SECOND QUANTIZATION 3 use Equation (1.41)for A.When taking e=ie (1.42) Vx【e±ire(k=土kx[eire,kl (1.43) Therefore E=否工受an侧-侧 (1.44) B=iVV 内-e(.4的 1 because in the expression for A the same is true and E and B are derived from A.Classically the E2 and B2 in Equation (1.20)contribute equally to the energy.We might therefore expect that to be the case here too,so only one of two terms would need to be ctati s not orrectas will be shown below by calculating both terms the simpler E2term using Equation (1.44) IE即=EE=g∑kuV受[(k)exex(k)-a威ke-ek ∑r√学[a或ke-re克k)-ake-ek] (1.46 When working out this expression we get four terms,each summed over kThere are two types of exponentials among the four terms:those with the difference and th E2= ,Vakd或ke-e,k-ik的+i 4π 2V. -0∑，w[a.(kw.(k)e内：erk+h 2Vax (1.47) The symbol .indicates the Hermitian conjugate of the previous term but with the proviso,indicated by the line over h.c.,that in a product of operators their original order is preserved.Thus the h.c.of ax(k)a,(k')is
SECOND QUANTIZATION 13 from Equation (1.17) where we use Equation (1.41) for A. When taking derivatives of A we will use the relations ∂ ∂t e±ikx = ∓iωke±ikx (1.42) ∇ × [e±ikxελ(k)] = ±ik × [e±ikxελ(k)] (1.43) Therefore E = i 4π V k,λ ωk 2 aλ(k)eikxελ(k) − a † λ(k)e−ikxε∗ λ(k) (1.44) B = i 4π V k,λ 1 2ωk aλ(k)eikxk × ελ(k) − a † λ(k)e−ikxk × ε∗ λ(k) (1.45) Notice that the second terms in Equation (1.44) and Equation (1.45) are the Hermitian conjugates of their respective first terms, as they should be, because in the expression for A the same is true and E and B are derived from A. Classically the E2 and B2 in Equation (1.20) contribute equally to the energy. We might therefore expect that to be the case here too, so only one of two terms would need to be calculated. This expectation is not correct as will be shown below by calculating both terms. We begin with the simpler |E| 2 term using Equation (1.44) |E| 2 = E · E† = 4π V k,λ ωk 2 aλ(k)eikxελ(k) − a † λ(k)e−ikxε∗ λ(k) · k,λ ω k 2 a† λ(k )e−ik xε∗ λ(k ) − aλ(k )eik xελ(k ) (1.46) When working out this expression we get four terms, each summed over k, λ, k , λ . There are two types of exponentials among the four terms: those with the difference and those with the sum of k and k in the exponent. We reorder Equation (1.46) with the terms involving the difference k − k first |E| 2 = 4π 2V k,kλλ ωkω k aλ(k)a † λ(k )ei(k−k )xελ(k) · ε∗ λ(k ) + h.c. − 4π 2V k,k,λ,λ ωkω k aλ(k)aλ(k )ei(k+k )xελ(k) · ελ(k ) + h.c. (1.47) The symbol h.c. indicates the Hermitian conjugate of the previous term but with the proviso, indicated by the line over h.c., that in a product of operators their original order is preserved. Thus the h.c. of aλ(k)a† λ(k ) is
4 FLECTROMAGNETIC RADIATION AND MATTER a(k')a(k)while the h.c.of ax(k)a,(k')is a(k)a(k').That these are not the same for all k,k,,will be obvious from the commutation relations integrals first.We use that d'xelkek)x=Vowk (1.48) have twe -k.In the @t)and exp(+)factors cancel,wh in the latter we get factors exp(-2iot)and exp(+2i@t).Note that is defined just below Equation(1.23)as =lkl so is positive,irrespective of the sign of k. We get dxE=2r∑o%[aka(k){e(k)·e克(k}+h.c.】 k -2ok [ax(k)ax(-k)e-2io'[ex(k).ex(-k))+h.c.] k入 (1.49) In the first sum the double sum over and reduces to a single sum overλbecause ek)·e,k)=6 (1.50 This relation holds for linear polarization as well as for circular and ica polanization.Nor that one and only one factor in the dop has a comp onjugatio n sign and that the tw factors have argument(k).The complex conjugation of one of the twoein E (1.50)is necessary when e is complex and it conforms with the definition of the dot product of complex vectors (compare the scalar product of two 'vectors'in Hilbert space).The second sum in Equation (1.49)will cancel a similar term in the expression for Bl2 so we do not have to deal with the dot products of tw (k)or two arguments Equation(1 28) We calculate the B2=B.Bt term using Equation (1.45) 1 -[a(k)eikxk x e(k)-aj(k)e-iexk x ei(k)] 1
