1 Electromagnetic Radiation and Matter 1.1 HAMILTONIAN AND VECTOR POTENTIAL The classical Hamiltonian describing the interaction of a particle with mass mand charge e with an electromagnetic field with vector potential A and scalar potentials H=2m(p-cA)+ (1.1) where p is the momentum of the particle.For example,A could be the vector potential of an external magnetic field while could be the Coulomb potential due to the presence of another charge.We do follow the 。 or the absolute value our case e= by casting the Lorentz force in the Hamiltonian formalism.It can also be obtained from the Hamiltonian of a free particle 发 (1.2) with the substitutions p→p-eA H→H-eφ (1.3) This is called the 'Minimal Substitution'.The substitutions(1.3)can be written in covariant form as pu→pu-eAu (1.4) ycHan P.Par
1 Electromagnetic Radiation and Matter 1.1 HAMILTONIAN AND VECTOR POTENTIAL The classical Hamiltonian describing the interaction of a particle with mass m and charge e with an electromagnetic field with vector potential A and scalar potential φ is H = 1 2m (p − eA) 2 + eφ (1.1) where p is the momentum of the particle. For example, A could be the vector potential of an external magnetic field while φ could be the Coulomb potential due to the presence of another charge. We do not follow the convention where e stands for the absolute value of the electron charge; in our case e = −1.602 × 10−19 C for an electron. This Hamiltonian is derived by casting the Lorentz force in the Hamiltonian formalism. It can also be obtained from the Hamiltonian of a free particle H = p2 2m (1.2) with the substitutions p → p − eA H → H − eφ (1.3) This is called the ‘Minimal Substitution’. The substitutions (1.3) can be written in covariant form as pµ → pµ − eAµ (1.4) An Introduction to Advanced Quantum Physics Hans P. Paar 2010 John Wiley & Sons, Ltd
4 ELECTROMAGNETIC RADIATION AND MATTER with Pu=(p,iE)and A=(A,i),c=1.The word 'minimal'indic cate that no additional terms that would in principle be allowed are included (experiment is the arbiter).An example would be a term that accounts for the intrinsic magnetic moment of the particle.As is known from the study of etic field.the Hamilt ian Fo a n nagnetic me ment a sociated momentum.To obtain the Hamiltonian of the quantized system we use as always the replacements p→ Ha固+-品 (1.5) or in covariant form 18 (1.6 The requirement that the substitutions in Equation(1.5)be covariant,that is,that they can be written in the form of Equation (1.6),explains the minus sign in Equation(1.5).The substitutions are a manifestation of 'First Quantization'in which momenta and energies,and functions dependent upon these,become operators.They lead to the commutation relations (1.7刀 where is the Kronecker delta. The stion a ses sas to how to We know from the s of p lanck and tein in the treatment of black body radiation and the photo-electric effect respectively that electromagnetic energy is quantized with quanta equal ho. The replacements Equation(1.6)are of no use for an explanation.We will address this issue fully in the next subsection when we introduce'Second which the vector and scaar poten ntials and thus the elec and magn fields become operators.This is a prototype of relativistic In preparation for quantization of the electromagnetic field,we will briefly review the arguments that lead to the wave equation for A with =0(the Coulomb Gauge).Recall from classical electromagnetism that when the vector and scalar r potentials are transformed into new ones by the Gauge transformation A→A'=A+VX 中→=0-船 (1.8)
