ELECTROMAGNETIC RADIATION AND MATTER So finally we have dhw%=2πf2Vl8(E(-E±o) (1.103) This result is to first order in V and is called Fermi's Golden Rule.For some applications the first order calculation is not precise enough,especially if it is zero,and we must go to second order.We will leave that calculation until state as (1.104) This can be seen as follows.If a system has a probability Pi(t)to be in the state i at time t,then a time dt later that probability will have on(1.104 0银nEquation (1.77 and Equation在78 we sce that Ho cons ists of two parts,one for a charged particle in an electrostatic electric field (the first two terms)and one for the electromagnetic field(the third term).These two parts in Ho are independent of each other.We assume that we know the eigenkets of each part so we can form the eigenkets of Ho as product eigenkets as parts.The pre pcopiateocahmltonianthatcomsisofwOnaie nden ct eigenke ts are formed fro m the B，…ofth charge cle part of Ho anc the eigenkets k,)of the part of Ho.The eigenvalues of Ho are the sum of the eigenvalues of the two parts.Consider as an example a system that decays from state A to state B under photon emission A→B+ (1.105) The initial state is r esented by A0)and Bk,)or more compactly A:0)and the final st B:k. We will now apply this ormalism to the case of spontaneous er a photon.This was first discussed by Einstein in 1917(having finished his theory of General Relativity)when he introduced his A and B coefficients that describe spontaneous and induced emission respectively.Examples of spontaneous emission can be found in atomic,molecular and nuclear physics.It plays an important role in black body radiation. 1.4 SPONTANEOUS EMISSION 1.4.1 First Order Result We consider the process A→B+y (1.106)
28 ELECTROMAGNETIC RADIATION AND MATTER So finally we have dwfi = 2π | f λV i | 2 δ(Ef − Ei ± ω) (1.103) This result is to first order in V and is called Fermi’s Golden Rule. For some applications the first order calculation is not precise enough, especially if it is zero, and we must go to second order. We will leave that calculation until we need its result in the subsection on photon scattering. The transition probability (per unit time) is related to the lifetime τ of a state as τ = 1 wfi (1.104) This can be seen as follows. If a system has a probability Pi(t) to be in the state i at time t, then a time dt later that probability will have changed by dPi = −wfiPidt. We can integrate this relation to get Pi(t) = Pi(0) e−wfit . Using the definition of lifetime τ from Pi(t) = Pi(0) e−t/τ we find Equation (1.104). In using Equation (1.77) and Equation (1.78) we see that H0 consists of two parts, one for a charged particle in an electrostatic electric field (the first two terms) and one for the electromagnetic field (the third term). These two parts in H0 are independent of each other. We assume that we know the eigenkets of each part so we can form the eigenkets of H0 as product eigenkets as appropriate for a Hamiltonian that consists of two independent parts. The product eigenkets are formed from the eigenkets A , B , ··· of the charged particle part of H0 and the eigenkets k, λ of the electromagnetic part of H0. The eigenvalues of H0 are the sum of the eigenvalues of the two parts. Consider as an example a system that decays from state A to state B under photon emission A → B + γ (1.105) The initial state is represented by A 0 and B k, λ or more compactly A; 0 and the final state by B; k, λ . We will now apply this formalism to the case of spontaneous emission of a photon. This was first discussed by Einstein in 1917 (having finished his theory of General Relativity) when he introduced his A and B coefficients that describe spontaneous and induced emission respectively. Examples of spontaneous emission can be found in atomic, molecular and nuclear physics. It plays an important role in black body radiation. 1.4 SPONTANEOUS EMISSION 1.4.1 First Order Result We consider the process A → B + γ (1.106)
