LQMCh15RadiationTheory

We are interested mainly in developing a formula for the transition probability per unit of time for an atom to go from state n_i to state n_j . There are at least two aspects , absorption and emission of radiation by the atom. The emission of radiation or decay of the atom can be of two kinds , induced by external radiation or spontaneous, in the absence of a radiation filed.

a) Electric dipole radiation

A first approach to quantum radiation theory comes from the classical theory of electromagnetism.

The most elementary radiation is attributed to an oscillating electric dipole with frequency ω radiating with average power ( see ref. 6)

dE/dt = ω⁴ /p/² (12π ε_o c³ )^-1 , energy/time ~ watts~ ( SI units ) (1)

where p =q r is the dipole moment C-m , ε_o is the electrical permittivity of empty space , c is the speed of light.

The dipole moment is a vector proportional to the position vector r = x i + y j + z k .

Before proceeding further , a quick check of formula (1) for hydrogen, tells us that we are within the ballpark.

Let's say that optical transitions in the hydrogen atom are of the order E ~ 1eV =1.6E-19 J . The frequencies ω = 2π f and

f ~ 1.0E15 hertz . The speed of light is c=3.0E8 m/s and ε_o = 8.84E-12 C² /N-m² .

The electric dipole p ~ q r_atomic ~ 1.6E-19 * (1.E-10m) = 1.6E-29 C-m .

Equation (1) suggests that the lifetime of an excited state in hydrogen that emits a photon in the optical region has a lifetime ,

t_{lifetime ~} E (12π ε_o c³ ) /( ω⁴ /p/² ) ~ 3.6E-9 seconds

t:e*12*%pi*epsi*c^3/(w^4*p^2),numer;

3.6091343864124633E-9

No external field is present in the derivation of (1) hence it should correspond to the spontaneous emission of radiation by an excited atomic electron.

We can generalize (1) to quantum mechanics defining the transition element

< p_i,j > = ∫ Ψ^*_i (q r ) Ψ_j dτ . (2)

In spherical coordinates the electric dipole moment has components p_x =qr sin(θ) cos(φ) , p_y =qr sin(θ) sin (φ) , p_z =qr cos(θ) .

The dipole selection rules follow from the integration of eq.(2).

The magnetic quantum numbers must remain the same or change by one unit

∆m = m_j - m_i = 0 , (+/-) 1 . (3)

The rule on the angular momentum l is

∆ l = l_j -l_i = (+/- ) 1 , (4)

it is said that the parity of the wave function changes in the transition.

To prove eq .(3) consider first the z component p_z .

p_z ~ sin(θ) and has no dependence on the angle φ . The expectation value has a factor coming from the φ integral

<p_z > ~ ∫₀^2π exp (m_i i ) exp (-m_j i ) dφ = 0 , unless m_i = m_j . (5)

Consider now the expectation value of the component p_x .

The component p_x ~ cos (φ) = (1/2){exp( iφ) + exp(-iφ) } .The matrix element is

< p_x > ~ ∫₀^2π exp [(m_i - m_j ) i ] {exp( iφ) + exp(-iφ) } dφ .

The conditions for a non vanishing integral are

m_i - m_j +1 = 0 or m_i - m_j -1 = 0 , thus ∆m = (+/-) 1 . (6)

As for the rule of the angular momentum eq.( 4) consider first the following, the components of the dipole moment are odd functions , it has odd parity.

Consider the wave functions in a more explicit manner. The wave function carries at least three quantum numbers ( spin is excluded). Therefore

Ψ ~ Ψ_n,l,m (r,θ,φ ) = R_n,l (r) Y_l,m ( θ,φ ) and the spherical harmonics are given by

. (7)

The matrix element < Ψ_n,l,m /p_z / Ψ_n',l',m' > ~ < P_l^m (cos(θ)) / cos ( θ) / P_l'^m' (cos(θ)) > , or in terms of the variable x=cos(θ) ,

sin (θ) = (1-x² )^1/2 the integrals have the form

I = ∫_-1⁺¹ P_l^m (x) x P_l'^m' (x) dx . (8)

Any integral of the form odd * odd * odd is automatically zero. Also zero are the integrals even*odd*even .