4 ELECTROMAGNETIC RADIATION AND MATTER with pµ = (p, iE) and Aµ = (A, iφ), c = 1. The word ‘minimal’ indicates that no additional terms that would in principle be allowed are included (experiment is the arbiter). An example would be a term that accounts for the intrinsic magnetic moment of the particle. As is known from the study of an atom in an external static magnetic field, the Hamiltonian Equation (1.1) already accounts for a magnetic moment associated with orbital angular momentum. To obtain the Hamiltonian of the quantized system we use as always the replacements p → 1 i ∇ H (or E) → −1 i ∂ ∂t (1.5) or in covariant form pµ → 1 i ∂ ∂xµ = 1 i ∂µ (1.6) The requirement that the substitutions in Equation (1.5) be covariant, that is, that they can be written in the form of Equation (1.6), explains the minus sign in Equation (1.5). The substitutions are a manifestation of ‘First Quantization’ in which momenta and energies, and functions dependent upon these, become operators. They lead to the commutation relations [pµ, xν ] = 1 i δµν (1.7) where δµν is the Kronecker delta. The question arises as to how to quantize the electromagnetic field. We know from the hypothesis of Planck and its extension by Einstein in the treatment of black body radiation and the photo-electric effect respectively that electromagnetic energy is quantized with quanta equal ω. The replacements Equation (1.6) are of no use for an explanation. We will address this issue fully in the next subsection when we introduce ‘Second Quantization’ in which the vector and scalar potentials and thus the electric and magnetic fields become operators. This is a prototype of relativistic quantum field theory. In preparation for quantization of the electromagnetic field, we will briefly review the arguments that lead to the wave equation for A with φ = 0 (the Coulomb Gauge). Recall from classical electromagnetism that when the vector and scalar potentials are transformed into new ones by the Gauge transformation A → A = A + ∇χ φ → φ = φ − ∂χ ∂t (1.8)
HAMILTONIAN AND VECTOR POTENTIAL 5 that the values of electric and magnetic fields Eand Bdo not change .This is so because under the Gauge transformation in Equation(1.8)we have E=-- →E=--A =E (1.9) at at and B=V×A→B=V×A=A (1.10) where we used in Equation (1.9)that the order of V and a/at can be exchanged while we used in Equation (1.10)x.The function x is arbitrary (uncor strained). ssign in Equatic 118) necessary for emain unchanged und the Gauge tra The minus sign also follows if we require that the relations Equation(1.8)can be written in a covariant form Au→A=Au+uX (1.11) Writing compone nt of A=(A,i)in Equation (1.11), on Fnds th eonrelation of Equation (1..One c transformation to find new A'and 'such that they satisfy the Lorentz condition A+==0 at (1.12) inhomogeneous wave equation 2x 2x一tr =-f(x,t) (1.13) The Lorentz condition is seen to be covariant as well.The Lorentz condition simplifies the differential equations for the vector and scalar potentials to 92A 72A- =-4xj -器 =-4rp (1.14) or uduAa =32Aa=0 (1.15) showing that the ave equations for the potentials are covariant as they should be.This is the Lorentz Gauge
HAMILTONIAN AND VECTOR POTENTIAL 5 that the values of electric and magnetic fields E and B do not change. This is so because under the Gauge transformation in Equation (1.8) we have E = −∇φ − ∂A ∂t → E = −∇φ − ∂A ∂t = E (1.9) and B = ∇ × A → B = ∇ × A = A (1.10) where we used in Equation (1.9) that the order of ∇ and ∂/∂t can be exchanged while we used in Equation (1.10) that ∇ × ∇χ = 0 for all χ. The function χ is arbitrary (unconstrained). The minus sign in Equation (1.8) is necessary for E to remain unchanged under the Gauge transformation. The minus sign also follows if we require that the relations Equation (1.8) can be written in a covariant form Aµ → A µ = Aµ + ∂µχ (1.11) Writing out the fourth component of Aµ = (A, iφ) in Equation (1.11), one finds the second relation of Equation (1.8). One can make a Gauge transformation to find new A and φ such that they satisfy the Lorentz condition ∇ · A + ∂φ ∂t = ∂µA µ = 0 (1.12) as follows. If ∇ · A + ∂φ/∂t = f(x, t) = 0 then the new A and φ will satisfy the Lorentz condition in Equation (1.12) if χ is required to satisfy the inhomogeneous wave equation ∇2χ − ∂2χ ∂t2 = −f(x, t) (1.13) The Lorentz condition is seen to be covariant as well. The Lorentz condition simplifies the differential equations for the vector and scalar potentials to ∇2A − ∂2A ∂t2 = −4πj ∇2φ − ∂2φ ∂t2 = −4πρ (1.14) or ∂µ∂µAα = ∂2Aα = 0 (1.15) showing that the wave equations for the potentials are covariant as they should be. This is the Lorentz Gauge.
