Solutions to Problem Set #2 in Classical Electrodynamics, Assignments of Physics

Download Solutions to Problem Set #2 in Classical Electrodynamics and more Assignments Physics in PDF only on Docsity! PHY–396 K. Solutions for problem set #2. Problem 1(a): As discussed in class for the massless case (EM), ∂L/∂(∂µAν) = −Fµν . Clearly, ∂L/∂(Aν) = +m2Aν − Jν . Hence, the Euler–Lagrange field equation is −∂µ ∂L ∂(∂µAν) + ∂L ∂(Aν) ≡ ∂µFµν + m2Aν − Jν = 0, (S.1) or in terms of Aν and their explicit derivatives, ∂2Aν − ∂ν(∂µAµ) + m2Aν − Jν = 0. (S.2) Problem 1(b): Take the divergence ∂ν of the field equation (S.2); the first two terms cancel out while the rest becomes m2 ∂νA ν − ∂νJν = 0. (S.3) In the massless case, this equation enforces the current conservation ∂νJ ν = 0 regardless of the 4–vector potential Aν(x), but there is no such constraint in the massive case at hand. Instead, eq. (S.3) simply relates the current divergence to the 4–potential divergence. In particular, if the current happens to satisfy ∂νJ ν , then — and only then — eq. (S.3) requires ∂νA ν = 0 as well. Consequently, the field equation (S.2) simplifies to (∂2 + m2)Aν = Jν . Q.E .D. Problem 1(c): As in (a), ∂L/∂(∂µAν) = −Fµν . In particular, ∂L/∂(∂0Ai) = −F0i = −Ei while ∂L/∂(∂0A0) = −F00 = 0. 1 Problem 1(d): In terms of the Hamiltonian and Lagrangian densities, eq. (3) means H = −Ȧ · E − L. (3′) In terms of A, E and A0, Ȧ = −E − ∇A0, −Ȧ · E = E2 + E · ∇A0, L = 12 ( E2 − (∇×A)2 ) + 12m 2(A20 −A2) − (A0J0 −A · J). (S.4) Consequently, H = 12E 2 + E · ∇A0 − 12m 2A20 + A0J0 + 1 2(∇×A) 2 + 12m 2A2 − A · J, (S.5) which immediately leads to eq. (4) (after one integration by parts). Q.E .D. Problem 1(e): Expanding the left hand side of eq. (5) gives us δH δA0(x) ≡ ∂H ∂(A0) − ∇ i ∂H ∂(∇ i A0) = −m2A0 + J0 − ∇iE i, hence A0(x, t) = J0 − ∇ · E m2 (S.6) everywhere in spacetime. Please note that there are no derivatives at the left hand side of this formula. 2 Problem 2(b): According to eqs. (8), at equal times [B̂i, 12B̂ 2(x′)] = 0 while [B̂i, 12Ê 2(x′)] = Êj(x′)× [B̂i(x), Êj(x′)] = Êj(x′)× (+ih̄c)jik ∂ ∂x′k δ(3)(x′ − x), (S.18) hence [B̂i(x), Ĥ] = ∫ d3x′ (+ih̄c)jik Êj(x′)× ∂ ∂x′k δ(3)(x′ − x) = −ih̄c jik ∇ k Êj(x), (S.19) or in vector notations, [B̂(x), Ĥ] = −ih̄c∇× Ê. (S.20) Consequently, in the Heisenberg picture 1 c ∂ ∂t B̂(x, t) = −∇× Ê(x, t). (9.a) Likewise, [Êi, 