Quantum Field Theory - Assignment 7 Solutions | PHY 396K, Assignments of Physics

Download Quantum Field Theory - Assignment 7 Solutions | PHY 396K and more Assignments Physics in PDF only on Docsity! PHY–396 K. Solutions for problem set #7. Problem 1(a): γµγν = ±γνγµ where the sign is ‘+’ for µ = ν and ‘−’ otherwise. Hence for any product Γ of the γ matrices, γµΓ = (−1)nµΓγµ where nµ is the number of γ ν 6=µ factors of Γ. For Γ = γ5 ≡ iγ0γ1γ2γ3, nµ = 3 for any µ = 0, 1, 2, 3; thus γ µγ5 = −γ5γµ. Problem 1(b): First, ( γ5 ≡ iγ0γ1γ2γ3 )† = −i(γ3)†(γ2)†(γ1)†(γ0)† = +iγ3γ2γ1γ0 = +i((γ3γ2)γ1)γ0 = (−1)3i γ0((γ3γ2)γ1) = (−1)3+2i γ0(γ1(γ3γ2)) = (−1)3+2+1i γ0(γ1(γ2γ3)) = +iγ0γ1γ2γ3 ≡ +γ5. (S.1) Second, (γ5)2 = γ5(γ5)† = (iγ0γ1γ2γ3)(iγ3γ2γ1γ0) = −γ0γ1γ2(γ3γ3)γ2γ1γ0 = +γ0γ1(γ2γ2)γ1γ0 = −γ0(γ1γ1)γ0 = +γ0γ0 = +1. (S.2) Problem 1(c): Any four distinct γκ, γλ, γµ, γν are γ0, γ1, γ2, γ3 in some order. They all anticommute with each other, hence γκγλγµγν = ²κλµνγ0γ1γ2γ3 ≡ −i²κλµνγ5. The rest is obvious. Problem 1(d): i²κλµν γκγ 5 = γκ γ [κγλγµγν] = 14 γκ ( γκγ[λγµγν] − γ[λ)γκγ(µγν] + γ[λγµ)γκγ(ν] − γ[λγµγν]γκ ) = 14 ( 4γ[λγµγν] + 2γ[λγµγν] + 4g[λµγν] + 2γ[νγµγλ] ) = 14(4 + 2 + 0− 2)γ [λγµγν] = γ[λγµγν]. (S.3) 1 Problem 1(e): Proof by inspection: In the Weyl basis, the 16 matrices are 1 = ( 1 0 0 1 ) , γµ = ( 0 σµ σ̄µ 0 ) , γ[µγν] = ( σ[µσ̄ν] 0 0 σ̄[µσν] ) , γ5γµ = ( 0 −σµ σ̄µ 0 ) , γ5 = ( −1 0 0 +1 ) , (S.4) 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.4). Q.E .D. Algebraic Proof: Without making any assumption about the matrix form of the γµ operators, let us consider the Clifford algebra γµγν+γνγµ = 2gµν . Because of these anticommutation relations, one may re-order any product of the γ’s as ±γ0 · · · γ0 γ1 · · · γ1 γ2 · · · γ2 γ3 · · · γ3 and then further simplify it to ±(γ0 or 1) × (γ1 or 1) × (γ2 or 1) × (γ3 or 1). The net result is (up to a sign or ±i factor) one of the 16 operators 1, γµ, γ[µγν], γ5γµ (cf. (d)) or γ5 (cf. (c)). Consequently, any operator Γ algebraically constructed of the γµ’s is a linear combination of these 16 operators. Incidentally, the algebraic argument explains why the γµ (and hence all their products) should be realized as 4 × 4 matrices since any lesser matrix size would not accommodate 16 independent products. That is, the γ’s are 4× 4 matrices in four spacetime dimensions; different dimensions call for different matrix sizes. Specifically, in spacetimes of even dimensions d, there are 2d independent products of the γ operators, so we need matrices of size 2d/2 × 2d/2: 2× 2 in two dimensions, 4× 4 in four, 8× 8 in six, 16× 16 in eight, 32× 32 in ten, etc.,etc.. In odd dimensions, there are only 2d−1 independent operators because γd+1 ≡ (i)γ0γ1 · · · γd−1 — the analogue of the γ5 operator in 4d — commutes rather than anticommutes with all the γµ and hence with the whole algebra. Consequently, one has two distinct representations of the Clifford algebra — one with γd+1 = +1 and one with γd+1 = −1 — but in each repre- sentation there are