Dirac Equation Homework 4

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 {γµ, γν} = + 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 {γ , γ } = 2g × 14

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4.1.2: Boost and rotation generators We can write i i i Sµν = [γµ, γν] = ({γµ, γν} − 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

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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] = ({γµ, γν} − 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 ]

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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)

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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 = −i{γσ(gµρ − γργµ) + (gµσ − γσγµ)γρ} = −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.

{γ5, γµ} = 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.

{γ5, γ1} = i(γ0γ1γ2γ3γ1 + γ1γ0γ1γ2γ3) (29) = i((−1)2γ0γ1γ1γ2γ3 + (−1)γ0γ1γ1γ2γ3) (30) = 0 (31)

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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)

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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.

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

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Part (ii) We have that

† † † ψ → ψ Λ 1 2 In inﬁnitesimal form, this reads

† † i µν † ψ → ψ 1 + 2 ωµν(S ) If we deﬁne

ψ¯ := ψ†γ0

we have the transformation

¯ † i µν † 0 ψ → ψ 1 + 2 ωµν(S ) γ Since Sij is given by 1 σk 0 Sij = ijk 2 0 σk we have (Sij)† = (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 † 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)†γ0 = γ0(Sµν)

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and the transformation for ψ¯ becomes

† 0 † 0 i µν ψ γ → ψ γ 1 + 2 ωµν(S ) For ﬁnite rotations, this is just

¯ ¯ −1 ψ → ψΛ 1 2 From this relationship, we immediately see that

¯ ¯ −1 ψψ → ψΛ 1 Λ 1 ψ 2 2 so ψψ¯ → ψψ¯

that is, ψψ¯ transforms like a Lorentz scalar.

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4.4 - Gordon identity We begin with the following two equations,

(p/ − m)u(p) = 0 (38) u¯(p0)(p/0 − m) = 0 (39) Multiplying with appropriate spinors and adding these two together,

u¯(p0)γµ(p/ − m)u(p) +u ¯(p0)(p/0 − m)γµu(p) = 0 (40) 0 µ α α µ 0 0 µ ⇒ u¯(p )(γ γ pα + γ γ pα)u(p) − 2mu¯(p )γ u(p) = 0 (41) We can write the ﬁrst term as, 1 1 u¯(p0)(γµγα ((p + p0 ) + (p − p0 )) + γαγµ ((p + p0 ) − (p − p0 )))u(p) (42) 2 α α α α 2 α α α α 1 1 = (p + p0 )¯u(p0)({γµ, γα})u(p) + (p − p0 )¯u(p0)[γµ, γα]u(p) (43) 2 α α 2 α α 0 0 µα 0 0 = (pµ + pµ)¯u(p )u(p) − iσ (pα − pα)¯u(p )u(p) (44) Therefore, Gordon identity,

0 0 µα 0 0 µ u¯(p ) (pµ + pµ) − iσ (pα − pα) u(p) − 2mu¯(p )γ 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.

0 0 µα 0 0 µ u¯(p ) (pµ + pµ) − iσ (pα − pα) u(p) + 2mu¯(p )γ u(p) = 0 (46)

4.5 - 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,

(p/ − m)u(p) = 0 (47) u¯(p)(p/ − m) = 0 (48) Proceeding as suggested in the problem,

u¯(p)γµ(p/ − m)u(p) +u ¯(p)(p/ − m)γµu(p) = 0 (49) µ α α µ µ ⇒ pαu¯(p)(γ γ + γ γ )u(p) − 2mu¯(p)γ u(p) = 0 (50) ⇒ pµu¯(p)u(p) − m u¯(p)γµu(p) = 0 (51)

The same calculation with the v spinors yields,

pµv¯(p)v(p) + m v¯(p)γµv(p) = 0 (52)

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If we put µ = 0 in the u equation, we get,

0 0 p u¯s(p)ur(p) − m u¯s(p)γ ur(p) = 0 (53) † 0 0 ⇒ Eu¯s(p)ur(p) − m us(p)γ γ ur(p) = 0 (54) † ⇒ Eu¯s(p)ur(p) − m us(p)ur(p) = 0 (55)

⇒ Eu¯s(p)ur(p) − 2m E δrs = 0 (56)

⇒ u¯s(p)ur(p) = 2m δrs (57)

p0v¯(p)v(p) + m v¯(p)γ0v(p) = 0 (58)

⇒ v¯s(p)vr(p) = −2m δrs (59)

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Part (ii) Ex 4.10 Lahiri and Pal

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4.6 - Helicity and chirality projection operators