Phys624 Dirac Equation Homework 4 Homework 4 Solutions 4.1 - Weyl or Chiral representation for γ-matrices 4.1.1: Anti-commutation relations We can write out the γµ matrices as 0 σµ γµ = σ¯µ 0 where µ µ σ = (1; σ); σ¯ = (12; − σ) The anticommutator is 0 σµ 0 σν 0 σν 0 σµ fγµ; γνg = + σ¯µ 0 σ¯ν 0 σ¯ν 0 σ¯µ 0 σµσ¯ν 0 σνσ¯µ 0 = + 0σ ¯µσν 0σ ¯νσµ σµσ¯ν + σνσ¯µ 0 = 0σ ¯µσν +σ ¯νσµ Consider the upper-left component, σµσ¯ν + σνσ¯µ. For µ = ν = 0, 0 0 0 0 σ σ¯ + σ σ¯ = 2 × 12 For µ = 0 and ν 6= 0, σ0σ¯i + σiσ¯0 = 0 For µ 6= 0 and ν 6= 0, we get σiσ¯j + σjσ¯i = −σiσj − σjσi = −σi; σj = −2δij Putting all of these together, we get µ ν ν µ µν σ σ¯ + σ σ¯ = 2g × 12 In exactly the same way, µ ν ν µ µν σ¯ σ +σ ¯ σ = 2g × 12 so µ ν µν fγ ; γ g = 2g × 14 1 Phys624 Dirac Equation Homework 4 4.1.2: Boost and rotation generators We can write i i i Sµν = [γµ; γν] = (fγµ; γνg − 2γνγµ) = (gµν − γνγµ) (1) 4 4 2 And, 0 σν 0 σµ γνγµ = σ¯ν 0 σ¯µ 0 σνσ¯µ 0 = 0σ ¯νσµ i gµν − σνσ¯µ 0 Sµν = (2) 2 0 gµν − σ¯νσµ Then, i −σi 0 Ki = S0i = (3) 2 0 σi 1 J k = ijkSij (4) 2 1 i gij − σjσ¯i 0 = ijk (5) 2 2 0 gij − σ¯jσi i σjσi 0 = ijk (6) 4 0 σjσi i [σj; σi] 0 = ijk (7) 8 0 [σj; σi] i 2ijimσm 0 = ijk (8) 8 0 2ijimσm 1σk 0 = (9) 2 0 σk 2 Phys624 Dirac Equation Homework 4 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, [Sµν;Sρσ] (10) We look at this because the Lorentz generators are made out of Sµν,and their commutation will follow from the quantity above. We can write i i i Sµν = [γµ; γν] = (fγµ; γνg − 2γνγµ) = (gµν − γνγµ) (11) 4 4 2 Now, since gµν is implicitly multiplied with the identity spinor space, the commutator we are after is µν ρσ 1 µν ν µ ρσ σ ρ 1 σ ρ ν µ [S ;S ] = − 4 [g − γ γ ; g − γ γ ] = 4 [γ γ ; γ γ ] 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 gµνγργσ. We collect all these terms in the end, and rewrite them in terms of Sµν. Carrying out the calculation, µν ρσ 1 σ ρ ν µ ν µ σ ρ [S ;S ] = 4 (γ γ γ γ − γ γ γ γ ) 1 σρ ρ σ ν µ ν µ σ ρ = 4 ((2g − γ γ )γ γ − γ γ γ γ ) 1 σρ ν µ ρ σ ν µ ν µ σ ρ = 4 (2g γ γ − γ γ γ γ − γ γ γ γ ) 1 σρ ν µ ρ σν ν σ µ ν µ σ ρ = 4 (2g γ γ − γ (2g − γ γ )γ − γ γ γ γ ) 1 σρ ν µ σν ρ µ ρ ν σ µ ν µ σ ρ = 4 (2g γ γ − 2g γ γ + γ γ γ γ − γ γ γ γ ) 1 σρ ν µ σν ρ µ ρ ν σµ µ σ ν µ σ ρ = 4 (2g γ γ − 2g γ γ + γ γ (2g − γ γ ) − γ γ γ γ ) 1 σρ ν µ σν ρ µ σµ ρ ν ρ ν µ σ ν µ σ ρ = 4 (2g γ γ − 2g γ γ + 2g γ γ − γ γ γ γ − γ γ γ γ ) 1 σρ ν µ σν ρ µ σµ ρ ν ρν ν ρ µ σ ν µ σ ρ = 4 (2g γ γ − 2g γ γ + 2g γ γ − (2g − γ γ )γ γ − γ γ γ γ ) 1 σρ ν µ σν ρ µ σµ ρ ν ρν µ σ ν ρ µ σ ν µ σ ρ = 4 (2g γ γ − 2g γ γ + 2g γ γ − 2g γ γ + γ γ γ γ − γ γ γ γ ) 1 σρ ν µ σν ρ µ σµ ρ ν ρν µ σ = 4 (2g γ γ − 2g γ γ + 2g γ γ − 2g γ γ + γν(2gρµ − γµγρ)γσ − γνγµγσγρ) 1 σρ ν µ σν ρ µ σµ ρ ν ρν µ σ ρµ ν σ = 4 (2g γ γ − 2g γ γ + 2g γ γ − 2g γ γ + 2g γ γ − γνγµγργσ − γνγµγσγρ) 1 σρ ν µ σν ρ µ σµ ρ ν ρν µ σ ρµ ν σ = 4 (2g γ γ − 2g γ γ + 2g γ γ − 2g γ γ + 2g γ γ − γνγµ(2gρσ − γσγρ) − γνγµγσγρ) 1 σρ ν µ σν ρ µ σµ ρ ν ρν µ σ ρµ ν σ ρσ ν µ = 4 (2g γ γ − 2g γ γ + 2g γ γ − 2g γ γ + 2g γ γ − 2g γ γ + γνγµγσγρ − γνγµγσγρ) 1 νσ ρ µ µσ ρ ν νρ µ σ µρ ν σ = − 2 (g γ γ − g γ γ + g γ γ − g γ γ ) Now we can add gνσgρµ − gµρgσν and gνρgσµ − gµσgνρ: µν ρσ i µρ ρ µ νσ i νρ ρ ν µσ [S ;S ] = i[− 2 (g − γ γ )g + 2 (g − γ γ )g i σµ µ σ νρ i σν ν σ µρ − 2 (g − γ γ )g + 2 (g − γ γ )g ] 3 Phys624 Dirac Equation Homework 4 Using the above and the fact that Sµν is antisymmetric, we get [Sµν;Sρσ] = i(gνρSµσ − gµρSνσ − gνσSµρ + gµσSνρ) 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 Sµν satisfies the same commutation relation as Lorentz transformation generator. However, let us calculate the commutators explicitly in terms of J i;Ki etc. 1 [J i;J j] = mnipqj[Smn;Spq] (12) 4 i = mnipqj(gnpSmq − gmpSnq − gnqSmp + gmqSnp) (13) 4 = −imnipqjgmpSnq (14) = imnimqjSnq (15) nq = i (δnqδij − δjnδiq) S (16) nq = −i δjnδiq S (17) i = − (δ δ − δ δ )Snq (18) 2 jn iq jq in i = ijknqkSnq (19) 2 = iijkJ k (20) where we have used the -tensor contraction identity ijk imn = δjmδkn − δjnδkm (21) and the anti-symmetry of Sµν in the above derivation. The other two commutation relations follow from similar manipulations. [Ki;Kj] = [S0i;S0j] = −iSij = −iijkJ k (22) 1 [Ki;J j] = mnj[S0i;Smn] (23) 2 i = mnj(gimS0n − ginS0m) (24) 2 = −iinjS0n (25) = iijkJ k (26) 4 Phys624 Dirac Equation Homework 4 4.2.2: Meaning of µ index on γµ We have the commutator i [γµ;Sρσ] = [γµ; gρσ − γσγρ] 2 i = − [γµ; γσγρ] 2 i = − γσ[γµ; γρ] + [γµ; γσ]γρ 2 = −ifγσ(gµρ − γργµ) + (gµσ − γσγµ)γρg = −igµργσ − igµσγρ + iγσγργµ + iγσγµγρ = −igµργσ − igµσγρ + 2igµργσ − iγσγµγρ + iγσγµγρ = igµργσ − igµσγρ We can write the right-hand side down in the same form by substituting the explicit repre- ρσ µ sentation of (J )ν : ρσ µ ν µα ρ σ σ ρ ν (J )ν γ = ig (δαδν − δαδν )γ µα ρ σ ν µα σ ρ ν = ig δαδν γ − ig δαδν γ = igµργσ − igµσγρ Thus, µ ρσ ρσ µ ν [γ ;S ] = (J )ν γ 4.2.3: Chirality projection operator Following the steps in Problem 4.2.1, we can write, i Sµν = (gµν − γµγν) (27) 2 Note that γ5 anti-commutes with all the γµ. fγ5; γµg = i(γ0γ1γ2γ3γµ + γµγ0γ1γ2γ3) (28) For a given value of µ, we can anti-commute the γµ in each term all the way to the corre- sponding γ in γ5. 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. fγ5; γ1g = i(γ0γ1γ2γ3γ1 + γ1γ0γ1γ2γ3) (29) = i((−1)2γ0γ1γ1γ2γ3 + (−1)γ0γ1γ1γ2γ3) (30) = 0 (31) 5 Phys624 Dirac Equation Homework 4 Therefore, i [γ5;Sµν] = [γ5; (gµν − γµγν)] (32) 2 i = − [γ5; γµγν] (33) 2 i = − [γ5; γµ]γν + γµ[γ5; γν] (34) 2 i = − [γ5; γµ]γν + γµ[γ5; γν] (35) 2 = −i(γ5γµγν + γµγ5γν) (36) = −i(γ5γµγν − γ5γµγν) = 0 (37) 6 Phys624 Dirac Equation Homework 4 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 Phys624 Dirac Equation Homework 4 8 Phys624 Dirac Equation Homework 4 4.3.2: Bilinears Part (i) Ex 4.5 Lahiri and Pal 9 Phys624 Dirac Equation Homework 4 Part (ii) We have that y y y ! Λ 1 2 In infinitesimal form, this reads y y i µν y ! 1 + 2 !µν(S ) If we define ¯ := yγ0 we have the transformation ¯ y i µν y 0 ! 1 + 2 !µν(S ) γ Since Sij is given by 1 σk 0 Sij = ijk 2 0 σk we have (Sij)y = (Sij). Also, 1 0 1σk 0 σk 0 0 1 [γ0;Sij] = ijk − 2 1 0 0 σk 0 σk 1 0 1 0 σk 0 σk = ijk − 2 σk 0 σk 0 = 0 In terms with µ or ν zero, we have i σi 0 S0i = − ; 2 0 −σi i y i σ 0 S0i = = −S0i 2 0 −σi Now consider the anti-commutator, i 0 1σi 0 σi 0 0 1 γ0;S0i = − + 2 1 0 0 −σi 0 −σi 1 0 i 0 −σi 0 σi = + 2 σi 0 −σi 0 = 0 Therefore, (Sµν)yγ0 = γ0(Sµν) 10 Phys624 Dirac Equation Homework 4 and the transformation for ¯ becomes y 0 y 0 i µν γ ! γ 1 + 2 !µν(S ) For finite rotations, this is just ¯ ¯ −1 ! Λ 1 2 From this relationship, we immediately see that ¯ ¯ −1 ! Λ 1 Λ 1 2 2 so ¯ ! ¯ that is, ¯ transforms like a Lorentz scalar.