Chapter 15 – The Quantum Theory of Radiation - (unfinished )

by Reinaldo Baretti Machín

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References:

1. Principles of quantum mechanics; nonrelativistic wave mechanics with illustrative applications.    

by    W. V. Houston

 

2. The Quantum Theory of Radiation: Third Edition by Walter Heitler

 

3. Theoretical Physics by A S Kompaneyets and George Yankovsky , Part III

4. Fundamentals of Modern Physics by Robert Martin Eisberg (Hardcover - June 1961) , chapter 9.

5. http://en.wikipedia.org/wiki/Dipole

 

 

We are interested mainly in developing a formula for the transition probability per unit of time for an atom to go from state ni to state nj . 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 εo c3 )-1      , energy/time ~ watts~ ( SI units )                              (1)

where      p =q  is the dipole moment C-m , εo 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  εo = 8.84E-12  C2 /N-m2 .

The electric dipole   p ~ q ratomic ~ 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 ,

                                        

                             tlifetime ~  E (12π εo 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

 

                                                  < pi,j >  = ∫ Ψ *i (q r ) Ψ j dτ                     .        (2)

In spherical coordinates the electric dipole moment has components px =qr sin(θ) cos(φ)  ,  py =qr sin(θ) sin (φ) ,  pz =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 =  mj - mi = 0 , (+/-) 1                   .        (3)  

The rule on the angular momentum l is

                                                ∆ l = lj -li = (+/- ) 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  pz .

 

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

 

                               <pz > ~ ∫0 exp  (mi  i ) exp (-mj i ) dφ  = 0  , unless  mi = mj   . (5)

 

Consider now the expectation value of the  component  px .

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

 

               < px > ~  ∫0 exp  [(mi  - mj ) i ]  {exp( iφ) + exp(-iφ) }  dφ .

The conditions for a non vanishing integral  are

 

                         mi - mj +1 = 0   or    mi - mj -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 ) = Rn,l (r) Yl,m ( θ,φ )   and the spherical harmonics are given by            

 

                 

                                .  (7)

 

 

The matrix element      < Ψ n,l,m /pz / Ψ n',l',m' > ~    < Plm (cos(θ)) / cos ( θ) / Pl' 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 Plm (x) x Pl' 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

 

                          Ilm ; l',m  =      -1+1 Plm (x) x Pl' m  (x) dx             ∆m =0                           (9)

or                       Ilm ; l',m+1  =      -1+1 Plm (x) (1-x2 )1/2 Pl' 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  Plm (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  pz expectation vanishes unless ∆m =0.

<0,0/ pz / 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/ pz / 2,0 >

example  with ∆m =0    ,  ∆l =2

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

0

 

 

 

c)    <0,0/ pz / 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 py component. As shown above (see eq.(6) )  it requires ∆m=1 .

d)  <0,0/ py / 1,1 >

example with ∆m =1    ,  ∆l =1

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

 

 

e)  <0,0/ py / 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/ py / 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  qs (t) sin (s π x/L )                                  . (15)

The Fourier  coefficients   are given by

                                            

                                                qs (t) = (2/L) ∫0L  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) =dqs /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

                                                    = τy (∆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  ps conjugate to qs is

 

                                                         ps = ∂ T/ ∂ qs' = (M/2) qs (t) '        .  (20)

We have the hamiltonian

 

 

                                  H(qs , ps ) = ∑s { ps2 /M   +   (1/4) (τ  π2 s 2  /L)  q s 2 }       . (21)

 

                     

 

 

A big leap is taken by considering the Fourier coefficients qs as coordinates of the system. Eq.(21) is very much like the hamiltonian of the one dimensional harmonic oscillator  h (x,p) = px2 /(2m)   + (1/2) k x2 .

The hamiltonian   ps2 /M   +   (1/4) (τ  π2 s 2  /L)  q s 2  differs slightly from  px2 /(2m)   + (1/2) k x2 . Compare ps2 /M with px2 /(2m) .

Corresponding to the hamiltonian ,  ps2 /M   +   (1/4) (τ  π2 s 2  /L)  q s 2 , there is the Schrodinger equation

   

                                   - ( h'2 /M) ∂2 Ψ(qs )/ ∂ qs 2   +  ks  q s 2 Ψ(qs ) = Es  Ψ(qs )               ,       (22)

with ks = (1/4) (τ  π2 s 2  /L) .

