《高等量子力学》课程参考教材：An Introduction to Advanced Quantum Physics（Hans P. Paar）.pdf_P26-P30

18 ELECTROMAGNETIC RADIATION AND MATTER with the terms involving the difference k − k first and insert the four products of the type Equation (1.60) in their appropriate terms to get E × B† = − 4π 2V k,kλλ aλ(k)a† λ(k )ei(k−k )x{ελ(k) · ε∗ λ(k )} k + h.c. − 4π 2V k,k,λ,λ aλ(k)aλ(k )ei(k+k )x{ελ(k) · ελ(k )} k + h.c. (1.61) 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 d3x E × B† = 2π k,λ,λ aλ(k)a† λ(k){ελ(k) · ε∗ λ(k)} k + h.c. k +2π k,λ,λ aλ(k)aλ(−k) e−2iωkt {ελ(k) · ελ(−k)} k + h.c. (1.62) Note that all terms are proportional to k as they should be because E × B† should be parallel to k for each value of k. The second sum is zero because its terms are antisymmetric under the exchange k → −k, while positive and negative values of k appear equally. This is good because the time dependence of Pem is removed as expected because the momentum of an electromagnetic field in a vacuum is conserved. We also get 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. To show that the second sum in Equation (1.62) is zero, we split the sum over k into two equal pieces, each with a factor 1 2 in front, and then replace k → −k in the first sum k,λ,λ aλ(k)aλ(−k) e−2iωkt {ελ(k) · ελ(−k)} k = 1 2 k,λ,λ aλ(k)aλ(−k) e−2iωkt {ελ(k) · ελ(−k)} k + 1 2 k,λ,λ aλ(k)aλ(−k) e−2iωkt {ελ(k) · ελ(−k)} k = 1 2 k,λ,λ aλ(−k )aλ(k ) e−2iωkt {ελ(−k ) · ελ(k )}(−k ) + 1 2 k,λ,λ aλ(k)aλ(−k) e−2iωkt {ελ(k) · ελ(−k)} k (1.63) Inspection shows that the last two sums cancel each other because of the minus sign in front of k in the first of the two sums. The factors multiplying −k and k in the last two sums are equal because aλ(k) and aλ(−k) commute

SECOND OUANTIZATION 19 for all k,k',,see Equation(1.32),and in executing the double sum over andit does not matter in which order one does this.The h.c.term in the second sum in Equation (1.62)is also zero,of course,because it is the Hermi- tian conjugate of the one we just considered apart from the order of the oper- ators,but they commute as well.The symmetry argument(luckily)does not hold for the first sum in Equation(1.62)because that sum involves the prod- nute,see Equation(1.31) ed in Equa n1.63 oes n t work for the first sum Using Equation(1.50)for the dot product of the polarization vectors and inserting the factor 1/(4x)from Equation(1.58)we get Pem=ax(k)ai(k)+ai(k)ax(k)k (1.64) =∑kaj(k)a(k)+1]k (1.65) =∑[Nk)+]k (1.66) k =∑Nkk (1.67 11 where we have used the commutation relations Equation(1.31)in Equation (1.64)and introduced the number operator Na(k)from Equation(1.35)in Equation(1.65).The 'zero point momentum'evident in Equation (1.66) sums to zero if k has no preferred direction for an electromagnetic field in a vacuum,a reasonable assumption.We evade the zero point momentum, case for the en ,see Equation(1.57列 remarkable as the result in Equation (1.57)for th same reasons.It shows that the momentum of the electromagnetic field is quantized,confirming that we may speak of 'photons'.The quantum of momentum is k(hk really)where k was first introduced in the plane wave solutions of the wave Equation (1.21).The vector k derives its physical meaning from that exp ression:Ikl is the wave vector with gnitude /1.We nclude tha m bk of ohoton is e the deBroglie relat e can us -p to calculate the mass of the photon.Using E=hok and p=hk we find m2=0.This is as expected because a'particle can travel with the speed of light(in a vacuum)only if it is massless. Also here we see that the choice of pre-factor in Equation (1.29)and Equation(1.30)is a good one because it leads again to a result in agreement with experiment 1.2.4 Polarization and Spin The polarization factors ea(k)and its complex conjugate define the polariza- tion as discussed earlier.A photon with momentum k along the z-axis that

