Quantization of the E-M field

Transcription

1 April 6, 20 Lecture XXVI Quantization of the E-M field 2.0. Electric quadrupole transition If E transitions are forbidden by selection rules, then we consider the next term in the expansion of the spatial dependence of the field operator and the magnetic term in the interaction Hamiltonian. Recall that the e mcp A term in the interaction, that can connect initial and final states with a difference of a single photon f; n kλ H i; n kλ = e f e i(k x) p ɛ kλ i e iωt m 2ωV Expanding the exponent to first order, where k x is the small parameter (long wavelength and small atom) gives us f; n kλ H i; n kλ = e m 2ωV f ( + i(k x))p ɛ kλ i e iωt Typically the first order term contributes only if the zeroth order (dipole) is forbidden. Then f; n kλ H i; n kλ = i e m 2ωV f (k x)p ɛ kλ i e iωt Consider evaluation of the matrix element We can expand f (k x)p ɛ kλ i f (k x)(p ɛ kλ ) i = 2 f (k x)(p ɛ kλ) + (k p)(ɛ kλ x) i + 2 f (k x)(p ɛ kλ) (k p)(ɛ kλ x) i The first term 2 ((k x)(p ɛ kλ) + (k p)(ɛ kλ x)) = 2 k (xp + px) ɛ kλ As before, we can write the operator p = im [H 0, x] so that Finally xp + px = im [H 0, xx]. 2 k f xp + px i ɛ kλ = imω 2 k f xx i ɛ kλ.

2 This transition is the electric quadrupole (E2) transition. Since k ɛ = 0, where k i x i x j ɛ j = k i T ij ɛ j T ij = x i x j δ ij x 2. has zero trace and 5 independent components that can be written as a linear combination of Y2 m (spherical tensor operator). The the WE theorem tells use that the total angular momentum of initial and final states can change by at most 2. Spherical tensor operator Let s examine the tensor operator T ij in more detail. Recall that a cartesian tensor operator can be assembled from cartesian vector operators U and V according to W ij = U i V j As long as U and V transform as vectors under rotations then so will W ij. But the cartesian tensor does not transform irreducibly. We can write U i V j = U V δ ij + 2 (U iv j U j V i ) + 2 (U iv j + U j V i ) U V δ ij The first term transforms as a scalar, the second as a vector (namely U V) and the third term is a X symmetric traceless tensor. The 9 real parameters of the X cartesian tensor correspond to one parameter for the scalar, for the vector and 5 for the symmetric zero trace tensor. We can identify a correspondence between the spherical tensors with spherical harmonics. Y 0 = 4π cos θ = z 4π r T 0 = 4π V z Y ± = 8π sin x ± iy θe±iφ = T± V x ± iv = y 4π 2r 4π 2 ( ) 2 ( ) 2 x ± iy Y 2 ±2 = 2π sin2 θe ±2iφ = T±2 2 Vx ± iv = y 8π 2r 8π 2 Y 2 ± = 8π sin θ cos (x ± iy)z θe±iφ = T 2 (V x ± iv y ) 4π 2r 2 ± = V z 4π 2 Y2 0 = 6π ( (x + iy)/ 2(x iy)/ 2 cos2 θ ) = 6π 2z2 r 2 T0 2 = 6π 2 ( Vz 2 (V x + iv y )(V x iv y )/2 ) The above is obviously a special case, namely U = V but at least we see that we can write a cartesian tensor in spherical basis. More generally 2

