Solutions for Problem Set 6 - Quantum Field Theory I | PHY 396K, Assignments of Physics

Download Solutions for Problem Set 6 - Quantum Field Theory I | PHY 396K and more Assignments Physics in PDF only on Docsity! PHY–396 K. Solutions for problem set #6. Problem 1(a): By definition, Ŝµν def = i4 [γ µ, γν ] = i2(γ µγν − gµν). Using this formula, we saw in class that[ γλ, Sµν ] = i ( gλµγν − gλνγµ ) . Consequently, by Leibniz rule, [ γκγλ, Sµν ] = γκ [ γλ, Sµν ] + [ γκ, Sµν ] γλ = γκ ( igλµγν − igλνγµ ) + ( igκµγν − igκνγµ ) γλ = igλµγκγν − igκνγµγλ − igλνγκγµ + igκµγνγλ = igλµ ( γκγν − gκν ) − igκν ( γµγλ − gλµ ) − igλν ( γκγµ − gκµ ) + igκµ ( γνγλ − gλν ) = 2gλµSκν − 2gκνSµλ − 2gλνSκµ + 2gκµSνλ, and therefore, [ Sκλ, Sµν ] = i2 [ γκγλ, Sµν ] = igλµSκν − igκνSµλ − igλνSκµ + igκµSνλ = igλµSκν + igκνSλµ − igλνSκµ − igκµSλν . (S.1) Q.E .D. Problem 1(b): Let F = − i2θαβS αβ, thus M = eF and M−1 = e−F . We shall use the multiple- commutator formula for the e−Fγµe+F , so we begin by evaluating the single commutator [ γµ, F ] = − i2θαβ [ γµ, F ] = 12θαβ ( gµαγβ − gµβγα ) = 12θ µ βγ β − 12θ µ α γ α = θµνγ ν . (S.2) The multiple commutators follow immediately from this formula, [[ γµ, F ] , F ] = θµλθ λ νγ ν ,[[[ γµ, F ] , F ] , F ] = θµλθ λ ρθ ρ νγ ν , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (S.3) 1 Consequently, M−1γµM = e−Fγµe+F = γµ + [ γµ, F ] + 12 [[ γµ, F ] , F ] + 16 [[[ γµ, F ] , F ] , F ] + · · · = γµ + θµνγ ν + 12 θ µ λθ λ νγ ν + 16 θ µ λθ λ ρθ ρ νγ ν + · · · = Lµνγ ν . (S.4) Q.E .D. Problem 1(c): { γρ, γλγµγν } = 2gρλ γµγν − 2gρµ γλγν + 2gρν γλγµ,[ γρ, γκγλγµγν ] = 2gρκ γλγµγν − 2gρλ γκγµγν + 2gρµ γκγλγν − 2gρν γκγλγµ,[ Sρσ, γλγµγν ] = igσλ γργµγν + igσµ γλγργν + igσν γλγµγρ − igρλ γσγµγν − igρµ γλγσγν − igρν γλγµγσ. The algebra is straigtforward. Problem 1(d): γα γα = 1 2{γ α, γβ}gαβ = gαβgαβ = 4; γαγνγα = 2g ανγα − γν γαγα = 2γν − γν(4) = −2γν ; γαγµγνγα = 2g αµγνγα − γµ γαγνγα = 2γνγµ − γµ (−2γν) = 2{γν , γµ} = 4gµν ; γαγλγµγνγα = 2g αλγµγνγα − γλ γαγµγνγα = 2γµγνγλ − γλ (4gµν) = 2 ( γµγν − 2gµν ) γλ = −2γνγµγλ. (S.5) Problem 1(e): Gauge-covariant derivatives Dµ do not commute with each other: [Dµ, Dν ] = iqFµν . Therefore( γµDµ )2 6= D2 but rather ( γµDµ )2 = γµγν DµDν = (g µν − 2iSµν) DµDν = D2 − iSµν [Dµ, Dν ] = D2 + qFµν Sµν (S.6) 2 Next consider the determinant det ( Xµσ µ ) = det ( X0 +X3 X1 − iX2 X1 + iX2 X0 −X3 ) = (X0) 2 − (X3)2 − (X1)2 − (X2)2 ≡ X2. (S.16) According to eq. (4), X ′ 2 = det ( X ′µσ µ ) = |det(U)|2 det ( Xµσ µ ) = X2 (S.17) because the SL(2,C) matrices U have det(U) = 1. Consequently, for any 4–vector Xµ, we have (LX)2 = X2, which means that Xµ → LµνXν is indeed a Lorentz transform. Finally, consider the group law: σλ L λ µ(U2U1) = (U2U1)σµ(U2U1) † = U2 ( U1σµU † 1 = σν L ν µ(U1) ) U †2 = ( U2 σν U † 2 ) Lνµ(U1) = σλ L λ ν(U2)L ν µ(U1) (S.18) and hence Lλµ(U2U1) = L λ ν(U2)L ν µ(U1) i.e., L(U2U1) = L(U2)L(U1). Q.E .D. Problem 2(d): For