SECOND QUANTIZATION 19 for all k, k , λ, λ , see Equation (1.32), and in executing the double sum over λ and λ it does not matter in which order one does this. The h.c. term in the second sum in Equation (1.62) is also zero, of course, because it is the Hermitian conjugate of the one we just considered apart from the order of the operators, but they commute as well. The symmetry argument (luckily) does not hold for the first sum in Equation (1.62) because that sum involves the product of aλ(k)a † λ(k) and its h.c. These do not commute, see Equation (1.31), so the trick employed in Equation (1.63) does not work for the first sum. Using Equation (1.50) for the dot product of the polarization vectors and inserting the factor 1/(4π) from Equation (1.58) we get Pem = 1 2 k,λ aλ(k)a † λ(k) + a† λ(k)aλ(k) k (1.64) = 1 2 k,λ a † λ(k)aλ(k) + 1 k (1.65) = k,λ Nλ(k) + 1 2 k (1.66) = k,λ Nλ(k) k (1.67) where we have used the commutation relations Equation (1.31) in Equation (1.64) and introduced the number operator Nλ(k) from Equation (1.35) in Equation (1.65). The ‘zero point momentum’ evident in Equation (1.66) sums to zero if k has no preferred direction for an electromagnetic field in a vacuum, a reasonable assumption. We evade the zero point momentum, contrary to the case for the energy, see Equation (1.57). The result in Equation (1.67) is equally remarkable as the result in Equation (1.57) for the same reasons. It shows that the momentum of the electromagnetic field is quantized, confirming that we may speak of ‘photons’. The quantum of momentum is k (k really) where k was first introduced in the plane wave solutions of the wave Equation (1.21). The vector k derives its physical meaning from that expression: |k| is the wave vector with magnitude 2π/λ. We conclude that the momentum k of a photon is equal to 2π/λ or p = h/λ, the deBroglie relation. We can use the Einstein relation m2 = E2 − p2 to calculate the mass of the photon. Using E = ωk and p = k we find m2 = 0. This is as expected because a ‘particle’ can travel with the speed of light (in a vacuum) only if it is massless. Also here we see that the choice of pre-factor in Equation (1.29) and Equation (1.30) is a good one because it leads again to a result in agreement with experiment. 1.2.4 Polarization and Spin The polarization factors ελ(k) and its complex conjugate define the polarization as discussed earlier. A photon with momentum k along the z-axis that

ELECTROMAGNETIC RADIATION AND MATTER is linearly polarized along the x-axis is represented by af(k)0),in analogy with a plane wave in classical electromagnetism.Likewise a photon linearly polarized along the yaxis is represented by(the unit vector e1 and e2 are chosen along the x-axis and the y-axis respectively and they form with k/k a right-handed triplet of (real)unit vectors in the order (e1,e2,k/@k).When discussing linearly polarized photons,we can and do omit the complex conjugate symbol over e(k).Such photons are called transverse photons. To represent circularly polarized photons we introduce aR(k)=a1(k)+ia2(k) aL(k)=a(k)-ia(k) (1.68) We will show that a right-handed (RH)photon is created by k,R)= ak(k)o)and that a left-handed (LH)photon is created by k,L)=af(k)0). Creation operators in the expression for A(x,t)are accompanied by a factor exp(-ikx).If we fix the position x and let t vary we get a factor exp(ict). We separate the real and imaginary parts of the resulting expression ak(k)e)=(a1+ia2)te =(af-ia)(cosot +isin@t)o) =(af cosot+a sin ot)0)+i(af sinwt-a cosot)0)(1.69) Consider first the real part of Equation(1.69).We plot in Figure 1.1 on an x,y coordinate system the coefficient of a plotted along the x-axis and the coefficient of a plotted along the y-axis(the respective directions of polarization related to af and a).It is seen that the point defined by these two coefficients lies on a circle and that the point moves in a counter-clockwise direction (in the direction from the x-axis to the y-axis) round the circle as time in es.The latter statement can be checked by substitut (A in Figure 1.1)and ot=π/2(Bi Figure 1. .1)in Equation(1.69)and following the position of the point along the circle from A to B.The same counter-clockwise direction of movement of the point is specified by the imaginary part of Equation (1.69).This is consistent with the subscript R of the coefficient ag.The photon moves in the direction of k, in the figu out of the paper toward the and the 0 -clockwise movement e po orrespon to a RH ovement relative tok in the manner defined by the right-hand rule of the e cross product of two vectors.In a similar manner we can show that a LH photon is represented by k,L)=af(k)0).For our discussion we have selected the term in A that creates a photon.One can (and should)verify that the arguments also hold for the terms that t annihilate a phot