3 T0 0 = U V = (U +V + U V + U 0 V 0 ) T q = (U V) q i 2 T 2 ±2 = U ± V ± T 2 ± = U ±V 0 + U 0 V ± 2 T 2 0 = U +V + 2U 0 V 0 + U V + 6 Indeed if X k q and Y k2 q 2 are irreducible spherical tensors under rotations then Z k q = q,q 2 kq k q q k 2 q 2 X k q Y k2 q 2 wnere kq k q k 2 q 2 is a Clebsch Gordon coefficient. We see that spherical tensors can be combined in the same way as angular momentum states as we can write analogously kq = q,q 2 kq k q q k 2 q 2 k q k 2 q 2 Back to the quadrupole operator Wigner Ekhart Theorem T ij T 2 ±2,±,0 = x i x j δ ij T k=2 q=±2,±,0 r 2 Y m=±2,±,0 l=2 We take advantage of the spherical tensor formalism to evaluate the matrix element for the electric quadrupole transition. Suppose the quantum numbers of initial and final state are α, j, m and α, j, m respectively. Then we need to evaluate α, j, m Tq 2 α, j, m The Wiger Ekhart theorem states that α, j, m T k q α, j, m = j, m k, q, j, m α j T k αj 2j + The Clebsch Gordon coefficient is zero unless j = m + q and k + j j k j We see that l 2. Parity is conserved in electromagnetic interactions. The quadrupole operator is even under the parity operation. Therefore, initial and final states must have the same parity so l = 2, 0. The wave functions of initial and final atomic states are R i (r)yl m (θ, φ) and R f ()Yl m (θ, φ) respectively. And we saw above that Tq k = r 2 Y q k (θ, φ). Therefore f T k q i = R f (r)r i (r)r 2 r 2 dr Yl m Y q dω k Y l m

4 We can use the triple integral for spherical harmonics dωyl m (2l (θ, φ)y m l (θ, φ)y m2 + )(2l 2 + ) l 2 (θ, φ) = l0 l 0l 2 0l m l 2 m 2 lm 4π(2l + ) The last Clebsch Gordon coefficient enforces the Wigner Ekhart selection rules. Note that for any given initial and final state, there will be a non-zero for only one of the five T k q. All that remains is the radial integrals Magnetic dipole transition The second term can be written 2 f (k x)(p ɛ kλ) (k p)(ɛ kλ x) i = 2 f (k ɛ kλ) (x p) i k ɛ kλ is the leading term in the plane-wave expansion of the magnetic field B and x p is the orbital angular momentum. This term contributes to magnetic dipole M transitions. The operator is constructed from Y m and corresponds to transitions between states with l. The M transition will be relevant when the E transition is zero, perhaps because initial and final states have δl even so no parity change. This might correspond to a spin flip. As stated above, the intrinsic parity of the photon is odd, so if initial and final states of the emitting atom have the same parity, it must be that the photon has some orbital angular momentum with respect to the atomic coordinate system. If the orbital angular momenum is kr = l =, then the photon would have had to be emitted at r = /k. But as we determined earlier, kr for typical energy differences between levels and so the rate for such a process is low. The next term in the interaction Hamiltonian is ( ) µ B = e σ (k ɛ) 2mc 2ωV to be compared to ( ) e L (k ɛ) 2mc 2ωV Clearly of the same order. 4

5 2.. PLANK RADIATION LAW 2. Plank radiation law Suppoe we have atoms that make transitions between state A and B as follows A γ + B The higher energy state A decays to state B with emission of a photon. Then state B absorbs a photon and transitions to A. If the system is in equilibrium then N(B)w abs = N(A)w emis where N(B), N(A) are the numbers of atoms in states A and B respectively. wabs is the probablity that an atom in the state B absorbs a photon and w emis is the probablity that an atom in state A emits a photon. If the atoms are in thermal equilibrium then N(B) N(A) = w emis = e EB/kT w abs e = E A/kT e ω/kt (2.) The transition probability from state A and n photons to the state B with n + photons w emis B, n γ + e ik x ɛ α pa k A, n γ 2 = (n γ + ) B e ik x ɛ α p A 2 The probability for transition from state B with n photons to state A with n is w abs A, n γ e +ik x ɛ α pa k A, n γ 2 = (n γ ) A e +ik x ɛ α p B 2 The ratio w emis = n + w abs n Together with Equation 2. we find that the number of photons in thermal equilibrium with wave number k and polarization ɛ is n = e ω/kt That is, we imagine that there is an oscillator with omega = c k and that it is in equilibrium at temperature T when it is in the n th energy level. Now imagine a box with walls that absorb and emit photons at all wavelengths and polarizations. It is filled with oscillators at every possible frequency. Each oscillator will be at energy E n = ω n e ω/kt The total number of oscillators per unit frequency is the density of states dn = 2 V 4πk2 dk (2π) = 2 V 4πω2 dω (2π) c The number of states per unit frequency per unit volume is ρ(ω) = 8πω2 (2π) c The total energy per unit frequency per unit volume is U(ω) = ρ(ω)e n = 8π ω (2π) c (e ω/kt ) 5