Solutions to Problem Set #3 in Quantum Mechanics and Quantum Field Theory, Assignments of Physics

Download Solutions to Problem Set #3 in Quantum Mechanics and Quantum Field Theory and more Assignments Physics in PDF only on Docsity! PHY–396 K. Solutions for problem set #3. Problem 1(a): Use product-of-exponentials formula ∀Â, B̂ : eÂeB̂ = exp ( Â+ B̂ + 12 [Â, B̂] + 1 12 [(Â− B̂), [Â, B̂]] + · · · ) . (S.1) In particular, eαâ † e−α ∗â = exp ( α↠− α∗â + 12αα ∗ + 0 ) , hence |α〉 def= eαâ †−α∗â |0〉 = e−|α| 2/2eαâ † e−α ∗â |0〉 = e−|α| 2/2eαâ † |0〉 . (S.2) (Note e−α ∗â |0〉 = |0〉 since â |0〉 = 0.) Next, [â, â†] = 1 implies that for any function f(â†), [â, f(â†)] = f ′(â†). In particular, [â, eαâ † ] = αeαâ † or in other words, (â−α)eα↠= eα↠â and hence (â−α) |α〉 ∝ eα↠â |0〉 = 0. Q.E .D. Problem 1(b): For any normal-ordered product of creation and annihilation operators — i.e., a product in which all creation operators are to the right of all annihilation operators — one has 〈α| (â†)k(â)` |α〉 = (α∗)kα`, simply because â |α〉 = α |α〉 and 〈α| ↠= α∗ 〈α|. In particu- lar, 〈α| (n̂ = â†â) |α〉 = α∗α. On the other hand, n̂2 = â†ââ†â = â†â†ââ + â†â =⇒ 〈α| n̂2 |α〉 = (α∗)2α2 + α∗α = n̄2 + n̄ (S.3) hence ∆n = √ 〈n̂2〉 − n̄2 = √ n̄. In a similar manner, q̂ = √ h̄ 2mω (â + â†), q̂2 = h̄ 2mω ( (â)2 + (â†)2 + 2â†â + 1 ) =⇒ 1 〈α| q̂2 |α〉 = h̄ 2mω ( (α + α∗)2 + 1 ) = 〈α| q̂ |α〉2 + h̄ 2mω and likewise 〈α| p̂2 |α〉 = mωh̄ 2 ( (−iα + iα∗)2 + 1 ) = 〈α| p̂ |α〉2 + mωh̄ 2 , thus ∆q = √ h̄ 2mω , ∆p = √ mωh̄ 2 , ∆q∆p = h̄ 2 . (S.4) Q.E .D. Problem 1(c): In the Schrödinger picture, ↠is time independent, hence (d/dt)eαâ † = (dα/dt)â†eαâ † . Using time independence of the magnitude |α|, we then have d dt ( |α〉 = e−|α| 2/2eαâ † |0〉 ) = dα dt ↠|α〉 = 1 α dα dt â†â |α〉 . (S.5) Specifically, for α = α0e −iωt, (d/dt) |α〉 = −iωâ†âα and consequently, for the state vector (2), ih̄ d dt |ψ〉 = h̄ωâ†â |ψ〉 + 12 h̄ωψ = Ĥ |ψ〉 . (S.6) Q.E .D. Problem 1(d): First, quantum overlap with the ground state, 〈0|α〉 = e−|α|2/2 〈0| eα↠|0〉 = e−|α|2/2 because 〈0| eα↠= 〈0| (note 〈0| ↠= 0). Now, for two different coherent states, 〈α|beta〉 = e−|α| 2/2 〈0| eα ∗â |β〉 = e−|α| 2/2 〈0| eα ∗β |β〉 = e−|α| 2/2 eα ∗β e−|β| 2/2 or in terms of the probability overlap, |〈α|β〉|2 = e−|α−β| 2 . (S.7) 2 Problem 2(a): Classical equation of motion for a complex field follow from zero variational derivative of the action with respect to both real and imaginary parts if δΦ(x). Equivalently, we treat Φ(x) and Φ∗(x) as independent, thus ∂L ∂(∂µΦ) = ∂µΦ∗, ∂L ∂(∂µΦ∗) = ∂µΦ, ∂L ∂(Φ) = −m2Φ∗−λ2 (Φ ∗)2Φ, ∂L ∂(Φ∗) = −m2Φ−λ2 Φ ∗Φ2, and the Euler–Lagrange field equations ∂2Φ∗ +m2Φ∗ + λ2 (Φ ∗)2Φ = 0, ∂2Φ +m2Φ + λ2 Φ ∗Φ2 = 0. (S.16) Consequently, the divergence of the current (7) evaluates to ∂µJ µ = i∂µ ( Φ∗(∂µΦ)− (∂µΦ∗)Φ ) = iΦ∗(∂2Φ) − i(∂2Φ∗)Φ = −iΦ∗(m2Φ + λ2 Φ ∗Φ2) + i(m2Φ∗ + λ2 (Φ ∗)2Φ)Φ = 0. (S.17) Q.E .D. Problem 2(b): 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.18) The canonical conjugation implies canonical Poisson brackets between the classical fields Φ, Φ∗, Π and Π∗ and hence the canonical commutation relation between their quantum counter- parts: [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.19) 5 [Φ̂(x), Π̂(x′)] = [Φ̂†(x), Π̂†(x′)] = iδ(3)(x− x′). (In the Heisenberg picture for the quantum fields, these commutation relations hold for equal times t = t′ only.) Classically, the Hamiltonian density is H = Π∂0Φ + Π∗∂0Φ∗ − L = Π∗Π + ∇Φ∗ · ∇Φ + m2 Φ∗Φ + 14λ ( Φ∗Φ )2 , (S.20) so the quantum theory’s Hamiltonian is obviously (8) (modulo operator-ordering ambiguity). Q.E .D. Problem 2(c): Fourier transforming the canonical commutation relations (S.19) results in [Φ̂p, Φ̂p′ ] = [Φ̂p, Φ̂ † p′ ] = [Φ̂ † p, Φ̂ † p′ ] = 0, [Π̂p, Π̂p′ ] = [Π̂p, Π̂ † p′ ] = [Π̂ † p, Π̂ † p′ ] = 0, [Φ̂p, Π̂ † p′ ] = [Φ̂ † p, Π̂p′ ] = 0, (S.21) [Φ̂p, Π̂p′ ] = [Φ̂ † p, Π̂ † p′ ] = (2π) 3iδ(3)(p + p′). Consequenly, [âp, âp′ ] = [âp, b̂ † p′ ] = [b̂ † p, b̂ † p′ ] = 0 (S.22) because all Φp and all Π † p commute with each other. Similarly, [b̂p, b̂p′ ] = [b̂p, â † p′ ] = [â † p, â † p′ ] = 0 (S.23) because all Φ†p and all Πp commute with each other too. Less obviously [âp, b̂p′ ] = i 2 √ Ep Ep′ (2π)3iδ(3)(p + p′) + i2 √ Ep′ Ep (2π)3(−i)δ(3)(−p− p′) = 0 (S.24) 6 and likewise [â†p, b̂ † p′ ] = 0. Finally, [âp, â † p′ ] = [b̂p, b̂ † p′ ] = −i 2 √ Ep Ep′ (2π)3iδ(3)(p− p′) + i2 √ Ep′ Ep (2π)3(−i)δ(3)(p′ − p) = (2π)3δ(3)(p− p′). (S.25) Q.E .D. Problem 2(d): First, Fourier-transforming the free Hamiltonian gives us Ĥfree = ∫ d3p (2π)3 ( Π̂†pΠ̂p + E 2 p Φ̂ † pΦ̂p ) . (S.26) Second, we reverse the definitions (9) to obtain Φ̂p = âp + b̂ † −p√ 2Ep and Π̂p = √ 1 2Ep ( −ib̂p + iâ † −p ) . (S.27) Third, we calculate E2p Φ̂ † pΦ̂p + Π̂−pΠ̂ † −p = 1 2Ep ( â†p + b̂−p ) ( âp + b̂ † −p ) + 12Ep ( iâ†p − ib̂−p ) ( −iâp + ib̂ † −p ) = Ep ( â†pâp + b̂−pb̂ † −p ) . (S.28) Finally, we put eqs. (S.28) and (S.26) together and derive Ĥfree = ∫ d3p (2π)3 ( E2p Φ̂ † pΦ̂p + Π̂−pΠ̂ † −p ) = ∫ d3p (2π)3 Ep ( â†pâp + b̂−pb̂ † −p ) = ∫ d3p (2π)3 Ep ( â†pâp + b̂ † pb̂p ) + const. (S.29) Q.E .D. 7