14 ELECTROMAGNETIC RADIATION AND MATTER aλ(k )a † λ(k) while the h.c. of aλ(k)a † λ(k ) is a† λ(k)aλ(k ). That these are not the same for all k, k , λ, λ will be obvious from the commutation relations Equation (1.31). We must integrate the four terms over all of space. We exchange the order of the integration and the summations so as to do the integrals first. We use that d3x ei(k±k )·x = Vδkk (1.48) Thus the double sum over k and k reduces to a single sum over k and we have two types of terms: those with k = k and those with k = −k. In the former the exp(−iωkt) and exp(+iωkt) factors cancel, while in the latter we get factors exp(−2iωkt) and exp(+2iωkt). Note that ωk is defined just below Equation (1.23) as ωk = |k| so ωk is positive, irrespective of the sign of k. We get d3x |E| 2 = 2π k,λ,λ ωk aλ(k)a† λ(k){ελ(k) · ε∗ λ(k)} + h.c. −2π k,λ ωk aλ(k)aλ(−k) e−2iωkt {ελ(k) · ελ(−k)} + h.c. (1.49) In the first sum the double sum over λ and λ reduces to a single sum over λ because ελ(k) · ε∗ λ(k) = δλλ (1.50) This relation holds for linear polarization as well as for circular and elliptical polarization. Note that one and only one factor in the dot product has a complex conjugation sign and that the two factors have the same argument (k). The complex conjugation of one of the two ε in Equation (1.50) is necessary when ε is complex and it conforms with the definition of the dot product of complex vectors (compare the scalar product of two ‘vectors’ in Hilbert space). The second sum in Equation (1.49) will cancel a similar term in the expression for |B| 2 so we do not have to deal with the dot products of two ελ(k) or two ε∗ λ(k) that also have different arguments (k). Compare with the comments below Equation (1.27) and Equation (1.28). We calculate the |B| 2 = B · B† term using Equation (1.45) |B| 2 = 4π V k,λ 1 2ωk aλ(k)eikxk × ελ(k) − a † λ(k)e−ikxk × ε∗ λ(k) · k ,λ 1 2ω k a† λ(k )e−ik xk × ε∗ λ(k ) − aλ(k )eik xk × ελ(k ) (1.51)
SECOND QUANTIZATION ssion we get dot products of two cross [k×ek][K×e克(k]=e克(k)[k×ek》×K] =-e(k)-[k×k×e(k =-e克k)-[k·e(kk-(k-k)e:k =(k.k)ex(k)-ei(k)] (1.52) The first step in the above equation is a (counter-clockwise)cyclic per- expands the triple (ab)e.In the fourth step we use =士k and thusk,ek)=±keak=0 according to Equation(1.19).The other three dot products of two cross products can be evaluated in a manner similar to Equation(1.52). When working out Equation (1.51)we notice some similarity between this calculation and the one done earlier for E,especially in the way summed over k the products of the e nentials are structured.We e get four terms,each entials four terms:those with the difference an among the h the sum of k and in the exponent.We reorder Equation(1.51)with the terms involving the difference k-k'first and insert the four dot products of two cross products as evaluated in Equation(1.52)in their appropriate terms = 2可aet-xk.K)e.-e防k+ic】 1 4 1 -[z(k)ax(k)et+(k-K)ex(k)·erK月+h.c] (1.53 We exchange the integration over all of space with the summations, use Equation (1.48)as before to find that the double sum over k and k reduces to a single sum over k.The time-dependent exponentials cancel in the two terms with k'=k but not in the two terms with k'=-k.Note that the factor k.k'that appears in Equation (1.53)is equal to +k2= depending upon the relative signs of k and k'.We get 72应akee,因e+5c]+ xB2=2m∑1 2a.(k)ow(-k)e-e(k)v-k+:j) k,4 (1.54) Note the factor in the first sum and-in the second sum
SECOND QUANTIZATION 15 When working out this expression we get dot products of two cross products. They can be evaluated as k × ελ(k) · k × ε∗ λ(k ) = ε∗ λ(k ) · {k × ελ(k)} × k = −ε∗ λ(k ) · k × {k × ελ(k)} = −ε∗ λ(k ) · {k · ελ(k)}k − (k · k )ελ(k) = (k · k ) ελ(k) · ε∗ λ(k ) (1.52) The first step in the above equation is a (counter-clockwise) cyclic permutation as in the triple product a · (b × c) = c · (a × b). The third step expands the triple product as in a × (b × c) = (a · c)b − (a · b)c. In the fourth step we use that k = ±k and thus k · ελ(k) = ±k · ελ(k) = 0 according to Equation (1.19). The other three dot products of two cross products can be evaluated in a manner similar to Equation (1.52). When working out Equation (1.51) we notice some similarity between this calculation and the one done earlier for |E| 2, especially in the way the products of the exponentials are structured. We get four terms, each summed over k, λ, k , λ . There are two types of exponentials among the four terms: those with the difference and those with the sum of k and k in the exponent. We reorder Equation (1.51) with the terms involving the difference k − k first and insert the four dot products of two cross products as evaluated in Equation (1.52) in their appropriate terms |B| 2 = 4π 2V k,kλλ 1 ωkω k aλ(k)a† λ(k )ei(k−k )x(k · k ){ελ(k) · ε∗ λ(k )} + h.c. − 4π 2V k,k,λ,λ 1 ωkω k aλ(k)aλ(k )ei(k+k )x(k · k ){ελ(k) · ελ(k )} + h.c. (1.53) We exchange the integration over all of space with the summations, and use Equation (1.48) as before to find that the double sum over k and k reduces to a single sum over k. The time-dependent exponentials cancel in the two terms with k = k but not in the two terms with k = −k. Note that the factor k · k that appears in Equation (1.53) is equal to ±k2 = ±ω2 k depending upon the relative signs of k and k . We get d3x B2 = 2π k,λ,λ 1 ωk aλ(k)a† λ(k){ελ(k) · ε∗ λ(k)} + h.c. (+ω2 k) −2π k,λ,λ 1 ωk aλ(k)aλ(−k) e−2iωkt {ελ(k) · ελ(−k)} + h.c. (−ω2 k) (1.54) Note the factor +ω2 k in the first sum and −ω2 k in the second sum.
