Solutions for Problem Set 5 - Quantum Field Theory I | PHY 396K, Assignments of Physics

Download Solutions for Problem Set 5 - Quantum Field Theory I | PHY 396K and more Assignments Physics in PDF only on Docsity! PHY–396 K. Solutions for problem set #5. Problem 1(a): Let ∆Tµν = ∂λK[λµ]ν . Regardless of the specific form of the K[λµ]ν(φ, ∂φ) tensor, its anti- symmetry with respect to its first two indices implies ∂µ∆T µν = ∂µ∂λK[λµ]ν = 0 (S.1) and hence the first eq. (3). Furthermore,∫ d3x ( ∆T 0ν = ∂iKi0ν ) = ∮ boundary of space d2Area iKi0ν −→ 0 (S.2) when the integral is taken over the whole space, hence the second eq. (3). Problem 1(b): In the Noether’s formula (1) for the stress-energy tensor, φa stand for the independent fields, however labelled. In the electromagnetic case, the independent fields are components of the 4–vector Aλ(x), hence TµνNoether(EM) = ∂L ∂(∂µAλ) ∂νAλ − gµν L = −Fµλ ∂νAλ + 14g µν FκλF κλ. (S.3) While the second term here is clearly both gauge invariant and symmetric in µ↔ ν, the first term is neither. Problem 1(c): Clearly, one can easily restore both symmetry and gauge invariance of the electromagnetic stress-energy tensor by replacing ∂νAλ in eq. (S.3) with F ν λ, hence eq. (5). The correction amounts to Tµν − TµνNoether = −F µλ (F νλ − ∂νAλ = −∂λAν)) = ∂λ ( FµλAν def = K[λµ]ν ) − Aν ( ∂λF µλ ) . (S.4) Since the free electromagnetic field satisfies ∂λF µλ = 0, the second term on the right hand side here vanishes — and the remaining correction indeed has form (2). 1 Problem 1(d): As explained in class, the Lagrangian L = −14FκλF κλ = 12(E 2 − B2). Combining this fact with eq. (5), we have the energy density H = T 00 = −F 0iF 0i − L = +E2 − 12 ( E2 −B2 ) = 12 ( E2 + B2 ) (S.5) in agreement with the standard electromagnetic formulæ (note the c = 1, rationalized units here). Likewise, the energy flux and the momentum density are Si = T i0 = T 0i = −F 0jF ij = −(−Ej)(ijkBk) = +ijkEjBk = (E×B)i, (S.6) in agreement with the Poynting vector S = E × B (again, in the c = 1, rationalized units). Finally, the (3–dimensional) stress tensor is T ijEM = −F iλF jλ − g ijL = −F i0F j0 − F ikF jk + δ ijL = −EiEj + ik`B` jkmBm + 12δ ij ( E2 −B2 ) = −EiEj − BiBj + 12δ ij ( E2 + B2 ) . (S.7) Problem 2(a): In a sense, eq. (6) follows from eq. (S.4), but it is just as easy to derive it directly from Maxwell equations. Starting with eq. (5), we immediately have ∂µ T µν EM = − ( ∂µF µλ ) F νλ − Fµλ ( ∂µF ν λ ) + 12Fκλ ( ∂νF κλ ) . (S.8) Using the antisymmetry Fµλ = −F λµ, the second term on the right hand side here becomes −Fµλ ∂µF νλ = +Fµλ ∂µF λν = +Fµλ ∂λF νµ = 12Fµλ ( ∂λF νµ + ∂µF λν ) = −12Fµλ ( ∂νFµλ ) (S.9) (using Maxwell equation ∂λF νµ + ∂µF λν + ∂νFµλ = 0) and thus precisely cancels the third term on the right hand side of eq. (S.8). Consequently, ∂µ T µν EM = − ( ∂µF µλ ) F νλ = − Jλ F νλ (S.10) by the other Maxwell equation ∂µF µλ = Jλ. Q.E .D. 2 cancellation of all terms except those containing the gauge field strength tensor Fµν . Hence, ∂µT µν mat = iqF µν (ΦDµΦ ∗ − Φ∗DµΦ) = Fµν Jν . (S.20) Q.E .D. Problem 3(a): In terms of momentum-and-polarization modes Ek,λ and Bk,λ of the electric and magnetic fields, L = ∫ d3x ( L = 12E 2 − 12B 2 ) = ∫ d3k (2π)3 ∑ λ=−1,0,+1 ( 1 2E ∗ k,λEk,λ − 12B ∗ k,λBk,λ ) . (S.21) As explained in the solutions to the previous homework, Bk,λ = ikλAk,λ and Ek,λ = −Ȧk,λ − ikδλ,0A0k . (S.22) Substituting these values into eq. (S.21) and separating the transverse modes λ = ±1 from the longitudinal modes λ = 0, we arrive at eqs. (12–14), Q.E .D. Problem 3(b): In 3D notations, gauge transforms Aµ(x)→ Aµ(x)− ∂µΛ(x) look like A(x, t) 7→ A(x, t) + ∇Λ(x, t), A0(x, t) 7→ A0(x, t) − Λ̇(x, t). (S.23) In terms of momentum-and-polarization modes, a generic gauge transform (S.23) for an arbi- trary Λ(x, t) becomes Ak,λ(t) 7→ Ak,λ(t) + ikδλ,0Λk(t), A0k(t) 7→ A0k(t) − Λ̇k(t) (S.24) for arbitrary Λk(t), hence gauge invariance of the transverse Ak,λ(t) (λ = ±1) and eqs. (15) for the longitudinal and scalar modes. 