12Ê 2(x′)] = 0 while [Êi, 12B̂ 2(x′)] = B̂j(x′)× [Êi(x), B̂j(x′)] = B̂j(x′)× (−ih̄c)ijk ∂ ∂xk δ(3)(x′ − x), (S.21) hence [Êi(x), Ĥ] = ∫ d3x′ (−ih̄c)ijk B̂j(x′)× ∂ ∂xk δ(3)(x′ − x) = −ih̄c ijk ∇ k B̂j(x), (S.22) or in vector notations, [Ê(x), Ĥ] = +ih̄c∇× B̂. (S.23) Consequently, in the Heisenberg picture 1 c ∂ ∂t B̂(x, t) = +∇× Ê(x, t). (9.b) Q.E .D. 5 Problem 3(a): In the Hamiltonian formalism for the classical fields Φ(x) and Φ∗(x), the canonical conjugate fields are Π(x) = ∂L ∂(∂0Φ∗) = ∂0Φ(x) and Π ∗(x) = ∂L ∂(∂0Φ) = ∂0Φ ∗(x). (S.24) The canonical conjugation implies canonical Poisson brackets between the classical fields Φ(x) and Π∗(x), and likewise Φ∗(x) and Π(x) and hence the canonical commutation relation between their quantum counterparts: In the Schrödinger picture [Φ̂(x), Φ̂(x′)] = [Φ̂(x), Φ̂†(x′)] = [Φ̂†(x), Φ̂†(x′)] = 0, [Π̂(x), Π̂(x′)] = [Π̂(x), Π̂†(x′)] = [Π̂†(x), Π̂†(x′)] = 0, [Φ̂(x), Π̂(x′)] = [Φ̂†(x), Π̂†(x′)] = 0, (S.25) [Φ̂(x), Π̂†(x′)] = [Φ̂†(x), Π̂(x′)] = 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∂0Φ + Π∗∂0Φ∗ − L = Π∗Π + ∇Φ∗ · ∇Φ + m2 Φ∗Φ, (S.26) so the quantum theory’s Hamiltonian is obviously as in eq. (13). Q.E .D. Problem 3(b): Fourier transforming the canonical commutation relations (S.25) results in [Φ̂p, Φ̂p′ ] = [Φ̂p, Φ̂ † p′ ] = [Φ̂ † p, Φ̂ † p′ ] = 0, [Π̂p, Π̂p′ ] = [Π̂p, Π̂ † p′ ] = [Π̂ † p, Π̂ † p′ ] = 0, [Φ̂p, Π̂p′ ] = [Φ̂ † p, Π̂ † p′ ] = 0, [Φ̂p, Π̂ † p′ ] = [Φ̂ † p, Π̂p′ ] δp,p′ . (S.27) 6 Consequently, [âp, âp′ ] = [âp, b̂ † p′ ] = [b̂ † p, b̂ † p′ ] = 0 (S.28) because all the Φ̂p and all the Π̂p operators commute with each other, and likewise Similarly, [b̂p, b̂p′ ] = [b̂p, â † p′ ] = [â † p, â † p′ ] = 0 (S.29) because all the Φ̂†p and all Π̂ † p operators commute with each other too. Less obviously [âp, b̂p′ ] = 1 2 √ EpEp′ ( EpEp′(0) + iEp(iδp,−p′) + iEp′(−iδp,−p′) − (0) ) = δp,−p′ × Ep′ − Ep 2 √ EpEp′ = 0, (S.30) and similarly [â†p, b̂ † p′ ] = 0. Finally, [âp, â † p′ ] = [b̂p, b̂ † p′ ] = 1 2 √ EpEp′ (EpEp′(0) − iEp(iδp,p′) + iEp′(−iδp,p′) + (0)) = δp,p′ × Ep′ + Ep 2 √ EpEp′ = δp,p′ . (S.31) Q.E .D. Problem 3(c): First, Fourier-transforming the free Hamiltonian (13) gives us Ĥfree = ∑ p ( Π̂†pΠ̂p + E 2 p Φ̂ † pΦ̂p ) . (S.32) Second, we reverse the definitions (15) to obtain Φ̂p = 1√ 2Ep ( âp + b̂ † −p ) and Π̂p = √ 2Ep 2i ( b̂p − â † −p ) . (S.33) 7