Problem 1(a): As discussed in class, Euler Lagrange equations for charged fields can be written in a manifestly covariant form as L (D µ φ) L
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1 PHY 396 K. Solutions for problem set #. Problem a: As discussed in class, Euler Lagrange equations for charged fields can be written in a manifestly covariant form as D µ D µ φ φ = 0. S. In particularly, for φ = Φ, we have D µ Φ = Dµ Φ, Φ = m Φ, which gives us D µ D µ Φ + m Φ = 0. S. Likewise, for φ = Φ we have D µ Φ = Dµ Φ, Φ = m Φ, and therefore D µ D µ Φ + m Φ = 0. S.3 As for the vector fields A ν, the Lagrangian depends on µ A ν only through F µν, which gives us the usual Maxwell equation µ F µν = J ν where J ν A ν. S.4 To obtain the current J ν, we notice that the covariant derivatives of the charged fields Φ and
2 Φ depend on the gauge field: D µ Φ A ν = iqδ ν µφ, D µ Φ A ν = iqδ ν µφ. S.5 Consequently, J ν = D ν Φ iqφ = iq ΦD ν Φ Φ D ν Φ. D ν Φ iqφ S.6 Note that all derivatives on the last line here are gauge-covariant, which makes the current J ν gauge invariant. In a non-covariant form, J ν = iqφ D ν Φ iqφ D ν Φ q Φ Φ A ν. S.7 To prove the conservation of this current, we use the Leibniz rule for covariant derivatives, D ν XY = XD ν Y + Y D ν X. Thus, µ Φ D µ Φ = D µ Φ D µ Φ = D µ Φ D µ Φ + Φ D µ D µ Φ, µ ΦD µ Φ = D µ ΦD µ Φ = D µ ΦD µ Φ + Φ D µ D µ Φ, S.8 and hence in light of eq. S.6for the current, νj ν = iq D ν ΦD ν Φ + ΦD ν D ν Φ = iqφ D Φ iqφ D Φ by equations of motion = iqφ m Φ iqφ m Φ = 0. + iq D ν Φ D ν Φ + Φ D ν D ν Φ S.9
3 Problem b: According to the Noether theorem, where T µν Noether = µ A λ ν A λ + µ Φ ν Φ + = T µν µν Noether EM + T Noether matter µ Φ ν Φ g µν L S.0 similar to the free EM fields, and T µν Noether EM = F µλ ν A λ + 4 gµν F κλ F κλ S. T µν Noether matter = Dµ Φ ν Φ + D µ Φ ν Φ g µν D λ Φ D λ Φ m Φ Φ. S. Both terms on the second line of eq. S.0 lack µ ν symmetry and gauge invariance and thus need λ K λµν corrections for some K λµν = K µλν. We would like to show that the same K λµν = F µλ A ν we used to improve the free electromagnetic stress-energy tensor will now improve both the T µν µν EM and T mat at the same time! Indeed, to improve the scalar fields stress-energy tensor we need T µν matter T µν µν true matter T Noether matter = D µ Φ D ν Φ ν Φ + D µ ΦD ν Φ ν Φ = D µ Φ iqa ν Φ + D µ Φ iqa ν Φ S.3 = A ν iqφ D µ Φ iqφd µ Φ = A ν J µ, while the improvement of the EM stress-energy requires cf. previous homework. T µν EM = F µλ F ν λ ν A λ = +F µλ λa ν = λ F λµ A ν + A ν J µ. S.4 Altogether, to symmetrize the whole stress-energy tensor, we need T µν tot T µν µν true total T Noether total = λ F µλ A ν K λµν. 3
4 Problem c: Because the fields Φx and Φ x have opposite electric charges, their product is neutral and therefore µ Φ Φ = D µ Φ Φ = D µ Φ Φ + Φ D µ Φ. Similarly, µ D µ Φ D ν Φ = D µ D µ Φ D ν Φ + D µ Φ D µ D ν Φ = m Φ D ν Φ + D µ Φ D ν D µ Φ + iqf µν Φ S.5 where we have applied the field equation D µ D µ + m Φ x = 0 to the first term on the right hand side and used [D µ, D ν ]Φ = iqf µν Φ to expand the second term. Likewise, µ D µ Φ D ν Φ = D µ D µ Φ D ν Φ + D µ Φ D µ D ν Φ = m Φ D ν Φ + D µ Φ D ν D µ Φ iqf µν Φ S.6 and µ [ g ] µν D λ Φ D λ Φ m Φ Φ = ν D λ Φ D λ Φ + m ν Φ Φ = D ν D µ Φ D µ Φ D µ Φ D ν D µ Φ + m Φ D ν Φ + m Φ D ν Φ. Together, the left hand sides of eqs. S.5, S.6 and S.7 comprise µ T µν mat S.7 cf. eq. 7. On the other hand, combining the right hand sides of these three equations results in massive cancellation of all terms except those containing the gauge field strength tensor F µν. Thus, µ T µν mat = D µ Φ iqf µν Φ + D µ Φ iqf µν Φ = F µν iqφ D µ Φ iqφ D µ Φ S.8 = F µν J ν. And since in the previous homework we have shown µ T µν EM = F µν J ν, it follows that the total stress tensor 4 is conserved, µ T µν = 0. 4
5 Problem a: According to the time-independent Maxwell equations 0, the divergences Ê and ˆB vanish as operators and therefore should commute with any other operator in the theory. On the other hand, the commutation relations of the Ê and ˆB follow directly from those of the Ê and ˆB field themselves: [ Êx, whateverx ] = [Êi x i x, same whateverx ], [ ˆBx, whateverx ] = [ ˆBi x i x, same whateverx ]. S.9 Thus to make sure that the fields commutation relations are consistent with eqs. 0, we must verify that x i [Êi x, Êj x ] = 0, x i [Êi x, ˆB j x ] = 0, x i [ ˆBi x, Êj x ] = 0, x i [ ˆBi x, ˆB j x ] = 0, Plugging in the commutation relations into these formulæ, we immediately see that S.0 x i [Êi x, Êj x ] = x i [ ˆBi x, ˆB j x ] = 0 S. while x i [Êi x, ˆB j x ] = x i i hc ɛ ijk x k δ3 x x = 0 S. because ɛ ijk x i x k 0. Likewise, x i [ ˆBi x, Êj x ] = x i +i hc ɛ jik = i hc ɛ jik x i x k δ3 x x x k δ3 x x = 0. S.3 5
6 Problem b: According to eqs., at equal times [ ˆB i, ˆB x ] = 0 while [ ˆB i, Ê x ] = Êj x [ ˆB i x, Êj x ] = Êj x +i hcɛ jik x k δ3 x x, hence [ ˆB i x, Ĥ] = d 3 x +i hcɛ jik Ê j x x k δ3 x x = i hc ɛ jik k Ê j x, S.4 S.5 or in vector notations, [ˆBx, Ĥ] = i hc Ê. S.6 Consequently, in the Heisenberg picture c t ˆBx, t = Êx, t..a Likewise, [Êi, Ê x ] = 0 while [Êi, ˆB x ] = ˆB j x [Êi x, ˆB j x ] = ˆB j x i hcɛ ijk x k δ3 x x, hence [Êi x, Ĥ] = d 3 x i hcɛ ijk ˆBj x x k δ3 x x = i hc ɛ ijk ˆBj k x, S.7 S.8 or in vector notations, [Êx, Ĥ] = +i hc ˆB. S.9 Consequently, in the Heisenberg picture c t ˆBx, t = + Êx, t..b 6
