SECOND OUANTIZATION 23 Table 1.1 Comparison of terminology for photons Classical Quantum Physical wave Photon city c RH and transverse polarization s +1 and-1 No longitudinal polarization No s:=0 The number operator Na(k)does not commute with A,E or B.This is because the numb pera es no mmute with ax and a a s can be seen from equations (1.36)and (1.37)and because Equation (1),Equation (1.44)and Equation(1.45)for A,E and B have terms proportional to either ax or af.The non-zero commutators lead to uncertainty relations between N,(k)and E and B.When the operators corresponding to two observables do not commute,an uncertainty relation results for their eigenvalues.The prototype s for this ar and position o energ and ri Equatio 1. eigenvalue of Na(k)is specified (so we know exactly how many photons are present in the system)the values of E and B(and A)are uncertain. This leads to the strange conclusion that the vacuum (which has exactly zero photons)has uncertain values of its electric and magnetic fields.One an sh that while (0E|0)=0 and (0B|0)=0 the val s of (10) and (B)are infini therefore the standard red a and ag are infinite and the electromagnetic energy Eem =oo according to Equation(1.20).Compare this with the zero point energy problem of the harmonic oscillator and the electromagnetic field.This means trouble!In the classical domain with well defined electric and magnetic fields the number of photons is extremely lar and average er fnis es of space lead ngful quantities.In the physical do nain we dea exactly known t very small number of photons (0,1,2,...)and the electri and magnetic held have uncertain values (but we do not care to know them). 1.2.5 Hamiltonian The Hamiltonian for a charged particle in an electromagnetic field is given by the sum of two parts,one given by Equation (1.1),the other by Equation(1.20)as H=2mp-eAP+e0+8元xE2+B的) (1.74) If we expand the first part we ge H= 易++3PxE+B的)-玩PA-玩Ap+ -A2(1.75)
SECOND QUANTIZATION 23 Table 1.1 Comparison of terminology for photons Classical Quantum Physical Wave Photon Velocity c Mass zero RH and LH transverse polarization Helicity is +1 and −1 No longitudinal polarization No sz = 0 The number operator Nλ(k) does not commute with A, E or B. This is because the number operator does not commute with aλ and a† λ as can be seen from equations (1.36) and (1.37) and because Equation (1.30), Equation (1.44) and Equation (1.45) for A, E and B have terms proportional to either aλ or a† λ. The non-zero commutators lead to uncertainty relations between Nλ(k) and E and B. When the operators corresponding to two observables do not commute, an uncertainty relation results for their eigenvalues. The prototypes for this are momentum and position or energy and time, see Equation (1.7), leading to the Heisenberg uncertainty relations. So if the eigenvalue of Nλ(k) is specified (so we know exactly how many photons are present in the system) the values of E and B (and A) are uncertain. This leads to the strange conclusion that the vacuum (which has exactly zero photons) has uncertain values of its electric and magnetic fields. One can show that while 0 E 0 = 0 and 0 B 0 = 0 the values of 0| E2 |0 and 0 B2 0 are infinite and therefore the standard deviations squared σ 2 E and σ 2 B are infinite and the electromagnetic energy Eem = ∞ according to Equation (1.20). Compare this with the zero point energy problem of the harmonic oscillator and the electromagnetic field. This means trouble! In the classical domain with well defined electric and magnetic fields the number of photons is extremely large, and averages over finite volumes of space lead to meaningful quantities. In the quantum physical domain we deal with an exactly known but very small number of photons (0, 1, 2, ...) and the electric and magnetic field have uncertain values (but we do not care to know them). 1.2.5 Hamiltonian The Hamiltonian for a charged particle in an electromagnetic field is given by the sum of two parts, one given by Equation (1.1), the other by Equation (1.20) as H = 1 2m (p − eA) 2 + eφ + 1 8π d3 x(E2 + B2) (1.74) If we expand the first part we get H = p2 2m + eφ + 1 8π d3 x(E2 + B2) − e 2m p·A − e 2m A·p + e2 2m A2 (1.75)