only 2d−1 independent operator products, which call for the matrix size of 2(d−1)/2 × 2(d−1)/2. For example, in three spacetime dimensions (two space, one time), can take (γ0, γ1, γ2) = (σ3, iσ1, iσ2) for γ 4 ≡ iγ0γ1γ2 = +1 or (γ0, γ1, γ2) = (σ3, iσ1,−iσ2) for γ 4 = −1, but in both cases we have 2× 2 matrices. Problem 1(f): Under a continuous Lorentz symmetry x 7→ x′ = Lx, the Dirac spinor field and its conjugate 2 Problem 4(a): The overall statistics of the operator product B̂Ĉ corresponds to (−1)A(BC) = (−1)AB(−1)AC . Therefore, [Â, B̂Ĉ} def = ÂB̂Ĉ − (−1)AB(−1)ACB̂Ĉ = ( ÂB̂ − (−1)ABB̂ ) Ĉ + (−1)ABB̂ ( ÂĈ − (−1)ACĈ ) = [Â, B̂}Ĉ + (−1)ABB̂[Â, Ĉ}. (S.13) Likewise, [ÂB̂, Ĉ} = ÂB̂Ĉ − (−1)AC(−1)BCĈÂB̂ =  ( B̂Ĉ − (−1)BCĈB̂ ) + (−1)BC ( ÂĈ − (−1)ACĈ ) B̂ = Â[B̂, Ĉ} + (−1)BC [Â, Ĉ}B̂. (S.14) Problem 4(b): [ÂB̂, ĈD̂} = Â[B̂, ĈD̂} + (−1)BC(−1)BD[Â, ĈD̂}B̂ = Â[B̂, Ĉ}D̂ + (−1)BCÂĈ[B̂, D̂} + (−1)BC(−1)BD[Â, Ĉ}D̂B̂ + (−1)AC(−1)BC(−1)BDĈ[Â, D̂}B̂. (S.15) Problem 4(c): (−1)CA[Â, [B̂, Ĉ}} = (−1)CAÂB̂Ĉ − (−1)BC(−1)CAÂĈB̂ − (−1)ABB̂Ĉ + (−1)AB(−1)BCĈB̂Â, (−1)BC [Ĉ, [Â, B̂}} = (−1)BCĈÂB̂ − (−1)AB(−1)BCĈB̂ − (−1)CAÂB̂Ĉ + (−1)CA(−1)ABB̂ÂĈ, (−1)AB[B̂, [Ĉ, Â}} = (−1)ABB̂Ĉ − (−1)CA(−1)ABB̂ÂĈ − (−1)BCĈÂB̂ + (−1)BC(−1)CAÂĈB̂. (S.16) Upon adding these three equations together, their right hand sides cancel out while the left hand sides add up to the Jacobi identity (4). 5 Problem 5(a): Using the Leibniz rules (S.13) and (S.14) and the anticommutation relations (6), the calculation is straightforward. [â†αâβ, â † γ ] = δβγ â † α , [â†αâβ, âδ] = −δαδ âβ , [â†αâβ, â † γ âδ] = δβγ â † αâδ − δαδ â † γ âβ . (S.17) Problem 5(b): According to eq. (S.17), the commutator [â†αâβ, â † γ âδ] has exactly the same form as its bosonic counterpart. Hence, the proof of [Â, B̂] = Ĉ proceeds exactly as in the bosonic case, cf. homework set #2 (problem 3(b)). Problem 5(c): Using the Leibniz rules and eqs. (S.17), [↵âν , â † αâ † β âγ âδ] = δναâ † µâ † β âγ âδ + δνβ â † µâ † αâγ âδ − δµγ â † αâ † β âν âδ − δµδâ † αâ † β âγ âν . (S.18) Problem 5(d): Again, we have a fermionic analogue to the bosonic second-quantized operators we studied in homework set #3 (problem 1(d)). Given eqs. (7) and (S.18) (in which we exchange γ ↔ δ), we have [ Â, â†αâ † β âδâγ ] = ∑ µ,ν 〈µ| Â1 |ν〉 [ ↵âν , â † αâ † β âδâγ ] = ∑ µ 〈µ| Â1 |α〉 â † µâ † β âδâγ + ∑ µ 〈µ| Â1 |β〉 â † αâ † µâδâγ − ∑ ν 〈δ| Â1 |ν〉 â † αâ † β âν âγ − ∑ ν 〈γ| Â1 |ν〉 â † αâ † β âδâν (S.19) 6 and consequently, in light of eq. (8), [ Â, B̂ ] = ∑ α,β,γ,δ 〈α⊗ β| B̂2 |γ ⊗ δ〉 [∑ µ 〈µ| Â1 |α〉 â † µâ † β âδâγ + ∑ µ 〈µ| Â1 |β〉 â † αâ † µâδâγ − ∑ ν 〈δ| Â1 |ν〉 â † αâ † β âν âγ − ∑ ν 〈γ| Â1 |ν〉 â † αâ † β âδâν ] = ∑ µ,β,γ,δ 〈µ⊗ β| Â1(1 st)B̂2 |γ ⊗ δ〉 â † µâ † β âδâγ + ∑ α,µ,γ,δ 〈α⊗ µ| Â1(2 nd)B̂2 |γ ⊗ δ〉 â † αâ † µâδâγ − ∑ α,β,γ,ν 〈α⊗ β| B̂2Â1(2 nd) |γ ⊗ ν〉 â†αâ † β âν âγ − ∑ α,β,ν,δ 〈α⊗ β| B̂2Â1(1 st) |ν ⊗ δ〉 â†αâ † β âδâν 〈〈renaming indices〉〉 = ∑ α,β,γ,δ 〈α⊗ β| [( A1(1 st) + A1(2 nd) ) , B̂2 ] |γ ⊗ δ〉 × â†αâ † β âδâγ = ∑ α,β,γ,δ 〈α⊗ β| Ĉ2 |γ ⊗ δ〉 × â † αâ † β âδâγ ≡ Ĉ. Q. E . D. 7