Photons Where We Consider a Radiation ﬁeld Whose Vector Potential a Ω = |K|C

Home , Wave vector

Photons where We consider a radiation field whose vector potential A ω = |k|c. satisfies the transversality condition Now ∇ · A = 0. A(x, t) 1 X X Because the electric and magnetic fields √ (α) ik·x ∗ (α) −ik·x = (ck,α(t)ˆ e + c (t)ˆ e ) V k,α 1 ∂ k α E = − A 1 X X c ∂t √ (α) ik·x ∗ (α) −ik·x = (ck,α(0)ˆ e + c (0)ˆ e ), B = ∇ × A V k,α k α satisfy the free space Maxwell equations where we have employed the four-vector notation

∇ · E = 0 k · x = k · x − ωt = k · x − |k|ct. ∂B ∇ × E = − ∂t The Hamiltonian function of the classical radiation field is ∇ · B = 0 Z 1 ∂E 1 2 2 3 ∇ × B = , H = (|B| + |E| ) d x c ∂t 2 1 Z = |∇ × A|2 + |(1/c)(∂A/∂t)|2 d3x. the vector potential satisfies the wave equation 2 1 ∂2A ∇2A − = 0. A straightforward calculation shows that c2 ∂t2 X X H = 2(ω/c)2c∗ c . We write the vector potential at the moment t = 0 as a k,α k,α α superposition of the periodically normalized plane waves k in an L-sided cube, Because the coefficients

u (x) = ˆ(α)eik·x, −iωt k,α ck,α(t) = ck,α(0)e like: satisfy the equation of motion

1 X X 2 √ c¨ = −ω c , A(x, t)|t=0 = (ck,α(0)uk,α(x) k,α k,α V k α=1,2 it would look like the classical radiation field were +c∗ (0)u∗ (x)). k,α k,α composed of independent harmonic oscillators. We define the variables Here V = L3 and ˆ(α), α = 1, 2 are real polarization 1 vectors. Q = (c + c∗ ) Due to the transversality condition we have k,α c k,α k,α iω P = − (c − c∗ ). ˆ(α) · k = 0, k,α c k,α k,α so the polarization can chosen so that the vectors With the help of these the Hamiltonian function can be (ˆ(1), ˆ(2), k/|k|) form a righthanded rectangular written as coordinate system. The Fourier components uk,α satisfy X X 1 H = (P 2 + ω2Q2 ). the orthogonality conditions 2 k,α k,α k α Z 1 3 ∗ d x u · u 0 = δ 0 δ 0 . V k,α k ,α0 kk αα Since ∂H Due to the periodicity conditions the wave vectors can ˙ = −Pk,α take the values ∂Qk,α ∂H = +Q˙ , kx, ky, kz = 2nπ/L, n = ±1, ±2,.... k,α ∂Pk,α At the moment t 6= 0 the vector potential is obtained the variables Pk,α and Qk,α are canonically conjugated simply by setting and the Hamiltonian function the sum of the total energies of the corresponding harmonic oscillators. c (t) = c (0)e−iωt k,α k,α Thus the classical radiation field can be thought to be a c∗ (t) = c∗ (0)eiωt, k,α k,α collection of independent harmonic oscillators. There • every oscillator is characterized by the wave vector k Thus we can write and the polarization ˆ(α), † a |nk,α i = c+|nk,α + 1i • the dynamic variables of every oscillator are k,α combinations of Fourier coefficients. ak,α|nk,α i = c−|nk,α − 1i.

