Physics 582, Problem Set 2 Solutions

Transcript

1 Physics 582, Problem Set 2 Solutions TAs: Hart Goldman and Ramanjit Sohal Fall 2018 Symmetries and Conservation Laws In this problem set we return to a study of scalar electrodynamics which has the Lagrangian L = D µ φ 2 m 2 0 φ 2 λ 4! φ F µνf µν, (0.1) where φ is a complex scalar field, A µ is a U(1) gauge field, and D µ is the covariant derivative D µ = µ + iea µ. (0.2) We will perform an analysis of this theory at the classical level, similar to what was done in the lecture notes for a Dirac fermion coupled to a gauge field. 1. Under an infinitesimal global U(1) rotation the fields transform as φ(x) e iθ φ(x) (1 + iθ)φ(x) = δφ(x) = iθφ(x) (0.3) The corresponding change in the Lagrangian is = δφ (x) = iθφ (x) (0.4) A µ (x) A µ (x) = (x) = 0. (0.5) = δφ + δφ δφ δφ + δ( µ φ) δ µφ + δ( µ φ ) δ µφ (0.6) [ = µ δ( µ φ) δφ + ] δ( µ φ ) δφ (0.7) = µ [D µ φ δφ + D µ φδφ ] (0.8) = θ µ [iφd µ φ iφ D µ φ]. (0.9) where we used the Euler-Lagrange equations in passing to the second line. = 0 under a U(1) rotation, we conclude that the current Since j µ = iφ(d µ φ) iφ D µ φ (0.10) is conserved.

2 2 2. We have that µ j µ = 0 = 0 j 0 = i j i. (0.11) Integrating the above expression over all space and using Gauss Law, we find 0 d 3 xj 0 = d 3 x i j i = 0, (0.12) where we have assumed that the currents vanish at spatial infinity. So, Q d 3 xj 0 = d 3 x [ iφ(d 0 φ) iφ D 0 φ ] (0.13) is a constant of motion. 3. Under a gauge transformation, while F µν F µν + µ ν Λ ν µ Λ = F µν, (0.14) D µ φ ( µ + iea µ + ie µ Λ)(e iθ φ) = e iθ D µ φ + e iθ (i µ θ + ie µ Λ)φ. (0.15) So, in order for this transformation to be a symmetry of the Lagrangian, we require 4. Under an infinitesimal gauge transformation, we have θ(x) = eλ(x). (0.16) φ(x) e ieλ φ(x) (1 ieλ)φ(x) = δφ(x) = ieλφ(x) (0.17) = δφ (x) = ieλφ (x) (0.18) A µ (x) A µ (x) + µ Λ(x) = (x) = µ Λ(x). (0.19) The variation of the Lagrangian under such a transformation is = δφ δφ + δ( µ φ) δ µφ + (φ φ ) + + [ = µ δ( µ φ) δφ + δ( µ φ ) δφ ] δ( µ A ν ) δ µa ν (0.20) + + δ( µ A ν ) δ µa ν (0.21) = µ ( eλj µ ) + µ Λ F µν µ ν Λ (0.22) [ = eλ µ j µ + µ Λ ej µ + ] (0.23)

3 3 where j µ (x) is the conserved current defined in Eq.(0.10) and the third term in the penultimate equation vanished as the anti-symmetric tensor F µν is contracted with a symmetric tensor. Since Λ can be chosen arbitrarily, both µ j µ and the quantity in square brackets must vanish. The vanishing of the first term just tells us, as before, that j µ is conserved. We can define the gauge current as J µ (x) = ej µ (x) = e [iφ(d µ φ) iφ D µ φ]. (0.24) This tells us that the current j µ couples to the gauge field A µ and carries gauge (electric) charge e. The corresponding constant of motion is simply Q gauge = eq = e d 3 x [ iφ(d 0 φ) iφ D 0 φ ] (0.25) which has the interpretation as the total electric charge. 5. The energy momentum tensor resulting from invariance of the Lagrangian due to space-time translations is given by T µν = g µν L + δ( µ φ) δ µφ + δ( µ φ ) δ µφ + Using the results from the previous sections we find δ( µ A ν ) δ µa ν. (0.26) T µν = µν µν T EM + T M (0.27) where T µν EM = 1 4 gµν F αβ F αβ F µλ ν A λ (0.28) [ T µν M = gµν D α φ 2 + m 2 0 φ 2 + λ ] 4! φ 4 + (D µ φ) ν φ + (D µ φ) ν φ. (0.29) Note that T µν is neither symmetric nor gauge invariant. In order to remedy this, we recall that we can define an improved energy-momentum tensor (known as the Belifante energy-momentum tensor) T µν = T µν + λ K λµν where K λµν is anti-symmetric in λ and µ. If we choose K λµν = F µλ A ν and use the equations of motion we find λ F µλ = e [iφ(d µ φ) iφ (D µ φ)], (0.30) λ K λµν = F µλ λ A ν + A ν (ieφ(d µ φ) ieφ (D µ φ)). (0.31)