6 ELECTROMAGNETIC RADIATION AND MATTER There is more freedom left in the choice of A and in that a further Gauge transformation as in Equation(1.11)can be made that results in the scalar potential being zero if we require that the new x satisfies 中at and 2x-ar 0-x =0 (1.16) compared with Equation(1.13).This is the Coulomb or Radiation Gauge. The electric and magnetic fields are now given by E=- aA B=V×A (1.17 at 】 n(1.10).The Lorentz condition in Equation(1.12)in the Coulomb Gauge is V.A=0 (1.18) We introduce the wave vector k and the corresponding angular frequency with k2=@2.It follows from Equation(1.18)for a traveling wave of the form A(x,t)=Aoexp(ik.x-@t)that k.A=0 (1.19) so k and A are perpendicular to each other.From the first relation in Equation (1.17)we find that E=iA,so E and A are parallel and thus k and E are perpe ndicular.The second relatic Equation (1.17) gives B=ik x A=k x E/@.This relation shows that B is in phase with E and is perpendicular to both k and E,so all three vectors E,B,k are mutually perpendicular.They form a right-handed triplet in that order because E x B=E x (k x E)/@=E2k/@where we used that E.k=0.We define the polarization of the electron agnetic field as the direction of the icfield.This is sc use the effect of the electri 山o of the magnetic in the exposure of photographi film.Because the electric field is perpendicular to its momentum we say that the electromagnetic field is transversely polarized.The cross product E x B equals the Poynting vector S,which is in the direction of k as it should be. Because Gauge transformations do not change the physical properties of the electro etic field the last conclusion about the ofE B nd k holds as well in the Lorentz Gauge and in deed i neral.N a by trans sforming away we have not removed a Coulomb potential th might be present,we only removed the scalar potential associated with the electromagnetic field described by the coupled E and B. The reader is urged to review this material from the text used in clas- sical electromagnetism.A problem about Gauge transformations and the Coulomb Gauge is provided at the end of this ch pter
6 ELECTROMAGNETIC RADIATION AND MATTER There is more freedom left in the choice of A and φ in that a further Gauge transformation as in Equation (1.11) can be made that results in the scalar potential φ being zero if we require that the new χ satisfies φ = ∂χ ∂t and ∇2χ − ∂2χ ∂t2 = 0 (1.16) compared with Equation (1.13). This is the Coulomb or Radiation Gauge. The electric and magnetic fields are now given by E = −∂A ∂t B = ∇ × A (1.17) compared with the relations for E and B used in Equation (1.9) and Equation (1.10). The Lorentz condition in Equation (1.12) in the Coulomb Gauge is ∇ · A = 0 (1.18) We introduce the wave vector k and the corresponding angular frequency ω with k2 = ω2. It follows from Equation (1.18) for a traveling wave of the form A(x, t) = A0exp(ik · x − ωt) that k · A = 0 (1.19) so k and A are perpendicular to each other. From the first relation in Equation (1.17) we find that E = iωA, so E and A are parallel and thus k and E are also perpendicular. The second relation in Equation (1.17) gives B = ik × A = k × E/ω. This relation shows that B is in phase with E and is perpendicular to both k and E, so all three vectors E, B, k are mutually perpendicular. They form a right-handed triplet in that order because E × B = E × (k × E)/ω = E2k/ω where we used that E · k = 0. We define the polarization of the electromagnetic field as the direction of the electric field. This is so because the effects of the electric field dominate those of the magnetic field, for example in the exposure of photographic film. Because the electric field is perpendicular to its momentum we say that the electromagnetic field is transversely polarized. The cross product E × B equals the Poynting vector S, which is in the direction of k as it should be. Because Gauge transformations do not change the physical properties of the electromagnetic field, the last conclusion about the orientation of E, B and k holds as well in the Lorentz Gauge and indeed in general. Note that by transforming away φ we have not removed a Coulomb potential that might be present, we only removed the scalar potential associated with the electromagnetic field described by the coupled E and B. The reader is urged to review this material from the text used in classical electromagnetism. A problem about Gauge transformations and the Coulomb Gauge is provided at the end of this chapter