So one must have the combination even*odd*odd or odd*odd*even to have a non zero integral.

Table 15.1 lists a few associated Legendre functions.

We will integrate a few expressions of the form

I_{lm ; l',m} =∫_-1⁺¹ P_l^m (x) x P_l'^m (x) dx ∆m =0 (9)

or I_{lm ; l',m+1} =∫_-1⁺¹ P_l^m (x) (1-x² )^1/2 P_l'^m+1 (x) dx ∆m = 1 . (10)

The results will justify the selection rule that says , I = 0 unless ∆l =1 .

MAXIMA code

define the functions P_l^m (x) and the electric dipole components

p00(x):=1 ; p10(x):=x ; p11(x): = -(1-x^2)^(1/2);

p20(x):=(1/2)*(3*x^2-1); p21(x):= -3*x*(1-x^2)^(1/2);

p30(x):=(1/2)*(5*x^3-3*x);

dipz(x):=x; dipy(x):=(1-x^2)^(1/2);

We already know that the p_z expectation vanishes unless ∆m =0.

<0,0/ p_z / 1,0 >

a) example with ∆m =0 , ∆l =1

integrate(p00(x)*dipz(x)*p10(x),x,-1,1); = 2/3

2/3

b) <0,0/ p_z / 2,0 >

example with ∆m =0 , ∆l =2

integrate(p00(x)*dipz(x)*p20(x),x,-1,1); =0

c) <0,0/ p_z / 3,0 >

integrate(p00(x)*dipz(x)*p30(x),x,-1,1); = 0

The examples from (a)-(c) show that the only non-zero case is (a) when ∆l =1 .

Now take the p_y component. As shown above (see eq.(6) ) it requires ∆m=1 .

d) <0,0/ p_y / 1,1 >

example with ∆m =1 , ∆l =1

integrate(p00(x)*dipy(x)*p11(x),x,-1,1); = -4/3

e) <0,0/ p_y / 2,1 > , ∆m =1 , ∆l =2

example with ∆m =1 , ∆l =1

integrate(p00(x)*dipy(x)*p21(x),x,-1,1); = 0

f) <0,0/ p_y / 3,1 > , ∆m =1 , ∆l =3

integrate(p00(x)*dipy(x)*p21(x),x,-1,1); = 0

Other trials will lead to the same conclusion. The conclusion is that the allowed dipole transitions rules are

∆m =0 , ∆l =(+/-) 1

and ∆m =(+/-)1 , ∆l = (+/-) 1 . (11)

A few associated Legendre functions , where x = cos(θ).

Table 15.1

b) Vibrating string as system of oscillators

The fundamental idea of this topic is to tie the energy of the electromagnetic field with the hamiltonian of a set of oscillators. The total energy of the radiation field is

U = (1/2)∫ ( E ² + B² ) d τ ~ ergs ~ (SI units) (12 )

On the other hand the hamiltonian of an oscillator is

H = (p² /2m + (1/2)k q² ) . ( 13 )

To establish such connection , it is illuminating to consider first the waves in a vibrating string of length L , as an analogy of the electromagnetic wave fields given by E and H. The vibrating string satisfies the wave equation

∂² y/ ∂ x² = (1/v² ) ∂² y/ ∂t² . (14)

Let the boundary conditions be y(0,t) =y(x=L,t) = 0 .

The Fourier expansion of the solution has the form

y(x,t) = ∑_s q_s (t) sin (s π x/L ) . (15)

The Fourier coefficients are given by

q_s (t) = (2/L) ∫₀^L y(x,t) sin ( s π x/L) dx . (16)

Take the string density to be a constant ρ = M /L . The kinetic energy of the string is the integral

T = (1/2) ρ ∫ (∂y/ ∂t )² dx , 0 ≤ x ≤ L .

= (1/2) ρ ∫ ∑_{s, n} q' _s (t) q' _n (t) sin (s π x/L ) sin (n π x/L ) dx

= (1/2) (M/L) ∑_s { q' _s (t) }² (L/2)

= (1/4) M ∑_s { q' _s (t) }² , (17)

where q' _s (t) =dq_s /dt .