4 4 Some rearrangement yields T µν = T µν EM + T µν M T µν EM = 1 4 gµν F αβ F αβ + F µλ F λ ν T µν M = gµν [ (D α φ) (D α φ) + m 2 0 φ 2 + λ 4! φ 4 ] + (D µ φ) (D ν φ) + (D µ φ)(d ν φ). (0.32) Hence T µν is manifestly symmetric and gauge-invariant. In the remainder of the problem set we will make use of the Belifante energy-momentum tensor, T µν. We will refer to T µν as the canonical energy-momentum tensor to distinguish between the two when necessary. Note that because we used the equations of motion to derive the Belifante energy-momentum tensor, strictly speaking these expressions only hold at the classical level. 6. We first recall that the conjugate momenta are given by Π = δ(( t φ) ) = Dt φ, Π = The Hamiltonian density 1 is given by δ( t φ) = (Dt φ), Π Aµ = δ( t A µ ) = F µt = (0, E i ). (0.33) H = T 00 (0.34) = 1 4 F αβ F αβ + F 0λ F λ 0 + (D 0 φ) (D 0 φ) (D i φ) (D i φ) + m 2 0 φ 2 + λ 4! φ 4 (0.35) = 1 2 (E2 + B 2 ) + ΠΠ (D i φ) (D i φ) + m 2 0 φ 2 + λ 4! φ 4. (0.36) The first term is the electromagnetic energy density, the second and third terms correspond to energy costs associated with fluctuations of φ in time and space (like a kinetic energy term), and the remaining terms are potential energy terms for φ. Due to our use of a modified energy-momentum tensor, the Gauss Law term which appeared in the previous problem set does not appear in this calculation. The momentum density 1 Strictly speaking, this is the energy density, not the Hamiltonian density (although often they are given by the same formal expression).

5 5 is given by P k = T 0k (0.37) = F 0λ F λ k + (D 0 φ) (D k φ) + (D 0 φ)(d k φ) (0.38) = (E B) k + ΠD k φ + Π (D k φ). (0.39) The first term the Poynting vector is the momentum carried by the electromagnetic field while the remaining terms represent the momentum carried by the scalar field. 7. Following the derivation in the notes, we have that under a general, infinitesimal coordinate transformation, x µ x µ = x µ + δx µ = x µ x ν = g µ ν + ν δx µ. (0.40) So, to first order the Jacobian is given by J = 1 + µ δx µ + O(δx 2 ). Using the this and the equations of motion, we can write the variation of the action as [ 0 = δs = d 4 x { µ δx µ L + δ( µ φ) δφ + δ( µ φ ) δφ + ]} δ( µ A ν ) δa ν [ = d 4 x { µ g µ νl δ( µ φ) νφ δ( µ φ ) νφ ] δ( µ A ν ) µa ν + µ [ δ( µ φ) δ T φ + δ( µ φ ) δ T φ + δ( µ A ν ) δ T A ν (0.41) (0.42) ]}, (0.43) where we have defined the total changes in the fields in terms of their functional changes and changes due to the coordinate transformation: δ T φ = δφ + µ φδx µ, δ T φ = δφ + µ φ δx µ, δ T A µ = + ν A µ δx ν. (0.44) Now, under an infinitesimal Lorentz transformation, we have that x µ x µ = x µ + ω µν x ν and so δx µ = ω µν, δφ = δφ = 0, = ω µν A ν (0.45) since φ and φ are scalars while A µ transforms as a four-vector. Plugging these expressions into the variation of the action and following the steps in the notes, we obtain 0 = δs = d 4 x { µ [ T ] } µ νx ρ F µ νa ρ ω νρ. (0.46)