HAMILTONIAN AND VECTOR POTENTIAL 7 We stress that the sequence of two Lorentz transformations has left the and magneticfields unchanged,so the physical propertie of the system have not been affected.This is in analogy with elementary classical mechanics where it is shown that the potential energy U(r)is defined up to a constant because adding a constant to U does not change the force F=-VU and does not change the physical properties of the system. The total Hamiltonian of the s consisting th no other charges and currents but the ones atedice n auation (.D n f Equation(1.1)and the energy of the electromagnetic field 1 Em8元 dx(E2+B2) (1.20) Because E or B may be complex we use absolute values in the integrand. The homogeneous wave Equation(1.14)for A becomes V2A-A =0 t2 (1.21) This equation describes free'electromagnetic fields,that is,fields in the absence of currents and charges.We now seek solutions of the homogen wave Equation (1.21).We in roduce the four-vector k is the wave vector and w the corresponding angular frequency.With x=(x,ict)we find that kx=k.x-wt.Therefore we can write a traveling wave A(x,t)=Aoexp(ik.x-@t)as A(x)=Aoexp(ikx).Traveling waves mem。ndmon e forcd the om including a factor (k)where8 is the Dirac function.The solution mus represent a three-dimensional vector so we make use of three mutually perpendicular unit vectors e(k)(=1,2,3).The Lorentz condition in the Coulomb Gauge Equation(1.19)requires that k and A are perpendicular to each other.If we choose e and e2 perpendicular to k (and to each other) and e3 parallel to k,such that e,e 3 form a right-handed set of unit vectors,terms proportio absent because of Eq ation(1.19の Because the wave Equation (1.2 1)is linear,its most general solution linear superposition of terms of the form a(k,@)exp [i(kx-t)e(k)(k2 =1,2.Our notation shows that the pre-factor ax(k,@depends upon k and and that the unit vectors s(k)depend upon(the direction of)k.The most general solution can thus be written as A=()广∫do2aeek (1.22) =1
HAMILTONIAN AND VECTOR POTENTIAL 7 We stress that the sequence of two Lorentz transformations has left the electric and magnetic fields unchanged, so the physical properties of the system have not been affected. This is in analogy with elementary classical mechanics where it is shown that the potential energy U(r) is defined up to a constant because adding a constant to U does not change the force F = −∇U and does not change the physical properties of the system. The total Hamiltonian of the system consisting of a charged particle in an electromagnetic field with no other charges and currents but the ones associated with the particle in Equation (1.1) consists of the part given in Equation (1.1) and the energy of the electromagnetic field Eem = 1 8π d3x(|E| 2 + |B| 2) (1.20) Because E or B may be complex we use absolute values in the integrand. The homogeneous wave Equation (1.14) for A becomes ∇2A − ∂2A ∂t2 = 0 (1.21) This equation describes ‘free’ electromagnetic fields, that is, fields in the absence of currents and charges. We now seek solutions of the homogeneous wave Equation (1.21). We introduce the four-vector kµ = (k, iω) where k is the wave vector and ω the corresponding angular frequency. With xµ = (x, ict) we find that kx = k · x − ωt. Therefore we can write a traveling wave A(x, t) = A0exp(ik · x − ωt) as A(x) = A0exp(ikx). Traveling waves with the vector potential of this form satisfy the wave equation provided that k2 − ω2 = k2 = 0. This condition can be enforced in the solution by including a factor δ(k2) where δ is the Dirac δ function. The solution must represent a three-dimensional vector so we make use of three mutually perpendicular unit vectors ελ(k) (λ = 1, 2, 3). The Lorentz condition in the Coulomb Gauge Equation (1.19) requires that k and A are perpendicular to each other. If we choose ε1 and ε2 perpendicular to k (and to each other) and ε3 parallel to k, such that ε1, ε2, ε3 form a right-handed set of unit vectors, terms proportional to ε3 must be absent because of Equation (1.19). Because the wave Equation (1.21) is linear, its most general solution is a linear superposition of terms of the form aλ(k, ω) exp [i(kx − ωt)]ελ(k)δ(k2) λ = 1, 2. Our notation shows that the pre-factor aλ(k, ω) depends upon k and ω and that the unit vectors ελ(k) depend upon (the direction of) k. The most general solution can thus be written as A(x, t) = 1 2π 2 d3k dω 2 λ=1 aλ(k) ei(k·x−ωt) ελ(k) δ(k2) (1.22)