Fig 1. Work done on a piece of string. τ is the tension in newtons.

The work done on a piece of string is calculated using various approximations.

∆W = vertical component of the tension *vertical diplacement of the midpoint

= τ_y (∆y/2) = τ sin (θ) ∆y/2

≈ τ tan (θ) ( ∂y/∂x )∆x /2

≈ τ ( ∂y/∂x )² ∆x /2 . (18)

The fundamental approximation is that the displacement is small, thus sin(θ) = tan(θ) = ∂y/∂x .

The potential energy for the complete string is the integral

V = (1/2) τ ∫₀^L ( ∂y/∂x )² dx

= (1/2) τ ∫ ∑_{s, n}(s π /L)(n π x/L) q _s (t) q _n (t) cos (s π x/L ) cos (n π x/L ) dx

= (1/2) τ ∑_s(s π /L)² q _s (t) ² (L/2)

= (1/4) (τ π²/L) ∑_s s ² q _s (t) ² . (19)

The momentum p_s conjugate to q_s is

p_s = ∂ T/ ∂ q_s^' = (M/2) q_s (t) ' . (20)

We have the hamiltonian

H(q_s , p_s ) = ∑_s { p_s² /M + (1/4) (τ π² s ² /L) q _s ² } . (21)

A big leap is taken by considering the Fourier coefficients q_s as coordinates of the system. Eq.(21) is very much like the hamiltonian of the one dimensional harmonic oscillator h (x,p) = p_x² /(2m) + (1/2) k x² .

The hamiltonian p_s² /M + (1/4) (τ π² s ² /L) q _s ² differs slightly from p_x² /(2m) + (1/2) k x² . Compare p_s² /M with p_x² /(2m) .

Corresponding to the hamiltonian , p_s² /M + (1/4) (τ π² s ² /L) q _s ² , there is the Schrodinger equation

- ( h'² /M) ∂² Ψ(q_s )/ ∂ q_s ² + k_s q _s ² Ψ(q_s ) = E_s Ψ(q_s ) , (22)

with k_s = (1/4) (τ π² s ² /L) .

Compare with the harmonic oscillator equation

-(1/2m) h'² ∂² Ψ(x)/ ∂x² + (1/2) k x² Ψ(x) = ε Ψ (x) . (23)

Multiply (22) by (1/2) and define ε_s= E_s /2 to get

- ( h'² /2M) ∂² Ψ(q_s )/ ∂ q_s ² + (1/2) k_s q _s ² Ψ(q_s ) = ε _s Ψ(q_s ) . (24)

The scale of "length" is L_scale = [ h'² /(Mk_s) ]^1/4 = [h'² 4 L /(M τ π² s² ) ]^1/4 .

The energy scale in eq.(24) is

ε_scale = k_s ( L_scale)² = h' (k_s /M)^1/2

= h' [ (τ π² s ² /(4ML) ]^1/2 ,

= h' ω_s . (25)

Hence the quantized energy eigenvalues of (24) are

ε_s = ( n_s + 1/2) h' ω_s . (26)

The corresponding eigenfunctions are,

Ψ(q_s ) = 1/( 2ⁿ n!)^1/2 ( α_s /π )^1/4 exp (- α_s q_s² /2) H_n ( α_s ^1/2 q_s ) , (27)

where α_s = 1/( L_scale ) ² = { M τ π² s² /(h'² 4 L) }^1/2 and H_n (u) is the hermitian polynomial of order n.

Example 1: A plane wave polarized in the + Y direction is traveling in the + X direction.

The fields are

E = e_y E₀ sin(kx-ωt) , B = e_z B₀ sin(kx-ωt) . ( 28 )

e_y is the unit vector along +Y and e_z is the unit vector along +Z.

Questions: What is the vector potential A ? What are the relations between E₀ , B₀ and the coefficients a_k,σ ?

The electric and magnetic fields are

E = - ∂A/ ∂t , B = del x A , and the vector A is also a plane wave.