24 ELECTROMAGNETIC RADIATION AND MATTER where we have written 'large'terms first.The value of e is given by e2= a1/137 so the last three terms can be considered'small'in comparison with the first three terms.We note that terms with p.A and A.p can be combined.Using a test function(x)we hav p.Av=v.Av=(V.A)v+A.Vv=A.py (1.76 equality,the product rule of calculus and Equation (1.18)an Equation (1.5)for the thir Again we see the advantage of using the Coulomb Gauge.Because the relation in Equation(1.76)is true for any we conclude that [p,A]=0 in the Coulomb Gauge and the Hamiltonian Equation(1.75)becomes with operators shown explicitly 2 H=- ++∑[N+引-品Av+ (1.77 The physical consequences of this Hamiltonian can be inferred by using time-dependent rturbation theory with the last two terms the time The first ree to an unperturbed v2 6=+的+∑N+引 (1.78) k We will review time-dependent perturbation theory in the next subsection 1.3 TIME-DEPENDENT PERTURBATION THEORY This will be a brief review because time-dependent perturbation the ory will have been discussed in detail in your favorite quantum physics book.Our treatment is not relativistically correct as is shown explicitly by the appear- ance of the time variable without the equivalent appearance of the three space variables.The time dependence leads to energy conservation explicit in the amplitude for a process without the equivalent three-momentum conservation.Special relativity requires that all four components of the energy-mon ntum four-vector appear without preference for any one of them.We will see that energy-momentum conservation can be violated in a process with an appeal to Heisenberg's uncertainty relations.In a relativis- and be off the r mass shell
24 ELECTROMAGNETIC RADIATION AND MATTER where we have written ‘large’ terms first. The value of e is given by e2 = α ≈ 1/137 so the last three terms can be considered ‘small’ in comparison with the first three terms. We note that terms with p · A and A · p can be combined. Using a test function ψ(x) we have p · A ψ = 1 i ∇ · A ψ = 1 i (∇ · A) ψ + 1 i A · ∇ψ = A · p ψ (1.76) We used Equation (1.5) in the first equality, the product rule of calculus in the second, and Equation (1.18) and Equation (1.5) for the third. Again we see the advantage of using the Coulomb Gauge. Because the relation in Equation (1.76) is true for any ψ we conclude that [p, A] = 0 in the Coulomb Gauge and the Hamiltonian Equation (1.75) becomes with operators shown explicitly H = − ∇2 2m + eφ + k,λ Nλ(k) + 1 2 ωk − e im A · ∇ + e2 2m A2 (1.77) The physical consequences of this Hamiltonian can be inferred by using time-dependent perturbation theory with the last two terms the timedependent perturbation. The first three terms correspond to an unperturbed Hamiltonian H0 with H0 = − ∇2 2m + eφ + k,λ Nλ(k) + 1 2 ωk (1.78) We will review time-dependent perturbation theory in the next subsection. 1.3 TIME-DEPENDENT PERTURBATION THEORY This will be a brief review because time-dependent perturbation theory will have been discussed in detail in your favorite quantum physics book. Our treatment is not relativistically correct as is shown explicitly by the appearance of the time variable without the equivalent appearance of the three space variables. The time dependence leads to energy conservation explicit in the amplitude for a process without the equivalent three-momentum conservation. Special relativity requires that all four components of the energy-momentum four-vector appear without preference for any one of them. We will see that energy-momentum conservation can be violated in a process with an appeal to Heisenberg’s uncertainty relations. In a relativistically correct formulation, energy-momentum conservation is respected in all parts of a process, but the photon can sometimes have non-zero mass and be off the mass shell