We quantize these oscillators by postulating that Pk,α Because the states |nk,α i are normalized we can calculate and Qk,α are not any more pure numbers but operators the coefficients as which satisfy the canonical commutation rules 2 2 |c+| = |c+| hnk,α + 1|nk,α + 1i [Q ,P 0 ] = i¯hδ 0 δ 0 k,α k ,α0 kk αα = hn |a a† |n i k,α k,α k,α k,α [Q ,Q 0 ] = 0 k,α k ,α0 = hn |N + [a , a† ]|n i [P ,P 0 ] = 0. k,α k,α k,α k,α k,α k,α k ,α0 = nk,α + 1, † We define dimensionless combinations a and a of 2 † k,α ,α |c−| = hn |a a |n i = n . k k,α k,α k,α k,α k,α the operators Pk,α and Qk,α as We choose the phase of the coefficients so that at the 1 √ moment t = 0 we have ak,α = (ωQk,α + iPk,α) 2¯hω q 1 a† |n i = n + 1|n + 1i † √ k,α k,α k,α k,α a = (ωQk,α − iPk,α). k,α 2¯hω p ak,α|nk,α i = nk,α |nk,α − 1i. It is easy to see that they satisfy the commutation relations Because † † n = hn |N |n i = hn |a a |n i [a , a 0 ] = δ 0 δαα0 k,α k,α k,α k,α k,α k,α k,α k,α k,α k ,α0 kk † † [a , a 0 ] = [a , a 0 ] = 0. and because the norm of a vectors is always non-negative k,α k ,α0 k,α k ,α0 we must have Note In these relations the operators must be evaluated n ≥ 0. † k,α at the same moment, i.e. [a , a 0 ] stands in fact for k,α k ,α0 From this we can deduce that the only possible † the commutator [a (t), a 0 (t)]. eigenvalues are k,α k ,α0 We further define the Hermitean operator nk,α = 0, 1, 2,.... N = a† a . We interprete k,α k,α k,α • the state |nk,α i to contain exactly nk,α photons, It is easy to see that each of which is characterized by a wave vector k and a polarization ˆ(α). [a ,N 0 ] = δ 0 δ 0 a k,α k ,α0 kk αα k,α † † † • the operator a to create a photon with the wave [a ,N 0 0 ] = −δ 0 δαα0 a . k,α k,α k ,α kk k,α vector k and the polarization ˆ(α). Due to the Hermiticity the eigenvalues nk,α of the • the operator a to destroy a photon with the wave operator Nk,α are real and the eigenvectors k,α vector k and the polarizartion ˆ(α). Nk,α|nk,α i = nk,α |nk,α i • the operator Nk,α to count the number of photons form an orthonormal complete basis. with the wave vector k and the polarization ˆ(α) in With the help of the commutation rule the state † † [a ,N 0 0 ] = −δ 0 δαα0 a k,α k ,α kk k,α The state composed of various kind of photons is a direct product of individual vectors |nk ,α i: we see that i i |n , n , . . . , n ,...i N a† |n i = (a† N + a† )|n i k1,α1 k2,α2 ki,αi k,α k,α k,α k,α k,α k,α k,α = |n i ⊗ |n i ⊗ · · · ⊗ |n i ⊗ · · · . k1,α1 k2,α2 ki,αi = (n + 1)a† |n i. k,α k,α k,α The vector |0i stands for the state that has no kind of Similarly we can show that photons, i.e.

N a |n i = (n − 1)a |n i. |0i = |0 i ⊗ |0 i ⊗ · · · . k,α k,α k,α k,α k,α k,α k1,α1 k2,α2 Application of any operator ak,α onto this results always where zero. We say that |0i represents the vacuum. ω = |k|c. It is easy to see that a general normalized photon state can be constructed applying operations a† When we choose the energy scale so that k,α consecutively: H|0i = 0,