20 ELECTROMAGNETIC RADIATION AND MATTER is linearly polarized along the x-axis is represented by a † 1(k) 0 , in analogy with a plane wave in classical electromagnetism. Likewise a photon linearly polarized along the y-axis is represented by a† 2(k) 0 . Here the unit vectors ε1 and ε2 are chosen along the x-axis and the y-axis respectively and they form with k/ωk a right-handed triplet of (real) unit vectors in the order (ε1, ε2, k/ωk). When discussing linearly polarized photons, we can and do omit the complex conjugate symbol over ελ(k). Such photons are called transverse photons. To represent circularly polarized photons we introduce aR(k) = a1(k) + ia2(k) aL(k) = a1(k) − ia2(k) (1.68) We will show that a right-handed (RH) photon is created by k, R = a † R(k) 0 and that a left-handed (LH) photon is created by k, L = a† L(k) 0 . Creation operators in the expression for A(x, t) are accompanied by a factor exp(−ikx). If we fix the position x and let t vary we get a factor exp(iωt). We separate the real and imaginary parts of the resulting expression a† R(k)eiωt 0 = (a1 + ia2) †eiωt 0 = (a† 1 − ia† 2)(cos ωt + isinωt) 0 = (a† 1 cos ωt + a† 2 sin ωt) 0 + i(a† 1 sin ωt − a † 2 cos ωt) 0 (1.69) Consider first the real part of Equation (1.69). We plot in Figure 1.1 on an x, y coordinate system the coefficient of a† 1 plotted along the x-axis and the coefficient of a † 2 plotted along the y-axis (the respective directions of polarization related to a† 1 and a† 2). It is seen that the point defined by these two coefficients lies on a circle and that the point moves in a counter-clockwise direction (in the direction from the x-axis to the y-axis) around the circle as time increases. The latter statement can be checked by substituting ωt = 0 (A in Figure 1.1) and ωt = π/2 (B in Figure 1.1) in Equation (1.69) and following the position of the point along the circle from A to B. The same counter-clockwise direction of movement of the point is specified by the imaginary part of Equation (1.69). This is consistent with the subscript R of the coefficient aR. The photon moves in the direction of k, in the figure out of the paper toward the reader, and the counter-clockwise movement of the point corresponds to a RH movement relative to k in the manner defined by the right-hand rule of the cross product of two vectors. In a similar manner we can show that a LH photon is represented by k, L = a† L(k) 0 . For our discussion we have selected the term in A that creates a photon. One can (and should) verify that the arguments also hold for the terms that annihilate a photon