SPONTANEOUS EMISSION 西 and use the Hamiltonian Equation(1.77).The Harmonic Perturbation is given by the last two terms =-Ap+ (1.107 with A given by Equation(1.41) A=Σ =ax(k)eike(k)+a(k)e-ike:(k)](1.108) which contains harmonic factors of the form exp(ti@t),just as we consid- ered in the previous section.We use Equation(1.94)to get 入9=-He- =-B:k-元Ap+2元AA:e-E(1.109 e- We assigned zero energy to the electromagnetic term in Ho so Ef=EB.The term linear in A creates or annihilates a photon through a(k)and ax(k)so it change s the numbe of photons s by 1.It can be seen that the Ate changes the numl er of photons by 0 or +2 so th s term does not c to Equation (1.106)because the product with the bra(k,would be zero In the present case we must create one photon so the product with (k, must be non-zero.So only the a(k)term in p.A contributes.We obtain A9=-√积（)@：k4减kpA-6 (1.110 where x is the coordinate of the charged particle(for example the electron in a H-atom)and p is its momentum,while k is the momentum of the photon.We know that af(k)0)=Tk,(the creation operator in action) so k,(k)0)=1 and one bra and one ket disappears from the product bra and ket.Thus we get 9-品-eip4- (1.111) If we compare this expression with the general expression in Equation(1.95) for a harn onic perturbation we find that 47 e-ikx g(k)·p (1.112)
SPONTANEOUS EMISSION 29 and use the Hamiltonian Equation (1.77). The Harmonic Perturbation is given by the last two terms λH1 = − e m A · p + e2 2m A2 (1.107) with A given by Equation (1.41) A(x) = 4π V k,λ 1 2ωk [aλ(k) eikxελ(k) + a † λ(k) e−ikxε∗ λ(k)] (1.108) which contains harmonic factors of the form exp(±iωt), just as we considered in the previous section. We use Equation (1.94) to get λ c˙ (1) f = −i f λH1 i e i(Ef −Ei)t = −i B; k, λ − e m A · p + e2 2m A2 A; 0 ei(EB−EA)t (1.109) We assigned zero energy to the electromagnetic term in H0 so Ef = EB. The term linear in A creates or annihilates a photon through a† λ(k) and aλ(k) so it changes the number of photons by ±1. It can be seen that the A2 term changes the number of photons by 0 or ±2 so this term does not contribute to Equation (1.106) because the product with the bra k, λ would be zero. In the present case we must create one photon so the product with k, λ must be non-zero. So only the a† λ(k) term in p · A contributes. We obtain λ c˙ (1) f = −i 4π V2ω − e m B; k, λ a† λ(k)e−ik·x ε∗ λ(k) · p A; 0 ei(EB−EA+ω)t (1.110) where x is the coordinate of the charged particle (for example the electron in a H-atom) and p is its momentum, while k is the momentum of the photon. We know that a † λ(k) 0 = √ 1 k, λ (the creation operator in action) so k, λ a† λ(k) 0 = 1 and one bra and one ket disappears from the product bra and ket. Thus we get λ c˙ (1) f = ie m 4π V2ωk B e−ik·x ε∗ λ(k) · p A ei(EB−EA+ω)t (1.111) If we compare this expression with the general expression in Equation (1.95) for a harmonic perturbation we find that λV = ie m 4π V2ωk e−ik·x ε∗ λ(k) · p (1.112)
30 ELECTROMAGNETIC RADIATION AND MATTER Substitution of this expression in Fermi's Golden Rule Equation(1.103) gives dwE=2πIBVA8(Eb-EA±o) =( -(Be-ik e(k)pA)8(EB -EA +(1.113) Note that the transition probability is of),that the-function requires that EA =EB+@and that the operator V must be such that(BVA)0. The expression in Equation (1.113)is rather complicated because of the exponential.We can expand the exponential in a Taylor series and see what we can learn,at least from the first few terms. 1.4.2 Dipole Transition Keeping for the moment only the first term of the expansion of the expo- nential in Equation(1.113)we get s=2a(原阳prE-+o (1.114) This is called the Dipole Approximation'for reasons that will become clear shortly.The matrix element(BpA)can be evaluated as follows (BPA）=m(BA =im(B[Ho,x]A) =im(EB-EA)(BxA) (1.115) dx 文=正=以，刘 (1.116 with H equal to the charged particle piece of Equation(1.78).The matrix element(BxA)can be calculated because we assumed we know the eigen- kets A)and The answer d its erenkets.For now we defne a quantity,then)Dipole de e on roble Moment,as D=Bex A) (1.117) and use it with Equation(1.115)in Equation(1.114)to get Tk|ek)·DP8(EB-EA+o) (1.118)
