QuantumPhysics.dvi

This is clearly a very powerful relation, because it impliesthat the details of the absorption and
emission process, encoded in the detailed common matrix elements, are irrelevant in determining
the photon occupation numbersnα(k).

Next, we denote the population numbers of the statesψAandψBbyN(A) andN(B) respec-
tively. Thermodynamic equilibrium requires that the detailed balance equation for the process,

N(A)wemi(α,k) =N(B)wabs(α,k) (20.39)

The ratio of these population numbers is given by the Boltzmann distribution formula,

N(A)

N(B)

=e−(EA−EB)/(kBT)=e− ̄hω/(kBT) (20.40)

herekBis the Boltzmann constant,Tis the temperature at equilibrium, andω=c|k|. Combining
(20.39) and (20.40), and eliminating the ratiosN(A)/N(B), andwemi/wabs, we find,

N(A)

N(B)

=

wabs(α,k)

wemi(α,k)

=

nα(k)

nα(k) + 1

=e− ̄hω/(kBT) (20.41)

The last equality may be readily solved fornα(k), and we find,

nα(k) =

1

e ̄hω/(kBT)− 1

(20.42)

This formula is an example of the occupation numbers for particles obeying Bose-Einstein statistics.

From this formula, it is a small extra step to derive the Planck formula for the black body
radiation spectrum as a function of temperature. The contributionU(ω)dωto the internal energy
density of the distribution of photons in an infinitesimal interval [ω,ω+dω] is given by

U(ω)dω=

1

L^3

× 2 × ̄hω×

1

e ̄hω/(kBT)− 1

×

[(

L

2 π

) 3

4 πk^2 dk

]

(20.43)

The factor 1/L^3 results from considering the energy density; the factor of 2results from the sum
over both polarizations of the photon; the factor ̄hωis the energy contribution of a single photon
with wave numberk; the denominator is the photon occupation number. Finally,the last factor, in
brackets, is the phase space factor giving the number of states in a spherical shell in wave-number
space of thicknessdk. It may be evaluated as follows. The allowed wave numbers in acubic box of
sizeL, with periodic boundary conditions, are

kx= 2πnx ky= 2πny kz= 2πnz (20.44)

The number of states in an infinitesimal volume element are then,

dnxdnydnz=

(

L

2 π

) 3

dkxdkydkz=

(

L

2 π

) 3

4 π k^2 dk (20.45)