U = exp ( − i2θ n~σ ) = cos θ2 − i sin θ 2 n~σ and U † = U−1 = cosθ2 + i sin θ 2 n~σ, Uσ0U † = 1 = σ0, which means that L(U) is merely a rotation of the 3d space. Specifically, ~σ · x′ = U(x~σ)U † = cos2 θ2 − i 2 sin θ [n~σ,x~σ] + sin 2 θ 2 (n~σ)(x~σ)(n~σ) = cos2 θ2 + sin θ (n× x) · ~σ + sin 2 θ 2 ( 2(nx)(n~σ)− (x~σ) ) = ~σ · ( cos θ ( x− n(nx) ) + sin θ n× x + n(nx) ) , thus x′ = cos θ ( x− n(nx) ) + sin θ n× x + n(nx), (S.19) which indeed describes a rotation through angle θ around axis n. On the other hand, for U = U † = exp ( − r2 n~σ ) = cosh r2 − sinh r 2 n~σ, U ( xµσµ ≡ t− x~σ ) U † = cosh2 r2 (t− x~σ) − 1 2 sinh r {n~σ, t− x~σ} + sinh2 r2 (n~σ)(t− x~σ)(n~σ) = cosh2 r2 (t− x~σ) − sinh r ( tn~σ − nx ) + sinh2 r2 ( t− 2(nx)(n~σ) + (x~σ) ) = ( cosh r t + sinh r nx ) − (~σn) ( sinh r t + cosh r nx ) − ~σ · ( x− n(nx) ) , (S.20) 5 and therefore, t′ = (cosh r) t + (sinh r) nx, x′ = n ( (sinh r) t + (cosh r) nx ) + ( x− n(nx) ) , (S.21) which is precisely the Lorentz boost of rapidity r in the direction n. (The rapidity r is related to the usual parameters of a Lorentz boost according to β = tanh r, γ = cosh r, γβ = sinh r. For several boosts in the same directions, the rapidities add up, rtot = r1 + r2 + · · ·.) Q.E .D. Problem 3(a): In light of eq. (5), Φ̂(x′ = Lx) = ∫ d3p′ (2π)3 1 2Ep′ [ e−ip ′x′ √ 2p′0 â(p′) + e+ip ′x′ √ 2p′0 â†(p′) ] p′0=Ep′ = ∫ d3p (2π)3 1 2Ep [ e−ipx √ 2(Lp)0 â(Lp) + e+ipx √ 2(Lp)0 â†(Lp) ] p0=Ep (S.22) where p′ = Lp and hence p′x′ = px and ∫ d3p′ 2Ep′ = ∫ d3p 2Ep . At the same time, eq. (7) implies Φ̂(Lx) = D̂(L) Φ̂(x) D̂†(L) = ∫ d3p (2π)3 1 2Ep [ e−ipx √ 2p0 D̂(L) â(p) D̂†(L) + e+ipx √ 2p0 D̂(L) â†(p) D̂†(L) ] p0=Ep (S.23) Since eqs. (S.22) and (S.23) should agree for all x, Fourier transforms of their respective right hand sides should agree for all p, hence √ 2p0 D̂(L) â(p) D̂†(L) = √ 2(Lp)0 â(Lp),√ 2p0 D̂(L) â†(p) D̂†(L) = √ 2(Lp)0 â†(Lp). (8.1–2) Consequently, D(L) |p〉 = D̂(L) (√ 2p0 â†(p) |0〉 ) = √ 2p0 D̂(L) â†(p) ( |0〉 = D̂†(L) |0〉 ) = (√ 2p0 D̂(L) â†(p) D̂†(L) = √ (Lp)0 â†(Lp) ) |0〉 = |Lp〉 , (8.3) 6 and likewise D(L) |p1, p2〉 = D̂(L) (√ (2p01)(2p 0 2) â †(p1)â †(p2) |0〉 ) = √ (2p01)(2p 0 2) D̂(L) â †(p1)â †(p2) ( |0〉 = D̂†(L) |0〉 ) = (√ 2p01 D̂(L) â †(p1) D̂†(L) )(√ 2p02 D̂(L) â †(p2) D̂†(L) ) |0〉 = (√ 2(Lp1)0 â †(Lp1) )(√ 2(Lp2)0 â †(Lp2) ) |0〉 = |Lp1, Lp2〉 , (8.4) etc., etc. Q.E .D. Problem 3(b): Using eq. (11) twice, we have D̂(L2)D̂(L1) φ̂a(x) ( D̂(L2)D̂(L1) )† = D̂(L2) ( D̂(L1) φ̂a(x) D̂†(L1) ) D̂†(L2) = D̂(L2) ( M ba (L −1 1 ) φ̂b(L1x) ) D̂†(L2) = M ba (L −1 1 ) ( D̂(L2) φ̂b(L1x) D̂†(L2) ) = M ba (L −1 1 )M c b (L −1 2 ) φ̂c(L2L1x). (S.24) On the other hand, D̂(L2L1) φ̂a(x) D̂†(L2L1) = M ca ( (L2L1) −1 = L−11 L −1 2 ) φ̂c(L2L1x), (S.25) which obviously agrees with (S.24) if and only if M ca (L −1 1 L −1 2 ) = M b a (L −1 1 )M c b (L −1 2 ) (S.26) i.e., if and only if the matrices M ba (L) form a representation of the Lorentz group. 7