7 Problem 3a: In the Hamiltonian formalism for the classical fields Φx and Φ x, the canonical conjugate fields are Πx = 0 Φ = 0Φx and Π x = 0 Φ = 0Φ x. S.30 The canonical conjugation implies canonical Poisson brackets between the classical fields Φx and Π x, and likewise Φ x and Πx and hence the canonical commutation relation between their quantum counterparts: In the Schrödinger picture [ˆΦx, ˆΦx ] = [ˆΦx, ˆΦ x ] = [ˆΦ x, ˆΦ x ] = 0, [ˆΠx, ˆΠx ] = [ˆΠx, ˆΠ x ] = [ˆΠ x, ˆΠ x ] = 0, [ˆΦx, ˆΠx ] = [ˆΦ x, ˆΠ x ] = 0, S.3 [ˆΦx, ˆΠ x ] = [ˆΦ x, ˆΠx ] = iδ 3 x x. In the Heisenberg picture, we have similar commutation relations for equal times t = t ; for un-equal times, the formulæ are much more complicated. Classically, the Hamiltonian density is H = Π 0 Φ + Π 0 Φ L = Π Π + Φ Φ + m Φ Φ, S.3 so the quantum theory s Hamiltonian is obviously as in eq. 5. Problem 3b: Fourier transforming the canonical commutation relations S.3 results in [ˆΦ p, ˆΦ p ] = [ˆΦ p, ˆΦ p ] = [ˆΦ p, ˆΦ p ] = 0, [ˆΠ p, ˆΠ p ] = [ˆΠ p, ˆΠ p ] = [ˆΠ p, ˆΠ p ] = 0, [ˆΦ p, ˆΠ p ] = [ˆΦ p, ˆΠ p ] = 0, S.33 [ˆΦ p, ˆΠ p ] = [ˆΦ p, ˆΠ p ] = δ p,p. 7
8 Consequently, [â p, â p ] = [â p, ˆb p ] = [ˆb p, ˆb p ] = 0 S.34 because all the ˆΦ p and all the ˆΠ p operators commute with each other, and likewise [ˆb p, ˆb p ] = [ˆb p, â p ] = [â p, â p ] = 0 S.35 because all the ˆΦ p and all ˆΠ p operators commute with each other too. Less obviously [â p, ˆb p ] = E p E p = δ p, p E p E p E p E p Ep E p 0 + ie p iδ p, p + ie p iδ p, p 0 = 0, S.36 and similarly [â p, ˆb p ] = 0. Finally, [â p, â p ] = [ˆb p, ˆb p ] = E p E p 0 ie p iδ p,p + ie p iδ p,p + 0 E p E p = δ p,p E p + E p E p E p = δ p,p. S.37 Problem 3c: First, Fourier-transforming the free Hamiltonian 5 gives us Ĥ free = p ˆΠ p ˆΠp + E p ˆΦ p ˆΦ p. S.38 Second, we reverse the definitions 7 to obtain ˆΦ p = â p + ˆb p Ep and ˆΠp = Ep i ˆbp â p. S.39 8
9 Third, we calculate E p ˆΦ p ˆΦ p + ˆΠ p ˆΠ p = E p â p + ˆb p âp + ˆb p = E p â pâ p + ˆb pˆb p = E p â pâ p + ˆb pˆb p + Finally, we put eqs. S.40 and S.38 together and derive Ĥ free = E ˆΦ p ˆΦ p p + ˆΠ p ˆΠ p p = E p â pâ p + ˆb pˆb p + p = E p â pâ p + E pˆb pˆbp + const. p + E p â p ˆb p âp ˆb p S.40 S.4 Problem 3d: We begin by Fourier transforming the charge operator ˆQ = d 3 x ˆρx = d 3 x {ˆΠ i x, ˆΦx } {ˆΠx, i ˆΦ x }, S.4 which yields ˆQ = p i {ˆΦ p, ˆΠ } p i {ˆΦp, ˆΠ } p. S.43 Next, we re-express the anticommutators here in terms of the creation and annihilator operators according to eqs. S.39. After simple algebra we find i {ˆΦ p, ˆΠ } p = â pâ p ˆb pˆb p + â pˆb p â pˆb p i, {ˆΦp, ˆΠ p} = â pâ p ˆb pˆb p â pˆb p + â pˆb p, S.44 and therefore ˆQ = p â pâ p ˆb pˆb p = p â pâ p ˆb pˆb p. 9
10 Problem 3e: According to eq., classically T 0i = 0 Φ i Φ + 0 Φ i Φ = Π i Φ Π i Φ, S.45 and hence in the quantum theory ˆP mech = d 3 x {ˆΠ, ˆΦ } {ˆΠ, ˆΦ }. S.46 Fourier-transforming this formula, we arrive at ˆP mech = p ip {ˆΠ p, ˆΦ } ip p + {ˆΠp, ˆΦ } p, S.47 and hence in light of eqs. S.44, ˆP mech = p p â pâ p p ˆb pˆb p = p p â pâ p + p ˆb pˆb p. 3 0
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