Solutions to Problem Set 4 - Quantum Field Theory I | PHY 396K, Assignments of Physics

Download Solutions to Problem Set 4 - Quantum Field Theory I | PHY 396K and more Assignments Physics in PDF only on Docsity! PHY–396 K. Solutions for problem set #4. Problem 1(a): The conjugacy relations †k,λ = −Â−k,λ, Ê † k,λ = −Ê−k,λ follow from hermiticity of the Â(x) and Ê(x) quantum fields and from the third eq. (6) for the polarization vectors: †k,λ = ∫ d3x e+ikxeλ(k) · †(x) = ∫ d3x e−i(−k)x(−e∗λ(−k)) ·  = −Â−k,λ , (S.1) and ditto for the ʆk,λ = −Ê−k,λ. The equal-time commutation relations follow from eqs. (1): Obviously, [Âk,λ, Âk′,λ′ ] = 0, [Êk,λ, Êk′,λ′ ] = 0. (S.2) Less obviously, [Âk,λ, Ê † k′,λ′ ] = ∫ d3x e−ikx ( e∗λ(k) )i∫ d3y e+ik ′y ( eλ′(k ′) )j [Âi(x), Êj(y)] = ∫ d3x e−i(k−k ′)x ( −ie∗λ(k) · eλ′(k′) ) = −i(2π)3δ(3)(k− k′)δλ,λ′ , (S.3) or equivalently, [Âk,λ, Êk′,λ′ ] = +i(2π) 3δ(3)(k + k′)δλ,λ′ . (S.4) Problem 1(b): There are four terms in the Hamiltonian density (2), so let us consider them one by one. Combining Fourier transform with decomposition into polarization modes it is easy to see that in light of eq. (4), ∫ d3x Ê2(x) = ∫ d3k (2π)3 ∑ λ ʆk,λÊk,λ (S.5) and likewise ∫ d3x Â2(x) = ∫ d3k (2π)3 ∑ λ †k,λÂk,λ . (S.6) 1 Furthermore, using eq. (5) we have ∇× Â(x) = ∫ d3k (2π)3 ∑ λ eikx ( ik× eλ(k) = λ|k|eλ(k) ) Âk,λ (S.7) and hence ∫ d3x ( ∇× Â(x) )2 = ∫ d3k (2π)3 ∑ λ λ2k2 †k,λÂk,λ . (S.8) Finally, the first eq. (6) gives us ∇ · Ê(x) = ∫ d3k (2π)3 ∑ λ eikx ( ik · eλ(k) = i|k|δλ,0 ) Êk,λ (S.9) and hence ∫ d3x ( ∇ · Ê(x) )2 = ∫ d3k (2π)3 k2ʆk,0Êk,0 . (S.10) In light of all these formulæ, we assemble the Hamiltonian (2) as Ĥ = ∫ d3k (2π)3 ∑ λ (( 1 2 + k2 2m2 δλ,0 = Ck,λ 2 ) ʆk,λÊk,λ + ( m2 + λ2k2 2 = ω2k 2Ck,λ ) †k,λÂk,λ ) . (8) Problem 1(c): Given eqs. (S.2) and (S.4), we have [âk,λ, âk′,λ′ = −i 2 √ ωkCk′,λ′ ωk′Ck,λ ( [Âk,λ, Êk′,λ′ ] = (+i)(2π) 3δ(3)(k + k′)δλ,λ′ ) + −i 2 √ ωk′Ck,λ ωkCk′,λ′ ( [Êk,λ, Âk′,λ′ ] = (−i)(2π)3δ(3)(k + k′)δλ,λ′ ) = 12(2π) 3δ(3)(k + k′)δλ,λ′ − 12(2π) 3δ(3)(k + k′)δλ,λ′ = 0 (S.11) 2 field governed by Hamiltonian (10), âk,λ(t) = e −iωtâk,λ(0) and â † k,λ(t) = e +iωtâ†k,λ(0) where ω ≡ ωk. Substituting this time dependence into eq. (S.18) and switching to relativistic notations, we immediately arrive at Â(x) = ∫ d3k (2π)3 1√ 2ωk ∑ λ √ Ck,λ ( e−ikxeλ(k) âk,λ(0) + e +ikxe∗λ(k) â † k,λ(0) ) k0=+ωk . (11) Q.E .D. Problem 1(f): The 3–scalar field Â0(x) is governed by eqs. (3) and (S.9), Â0(x, t) = ∫ d3k (2π)3 −i|k| m2 eikx Êk,0(t). (S.19) Reversing eqs. (9) for the Êk,λ operator, we have Êk,λ = i √ ωk 2Ck,λ ( âk,λ + â † −k,λ ) . (S.20) In particular, Êk,0 = im√ 2ωk ( âk,0 + â † −k,0 ) and hence Â0(x) = ∫ d3k (2π)3 |k| m e+ikx√ 2ωk ( âk,0 + â † −k,0 ) = ∫ d3k (2π)3 |k| m e+ikx√ 2ωk âk,0 + ∫ d3k (2π)3 |k| m e−ikx√ 2ωk â†+k,0 = ∫ d3k (2π)3 1√ 2ωk |k| m ( e+ikx âk,0 + e −ikx â†k,0 ) . (S.21) As to the time dependence, it works exactly as in eq. (11) for the vector field, thus Â0(x) = ∫ d3k (2π)3 1√ 2ωk |k| m ( e−ikx âk,0(0) + e +ikx â†k,0(0) ) k0=+ωk . (S.22) Similarity between eqs. (11) and (S.22) allows us to combine them into eq. (12) where f(k, λ) = Ck,λ eλ(k) and f 0(k, λ) = |k| m δλ,0 (S.23) or equivalently, (13). 5 Problem 1(g): According to eq. (12), (∂2 +m2)µ(x) = ∫ d3k (2π)3 1√ 2ωk ∑ λ ( (−k2 +m2)e−ikxfµ(k, λ) âk,λ(0) (S.24) + (−k2 +m2)e+ikxf∗µ(k, λ) â†k,λ(0) ) k0=+ωk , which vanishes because (−k2 +m2) = 0 for k0 = ωk. Likewise, ∂µÂ µ(x) = ∫ d3k (2π)3 1√ 2ωk ∑ λ ( e−ikx ( ikµf µ(k, λ) ) âk,λ(0) (S.25) + e+ikx ( −ikµf∗µ(k, λ) ) â†k,λ(0) ) k0=+ωk , which vanishes because kµf µ(k, λ) = 0 for all polarizations λ. Q.E .D. Problem 2(a): The simplest way to prove this lemma is by direct inspection, component by component: ∑ λ f i(k, λ)f∗j(k, λ) = ∑ λ eiλ(k)e ∗j λ (k) + k2 m2 ei0(k)e ∗j 0 (k) = δ ij + kikj m2 ; ∑ λ f i(k, λ)f∗0(k, λ) = f i(k, 0)f∗0(k, 0) = kiωk m2 ; ∑ λ f0(k, λ)f∗0(k, λ) = ∣∣f0(k, 0)∣∣0 = k2 m2 = −1 + ω2k m2 . (S.26) Alternatively, we may use the fact that the three four-vectors fµ(k, λ) (fixed k, λ = −1, 0,+1) are orthogonal to each other and also to the kµ = (ωk,k). Furthermore, each ( f(k, λ) )2 = −1. Consequently, the symmetric matrix (in Lorentz indices µ, ν) on the left hand side of eq. (14) has to be (minus) the projection matrix onto four-vectors orthogonal to the kµ, and that is precisely the matrix appearing on the right hand side of eq. (14) (note k2 = m2). Problem 2(b): The operator product µ(x)Âν(y) comprises ââ, â†â†, â†â and â↠terms. The first three 6 kinds of terms have zero matrix elements between vacuum states while 〈0| âk,λâ † k′,λ′ |0〉 = (2π)3δ(3)(k− k′)δλ,λ′ . Consequently, 〈0| µ(x)Âν(y) |0〉 = ∫ d3k (2π)3 1 2ωk ∑ λ [ e−ik(x−y) fµ(k, λ)f∗ν(k, λ) ] k0=+ωk = ∫ d3k (2π)3 1 2ωk [( −gµν + k µkν m2 ) e−ik(x−y) ] k0=+ωk = ( −gµν − 1 m2 ∂ ∂xµ ∂ ∂xν )∫ d3k (2π)3 1 2ωk [ e−ik(x−y) ] k0=+ωk ≡ ( −gµν − 1 m2 ∂ ∂xµ ∂ ∂xν ) D(x− y). (15) Problem 2(c): Actually, eq. (16) is not completely correct; there is a subtlety associated with time-ordering temporal components A0 of vector fields, so the correct statement is GµνF ≡ 〈0|T ∗µ(x)Âν(y) |0〉 = ( −gµν − ∂ µ∂ν m2 ) DF (x− y) = ∫ d4k (2π)4 ( −gµν + k µkν m2 ) ie−ik(x−y) k2 −m2 + i0 (S.27) where 〈0|T∗µ(x)Âν(y) |0〉 = 〈0|Tµ(x)Âν(y) |0〉 + iδµ0δν0δ(4)(x− y). (S.28) For the explanation of the T∗ modification of the time-ordered product, please see Quantum Field Theory by Claude Itzykson and Jean–Bernard Zuber. To derive eq. (S.27), we start with eq. (15) and immediately see that for the un-modified time-ordering, 〈0|Tµ(x)Âν(y) |0〉 = θ(x0 − y0) 〈0| µ(x)Âν(y) |0〉 + θ(y0 − x0) 〈0| Âν(y)µ(x) |0〉 = θ(x0 − y0) ( −gµν − 1 m2 ∂ ∂xµ ∂ ∂xν ) D(x− y) + θ(y0 − x0) ( −gµν − 1 m2 ∂ ∂xµ ∂ ∂xν ) D(y − x) (S.29) 7 while the second condition provides for [b̂i, b̂j ] = ∑ k ( uikvjk − vikujk ) ≡ ( uv> − vu> ) ij = 0 (and hence via hermitian conjugation, [b̂†i , b̂ † j ] = 0). For the problem at hand, the transform (18) is a special case of (S.37) with block-diagonal u and v matrices where each block comprises just two modes, namely +k and −k. Given t−k = t+k, each block looks like u = ( c 0 0 c ) , v = ( 0 s s 0 ) (S.39) where c = cosh(tk) and s = sinh(tk). The blocks (S.39) clearly satisfy both conditions (S.38) (note c2 − s2 = 1), hence the entire matrices u and v satisfy (S.38) as well and the operators b̂k and b̂ † k satisfy the bosonic commutation relations. Problem 3(c): According to eqs. (18), b̂†kb̂k = cosh 2(tk) ã † kãk + sinh 2(tk) ( ã−kã † −k = ã † −kã−k + V ) + cosh(tk) sinh(tk) ( ã−kãk + ã † kã † −k ) (S.40) and therefore ( b̂†+kb̂+k + b̂ † −kb̂−k ) = cosh(2tk) ( ã†+kã+k + ã † −kã−k ) + sinh(2tk) ( ã−kã+k + ã † +kã † −k ) + 2V sinh2(tk) (S.41) where V = [ãk, ã † k] = (2π) 3δ(3)(0) = volume of the superfluid. Consequently, as long as t−k = t+k and ω−k = ω+k,∫ d3k (2π)3 ( ωk cosh(2tk) ã † kãk + 1 2ωk sinh(2tk) ( ã−kã+k + ã † +kã † −k )) = ∫ d3k (2π)3 ωk b̂ † kb̂k + const. (S.42) Let us now take another look at the ‘free’ Hamiltonian (17) and expand it in terms of the 10 ãk and ã † k modes of the shifted fields: Ĥ2 = ∫ d3k (2π)3 (( k2 2M + λn ) ã†kãk + 1 2λn ( ã−kã+k + ã † +kã † −k )) , (S.43) and therefore, in light of eq. (S.42), Ĥ2 = ∫ d3k (2π)3 ωkb̂ † kb̂k + const (19) where ωk cosh(2tk) = k2 2M + λn, ωk sinh(2tk) = λn. (S.44) Solving these equations for ωk and tk gives us ω2k = ( k2 2M + λn )2 − (λn)2 = k 2 2M ( k2 2M + 2λn ) , (S.45) cf. eq. (20), and tanh(2tk) = 2Mλn k2 + 2Mλn i.e., tk = 1 4 log k2 + 4Mλn k2 . (S.46) Aside: Note that at this level of analysis (i.e., ignoring the quasiparticle interactions following from the Ĥint), the frequencies (20) are simply the frequencies of the plane-wave