Donald Gary Swanson
In Section 5.4.2 where we treated time-dependent perturbation theory, we took the perturbation to be of the general form Hˆ(1)(r, t) = 2Hˆ(1)(r) cosωt (9.1) but for absorption and emission of photons, we want the perturbation to be an electromagnetic field of the form E = E0 cosωt eˆz (9.2) so that the perturbing potential is V (r, t) = − ∫ E dz = −qE0z cosωt (9.3) and because of the factor of 2 in Equation (9.1), the matrix element is Hˆ(1)kj = − 12E0〈k|℘z|j〉 (9.4) where ℘z ≡ qz is the z-component of the dipole moment. In the derivation of Equation (5.48) from Equation (5.47), we integrated over the energy E = ~ω, whereas if we instead integrate over ω, the result is Rj→k = 2pi ~2 |Hˆ(1)kj |2 , (9.5) where the δ-function in Equation (5.48) tells us to evaluate the expression at ω = ωkj . The energy density in an electric field is given by and we let u→ ρ(ω)dω so that Rj→k = pi ²0~2 |〈k|℘z|j〉|2ρ(ω) , (9.6) and There is nothing special about the z-direction in evaluating the matrix element, so it is common to use the magnitude instead of the single component, where by symmetry, |℘|2 = |℘x|2 + |℘y|2 + |℘z|2 = 3|℘z|2 so the final result is expressed as Rj→k = pi 3²0~2 |〈k|℘|j〉|2ρ(ω) . (9.7) If we imagine a closed system in equilibrium, where there are atoms in both the lower state (ψk) and in the upper state (ψj), we can be assured that there is a balance between the rates of emission and absorption. If we define the spontaneous emission rate to be A∗ so that if there are Nj atoms in the upper state, there will be NjA transitions per second from state j to state k, as illustrated in Figure 5.2c. There are also transitions from state j to state k by stimulated emission as illustrated in Figure 5.2a. This rate is proportional to the energy density in the electromagnetic field of the wave and is given by Bjkρ(ω), where Bjk is proportional to Rk from Equation (5.48) and ρ(ω) is the energy density in the electric field of the wave at frequency ω. The total number of transitions per second from state j to state k due to stimulated emission is therefore NjBjkρ(ω). If there is an equilibrium, there must be an equal number of transitions from state k to state j by absorbing photons as illustrated in Figure 5.2b. The total number of transitions per second from state k to state j is then given by NkBkjρ(ω) where Nk is the number of atoms in the lower state (ψk) and Bkj is the transition rate for the absorption process. The rate equation is therefore dNj dt = −NjA−NjBjkρ(ω) +NkBkjρ(ω) . (9.8) In equilibrium, there is a steady state, so dNj/dt = 0, so that the energy density in the wave must be given by ρ(ω) = NjA NkBkj −NjBjk = A (Nk/Nj)Bkj −Bjk . (9.9) Equilibrium also implies, however, that there is a temperature T and that the relative probability of finding an atom in one state compared to another is given by the Boltzmann factor of Equation (7.26), which depends on the energies of the two states such that Nk Nj = e(Ej−Ek)/KT , (9.10) but with Ej − Ek = ~ω, this becomes ρ(ω) = A e~ω/KTBkj −Bjk . (9.11) We wish to relate this energy density expression to the black-body distribution since both are equilibrium expressions, but Equation (9.11) is in terms of the energy density of the electromagnetic field, while our black-body expression of Equation (1.3), which is the same as Equation (7.31) without the zero-point term, is given in terms of the intensity, I(λ, T ). In order to relate these two expressions, we revisit the expression for the density of states given by Equation (7.4) where dxdy dz = V , dpxdpydpz = 4pip2 dp, and p = ~k. Then we have N dxdy dz dpxdpydpz = 4piV h3 ~3k2 dk = V 2pi2 k2 dk . (9.12) Then, since k = ω/c, and since there are two independent photon states for each energy (spin 1 and spin −1), the density of photons is Nω V = ω2 pi2c3 dω e~ω/KT − 1 (9.13) where we have included the Bose-Einstein factor since photons have integral spin, and the energy density is therefore ρ(ω)dω = ~ω Nω V = ~ω3 pi2c3 dω e~ω/KT − 1 . (9.14) Comparing this result with Equation (9.11), we first conclude that Bkj = Bjk, which means that the transition rates are the same for the upward and downward transitions. Solving for A, we find for the spontaneous emission rate, A = ~ω3 pi2c3 Bkj , (9.15) and since, from Equation (9.7) Bkj = pi|〈k|℘|j〉|2 3²0~2 , (9.16) the spontaneous rate is given by A = ω3|〈k|℘|j〉|2 3pi²0~c3 . (9.17) and The lifetime in an excited state is the reciprocal of the spontaneous rate, although if there is more than one final state, the lifetime is the reciprocal of the sum of the various rates.
Taylor & Francis eBooks•Aug 9, 2006•0 citations