If we take A = - e_y (E₀ /ω) cos(kx-ωt) we can verify that

- ∂A/ ∂t = e_y (E₀ /ω) { - (-ω) sin(kx-ωt ) } = E and

del x B = (e_x ∂/∂x )[- e_y (E₀ /ω) cos(kx-ωt)] = - e_z (E₀ /ω) {- k sin(kx-ωt) }=

= e_z (E₀ k/ω) sin(kx-ωt) =e_z B₀ sin(kx-ωt) .

Which is true since B₀ = E₀ /c .

Rewriting A in complex notation

A= [-e_y (E₀ /(2ω) )exp(-iωt)] exp(+ikx) + [-e_y (E₀ /(2ω) )exp(+iωt)] exp(- ikx) ( 29 )

we can compare (31 ) with (29 ) and assign

a_k,σ = [-e_y (E₀ /(2ω) )exp(-iωt)] (30)

a^*_k,σ = [-e_y (E₀ /(2ω) )exp(+iωt)] . (31)

The instantaneous energy density of the electric field is (in empty space)

u_E = (1/2) ε₀ E² ~ ( Coul² /(N-m² ) ) N² /Coul² ~ N/m² =N-m/m³ ~ joules /m³ . (32)

Using B₀ = E₀ /c we find that the magnetic field has the same energy density as the electric field.

u_B = (1/2) B² /μ₀ = (1/2) E² /(c² μ₀ ) =(1/2) (ε₀ μ₀ )E² /μ₀ . Thus

u_B = u_E . (33)

The total energy density is given by

U_total = u_E + u_B = ε₀E² ~ joules/m³ . (34)

The aim of this section is to carry the steps that takes us from the total energy expression , ∫ ε₀E² dV , to a sum of energies of quantized harmonic oscillators of the form ∑_k ( n_k +1/2) h*ω_n . Moreover the Fourier coefficients become operators in a scheme called second quantization.

It was shown in the previous example ,that an electromagnetic radiation field can be obtained through the vector potential A from the equations

E = - ∂ A / ∂t , B = del x A , (35)

and the vector potential A is subject to the condition of zero divergence ; del ∙ A =0 . The fields E and B are called transverse or radiation fields .

The vector potential can be written in terms as a Fourier series. The radiation is enclosed in a cubical box of side L=(V)^1/3

The classical vector A , is a real vector. It may be assumed that the wave vector components are quantized according to

k_x = n_x π /L , k_y = n_y π /L , k_z = n_z π /L . If necessary the box size becomes infinite, L→ ∞ , and the k_i become continuous variables.

A(r,t ) = (1/V^1/2 ) ∑ _k,α=1,2 { c_k,σ ε^α exp(ik∙r-iωt) + c ^*_k,σ ε^α exp(-ik∙r+iωt ) } . ( 36 )

ε^α is the polarization vector .There are only two. It is a real unit vector e.g. ε^α = u_x , u_y or u_z the unit vectors along X and Y axis. Along with the unit vector u_k , ε¹ and ε² form an orthonormal triad.

Take a wave propagating along +X axis ,then one would have u_k = u_x , ε¹ = u_y, ε² = u_z .

To simplify the notation we reduce (30) to one dimension and one polarization. As in example #1 , let the wave propagate in the only along the X axis , back and forth. Then exp(ik∙r-iωt) reduces to exp(ikx-iωt) and exp(-ik∙r+iωt ) goes over into exp(-ikx+iωt ). Take ε^α = ε¹ = -u_y (minus the unit vector along Y). That is , we have only one polarization and the vector A is along the Y axis.

The vector potential A reduces to

A (x,t ) = (1/L^1/2 ) ∑ _k - u_y { c_k,σ exp(ikx-iωt) + c ^*_k,σ exp(-ikx+iωt ) } , (37-a)

≡ ∑_k A_k (x,t) . (37-b)

The Fourier series of E is

E = - ∂ A / ∂t = - u_y ∑_k{ -i ω_k c_k,σ exp(ikx-iωt) +i ω_k c^*k,σ exp(-ikx + iωt) } / L^1/2 . (38)

Using the orthogonality conditions we get for the total energy of the radiation filed a sum

U_total = ∫₀^L ε₀E² dx = (ε₀/ L) ∑_k,σ2 ω_k² c_k,σ c^* _k,σ (L)