† n (a ) ki,αi Y k ,α the Hamiltonian takes the form |n , n ,...i = i i |0i. k1,α1 k2,α2 q n ! X X k ,α ki,αi i i H = ¯hωNk,α. α † † k Note Since the operators a and a 0 commute the k,α k ,α0 order of operators does not matter. The many photon When it acts on a many photon state the result is states are symmetric with respect to the exchange of H|n , n ,...i photons. We say that the photons obey the Bose-Einstein k1,α1 k2,α2 statistics or that they are bosons. X = n ¯hωi|n , n ,...i. ki,αi k1,α1 k2,α2 Since the numbers nk,α tell us the number of photons of i type (k, α) in the volume under consideration we call them the occupation numbers of the state. The quantum mechanical momentum operator is exactly Correspondingly the space spanned by the state vectors is of the same form as the classical function (the Poynting called the occupation number space. vector): In the quantum theory the Fourier coefficients of a 1 Z classical radiation field must be replaced by the P = (E × B) d3x corresponding non-commuting creation and annihilation c X X 1 operators. Substituting = ¯hk(a† a + a a† ) 2 k,α k,α k,α k,α α c 7→ cp¯h/2ω a (t) k k,α k,α X X 1 p † = ¯hk(N + ). c∗ 7→ c ¯h/2ω a (t) k,α 2 k,α k,α k α we get Since here the summation goes over all wave vectors the r term associated with the factor 1/2 will not appear in the 1 X ¯h h √ (α) ik·x final result the terms ¯hk and −¯hk cancelling each other. A(x, t) = c ak,α(t)ˆ e V 2ω k,α For the momentum operator we get thus i +a† (t)ˆ(α)e−ik·x . X X k,α P = ¯hkNk,α. k α Note Here A is an operator defined at every point of the space whereas A of the classical theory is a three For one photon states we have component field defined at every point. The variables x and t are both in classical and quantum mechanical cases Ha† |0i =hωa ¯ † |0i k,α k,α variables parametrizing the fields. Fields like the operator P a† |0i =h ¯ka† |0i, A are called field operators or quantized fields. k,α k,α Also in the quantum theory the Hamiltonian is of the so form 1 Z H = (B · B + E · E) d3x. 2 ¯hω = ¯h|k|c = photon energy Substituting the field operator A into the equations ¯hk = photon momentum. 1 ∂ The photon mass will be E = − A c ∂t 1 B = ∇ × A (mass)2 = (E2 − |p|2c2) c4 1 and noting that this time the Fourier coefficients do not = [(¯hω)2 − (¯h|k|c)2] commute we get c4 = 0. 1 X X H = ¯hω(a† a + a a† ) 2 k,α k,α k,α k,α The photon state is also characterized by its polarization α k ˆ(α). Since ˆ(α) transforms under rotations like a vector X X 1 = (N + )¯hω, the photon is associated with one unit of angular k,α 2 k α momentum, i.e. the spin angular momentum of the photon is one. We define the circularly polarized • x and t in the expression for the field operator A are combinations not quantum mechanical variables but simply parameters which the operator A depends on. For 1 ˆ(±) = ∓√ (ˆ(1) ± iˆ(2)). example, it is not allowed to interprete the variables 2 x and t as the space-time coordinates of a photon. We rotate these vectors by an infinitesimal angle δφ • the quantized field A operates at every point (x, t) of around the progation direction k. Their changes are the space where it with a certain probability creates and annihilates excitation states called photons. δφ δˆ(±) = ∓√ (ˆ(2) ∓ iˆ(1)) Thus photons can be interpreted as the quantum 2 mechanical excitations of the radiation field. = ∓iδφ ˆ(±). We consider photon emission and and absorption of non We select the propagation direction k as the quantization relativistic atomic electrons. The relevant interaction axis and compare this expression with the transformation Hamiltonian is of the form properties of angular momentum eigenstates X e H = − A(x , t) · p int m c i i i i e |jmiR = 1 − Jzδφ |jmi = (1 − im δφ)|jmi. ¯h e2 + A(xi, t) · A(xi, t) , 2m2c2 We see that e where the transversality condition is taken into account • the spin components of the polarizations ˆ(±) are by replacing the operator pi · A with the operator A · pi. m = ±1. The summation goes over all electrons participating in • the state m = 0 is missing due to the transversality the process. The symbols xi stand for their position condition. coordinates. Note If we had to consider the interaction of spin and • our original linear polarization states are 50/50 radiation we should also include the term mixtures of m = 1 and m = −1 states. (spin) X e¯h H = − σi · [∇ × A(x, t)]| . int 2m c x=xi Hence the photon spin is always either parallel or i e antiparallel to the direction of the propagation. The operators a and a† are time dependent and so This time the Hamiltonian operator Hint operates not k,α k,α only on the atomic states but also on the photon states. they satisfy the Heisenberg equations of motion In the quantum theory of radiation i • the vector describing the initial state |ii is the direct a˙k,α = [H, ak,α] ¯h product of an atomic state A and a (many) photon i X X 0 state characterized by the occupation numbers n : = [¯hω N 0 0 , a ] k,α ¯h k ,α k,α k0 α0 |ii = |Ai ⊗ |nk,α i = |A; nk,α i. = −iωak,α like also • the vector describing the final state |fi is the direct a˙ † = iωa† . product of an atomic state B and a (many) photon k,α k,α state characterized by the occupation numbers n 0 : k ,α0 These equations have solutions

|fi = |Ai ⊗ |n 0 0 i = |A; n 0 0 i. −iωt k ,α k ,α ak,α = ak,α(0)e a† = a† (0)eiωt. k,α k,α Absorption Now The ﬁnal form of the ﬁeld operator is then