2 FLECTROMAGNETIC RADIATION AND MATTER momentum k.The subscript R on Eg also here correctly describes the polarization of A and E.In a similar manner one can show that a LH polarized A is given by A(x,t) (1.72 For our discussion we have selected the Hermitian conjugate terms of the pregsion for A.One can (and should)verity that the term too.We ion 1.6 The photon has spin 1.This follows from the fact that A is a four. vector whose fourth component(the scalar potential is identically zero in the Coulomb gauge.The remaining three components are constrained by the Lorentz condition in Equation(1.19)leaving only two independent odwoy oo noodo correspond to the two z-co mponents s. =+1 and of a spin 1 particle and not a s=particle as one might initially think These two possibilities correspond to RH and LH circular polarizations respectively Another characterization of the handedness of the circular polarization involves helicity.One defines the helicity b of a particle with momentum p and spin s as S.D h (1.73) It is seen that a RH photon with s:=+1 has b=+1,while a LH photon with s.=-1 hash=-1 if the axis of quantization is chosen along p. The abence of the s.component is ultimately a conseque of the fact that the photon ism ssless:t rentz an nd thu s the mb G aug can only be implemented for massless photons. Thus Equation(1.19)is true only for massless photons(classically:waves that in a vacuum travel at the speed of light)and only then is ex perpendicular to k.This can be seen by tracing the derivations of subsection 1.1.If the electromagnetic interaction is dese cribed in a relativistically manne (we will not do this her one finds that the phot y corre quires a mass in some cas We say that the photon mass she because oo the Einste E2-p2 =m2(with m=0 in this case)as an 'on the mass shell'particle will.When this happens it acquires a s:=0 component!Such photons are called longitudinal or scalar photons. There is another example of a massless particle,the mediator of the gravitational interaction,t e graviton whose spin is 2 with only two values for its z-component s:=+2. Note that when in classical electromagnetism a plane wave is called transversely polarized,it is called longitudinally polarized in quantum physics.This is confusing!In Table 1.1 we compare the classical and quantum physical properties of electromagnetic radiation

22 ELECTROMAGNETIC RADIATION AND MATTER momentum k. The subscript R on εR also here correctly describes the polarization of A and E. In a similar manner one can show that a LH polarized A is given by A(x, t) ∝ ε∗ Leiωt (1.72) For our discussion we have selected the Hermitian conjugate terms of the expression for A. One can (and should) verify that the arguments hold for the other term too. We will return to circular polarization in Section 1.6. The photon has spin 1. This follows from the fact that Aµ is a fourvector whose fourth component (the scalar potential φ) is identically zero in the Coulomb gauge. The remaining three components are constrained by the Lorentz condition in Equation (1.19) leaving only two independent components. Without getting too formal in proving this, we assert that these two components correspond to the two z-components sz = +1 and sz = −1 of a spin 1 particle and not a s = 1 2 particle as one might initially think. These two possibilities correspond to RH and LH circular polarizations respectively. Another characterization of the handedness of the circular polarization involves helicity. One defines the helicity h of a particle with momentum p and spin s as h = s · p |s||p| (1.73) It is seen that a RH photon with sz = +1 has h = +1, while a LH photon with sz = −1 has h = −1 if the axis of quantization is chosen along p. The absence of the sz = 0 component is ultimately a consequence of the fact that the photon is massless: the Lorentz and thus the Coulomb Gauge can only be implemented for massless photons. Thus Equation (1.19) is true only for massless photons (classically: waves that in a vacuum travel at the speed of light) and only then is ελ perpendicular to k. This can be seen by tracing the derivations of subsection 1.1. If the electromagnetic interaction is described in a relativistically correct manner (we will not do this here) one finds that the photon acquires a mass in some cases. We say that the photon is ‘off the mass shell’ because it does not satisfy the Einstein relation E2 − p2 = m2 (with m = 0 in this case) as an ‘on the mass shell’ particle will. When this happens it acquires a sz = 0 component! Such photons are called longitudinal or scalar photons. There is another example of a massless particle, the mediator of the gravitational interaction, the graviton whose spin is 2 with only two values for its z-component sz = ±2. Note that when in classical electromagnetism a plane wave is called transversely polarized, it is called longitudinally polarized in quantum physics. This is confusing! In Table 1.1 we compare the classical and quantum physical properties of electromagnetic radiation