solutions to the linearized classical field equations for the shifted fields. Indeed, expanding the classical Landau–Ginzburg field equations i ∂ ∂t Φ(x, t) = −∇ 2 2M Φ + λ(Φ∗Φ− n)Φ, −i ∂ ∂t Φ∗(x, t) = −∇ 2 2M Φ∗ + λ(Φ∗Φ− n)Φ, (S.47) to first order in Φ̃(x, t)Φ(x, t)− √ n, gives us i ∂ ∂t Φ̃(x, t) = −∇ 2 2M Φ̃ + λn ( Φ̃∗ + Φ̃ ) , −i ∂ ∂t Φ̃∗(x, t) = −∇ 2 2M Φ̃∗ + λn ( Φ̃∗ + Φ̃ ) , (S.48) 11 and the plane-wave solutions of these equations look like Φ̃(x, t) = A cosh(t) e+ikx−iωt + A∗ sinh(t) e−ikx+iωt, Φ̃∗(x, t) = A sinh(t) e+ikx−iωt + A∗ cosh(t) e−ikx+iωt, (S.49) where ω, t and k are related to each other exactly as in eqs. (20) and (S.46). Furthermore, in the free-shifted-fields approximation (i.e., ignoring the Ĥint), the time- dependent quantum fields expands into annihilation and creation operators according to Ψ̃(x, t) = ∫ d3k (2π)3 ( cosh(t) e+ikx−iωt b̂k(0) + sinh(t) e −ikx+iωt b̂†k(0) ) , Ψ̃†(x, t) = ∫ d3k (2π)3 ( sinh(t) e+ikx−iωt b̂k(0) + cosh(t) e −ikx+iωt b̂†k(0) ) , (S.50) where ω = ωk and t = tk, exactly as in classical solutions (S.49). Indeed, the free quantum fields (S.50) are precisely linear combinations of the free classical plane-wave solutions with operatorial coefficients — exactly as the free relativistic quantum fields we have studied in class and the relativistic vector fields discussed in problem 1 of this homework, cf eq. (12). Moreover, for both relativistic and non-relativistic quantum fields, the positive-frequency solutions (e−iωt with ω > 0) are accompanied by the annihilation operators (âk,λ for the vector fields or b̂k for Helium) while the negative-frequency solutions (e+iωt) are accompanied by the creation operators. These rules are universal in Quantum Field Theory, relativistic or otherwise. The only difference is the extra factor 1/ √ 2ωk,λ in the relativistic theories. Optional problem 3(*): First, note that the operator F̂ is anti-hermitian, F̂ † = −F̂ , hence eF̂ is a unitary operator, which means that the operator transform ãk → b̂k = e F̂ ãke −F̂ , ã†k → b̂ † k = e F̂ ã†ke −F̂ (S.51) is indeed consistent with the hermitian conjugation. The operators b̂k and b̂ † k defined by eqs. (S.51) obviously satisfy the same bosonic commutation relations as the ãk and ã † k op- erators, and just as obviously, the operators b̂k annihilate the ground state (quasi-particle vacuum) |Ω2〉 = eF̂ |coh〉 simply because the ãk operators annihilate the coherent state |coh〉. What is not so obvious is that eqs. (S.51) for the b̂k and b̂ † k operators agree with eqs. (18). 12