Solutions to Problem Set #7 in Quantum Field Theory, Assignments of Physics

Download Solutions to Problem Set #7 in Quantum Field Theory 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 γν 6=µ factors of Γ. For Γ = γ5 ≡ iγ0γ1γ2γ3, nµ = 3 for any µ = 0, 1, 2, 3; thus γµγ5 = −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 = κλµνγ0γ1γ2γ3 ≡ −iκλµνγ5. The rest is obvious. Problem 1(d): iκλµν 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 ) , γ0 = ( 0 1 1 0 ) , γi = ( 0 +σi −σi 0 ) , iγ[iγj] = ijk ( σk 0 0 σk ) , iγ[0γi] = ( −iσi 0 0 +iσi ) , (S.4) γ5γ0 = ( 0 −1 +1 0 ) , γ5γ1 = ( 0 −σi −σi 0 ) , γ5 = ( −1 0 0 +1 ) , 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, iγ[µγν], −iγ[λγµγν] = λµνργ5γρ (cf. (d)) or iγ[κγλγµγν] = κλµνγ5 (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 γ4 = −1, 2 Problem 2(d): In the Weyl convention for the Dirac matrices, the charge conjugation symmetry C acts on the Dirac field according to Ψ′(x) = ±γ2Ψ∗(x). Consequently Ψ′†(x) = ∓Ψ>(x)γ2 =⇒ Ψ′(x) = Ψ′†(x)γ0 = ∓Ψ>(x)γ2γ0, (S.11) and therefore for any Dirac bilinear, Ψ ′ ΓΨ′ = −Ψ>γ2γ0Γγ2Ψ∗ = +Ψ†(γ2γ0Γγ2)>Ψ = +Ψγ0γ2Γ>γ0γ2Ψ ≡ ΨΓcΨ. (S.12) The second equality of this formula follows by transposition of the Dirac “sandwich” Ψ> · · ·Ψ∗, which carries an extra minus sign because the fermionic fields Ψ and Ψ∗ anticommute with each other. Problem 2(e): By inspection, 1c ≡ γ0γ2γ0γ2 = +1. The γ5 matrix is symmetric and commutes with the γ0γ2, hence γc5 = +γ5. Among the four γµ matrices, the γ1 and γ3 are anti-symmetric and commute with the γ0γ2 while the γ0 and γ2 are symmetric but anti-commute with the γ 0γ2; hence, for all four γµ, γ c µ = −γµ. Finally, because of the transposition involved, (γµγν)c = γcνγcµ = +γνγµ, hence (γ[µγν])c = +γ[νγµ] = −γ[µγν]. Likewise, (γ5γµ)c = (γµ)c(γ5)c = −γµγ6 = +γ5γµ. Therefore, according to eq. (S.12), the scalar S, the pseudoscalar P and the axial vector Aµ are C–even while the vector Vµ and the tensor Tµν are C–odd. Problem 3(a): Given the anticommutation relations (2), we have â†αâβ âδ = −â † αâδâβ = −(δα,δ − âδâ † α)âβ = +âδâ † αâβ − δα,δâβ (S.13) and therefore [â†αâβ, âδ] = −δα,δâβ. Likewise, â†αâβ â † γ = â † α(δβ,γ − â†γ âβ) = δβ,γ â † α − â†αâ†γ âβ = δβ,γ â † α + â † γ â † αâβ (S.14) and therefore [â†αâβ, â † γ ] = +δβ,γ â † α. 5 Finally, by Leibniz rule [â†αâβ, â † γ âδ] = [â † αâβ, â † γ ]âδ + â † γ [â † αâβ, âδ] = +δβ,γ â † αâδ − δα,δâ † γ âβ . (S.15) Problem 3(b): According to eq. (S.15), 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 #3 (problem 2(b)). Problem 3(c): Using the Leibniz rules and eqs. (S.13) and (S.14), [↵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.16) Problem 3(d): Again, w have a fermionic analogue to the bosonic second-quantized operators we studied in homework set #3 (problem 2(d)). Given eqs. (4) and (S.16) (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〉 ↵⠆ β âδâγ + ∑ µ 〈µ| Â1 |β〉 â†α↵âδâγ − ∑ ν 〈δ| Â1 |ν〉 â†αâ † β âν âγ − ∑ ν 〈γ| Â1 |ν〉 â†αâ † β âδâν (S.17) 6 and consequently, in light of eq. (8), [ Â, B̂ ] = ∑ α,β,γ,δ 〈α⊗ β| B̂2 |γ ⊗ δ〉 [∑ µ 〈µ| Â1 |α〉 ↵⠆ β âδâγ + ∑ µ 〈µ| Â1 |β〉 â†α↵âδâγ − ∑ 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〉 ↵⠆ β âδâγ + ∑ α,µ,γ,δ 〈α⊗ µ| Â1(2nd)B̂2 |γ ⊗ δ〉 â†α↵âδâγ − ∑ α,β,γ,ν 〈α⊗ β| B̂2Â1(2nd) |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. Problem 4.