= ε₀ ∑_k2 ω_k² c_k,σ c^* _k,σ . (39)

In CGS units ε_{0 =1 ,} and E = -(1/c) ∂ A / ∂t where c is the sped of light. Thus

U_total =∑_k,σ2 ( ω_k /c ) ² c_k,σ c^* _k,σ . (40)

To make the connection with harmonic oscillators define

c_k,σ = (c/2) ( (i/ω_k) P_k,σ + Q_k,σ ) , (41)

c^*_k,σ = (c/2) ( -(i/ω_k) P_k,σ + Q_k,σ ) . (42)

Inserting in the expression for U_total gives

U_total =∑_k,σ (1/2) ( P_k,σ ² + ω_k² Q_k,σ ² ) . (43)

The radiation filed has been cast as a collection of independent harmonic oscillators.

Again we are reminded that initially ω_k = abs (k) c and abs(k) = [ (n_x π /L)² +(n_y π /L)² +(n_z π /L)² ]^1/2 . So there is some form of quantization from the boundary conditions.

(d) Creation and annihilation operators

We cross into another mathematical realm following Dirac suggestion that P_k,σ, Q_k,σ should be treated as non commuting operators.

We recall that [x,p_operator]ψ = (xp_operator -p_operator x)ψ = i h'ψ . By analogy

[Q_k,σ , P_k',σ' ] = ih' δ_{k k'} δ_σσ' . (44)

The Q 's conmute with themselves and so does the P's .

[Q_k,σ , Q_k',σ' ] = 0 , (45)

[P_k,σ , P_k',σ' ] = 0 . (46)

In Chapter 5 , http://www1.uprh.edu/rbaretti/LQMch5.htm , creation and annihilation operators were introduced in the context of the quantized oscillator.

We had the ground state with quantum number n=0 ; Ψ_n=0 (x) =(1/π)^1/4 exp(-x² /2) and characteristic energy E (n=0) = ( n+1/2)=1/2 .

The first excited state has n=1 , E= 3/2 and can be obtained with the rising or creation operator defined by

a+ = (1/2^1/2 ) ( -i p +x) = (1/2^1/2 ) ( - d/dx + x ) . (48)

Operating with with the rising operator a⁺ upon the ground state ,

a+ Ψ₀ = (1/2^1/2 ) ( - d/dx + x ) (1/π)^1/4 exp(-x² /2) = ( 2^1/2 / π^1/4 ) x exp(-x²/2) = Ψ_plus , (49)

The annihilation operator , a , is the conjugate of a⁺ ,

a ≡ (1/2^1/2 ) ( i p +x) = (1/2^1/2 ) ( d/dx + x ) . (50)

One readily verifies that

a Ψ₀ =(1/2^1/2 ) ( d/dx + x )(1/π)^1/4 exp(-x² /2)=0 . (51)

Conversely ,combining (48) and (50) we get

X = (1/2^1/2) ( a+ + a) , (52)

P = (1/2^1/2 )( a+ - a) . (53)

We are omitting the subscripts k,σ denoting the wave vector and the polarization degree of feedom.

Comparing the operators a and a⁺ with c_k,σ and c^*_k,σ one concludes they are the same by making h'=1, the speed of light c=1 and ω=1 and multiplying the a's operators by the factor 1/2^1/2. The correspondence between dimensionless a's and c's is

c = (1/2)^1/2 a . ( 54 )

Introducing the dimensions back ,from (40 ) we see that the coefficients c_k,σlead to non commuting operators a, a+

c_k,σ~ {energy (c/ω)² }^1/2 ~ {h' ω (c /ω)² /2}^1/2 → c (h'/2ω)^1/2a_operator , ( 55 )

c^* _k,σ → c (h'/2ω)^1/2a+_operator . ( 56 )

The properties of the operators a+ and a are summed up in abstract form by writing

a+ /n> = ( n+1)^1/2 /n+1> , ( 57 )

a /n> = n /n-1> , ( 58 )

along with the commutation relation

<n/aa+ - a+a/n> = δ_kk' δ_σσ' ( 59 )

It is some sort of an accounting procedure.