r |ii = |A; nk,α i 1 X ¯h h √ (α) ik·x−iωt A(x, t) = c ak,α(0)ˆ e |fi = |B; n − 1i. V 2ω k,α k,α i In the first order perturbation theory the amplitude of + a† (0)ˆ(α)e−ik·x+iωt . k,α the process |ii −→ |fi We should note that is the matrix element of the interaction operator HI • the operator A is Hermitean. between the states |ii and |fi. Up to this order • only a leads to a nonzero matrix element, is a† which adds one photon to the final state. The k,α k,α eventhough the field operator A is a linear relevant matrix element is now superposition of creation and annihilation operators † a and a , respectively. hB; nk,α + 1|Hint|A; nk,α i k,α k,α s • the term A · A is out of the question in this process e (nk,α + 1)¯h X = − hB|e−ik·xi p · ˆ(α)|Aieiωt. because it either changes the number of photons by m 2ωV i e i two or does not change it at all. If nk,α is very large then The first order transition matrix element is now q p nk,α + 1 ≈ nk,α , hB; nk,α − 1|Hint|A; nk,α i e and the semiclassical and quantum mechanical treatment = − hB; nk,α − 1| mec coincide. r X ¯h If n is small the semiclassical method fails completely. c a (0)eik·xi−iωtp · ˆ(α)|A; n i k,α 2ωV k,α i k,α In particular, the semicalssical treatment of the i spontaneous emission, n = 0, is impossible. The s k,α e nk,α ¯h X semiclassical method can be applied if we insert the atom = − hB|eik·xi p · ˆ(α)|Aie−iωt. m 2ωV i into the fictitious radiation field e i A(emis) = A(emis)e−ik·x+iωt, Comparing this with the matrix element of the 0 semiclassical perturbation potential where s eA (emis) (nk,α + 1)¯h † 0 i(ω/c)nˆ ·x A = c ˆ(α). Vni = − e ˆ · p 0 mec ni 2ωV (emis) we see that they both give exactly the same result The field A is not provided we use in the semiclassical theory the equivalent • directly proportional to the number of photons n , radiation field k,α (abs) (abs) (abs) ik·x−iωt • the complex conjugate of the field A . A = A0 e , Example Spontaneous emission from the state A to the where the amplitude is state B. s In the first order the transition rate is (abs) nk,α ¯h A = c ˆ(α). 0 2ωV wA→B 2π = |hB; 1 |H |A; 0i|2δ(E − E +hω ¯ ) Because the transition probability is ¯h k,α int B A 2 • according to the semiclassical theory directly 2π e2¯h X = hB|e−ik·xi ˆ(α) · p |Ai proportional to the intensity of the radiation, ¯h 2m2ωV i e i 2 |A0| ∝ nk,α , ×δ(EB − EA +hω ¯ ).

• according to the quantum theory directly Like in the photoelectric efect we can deduce that the proportional to the occupation number n , number of the allowed photon states ρ(E, dΩ) in the k,α energy interval (¯hω, ¯hω + d(¯hω)) and in the solid angle dΩ both the semiclassical and quantum mechanical results is 2 give equivalent results also at low intensities, i.e. when 2 V ω ρ(E, dΩ) = n dn dΩ = 3 3 d(¯hω) dΩ. nk,α is small. (2π) ¯hc The transition rate of photons emitting into a certain Emission solid angle is thus Now 2 2π e2¯h V ω2 dΩ |ii = |A; n i X −ik·xi (α) k,α wdΩ = hB|e ˆ · p |Ai , ¯h 2m2ωV i (2π)3¯hc3 e i |fi = |B; nk,α + 1i wherehω ¯ = E − E . and in the first order the only potential term of the field A B We consider only hydrogen like atoms so that only one r 1 X ¯h h electron participates in the process and we restrict to the √ (α) ik·x−iωt A(x, t) = c ak,α(0)ˆ e dipole approximation. Then V 2ω k,α 2 e ω (α) 2 † (α) −ik·x+iωti + a (0)ˆ e wdΩ = 2 2 3 |hB|p|Ai · ˆ | dΩ. k,α 8π me¯hc Earlier we saw that • in the term A · A creation and annihilation operators appear as quadratic, im (E − E ) hB|p|Ai = e B A hB|x|Ai ¯h only the quadratic term A · A contributes in the first = −imeωxBA. order perturbation theory. Only two of the terms of the form We let the symbol Θ(α) stand for the angle between the (α) † † † † vector xBA and the polarization direction ˆ , i.e. aa , a a, aa, a a