6 6 Since ω µν is anti-symmetric, the anti-symmetric part of the quantity in square brackets, namely, M µνρ T µν x ρ T µρ x ν + F µν A ρ F µρ A ν (0.47) must satisfy µ M µνρ = 0. (0.48) This conservation law implies, using µ T µν = 0 0 = µ M µνρ = T ρν T νρ + µ (F µν A ρ ) µ (F µρ A ν ) = T µν T νµ. (0.49) It is easy to see that both the canonical and Belifante energy-momentum tensors satisfy these equalities. Now, focusing on the spatial components, we can define L j 1 ] d 3 xɛ jkl M 0jk = d 3 x [ɛ jkl x k T 0l + (E A) j (0.50) 2 which almost looks like an angular momentum, were it not for the fact that this expression is not manifestly gauge invariant. We can instead define M µνρ = T µν x ρ T µρ x ν (0.51) which is manifestly gauge invariant and also clearly satisfies µ M µνρ = 0. We can then assign L j = 1 d 3 xɛ jkl M 0jk = d 3 xɛ jkl x k P l (0.52) 2 the meaning of a physical angular momentum. 8. As in the previous problem set, we write φ(x) = ρ(x)e iω(x), gauge fix ω(x) = 0, and take m 2 0 < 0 so that ρ acquires a non-zero vacuum expectation value ρ 0. We then expand ρ as ρ = ρ 0 + δρ where δρ represents the fluctuations of ρ about the ground state. The effective Lagrangian (as obtained in the previous problem set) is then L = 1 2 ( µρ) e2 ρ 2 A µ A µ m 2 0ρ 2 λ 4! ρ4 1 4 F µνf µν = 1 2 ( µδρ) e2 (ρ 0 + δρ) 2 A µ A µ m 2 0(ρ 0 + δρ) 2 λ 4! (ρ 0 + δρ) F µνf µν. (0.53) The conjugate momentum of δρ is therefore Π ρ = /δ( 0 δρ) = 0 ρ.

7 7 (a) Recall that we found the gauge current to be J µ (x) = ej µ (x) = e [iφ(d µ φ) iφ D µ φ]. (0.54) Plugging in ρ = ρ 0 + δρ, we find (b) Likewise, the Hamiltonian density 2 is H = 1 2 (E2 + B 2 ) + m 2 0(ρ 0 + δρ) 2 + λ 4! (ρ 0 + δρ) 4 J µ = 2e 2 A µ (ρ 0 + δρ) 2 (0.55) + ( 0 δρ) 2 + e 2 (A 0 ) 2 (ρ 0 + δρ) 2 i δρ i δρ + e 2 A i A i (ρ 0 + δρ) 2 = 1 2 (E2 + B 2 ) + m 2 0(ρ 0 + δρ) 2 + λ. (0.56) 4! (ρ 0 + δρ) 4 + Π 2 ρ + e 2 (A 0 ) 2 (ρ 0 + δρ) 2 i δρ i δρ + e 2 A i A i (ρ 0 + δρ) 2 (c) Finally, the momentum density is 9. Under a Wick rotation, P k = (E B) k + 2( 0 δρ)( k δρ) + 2e 2 A 0 A k (ρ 0 + δρ) 2 (0.57) = (E B) k + 2Π ρ ( k δρ) + 2e 2 A 0 A k (ρ 0 + δρ) 2. (0.58) x 0 ix D, 0 = x 0 i D. (0.59) As for the gauge potential, A µ, since it is a co-vector, it must transform in the same way as µ. Thus we have under a Wick rotation that A 0 ia D, D 0 id 0. (0.60) Using these relations we find that is = i dx 0 d d xl M d D xl E (0.61) where d D xl E = d D x [ φ 2 + m 2 0 φ 2 + λ 4! φ (E2 + B 2 ) (0.62) +e 2 A α A α φ 2 + ea α (iφ α φ iφ α φ)] (0.63) = d D x [ φ 2 + m 2 0 φ 2 + λ 4! φ (E2 + B 2 ) (0.64) +A α (iφ(d α φ) iφ D α φ)]. (0.65) 2 Again, strictly speaking, this is the energy density.

8 8 Here we have defined φ 2 D φ D φ + φ 2. The first three terms correspond to the energy density of a massive scalar field, the fourth term the energy density of the electromagnetic field, and the remaining terms correspond to the usual J A coupling of a conserved current to the electromagnetic field. Note that since we started with D = d + 1-dimensional Minkowski spacetime, our Wick rotated Euclidean theory has D spatial dimensions.