Introducing (56) in (40) we get the Hamiltonian operator of the radiation field writing

H = ∑_{k, σ} ( h'ω_k /2) ( a+a + a a+ )_k,σ . (60)

The expectation value is

H = < n_{k1 σ1} , n_{k2 σ2} , ..../ H / n_{k1 σ1} , n_{k2 σ2} , .... > . (61)

Each operator a or a+ acts exclusively upon the state labeled with subscripts k,σ.

Consider the state /n>,

( h'ω /2) <n/ a+a + a a+ /n> = ( h'ω /2) < n/ 2a⁺a +1 /n> , (62)

where we used (59). But <n/ a⁺ a/n> = n , thus

( h'ω /2) <n/ a+a + a a+ /n> = ( n+1/2 )h'ω . (63)

So (61) has an infinite number of terms

H = ∑_{k, σ} ( n_k,σ +1/2 )h'ω_k . (64)

The vacuum state where all n_k,σ = 0 , has infinite energy

H = ∑_{k, σ} h'ω_k /2 , (65)

called the zero point energy which is an infinite quantity.

This is no surprise if we recall the many assumptions and mathematical artifacts introduced in going from the classical expression

U_total = ∫₀^L ε₀E² dx , to the quantized H in (64) .

(e ) Time dependent perturbation theory

To deal with the question of the interaction of radiation and matter one has to cast Schrodinger equation in a time dependent form. Suppose the system has initially a Hamiltonian H₀ with characteristic functions ψ_n such that

H₀ ψ_n = E_nψ_n ( 66 )

with the time part separable by the complex exponential

ψ_n = u_n exp(-iE_n t/h') . ( 67 )

A perturbation H' is added to H₀ and they act upon a wave function

Ψ = ∑ _n a_n (t) u_n exp(-iE_n t/h') , (68 )

where the coefficients a_n (t) are unknown functions of time.

The time dependent Schrodinger equation is

(H₀ + H' ) Ψ = ih' ∂ Ψ / ∂t . (69)

The term H₀ Ψ = ∑ _n E_n a_n (t) u_n exp( -iE_n t/h') and the other term in (69) is

H' ∑ _m a_m (t) u_m exp( -iE_m t/h') = ih' ∑ _n (da_n/dt) u_n exp(-iE_n t/h') . (70)

Multiplying (70) by u_n' exp(iEt/h') and integrating over the space variables gives an equation for the a_n' coefficient

d a_n'/dt = (-i/h') ∑ _m <n' / H' / m > a_m (t) exp( -i(E_m - E_n' ) t/h') (71)

or dropping the prime over the index n ,

d a_n/dt = (-i/h') ∑ _m <n / H' / m > a_m (t) exp( -i(E_m - E_n ) t/h') , (72)

where <n / H' / m > = ∫ u_n * H' u_m dτ .

Eq (72) constitutes an infinite set of coupled differential equations. Many approximations are made starting here on a long winding road.

The first simplifying assumption is that initially there is only one state present in the sum (68) . First assume that at t=0 certain state labeled u₀ ( the subscript 0 need not be related to the ground state) is occupied with probability coefficient a₀ (t=0) =1 , all other

a_n(t=0) = 0 .

So (72) has only one term, setting a_m (t) = a₀ (t) ≈ 1 , E_m = E₀

d a_n/dt = (-i/h') <n / H' / 0 > exp( -i(E₀ - E_n ) t/h') . (73)

Integration of (73) gives

a_n (t) = <n / H' / 0 > { exp( -i(E₀ - E_n ) t/h') - 1}/(E₀ - E_n) . (74)

The square of a_n is

/ a_n (t) /² ≡ a_n (t) ( a_n (t) )* = 2 [<n / H' / 0 > /(E₀ - E_n) ]² (1- cos{(E₀ - E_n ) t/h'} ) .(75)

/ a_n /² is the probability that the system has made a transition from u₀ to u_n in a small time t . The time derivative is

d a_n ² /dt = (2/h') <n / H' / 0 >² /(E₀ - E_n) sin ( (E₀ - E_n ) t/h' ) ~ 1/time . (76)

Some manipulation is necessary with formula (75) if the energy spectrum forms a continuum.

Fermi's golden rule

Suppose that a continuum ( not a discrete spectrum) of states exists aroundE_n with a density ρ (E) ~ number of states /unit energy. For example this is the case in scattering problems. The exact expression for ρ (E) depends on the statistics obeyed by the particle , whether they are fermions (spin =1/2) or bosons ( spin =0 ,1 ) and indirectly on the boundary conditions.

Instead of (75) one writes

∫ a_n² ρ(E_n) dE_n ≈ 4 ρ(E_n) (H_n,0 )² ∫ [ sin {(E₀ - E_n ) t/2h'} /(E₀ - E_n) ]² dE_n . (77)

The integral is of the form ∫ {sin(α x) /x }² dx , see Fig 2 , α =t/(2h') . The maxima is equal to α² and for large α , the area under the curve is concentrated between x= - π/α and x = π/α .

We have

Area = (1/2)(2π/α)α² = α π . (78)

Inserting the area in (77)

∫ a_n² ρ(E_n) dE_n = (2π/h') ρ(E_n) (H_n,0 )² t ~ time (79)

A transition probability w(E₀ , E_n) is defined as

w(E₀ , E_n) = (2π/h') ρ(E_n) (H_n,0 )² ~ probability /time . (80)

The symbols E₀ and E_n lie within a continum of the electromagnetic radiation background.

Eq (80) is known as Fermi's golden rule.

Fig 2. Plot of f(α = 10 ,x) = {sin(αx) /x}² . The maxima at the origin is lim_x→0 f(α,x) = α² .

Examples :

Problem 15 from reference 1.

Consider a quantum oscillator initially in state n=0. At t=0 , the constant force F = c is applied i.e. H' = -cx . compute the probability of excitation to the first ( n=1) and second (n=2) states.

We use eq (75) . E₀ = 1/2 , E₁ = 3/2 , E₂ = 5/2 .

We can see that <1/H'/0> ≠ 0 , because the integrand in <1/H'/0> is even but <2/H'/0> = 0 because the integrand is odd.

H'_1,0 = <1/-cx/0> = ∫^+∞_-∞ ( 2^-12 (1/π)^1/4 2x exp(-x² /2) ) (-cx) ( (1/π)^1/4 exp(-x² /2) ) dx = -c/2^1/2 .

H' is small if c* Lscale << E₀ = 1/2 . But Lscale=1 , hence c << 1/2 . Inserting H'_1,0 in (75) yields

/ a₁ (t) /² ≡ a₁ (t) ( a₁ (t) )* = 2 [<n / H' / 0 > /(E₀ - E_n) ]² (1- cos{(E₀ - E_n ) t/h'} )

= 2 (c² /2) (1/(1/2 -3/2 )² ) (1 - cos (t) ) = 2 c² (sin² (t/2) )

**********

Formula (80) leads to the Einstein coefficients for absorption and induced emissions if one assumes that the atomic system in immersed in a blackbody reservoir of absolute temperature T.

Another possibility is that the atomic system interacts with the zero point square fluctuations ~ (E² + B² ) of the vacuum state. In this case it leads to the Einstein coefficient for spontaneous emission.

We start with spontaneous emission.

Let (see Ref. 3 )

ρ(E) = dN/(h'dω) ~ V ω² /(h' π² c³ ) number of states per unit energy (81)

To calculate the transition element , (H_i,f )² the simplest approximation is that of the dipole. See the previous example,

Suppose that H' = - eE ∙ r ~ force dot displacement ~ energy . It is called the dipole approximation since e r ( charge times displacement) is the classical electric dipole moment.

Then (H_i,f )² ~ e² <E² > < f/ r /i>² . The average of the squared field E² (ω) corresponds to fluctuations in the vacuum.

A quantized field theory expresses E and B in terms of non commuting creation (a⁺ ) and annihilation (a) operators, see ref. 8.

The operation of a+ on the vacuum state can "create" photons. We continue with our less rigorous approach.