 

Compare with the harmonic oscillator equation

 

 

                                        -(1/2m) h'22 Ψ(x)/ ∂x2  + (1/2) k x2 Ψ(x)  = ε Ψ (x)     .                   (23)

Multiply (22) by (1/2) and define  εs = Es /2   to get

 

 

                      - ( h'2 /2M) ∂2 Ψ(qs )/ ∂ qs 2   + (1/2) ks  q s 2 Ψ(qs ) = ε s  Ψ(qs )               .     (24)

 

 

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

The energy scale in eq.(24) is

 

                                                εscale = ks ( Lscale)2 = h' (ks /M)1/2                                

                                                         =  h' [ (τ  π2 s 2  /(4ML) ] 1/2  ,

                                                         =  h' ωs                                                                     . (25)

Hence the quantized energy eigenvalues of (24) are

 

                                                 εs = ( ns + 1/2) h' ωs            .                                               (26)

 

The corresponding eigenfunctions are,

 

                                                    

                          Ψ(qs ) = 1/( 2n n!)1/2 ( αs /π )1/4  exp (- αs qs2 /2) Hn ( αs 1/2 qs )    ,           (27)

where αs = 1/( Lscale ) 2 = { M τ π2 s2 /(h'2 4 L) }1/2  and Hn (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 =    ey E0  sin(kx-ωt)    ,   B = ez B0 sin(kx-ωt)     .                                                   ( 28 )

ey is the unit vector along  +Y and ez 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 = - ey (E0 /ω)  cos(kx-ωt)  we can verify that 

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

   del x B =  (ex ∂/∂x )[- ey (E0 /ω)  cos(kx-ωt)] = - ez (E0 /ω) {- k sin(kx-ωt) }=

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

 

Which is true since    B0 = E0 /c   .

 

Rewriting A in complex notation

 

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

 

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

 

a k,σ  = [-ey (E0 /(2ω) )exp(-iωt)]                                                                         (30)

 

a* k,σ = [-ey (E0 /(2ω) )exp(+iωt)]                                                                 .    (31)

 

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

 

               uE = (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.

 

  uB = (1/2)  B20 = (1/2) E2 /(c2 μ0 ) =(1/2) (ε0 μ0 )E20  . Thus

 

                                                                 uB = uE                                      .     (33) 

The total energy density is given by

                               Utotal   =   uE + uB = ε0E2  ~    joules/m3       .  (34) 

 

 


 

 

 

(c) Energy of the radiation field in empty space (see ref. 8)

 

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 ( nk +1/2) h*ωn . 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 

kx = nx π /L  , ky = ny π /L ,  kz = nz π /L . If necessary the box size becomes infinite, L→ ∞ , and the ki become continuous  variables.

 

 

                 

     A(r,t ) = (1/V1/2 )  ∑ 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 = ux , uy or uz the unit vectors along X and Y axis. Along with the unit vector uk , ε 1 and ε 2 form an orthonormal triad.

Take a wave propagating along +X axis ,then one would have uk = uxε 1 = uy , ε 2 = uz .

 

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 = ε 1 = -uy   (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/L1/2 )  ∑ k - uy  { c k,σ   exp(ikx-iωt) + c * k,σ  exp(-ikx+iωt ) }     , (37-a) 

 

               ≡  ∑k Ak (x,t)                                                                                     . (37-b)

The Fourier series of E is

 

       E = -  ∂ A / ∂t = - uyk{ - i ωk c k,σ   exp(ikx-iωt) + i ωk c* k,σ   exp(-ikx + iωt)  } / L1/2  . (38)

 

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

         Utotal  =   ∫0L  ε0E2 dx = (ε0 / L) ∑k,σ 2 ωk2  c k,σ c* k,σ (L)

 

                              =  ε0  ∑k 2 ωk2  c k,σ c* k,σ                                                        .    (39)

In CGS units  ε0 =1 ,  and  E = -(1/c)  ∂ A / ∂t  where c is the sped of light. Thus

                                          Utotal  =∑ k,σ 2 ( ωk /c ) 2  c k,σ c* k,σ                              . (40)

To make the connection with harmonic oscillators define

                                     ck,σ = (c/2) (   (i/ωk)  Pk,σ +   Qk,σ )                 ,                     (41) 

                                              c* k,σ = (c/2) (   -(i/ωk)  Pk,σ + Qk,σ )            .                   (42)

Inserting in the expression for Utotal gives

                                Utotal  =∑ k,σ  (1/2) ( Pk,σ 2  + ωk2 Qk,σ 2 )                     .    (43)

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

Again we are reminded  that initially ωk = abs (k) c  and  abs(k) = [ (nx π /L)2  +(ny π /L)2 +(nz 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 Pk,σ , Qk,σ should be treated as non commuting operators.