TIME-DEPENDENT PERTURBATION THEORY We assume that we can write the Hamiltonian in two pieces,a 'large time-independent on Ho and a'small time-dependent H1() Vexp with all time dependence in the exponential (Harmonic Perturbation')so V is a time-independent operator.The parameter A is introduced to keep track of the order of magnitude in the perturbation expansion of the solution to the Schrodinger equation.Thus we have H=Ho+H1()=H0+入Vei (1.79) We assume that the eigenkets)and eigenvalues E of Ho are known, that is we assume that the problem is solved for=0 Hon)=E0p】 (1.80 and that the kets are orthonormal 中m中n=dm (1.81) form a complete set.Let's simplify the notation by letting iEn and with is then 3 superposition of eekd of the st general eigenk full Hamiltonian H )=∑col网eE (1.82) where the c are constant.We find the most general eigenkets and eigen- values of the full Hamiltonian Equation (1.79)with 0 using a linear as in Equation (1.82)but with the coefficients functions |)=cn(me-正m (1.83) This should work because the eigenketsform a complete set.When this expression is substituted in the Schrodinger equation we get an equation for the coefficients c(t) ∑ca(t)He-t=-∑in(t))e- (1.84) In order to keep track of terms of the same order in A,we set cn(t)=c0+λc()+λ2c2(+… (1.85) Note that for A=0 the new c(t)are equal to the old time-independent c)of Equation (1.82)as they should be.Substitution of Equation (1.85)
TIME-DEPENDENT PERTURBATION THEORY 25 We assume that we can write the Hamiltonian in two pieces, a ‘large’ time-independent one H0 and a ‘small’ time-dependent one λH1(t) = λVexp(±iωt) with all time dependence in the exponential (‘Harmonic Perturbation’) so V is a time-independent operator. The parameter λ is introduced to keep track of the order of magnitude in the perturbation expansion of the solution to the Schrodinger equation. Thus we have ¨ H = H0 + λH1(t) = H0 + λVe±iωt (1.79) We assume that the eigenkets φn and eigenvalues E(0) n of H0 are known, that is we assume that the problem is solved for λ = 0 H0 φn = E(0) n φn (1.80) and that the kets φn are orthonormal φm φn = δmn (1.81) and form a complete set. Let’s simplify the notation by letting E(0) n → En and φn → n . The most general eigenket of H0 (and of the full Hamiltonian H with λ = 0) is then a superposition of eigenkets ψ = n c(0) n n e−iEnt (1.82) where the c (0) n are constant. We find the most general eigenkets and eigenvalues of the full Hamiltonian Equation (1.79) with λ = 0 using a linear superposition as in Equation (1.82) but with the coefficients functions of time ψ = n cn(t) n e−iEnt (1.83) This should work because the eigenkets n form a complete set. When this expression is substituted in the Schrodinger equation we get an equation for ¨ the coefficients cn(t) n cn(t)λH1 n e−iEnt = −1 i n c˙n(t) n e−iEnt (1.84) In order to keep track of terms of the same order in λ, we set cn(t) = c(0) n + λc(1) n (t) + λ2c(2) n (t) +··· (1.85) Note that for λ = 0 the new cn(t) are equal to the old time-independent c (0) n of Equation (1.82) as they should be. Substitution of Equation (1.85)
ELECTROMAGNETIC RADIATION AND MATTER in Equation (1.84)and equating terms with the same power of in the resulting equation(as the equation must be true if is varied)we get the iterative set of equations 0=-∑9网e-ifor20(1.86 ∑2He=-号∑eo (1.87) ∑cgH冰=-}∑岁efoc2 (1.88) (1.89列 o(m fond can only be solved if the preceding equation of order is solved first.This requires an iterative method that we might call a 'Bootstrap'.Fortunately, ra rely go beyond the first non- ro result and,with physicists never prove that the resulting series expan solve for the coefficients by multiplying Equation(1.86),Equation(1.87), Equation (1.88),...by (f from the left to find (t)=0 or co)constant (1.90) A4=-i∑0lHme,-E (1.91) 2=-∑c用lH网e-E (1.92) (1.93) Note that we multiplied the left and right side of each equation by an appropriate power of A in accordance with the power of found in the Equation (1.85).Thus far our results are genera We now specialize to the case e system in the stateat some time ti in the past,and we are asking for the amplitude for the system to be in the state f at a later time t2.We therefore set at ti all c(t1)=0 except that we set1.The relation in Equation (1.85)shows that atall except that ct)=1,the'initial co lition'.The rel cquation(1.82) shows that the system's state is=i as desired.To find the amplitude for the system to be in the state f at t2 we project out the ket f)from psi by calculating,using Equation (1.85),c(t2)=c(t)+(t)+c