We have ( E² (ω) + B ² (ω) ) V /(8π) = h'ω/2 , (82)

or since E² (ω) = B ² (ω)

E² (ω) V = 2π h'ω . (83)

However a plane wave will not have all three components of the field interacting with a dipole. For example a dipole in the z direction will only interact with E_z . So one may admit the plausibility of

E_z² V = E² (ω) V /3 = 2π h'ω/3 , (84)

and

( H_i,f )² = e² {2π h'ω/(3V) } < f/ r /i>² . (85)

Substitution of (81) and (85) in (80) yields the Einstein spontaneous emission coefficient labeled A_if

A_{i f} = (2π/h') V ω² /(h' π² c³ ) e² {2π h'ω/(3V) } < f/ r /i>²

= (4e² ω_if³ /(3h' c³ ) < f/ r /i>² ~ probabilty /time ~ 1/time . (86)

Remember that ω_{i f} = ( E_i - E_f ) /h' .

To check (86) we carry an order of magnitude calculation in CGS units.

The constants are e = 4.8E-10 esu , h'=1.05E-27erg-sec c=3E10 cm/s . Let < f/ r /i> ~ 1E-8 cm , and suppose we have an optical transition with λ = 5E-5 cm . Then ω= 2 π c/λ = 3.77E15/s.

Inserting these values in (86) gives

A_if = 5.80 E 7 /s , (87)

and a radiative lifetime of the order 1/A_if = 1.72 E-8 seconds. Compare this with the estimate given at the beginning of the chapter in section (a) Electric dipole radiation.

In the presence of radiation ,stimulated emission contributes to the decay augmenting the quantity given by (87) . To get the total probability for emission (i.e. decay) for example from i → f , the exponential Boltzman factor has to be introduced . The probabilities of finding a system in E_i or E_f are

P_i = C g_i exp(-E_i /kT) , ( 88 )

P_f = C g_f exp(-E_f /kT) , ( 89 )

where g_i , g_f are the degeneracies or weights of the initial and final states and C is a normalization constant.

We will suppose that E_i < E_f .

The sum of the probabilities per unit time for induced and spontaneous emission are

W^emision_i,f = P_f { u(ω_if) B_if + A _if } ~ 1/time . ( 90 )

The quantity u(ω) is given by Planck's formula

u(ω) = (hω³/(π² c³ ) ) { exp(h'ω/kT) -1 }^-1 , ( 91 )

where u(ω) ~energy-time/length³ ~ density of energy/frequency .Therefore the dimensions of B_if ~ length³ /time-h' .

The absorption probability per unit time would be

W^abs_fi = P_i u(ω_if) B_fi ~ 1/time . (92)

When the system is in equilibrium with its own radiation

W^abs_fi = W^emision_i,f ,

P_i u(ω_if) B_fi = P_f { u(ω_if) B_if + A _if } . (93)

Eq (93) gives a connection between the coefficients B_if and A_if . The g's and the B's must be related by

g_i B_fi = g_f B_if . (94)

One gets

{ ( P_i B_fi - P_f B_if)/P_f } u(ω_if) = A_if . (95)

But the left side of (95) is

{(g_i exp(-E_i/kT)B_fi - g_f exp(-E_f/kT)B_if )/g_f exp(-E_f/kT) }

= { ( exp(h'ω/kT) - 1) g_f B_if / g_f } = ( exp(h'ω/kT) - 1) B_if . (96)

The substitution of (96 ) in (95) cancels the factor ( exp(h'ω/kT) - 1) leaving

( h'ω_if³ /π² c³ ) B_if = A_if . (97)

The factor in front of B_if has a value ( asuming ω=3.77E15 rad/s)

( h'ω_if³ /π² c³ ) = 2.11E-14 ~joules-sec/m³ . (98)

Dividing

u(ω_if) B_if /A _if = (exp(h'ω/kT) -1)^-1 . (99)

In Fig 3 we plot (99) for ω = 3.77E15 rad/s ( λ= 5.0E-7 m in the visible region ) .The stimulated emission is a small fraction of the spontaneous emission.

Fig 3 .Plot of the factor (exp(h'ω/kT) -1)^-1 vs absolute temperature, with ω ~ visible region .

END OF CHAPTER