cos Θ(1) = sin θ cos φ in the operator A · A have non zero matrix elements cos Θ(2) = sin θ sin φ, provided that † 0 (α0) when θ and φ are the direction angles of the vector x. • a creates a photon of the type (k , ˆ ), Then (α 2 3 • a annihilates a photon of the type (k, ˆ ), e ω 2 2 (α) wdΩ = |xBA| cos Θ dΩ. 8π2¯hc3 and then † h1 0 |a a 0 |1 i = 1. The total transition rate is obtained by integrating over k ,α0 k,α k ,α0 k,α all propagation directions k/|k| and summing over both Now polarizations: e2ω3 2 hB; 1 0 |H |A; 1 i w = |xBA| . k ,α0 int k,α 3π¯hc3 2 e The life time of a state was obtained from the formula = hB; 1 0 A(x, t) · A(x, t) A; 1 i k ,α0 2 k,α 2mec 1 X e2 = wA→Bi , † † = hB; 1 0 (a a 0 + a 0 a ) τA k ,α0 2 k,α ,α0 ,α0 k,α i 2mec k k 2 c ¯h 0 0 0 where we have to sum also over the magnetic quantum √ (α) (α ) i(k−k )·x−i(ω−ω )t × ˆ · ˆ e A; 1k,αi numbers m. For example the life time of the hydrogen 2p 2V ωω0 2 2 state is e c ¯h (α) (α0) −i(ω−ω0)t −9 √ ˆ ˆ τ(2p −→ 1s) = 1.6 × 10 s. = 2 2 · e hB|Ai, 2mec 2V ωω0

±k·x Electron photon scattering where again the exponential functions e are replaced We consider the process by the constant 1 (the long wave length approximation). In the ﬁrst order we have thus

|1 i −→ |1 0 0 i, t k,α k ,α i Z 0 c(1)(t) = − eiωfit V (t0) dt0 ¯h fi i.e. t0 2 2 1 e c ¯h (α) (α0) • before the scattering the atom is in the state A, and = 2 √ 2δABˆ · ˆ (α) i¯h 2m c 0 k and ˆ are the wave vector and polarization of e 2V ωω t Z 0 0 the incoming photon. × ei(¯hω +EB −hω¯ −EA)t /h¯ dt0, 0 • after the scattering the atom is in the state B, k0 is 0 0 the wave vector and ˆ(α ) the polarization vector of where ω = |k|c and ω0 = |k |c. Now the outgoing photon. • in the transition amplitude c(1)(t) the interaction is The Hamiltonian of the interaction is in fact of second order: A · A. e e2 • in the second order correction c(2)(t) the term A · p Hint = − A(x, t) · p + 2 2 A(x, t) · A(x, t). is also of second order. mec 2mec Because To collect all contributions up to the second order in the interaction we have to consider also the correction c(2)(t), • the number of photons does not change in the into which we take all double actions of the operator scattering, A · p. Now