26 ELECTROMAGNETIC RADIATION AND MATTER in Equation (1.84) and equating terms with the same power of λ in the resulting equation (as the equation must be true if λ is varied) we get the iterative set of equations 0 = −1 i n c˙ (0) n n e−iEnt for λ0 (1.86) n c(0) n H1 n e−iEnt = −1 i n c˙ (1) n n e−iEnt for λ1 (1.87) n c(1) n (t)H1 n e−iEnt = −1 i n c˙ (2) n n e−iEnt for λ2 (1.88) ··· (1.89) and so on for larger powers of λ. The structure of Equation (1.87), Equation (1.88) and Equation (1.89) can be seen. An equation of order λn+1 can only be solved if the preceding equation of order λn is solved first. This requires an iterative method that we might call a ‘Bootstrap’. Fortunately, we rarely go beyond the first non-zero result and, with rare exceptions, physicists never prove that the resulting series expansion converges. We solve for the coefficients c˙ (f) n by multiplying Equation (1.86), Equation (1.87), Equation (1.88), ··· by f from the left to find c˙ (0) f (t) = 0 or c (0) f = constant (1.90) λc˙ (1) f (t) = −i n c(0) n f λH1 n e i(Ef −En)t (1.91) λ2c˙ (2) f (t) = −i n λc(1) n f λH1 n ei(Ef −En)t (1.92) ··· (1.93) Note that we multiplied the left and right side of each equation by an appropriate power of λ in accordance with the power of λ found in the terms in Equation (1.85). Thus far our results are general. We now specialize to the case where the system is in the state i at some time t1 in the past, and we are asking for the amplitude for the system to be in the state f at a later time t2. We therefore set at t1 all c (m) n (t1) = 0 except that we set c (0) i = 1. The relation in Equation (1.85) shows that at t1 all cn(t1) = 0 except that ci(t1) = 1, the ‘initial condition’. The relation in Equation (1.82) shows that the system’s state is ψ = i as desired. To find the amplitude for the system to be in the state f at t2 we project out the ket f from psi by calculating, using Equation (1.85), cf (t2) = c (0) f (t2) + λc (1) f (t2) + λ2c (2) f
TIME-DEPENDENT PERTURBATION THEORY 之 (tz)+...where c(tz)=0 from the initial condition.Equation (1.90) shows that cr (t2)=c(t),so c(t)=0.We find from Equation (1.91) ()=-iWHe-E加 (1.94) =-iff]avlie-E±g (1.95) We integrate this equation and obtain 广=- (1.96) =-hv明心d.两- (1.97) We know from the initial condition that (t)=0.We let t-o and t2→+oo and use 八e=sol (1.98) to get λc(∞)=2πfaVl8(Er-E:±o (1.99) We see that the amplitude for the system to be in the statefatis given by cf(oo)=c(oo)to first order in V,and is non-zero if fv and E=Ei ho.This corresponds to a selection rule and a change in the energy of the system by w.The meaning of this will become clear soon. We obtain the transition probability r unit time (or 'transition proba- bility'for short)wf from Equation(1.99)as xc(oo)2 dwf=Tim lertoo)2 T =▣号1 VIFWE-E±P (1.100) with T=t2-t1.We used that c(oo)=0;see the discussion just above Equation(1.94).The 8-function squared is evaluated as 6E=E)m2元-p (1.101) =8(Em2 (1.102)
TIME-DEPENDENT PERTURBATION THEORY 27 (t2) +··· where c (0) f (t2) = 0 from the initial condition. Equation (1.90) shows that c (0) n (t2) = c (0) n (t1), so c (0) f (t2) = 0. We find from Equation (1.91) λc˙ (1) f (t) = −i f λH1 i e i(Ef −Ei)t (1.94) = −i f λV i e i(Ef −Ei±ω)t (1.95) We integrate this equation and obtain t2 t1 λc˙ (1) f dt = λ[c (1) f (t2) − c (1) f (t1)] (1.96) = −i f λV i t2 t1 dt e i(Ef −Ei±ω)t (1.97) We know from the initial condition that c (1) f (t1) = 0. We let t1 → −∞ and t2 → +∞ and use 1 2π ∞ −∞ dt eiωt = δ(ω) (1.98) to get λc (1) f (∞) = 2πi f λV i δ(Ef − Ei ± ω) (1.99) We see that the amplitude for the system to be in the state f at t = ∞ is given by cf (∞) = c (1) f (∞) to first order in V, and is non-zero if f λV i = 0 and Ef = Ei ∓ ω. This corresponds to a selection rule and a change in the energy of the system by ∓ω. The meaning of this will become clear soon. We obtain the transition probability per unit time (or ‘transition probability’ for short) wfi from Equation (1.99) as dwfi = lim T→∞ cf (∞) 2 T = lim T→∞ λc (1) f (∞) 2 T = lim T→∞ 4π2 T f λV i 2 [δ(Ef − Ei ± ω)]2 (1.100) with T = t2 − t1. We used that c (0) f (∞) = 0; see the discussion just above Equation (1.94). The δ-function squared is evaluated as [δ(E)]2 = δ(E) lim T→∞ 1 2π +T/2 −T/2 dt eiEt (1.101) = δ(E) lim T→∞ T 2π (1.102)