30 ELECTROMAGNETIC RADIATION AND MATTER Substitution of this expression in Fermi’s Golden Rule Equation (1.103) gives dwfi = 2π B λV A 2 δ(Eb − EA ± ω) = 2π e m 2 4π V2ωk B e−ik·x ε∗ λ(k) · p A 2 δ(EB − EA + ω) (1.113) Note that the transition probability is of O(e2), that the δ-function requires that EA = EB + ω and that the operator V must be such that B V A = 0. The expression in Equation (1.113) is rather complicated because of the exponential. We can expand the exponential in a Taylor series and see what we can learn, at least from the first few terms. 1.4.2 Dipole Transition Keeping for the moment only the first term of the expansion of the exponential in Equation (1.113) we get dwfi = 2π e m 2 4π V2ωk B ε∗ λ(k) · p A 2 δ(EB − EA + ω) (1.114) This is called the ‘Dipole Approximation’ for reasons that will become clear shortly. The matrix element B p A can be evaluated as follows B p A = m B x˙ A = im B [H0, x] A = im (EB − EA) B x A (1.115) In the first step we used the non-relativistic relation between momentum and velocity, in the second we used x˙ = dx dt = i [H, x] (1.116) with H equal to the charged particle piece of Equation (1.78). The matrix element B x A can be calculated because we assumed we know the eigenkets A and B . The answer depends of course on the specific problem and its eigenkets. For now we define a quantity, the (transition) Dipole Moment, as D = B ex A (1.117) and use it with Equation (1.115) in Equation (1.114) to get dwfi = 4π2 V ωk | ε∗ λ(k) · D | 2 δ(EB − EA + ωk) (1.118)
SPONTANEOUS EMISSION 3 We have used the cor nstraint imposed by the delta function hen wer EB EA)2 by @p nowsee why thisis called the Dipole Approxir the transition probability depends upon the dipole moment(squared,as in classical radiation theory).The factor e(k).D implies that the probability for photon emission is a maximum when e(k)is parallel to D and thus k perpendicular to D,again in agreement with classical radiation theory for dipole raditiono se tha energy is conserved becuse ene in the final stare equal to energy in the in SaG thanks to the 8-function. ook over the ation to e where the comes from and verify that photon absorption would involve ax(k)and give 8(Eg -EA-k)(note the -sign in front of automatically and the transition probability would again conserve energy.As mentioned before,D is calculable for specific cases but one can sometimes see from symmetry that D has to be zero.An example is radiation in a H-atom when it transitions om a 2s state to a 1s state s state s have =0 re therefor spherically symmetric,ex is odd, therefore the transition d ipole moment is zero.Another way to see this is by looking at the integration over space to be done to calculate D.The wave functions are even functions but x is odd so the integral is zero. Experimentally we ask for the probability that a photon is emitted in a egion of phase spaceWe know from statistical physics that such a phase space volume is given by Vk'dikl Voi dor do (1.119) (2π3 (2π)3 (2π3 in wh Vo1©k)·D2VOC张经8EB-EA+E) (1.120) 2π5 We see that Vcancels as it should (so we can let V).We can integrate over the photon energy because it is constrained by the 8-function, enforcing energy conservation.If the experiment does not observe the polarization of the photon we must sum over A.We get d=∑ekD am=2玩 (1.121) We perform the sum over A using the three vectors e(k),e2(k),k/wk that form a triplet of (real)unit vectors.We can write D++D=(e1-DP+(e2.D+(kD 、 (1.122
SPONTANEOUS EMISSION 31 We have used the constraint imposed by the delta function when we replaced (EB − EA) 2 by ω2 k. We now see why this is called the Dipole Approximation: the transition probability depends upon the dipole moment (squared, as in classical radiation theory). The factor ε∗ λ(k) · D implies that the probability for photon emission is a maximum when ε∗ λ(k) is parallel to D and thus k perpendicular to D, again in agreement with classical radiation theory for dipole radiation. We also see that energy is conserved because the energy EB + ωk in the final state is equal to EA, the energy in the initial state, thanks to the δ-function. Look over the calculation to see where the + sign comes from and verify that photon absorption would involve aλ(k) and give δ(EB − EA − ωk) (note the − sign in front of ωk) automatically and the transition probability would again conserve energy. As mentioned before, D is calculable for specific cases but one can sometimes see from symmetry that D has to be zero. An example is radiation in a H-atom when it transitions from a 2s state to a 1s state. The s states have l = 0 and are therefore spherically symmetric, ex is odd, therefore the transition dipole moment is zero. Another way to see this is by looking at the integration over space to be done to calculate D. The wave functions are even functions but x is odd so the integral is zero. Experimentally we ask for the probability that a photon is emitted in a specific region of phase space d3k. We know from statistical physics that such a phase space volume is given by Vd3k (2π)3 = Vk2d|k| dk (2π)3 = Vω2 k dωk dk (2π)3 (1.119) The product of the transition probability wfi and the volume of phase space in which we find the photon is the desired quantity dwfi = 4π2 V ωk | ε∗ λ(k) · D | 2 Vω2 kdωkdk (2π)3 δ(EB − EA + ωk) (1.120) We see that V cancels as it should (so we can let V → ∞). We can integrate over the photon energy ωk because it is constrained by the δ-function, enforcing energy conservation. If the experiment does not observe the polarization of the photon we must sum over λ. We get dwfi dk = ω3 k 2π λ=1,2 | ε∗ λ(k) · D | 2 (1.121) We perform the sum over λ using the three vectors ε1(k), ε2(k), k/ωk that form a triplet of (real) unit vectors. We can write |D| 2 = D2 x + D2 y + D2 z = (ε1 · D) 2 + (ε2 · D) 2 + k · D ωk 2 (1.122)
ELECTROMAGNETIC RADIATION AND MATTER Define the angle 0 between k and D by k.D=@gIDI cos0.Then (exk)D)2=(e1D)}2+(e2D2 12 =1D12(1-cos20)=sin201D2 (1.123) We see that the probability is a maximum if =/2 or k LD as noted earlier.The integral over 0 is trivial and we get for the total transition probability 4 w=3021D2 (1.124) The luminosity C of the emitted radiation is obtained by multiplying wf by the energy of each photon so we get c=o (1.125) As an example we calculate the probability for spontaneous emission of a photon by nsition2p→1s We have that En= 12 ma-/n n,the fine-struct re constant and n the principal quantum number.For the 2p-1s transition we find that @=(3/8)ma2.We guess that the dipole moment is equal to the electron's charge times the Bohr radius ao =1/(am).Altogether we get wf (9/128)ma5,not very large because a21/137.To get a numerical value ast insert the c ers of hand c using dim nsional analy sis.We nd insert a factor c w=(9/128) h): 1.1x 10 he life r of a state is the inverse of th transition probability and we find that r =0.9ns,a very long time indeed (on an atomic scale).With the factor c2/h in Equation(1.124)and c2 in Equation (1.125),we see that the luminosity C does not contain h.This allows for the classical calculation of the lumin sity to be done with a result in agreement with Equation(1.125) 1.4.3 Higher Multipole Transition We now consider the next(second)term in the Taylor series'expansion of the exponential in Equation(1.113).The matrix element of that second term is (B:k,k·x(e*·pA) (1.126) The original order of the position coordinate x and the momentum operator p is kept as it was in the full matrix element Equation(1.113)because p acts
32 ELECTROMAGNETIC RADIATION AND MATTER Define the angle θ between k and D by k · D = ωk |D| cos θ. Then λ=1,2 (ελ(k) · D ) 2 = (ε1 · D) 2 + (ε2 · D) 2 = | D| 2 (1 − cos2 θ) = sin2 θ | D| 2 (1.123) We see that the probability is a maximum if θ = π/2 or k ⊥ D as noted earlier. The integral over θ is trivial and we get for the total transition probability wfi = 4 3 ω3 k |D| 2 (1.124) The luminosity L of the emitted radiation is obtained by multiplying wfi by the energy ω of each photon so we get L = 4 3 ω4 |D| 2 (1.125) As an example we calculate the probability for spontaneous emission of a photon by a H-atom in the transition 2p → 1s. We have that En = −(1/2) mα2/n2 with m the mass of the electron, α the fine-structure constant, and n the principal quantum number. For the 2p → 1s transition we find that ωk = (3/8) mα2. We guess that the dipole moment is equal to the electron’s charge times the Bohr radius a0 = 1/(αm). Altogether we get wfi = (9/128) mα5, not very large because α2 ≈ 1/137. To get a numerical value we must insert the correct powers of and c using dimensional analysis. We find that we need to insert a factor c2/ to get wfi = (9/128) mα5(c2/) = 1.1 × 109 s−1. The lifetime τ of a state is the inverse of the transition probability and we find that τ = 0.9 ns, a very long time indeed (on an atomic scale). With the factor c2/ in Equation (1.124) and c2 in Equation (1.125), we see that the luminosity L does not contain . This allows for the classical calculation of the luminosity to be done with a result in agreement with Equation (1.125). 1.4.3 Higher Multipole Transition We now consider the next (second) term in the Taylor series’ expansion of the exponential in Equation (1.113). The matrix element of that second term is B; k, λ (k · x)(ε∗ · p) A (1.126) The original order of the position coordinate x and the momentum operator p is kept as it was in the full matrix element Equation (1.113) because p acts