FLECTROMAGNETIC RADIATION AND MATTER When we sum Equation (1.49)and Equation (1.54)we see that the second sum in Equation (1.49)cancels the second sum in Equation (1.54) as mentioned earlier.This is good because the cancellation removes the time dependence present in E2 and Bl2 so that the energy Eem is time inder dent sitshould beforan n um.The olso gets rid of the troublesome tems invoingther of polarization vectors whose factors have too few or too many complex conjugations and are evaluated at different values of their momentum.The first sums in Equation (1.49)and Equation (1.54)are equal so we get a factor 2.Inserting the factor 1/(8)from Equation (1.20)we get,setting Ecm=H H=kaok [ax(k)af(k)+aj(k)az(k)] (1.55) =∑kao[2ak)a.(k)+1] (1.56) =>[N(k)+] (1.57) where we have used the commutation relation in equation (1.31)in eauation (1.56)and introduced the number operator Na(k)from Equation(1.35)in (1.57 a remarkable result for several reasons.First,it shows that the energy of the electromagnetic field is indeed quantized so that we may speak of 'photons'.Second,it shows that the quantum of energy is @k(hok really) where =lkl.The quantities where k and were first introduced in the plane wave solutions of the wave Equation(1.21)and they derive their sical m om n that expression: is the angula freque ho,the equation Pla nck wrote dowr for black ody radiation.And third,it has the remarkable consequence that even if there are no photons (Na(k)=0 for all k and A)the energy is not zero but in fact infinite!This is annoying to say the least and a sign of serious trouble ahead. In the meantime we give it a name:'zero point energy'and measure all ergies relative to it.This is the ond time that ncou rthis kind of r ense;the first time was in the treatment of the Harmonic Oscillato to which the current trouble can be traced.So it's really only one problem, some consolation！ The analogy between the above result for the energy spectrum of electro- magnetic energy and the eergy spectrum of the is the in an elect netic field wher n giving an e mentaryderivation of Black Body Radiation introductory physics.But it's no more than a manner of speak king. We now see why the choice of pre-factor in Equation(1.29),Equation (1.30)and Equation (1.41)is a good one even though it appeared to
16 ELECTROMAGNETIC RADIATION AND MATTER When we sum Equation (1.49) and Equation (1.54) we see that the second sum in Equation (1.49) cancels the second sum in Equation (1.54) as mentioned earlier. This is good because the cancellation removes the time dependence present in |E| 2 and |B| 2 so that the energy Eem is timeindependent as it should be for an electromagnetic field in vacuum. The cancellation also gets rid of the troublesome terms involving the dot product of polarization vectors whose factors have too few or too many complex conjugations and are evaluated at different values of their momentum. The first sums in Equation (1.49) and Equation (1.54) are equal so we get a factor 2. Inserting the factor 1/(8π) from Equation (1.20) we get, setting Eem = H H = 1 2 k,λ ωk aλ(k)a† λ(k) + a† λ(k)aλ(k) (1.55) = 1 2 k,λ ωk 2a† λ(k)aλ(k) + 1 (1.56) = k,λ Nλ(k) + 1 2 ωk (1.57) where we have used the commutation relation in Equation (1.31) in Equation (1.56) and introduced the number operator Nλ(k) from Equation (1.35) in Equation (1.57). This is a remarkable result for several reasons. First, it shows that the energy of the electromagnetic field is indeed quantized so that we may speak of ‘photons’. Second, it shows that the quantum of energy is ωk (ωk really) where ωk = |k|. The quantities where k and ω were first introduced in the plane wave solutions of the wave Equation (1.21) and they derive their physical meaning from that expression: ω is the angular frequency. Thus we find the relation E = ω, the equation Planck wrote down for black-body radiation. And third, it has the remarkable consequence that even if there are no photons (Nλ(k) = 0 for all k and λ) the energy is not zero but in fact infinite! This is annoying to say the least and a sign of serious trouble ahead. In the meantime we give it a name: ‘zero point energy’ and measure all energies relative to it. This is the second time that one encounters this kind of nonsense; the first time was in the treatment of the Harmonic Oscillator to which the current trouble can be traced. So it’s really only one problem, some consolation! The analogy between the above result for the energy spectrum of electromagnetic energy and the energy spectrum of the Harmonic Oscillator is the justification for speaking in terms of ‘oscillators’ present in an electromagnetic field when giving an elementary derivation of Black Body Radiation in introductory physics. But it’s no more than a manner of speaking. We now see why the choice of pre-factor in Equation (1.29), Equation (1.30) and Equation (1.41) is a good one even though it appeared to
SECOND QUANTIZATION 公 be rather arbitrary.With the choice of pre-factors in those relations we obtain the expression for the Hamiltonian Equation (1.57).Had we made a different choice then additional factors would have appeared in Equation(1.57)in disagreement with experiment,for example,black-body radiation. 1.2.3 Momentum The momentum of the electromagnetic field in a vacuum is given by the classical expression Pem二4 (1.58) The complex conjugate of B appears so that P is parallel to k.We replace the complex conjugate by the Hermitian conjugate but in the present case B is Hermitian so the t does not matter.We calculate the E x Bt using Equation (1.44)and Equation (1.45) Ex Bt=- [a(k)ex(k)-a(k)e-ikxej(k)] x∑ V2 [a减(ke-xK×ek的-ax (k)ek'x ex(k (1.59) When working out this expression we get a cross product of a vector and a cross product of two other vectors.They can be evaluated as e(k)×[k×e(k)]=ek)·e(k]k-[e(k)-k]e克 =ex(k)·e(k]k (1.60) The first step in the above equation expands the triple product as in a·(bxc)=c·(axb).In the second step we used that k=士k and thus e(k).ke(k=0 according to Equation (1.19).The other three cross produ vector nd ss produc t can evalua ed in a nner similar to Equation(1.60) Wh n working out Equation(1.60)w notice some si milarity between this calculation and the one done earlier for E2 and B2,especially in the way the products of the exponentials are structured.We get four terms,each summed over k,,k',A'.There are two oee iponentals among the four terms those with thdu the sum of k and k'in the exponent.We reorder 1159
SECOND QUANTIZATION 17 be rather arbitrary. With the choice of pre-factors in those relations we obtain the expression for the Hamiltonian Equation (1.57). Had we made a different choice then additional factors would have appeared in Equation (1.57) in disagreement with experiment, for example, black-body radiation. 1.2.3 Momentum The momentum of the electromagnetic field in a vacuum is given by the classical expression Pem = 1 4π d3x E × B∗ (1.58) The complex conjugate of B appears so that P is parallel to k. We replace the complex conjugate by the Hermitian conjugate but in the present case B is Hermitian so the † does not matter. We calculate the E × B† using Equation (1.44) and Equation (1.45) E × B† = −4π V k,λ ωk 2 aλ(k)eikxελ(k) − a † λ(k)e−ikxε∗ λ(k) × k,λ 1 2ω k a† λ(k )e−ik xk × ε∗ λ(k ) − aλ(k )eik xk × ελ(k ) (1.59) When working out this expression we get a cross product of a vector and a cross product of two other vectors. They can be evaluated as ελ(k) × k × ε∗ λ(k ) = [ελ(k) · ε∗ λ(k )] k − ελ(k) · k ε∗ λ = [ελ(k) · ε∗ λ(k )] k (1.60) The first step in the above equation expands the triple product as in a · (b × c) = c · (a × b). In the second step we used that k = ±k and thus ελ(k) · k = ελ · (±k) = 0 according to Equation (1.19). The other three cross products of a vector and a cross product can be evaluated in a manner similar to Equation (1.60). When working out Equation (1.60) we notice some similarity between this calculation and the one done earlier for |E| 2 and |B| 2, especially in the way the products of the exponentials are structured. We get four terms, each summed over k, λ, k , λ . There are two types of exponentials among the four terms: those with the difference and those with the sum of k and k in the exponent. We reorder Equation (1.59)