(a): The simplest form of the Dirac Lagrangian is L = Ψ(iγµ∂µ − m)Ψ, (S.18) which involves spacetime derivatives of the Ψ field but not of the Ψ. Consequently, by Noether theorem TµνN = ∂L ∂(∂µΨ) × ∂νΨ − gµνL = Ψiγµ∂νΨ − gµνΨ(iγλ∂λ −m)Ψ. (S.19) As usual for fields of non-zero spin, the Noether stress-energy tensor is not symmetric and the 7 To be precise, eq. (S.31) presumes “classical” fermionic fields which anticommute with each other, thus the charge density J0 can be written as either J0 − eΨγ0Ψ = −eΨ†Ψ ≡ −e ∑ α Ψ∗αΨα or J0 = +e ∑ α ΨαΨ ∗ α . (S.32) Alas, in the quantum theory Ψ̂†αΨ̂β is not equal to −Ψ̂βΨ̂ † α, and this gives rise to the operator ordering ambiguity in defining the quantum electric charge. Fortunately, this ambiguity amounts to a constant. Indeed, the quantum Dirac fields Ψ̂†α(x) and Ψ̂α(x) are linear combinations of the fermionic creation and annihilation operators (cf. eq. (5)), and the latter either anticommute with each other or have c-number anticommutators (cf. eqs. (6) and (7)). Therefore, the anticommutator {Ψ̂†α(x), Ψ̂β(y)} is a c-number function of (x− y). We shall calculate this function later in class, but for the moment all we need to know it’s a c-number, and therefore −e ∑ α Ψ̂†α(x)Ψ̂α(x) = +e ∑ α Ψ̂α(x)Ψ̂ † α(x) + a c-number constant. (S.33) Consequently, however we order the creation and annihilation operators in the quantized electric charge operators, it will give us the same result up to a c-number constant (which may be infinite). Hence, we may just as well take the simplest ordering and allow for an extra constant term, thus Q̂ = −e ∫ d3x Ψ̂†(x)Ψ̂(x) + a c-number constant. (10a) Next, let us expand the fields Ψ̂(x) and Ψ̂†(x) into creation and annihilation operators ac- cording to eqs. (S.23) and plug into the space integral (10a). Proceeding as in the previous 10 question, we have ∫ d3x Ψ̂†Ψ̂ = ∫ d3p (2π)3 1 (2Ep)2 ∑ s,s′ ( u†(p, s′)â†p,s′ + v †(−p, s′)b̂−p,s′ ) × × ( u(p, s)âp,s + v(−p, s)b̂ † −p,s ) = ∫ d3p (2π)3 1 2Ep ∑ s ( â†p,sâp,s + b̂−p,sb̂ † −p,s ) = ∫ d3p (2π)3 1 2Ep ∑ s ( â†p,sâp,s + b̂p,sb̂ † p,s ) = ∫ d3p (2π)3 1 2Ep ∑ s ( â†p,sâp,s − b̂†p,sb̂p,s + 2Ep(2π)3δ(3)(0) ) = ∫ d3p (2π)3 1 2Ep ∑ s ( â†p,sâp,s − b̂†p,sb̂p,s ) + infinite c-number constant, (S.34) and therefore Q̂ = ∫ d3p (2π)3 1 2Ep ∑ s ( −e â†p,sâp,s + e b̂†p,sb̂p,s ) + C (S.35) where C is the sum of c-number constants from eqs. (10.a) and (S.34). To determine the value of C, note that the vacuum state of the theory is invariant under the charge conjugation symmetry and therefore must have zero electric charge, Q̂ |0〉 = 0. On the other hand, the vacuum state |0〉 is annihilated by all the âp,s and b̂p,s operators and hence by the terms on the right hand side of eq. (S.35) except for the constant C. Consequently, Q̂ |0〉 = C |0〉 and the electric charge of the vacuum is C. And since this charge must vanish, we must have C = 0 — i.e., somehow the constant terms in eqs. (10a) and (S.34) must cancel each other — and therefore Q̂ = ∫ d3p (2π)3 1 2Ep ∑ s ( −e â†p,sâp,s + e b̂†p,sb̂p,s ) . (10b) Q.E .D. Problem 4(c): In this question, we start with eq. (11) for the net spin operator as a space integral of a Dirac 11 bilinear, so our first step is to expand the fields into momentum modes (S.23) and plug the expansion into eq. (11). This gives us Ŝnet = ∫ d3p (2π)3 1 2Ep S̃p (S.36) where S̃p = 1 2Ep ∑ s,s′ ( u†(p, s)â†p,s + v †(−p, s)b̂−p,s ) S ( u(p, s′)âp,s′ + v(−p, s ′)b̂†−p,s′ ) . (S.37) Next, we need to evaluate the Dirac “sandwiches” u†Su, v†Sv, etc., where S = ( 1 2 σ 0 0 12 σ ) . (S.38) For the non-relativistic modes (|p|  m) we approximate u(p, s) ≈ u(0, s) = √ m ( ξs ξs ) , v(−p, s) ≈ v(0, s) = √ m ( ηs ηs ) , (S.39) which gives us u†(p, s)Su(p, s′) ≈ m ξ†sσξs′ , v†(−p, s)Sv(−p, s′) ≈ m η†sσηs′ , u†(p, s)Sv(−p, s′) = O(|p|) ≈ 0, v†(−p, s)Su(p, s′) = O(|p|) ≈ 0, (S.40) and consequently S̃p ≈ ∑ s,s′ ( ξ†s σ 2 ξs′ × â†p,sâp,s′ + η † s σ 2 ηs′ × b̂−p,sb̂ † −bp,s′ ) + O(|p|/m). (S.41) At this point, let us separate the â†â terms from the b̂b̂† terms in the momentum integral (S.36), and then in the b̂b̂† part change the sign of the integration variable p. Putting the two parts 12