 

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

 

                                      [Qk,σ , Pk',σ' ] =  ih' δk k' δσσ'                 .   (44) 

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

                                         [Qk,σ , Qk',σ' ] = 0                            ,  (45)

                  

                                                  [Pk,σ , Pk',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 ;  Ψn=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+ = (1/21/2 ) ( -i p +x) = (1/21/2 ) ( - d/dx + x )                                        . (48)

 

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

 

a+ Ψ0 = (1/21/2 ) ( - d/dx + x ) (1/π)1/4 exp(-x2 /2) = (  21/2 / π1/4 )  x exp(-x2/2) = Ψplus  ,  (49)

 

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

                                                               

                                a ≡ (1/21/2 ) ( i p +x) = (1/21/2 ) (  d/dx + x )              . (50) 

 

One readily verifies that

                                 a Ψ0 =(1/21/2 ) (  d/dx + x )(1/π)1/4 exp(-x2 /2)=0   .   (51)

 

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

 

                              X  = (1/21/2) ( a+ +  a)                                            , (52)                                                     

                              P = (1/21/2 )( 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 ck,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 1/21/2 .  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,σ  lead 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)1/2 aoperator     ,   (   55   )  

                  

              

                c* k,σ   →  c (h'/2ω)1/2 a + 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> = δkk'  δσσ'                                                           ( 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'ωk /2) ( a+a   +   a a+ )k,σ                          .   (60)

The expectation value is

  

                       H   = < nk1 σ1  , nk2 σ2 , ..../ H  /  nk1 σ1  , nk2 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 ( nk,σ  +1/2 )h'ωk                                                   .  (64)

 

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

 

                                   H   = k, σ h'ωk /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

 

Utotal  =   ∫0L  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 ψn such that

 

                                                                       H0  ψn  = En ψn                                 ( 66  )   

with the time part separable by the complex exponential                                                                  

                                                                     ψn  = un exp(-iEn t/h')      .                  ( 67  )    

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

                                                      

                                                                      Ψ = ∑ n an (t) un exp(-iEn t/h')           ,    (68   )

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

 

The time dependent Schrodinger equation is

                                             

                                                                       (H0  + H' ) Ψ =  ih' ∂ Ψ / ∂t           .   (69)

The term H0 Ψ = ∑ n En an (t) un  exp( -iEn t/h')  and the other term in (69) is

 

                                                H' ∑ m  am (t) um  exp( -iEm t/h')  = ih' ∑ n  (da n/dt) un exp(-iEn t/h')     .  (70)

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

 

                                                              d an' /dt = (-i/h')   ∑ m <n' / H' / m > am (t)  exp( -i(Em - En' ) t/h')   (71)

or dropping the prime over the index n ,

 

                                                           d an /dt = (-i/h')   ∑ m <n / H' / m > am (t)  exp( -i(Em - En ) t/h')     , (72)

where    <n / H' / m >   = ∫ un * H' um 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

an(t=0) = 0 .   

 

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

                                                 

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

Integration of (73) gives 

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

The square of an is

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

 

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

            

                d  an 2 /dt =  (2/h') <n / H' / 0 >2  /(E0 - En)   sin ( (E0 - En ) 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 around  En 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 

 

                       ∫ an2 ρ(En) dEn  ≈    4 ρ(En)  (Hn,0  )2  ∫ [ sin {(E0 - En ) t/2h'} /(E0 - En) ]2 dEn     . (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)

 

                                            ∫ an2 ρ(En) dEn =  (2π/h') ρ(En)  (Hn,0  )2 t      ~  time                        (79)

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

                                      

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

The symbols E0 and En  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   limx→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/21/2 .

 

 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 - En) ] 2 (1- cos{(E0 - En ) 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 , (Hi,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   (Hi,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 Ez . So one may admit the plausibility of

                                                      Ez2 V =  E2 (ω) V /3 = 2π h'ω/3                             ,   (84)

and

                                                   ( Hi,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 Aif

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

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

Remember  that ωi f = ( Ei - Ef ) /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

                                                         Aif  = 5.80 E 7 /s                                     ,     (87)

and a radiative lifetime of the order 1/Aif  = 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 Ei or Ef are

                                                                            Pi = C gi exp(-Ei /kT)                          ,              ( 88  )

                                                                            Pf = C gf exp(-Ef /kT)                            ,            ( 89  )

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

We will suppose that   Ei < Ef .

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

                                                      Wemision i,f = Pf { u(ωif ) Bif +  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 Bif ~ length3 /time-h' .

The absorption probability per unit time would be

                                                  Wabsfi = Pi u(ωif ) Bfi  ~ 1/time                                       .     (92)

When the system is in equilibrium with its own radiation

                              Wabsfi  =     Wemision i,f        ,

                              Pi u(ωif ) Bfi  =   Pf { u(ωif ) Bif +  A if }                                             .   (93)

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

                                                                 gi Bfi = gf Bif                                                    .  (94)

One gets

                                          { ( Pi Bfi - Pf Bif )/Pf }  u(ωif )   =  Aif                                       .   (95)

But  the left side of (95) is

        {(gi exp(-Ei/kT)Bfi - gf exp(-Ef/kT)Bif )/gf exp(-Ef/kT) }

           =  { ( exp(h'ω/kT) - 1) gf Bif  / gf }  =   ( exp(h'ω/kT) - 1) Bif                             .  (96) 

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

                                            

                                                                    ( h'ωif32 c3 ) Bif  = Aif                         .  (97)

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

                                            

                                                ( h'ωif32 c3 ) = 2.11E-14    ~joules-sec/m3          .    (98)  

Dividing 

                                             u(ωif ) Bif /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