (p) = 0 (54) Eū s (p)u r (p) m u s(p)u r (p) = 0 (55) Eū s (p)u r (p) 2m E δ rs = 0 (56) ū s (p)u r (p) = 2m δ rs (57) p 0 v(p)v(p) + m v(p)γ 0 v(p) = 0 (58) v s (p)v r (p) = 2m δ rs (59) 13

14 Part (ii) Ex 4.10 Lahiri and Pal 14

15 4.6 - Helicity and chirality projection operators 15

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