2 t t0 • in order to be non zero the matrix element of the i Z Z 0 (2) X 0 00 iωfmt 0 interaction must contain products of photon creation c (t) = − dt dt e Vfm(t ) ¯h t t and annihilation operators, m 0 0 00 iωmit 00 ×e Vmi(t ). • in the term A · p creation and annihilation operators appear as linear, Thus there are two possibilities: the interaction A · p can • at the moment t1 annihilate the incoming photon photons c/V , so the diﬀerential cross section is (α) (k, ˆ ) and at some later time t2 create the 0 0 (α0) dσ 2 ω (α) (α0) outgoing photon (k , ˆ ) or = r δABˆ · ˆ dΩ 0 ω 0 • at the moment t1 create the outgoing photon 1 (p · ˆ(α )) (p · ˆ(α)) 0 (α0) X BI IA (k , ˆ ) and at some later time t2 annihilate the − me EI − EA − ¯hω incoming photon (k, ˆ(α)). I (α) (α0) 2 (p · ˆ )BI (p · ˆ )IA Between the moments t and t the atom is in the state I, + , 1 2 E − E +hω ¯ 0 which normally is neither of the states A and B. I A −13 In the intermediate state I there are thus two where r0 ≈ 2.82 × 10 cm is the classical radius of possibilities: either there are no photons present or both electron. This expression is known as the incoming and outgoing photons are present. We get thus Kramers-Heisenberg formula. (in the dipole approximation) Example Elastic scattering. Now A = B jahω ¯ =hω ¯ 0. Using the commutation 2 2 Z t Z t2 (2) 1 c ¯h e relations of the operators x and p, the completeness of c (t) = 2 √ − dt2 dt1 (i¯h) 2V ωω0 mec 0 0 the intermediate states and the relation X (α0) i(E −E +¯hω0)t /h¯ × hB|p · ˆ |Iie B I 2 pAB = imeωABxAB I we can write ×hI|p · ˆ(α)|Aiei(EI −EA−hω¯ )t1/h¯ (α) (α0) 1 X h (α) (α0) X (α) i(E −E +¯hω)t /h¯ ˆ · ˆ = (x · ˆ ) (p · ˆ ) + hB|p · ˆ |Iie B I 2 i¯h AI IA I I (α) (α0) i 0 0 −(p · ˆ )AI (x · ˆ )IA ×hI|p · ˆ(α )|Aiei(EI −EA−hω¯ )t1/h¯ 1 X 2 (α) (α0) 2 2 = (p · ˆ )AI (p · ˆ )IA, c ¯h e me¯h ωIA = − √ I i¯h2V ωω0 mec (α0) (α) where ωIA = (EI − EA)/¯h. X (p · ˆ )BI (p · ˆ )IA × We see that E − E − ¯hω i I A (α) (α0) δAAˆ · ˆ (α) (α0) ! (p · ˆ ) (p · ˆ ) " (α0) (α) BI IA 1 (p · ˆ ) (p · ˆ ) + 0 X AI IA EI − EA +hω ¯ − me¯h ωIA − ω I t Z 0 0 i(EB −EA+¯hω −hω¯ )t2/h¯ (α) (α ) # × dt2 e . (p · ˆ )AI (p · ˆ )IA 0 + ωIA + ω (1) For the transition rate we get combining the terms c (t) " (α0) (α) 1 X ω(p · ˆ )AI (p · ˆ )IA and c(2)(t) and taking into account the relation = − me¯h ωIA(ωIA − ω) I 2 Z t 0 # ixt0 0 ω(p · ˆ(α)) (p · ˆ(α )) lim e dt = 2πtδ(x) − AI IA . t→∞ 0 ωIA(ωIA + ω) the expression If ω is small then Z (1) (2) 2 1 1 ± (ω/ωIA) wdΩ = (|c + c | /t)ρ(E, dΩ) dE ≈ . ωIA ∓ ω ωIA 2 2 2 2 02 2π c ¯h e V ω Then = √ 2 3 3 dΩ ¯h 2V ωω0 mec (2π) ¯hc X 1 h (α0) (α) (p · ˆ )AI (p · ˆ )IA (α) (α0) ω2 × δABˆ · ˆ I IA

(α) (α0) i (α0) (α) −(p · ˆ )AI (p · ˆ )IA 1 X (p · ˆ )BI (p · ˆ )IA − X h (α0) (α) me EI − EA − ¯hω 2 I = me (x · ˆ )AI (x · ˆ )IA (α) (α0) 2 I (p · ˆ )BI (p · ˆ )IA 0 + . (α) (α ) i 0 −(x · ˆ )AI (x · ˆ )IA EI − EA +hω ¯ 2 (α0) (α) Because in the initial state there was exactly one photon = me([x · ˆ , x · ˆ ])AA in the volume V and the ﬂux density of the incoming = 0. The diﬀerential cross section is now 2 3 dσ r0 4 X 1 = ω dΩ me¯h ωIA I (α0) (α) ×[(p · ˆ )AI (p · ˆ )IA 2 (α) (α0) + (p · ˆ )AI (p · ˆ )IA]

2 r0me 4 X 1 = ω ¯h ωIA I (α0) (α) ×[(x · ˆ )AI (x · ˆ )IA 2 (α) (α0) +(x · ˆ )AI (x · ˆ )IA] .

At long wave lengths the diﬀerential cross section obeys the Rayleigh law or dσ 1 ∝ . dΩ λ4 Now

• for ordinary colourless gases ωIA corresponds to wave lengths in the ultraviolet,

• for the visible light we have then ω ωIA, so our approximations are valid in the atmossphere. The theory explains why the sky is blue and the sunset red.