5 Problem 3(c): According to eq. (13), canonically conjugate variables to the transverse Ak,λ are Πk,λ = A ∗ k,λ = −E∗k,λ. (Mind the double-counting due to A∗k,λ = −A−k,λ.) Consequently, upon quantization, we have canonical equal-time commutation relations [Âk,λ,  † k′,λ′ ] = 0, [Êk,λ, Ê † k′,λ′ ] = 0, [Âk,λ, Ê † k′,λ′ ] = −iδλ,λ′(2π) 3δ(3)(k− k′), (S.25) where †k,λ = −Â−k,λ and likewise Ê † k,λ = −Ê−k,λ. The Hamiltonian for the transverse modes follows from the Lagrangian (13) in an obvious way; upon quantization, we have Ĥ> = ∫ d3k (2π)3 ∑ λ=±1 ( 1 2Ê † k,λÊk,λ + 1 2k 2 †k,λÂk,λ ) . (S.26) Problem 3(d): Proceeding exactly as in the previous homework about the massive vector fields, we define annihilation and creation operators âk,λ = ωkÂk,λ − iÊk,λ√ 2ωk , â†k,λ = ωk † k,λ + iÊ † k,λ√ 2ωk , (S.27) where ωk = k ≡ |k|, as appropriate for the massless particles. The bosonic commutation relations between these operators and eq. (16) for the Hamiltonian (S.26) follow in exactly the same way as for the massive vector particles; see solutions for the previous homework for details. Problem 3(e): In the transverse gauge Ak,0 ≡ 0, the longitudinal Lagrangian (14) reduces to L‖ = ∫ d3k (2π)3 k2 2 ∣∣A0k∣∣2 , (S.28) which contains no time derivatives whatsoever and thus yields a time-independent equation of “motion” for the A0k modes, namely k 2A0k = 0. 6 The electric current Jµ(x) couples to the EM fields via the Lagrangian term L ⊃ −JµAµ = J ·A− J0A0; in terms of the momentum-and-polarization modes, this interaction term is Lint = ∫ d3k (2π)3 (∑ λ Ak,λJ ∗ k,λ − A0k,λ(J0k,λ)∗ ) . (S.29) Quantizing the transverse part of this interaction gibes us the interaction Hamiltonian for the photons Ĥint = ∫ d3k (2π)3 ∑ λ=±1 ( −Âk,λĴ † k,λ ) = ∫ d3k (2π)3 −1√ 2ωk ∑ λ=±1 ( âk,λĴ † k,λ + â † k,λĴk,λ ) , (S.30) which is responsible for emission and absorption of photons by matter (via the current oper- ators Ĵk,λ = Ĵ † −k,λ). The longitudinal part of the interaction (S.29) together with the free longitudinal La- grangian (14) have particularly simple form in the transverse gauge, namely L‖ = ∫ d3k (2π)3 1 2 ( k2 ∣∣A0k∣∣2 − A0k(J0k)∗ − (A0k)∗J0k) . (S.31) Again, there are no time derivatives in this Lagrangian, hence time-independent equations of “motion” k2A0k = J 0 k, or in spacetime terms, ∇2A0(x, t) = −J0(x, t), i.e., A0(x, t) is the instantaneous Coulomb potential for the charge density J0(x′, same t). Q.E .D. Since the A0 field is non-dynamical, we may substitute the solution of its time-independent equation of “motion” back into the Lagrangian (S.31). The result is ∆LCoulomb = ∫ d3k (2π)3 − ∣∣J0k∣∣2 2k2 = ∫∫ d3xd3y −1 8π|x− y| J0(x) J0(y), (S.32) i.e., (minus) the Coulomb energy due to the electric charge density J0(x). Physically, it is responsible for the Coulomb forces between charged particles giving rise to the Jµ(x). In the quantum theory of radiation, (the quantum analogue) of ∆HCoulomb = −∆LCoulomb is considered a part of the charged particles’ Hamiltonian (e.g., the Coulomb potential for the electrons in an atom) while (S.30) is treated as a perturbation giving rise to emission, absorption and scattering of photons by matter. 7