Spontaneous Emission (8-325)

2. Spontaneous Emission: Einstein A coefficient

a) Starting from the energy density per unit volume $u(ω)$ of the photon field [which Griffiths calls $ρ(ω)$] determine the mean value of the mean squared electric field $|~E (ω)|^2$ at this energy density, assuming the mean photon occupation number of a mode of frequency $ω$ is $n(ω)$.

b) Evaluate the electromagnetic energy density $u0(ω)$ in the absence of light $n(ω) = 0$ and hence the zero-point mean squared electric field $|~E(ω)|^2$ in the vacuum.

c) From your result for b) determine the transition rate for spontaneous emission in the absence of light, and hence the Einstein A coefficient for spontaneous emission due to

dipole transitions.

d) Using Griffiths Equation 9.54, etc. estimate the lifetime $τ = 1/A$ of an electron in the $|210>$ state of Hydrogen.


In Griffiths (and elsewhere) the volumic density of energy is

$u=ϵ/2*E^2$    if $u→ρ(ω)dω$  then $ρ(ω)=ϵ/2*|E(ω)|^2$    (in the question $ρ(ω)$  is called $u(ω)$)
If you are given the average occupancy  with photons of interval $dω$, $n(ω)$ then



$<|E(ω) |^2> =(∫_0^∞ n(ω)*|E(ω) |^2 dω)/(∫_0^∞ n(ω)dω)=$



The volume density as function of frequency $ρ(ω)$  is given by the Planck equation for the Black-body radiation.

$ρ(ω)=ℏ/(π^2 c^3 )*ω^3/(exp⁡(ℏω/k T)-1)$

Which is found by taking

$n(ω)=d_k/(exp⁡(ℏω/k T)-1)$   with $d_k=V/(π^2 c^3 ) ω^3$  being the photon “degeneracy” of the volume element

So that normally (in the presence of light)


When light is absent:

$n(ω)=d_k$  so that $ρ_0 (ω)=(ℏω^3)/(π^2 c^3 )$

Since in spectrum there is no frequency (you can approximate with only one value of $ω=ω_0$) one can write

$ρ_0 (ω)*δ(ω-ω_0 )=ϵ_0/2 |E(ω_0)|^2$   so that $|E(ω_0 ) |^2=(2/ϵ_0) *(ℏω_0^3)/(π^2 c^3 )$


$A=π/(3ϵ_0 ℏ^2 ) |p|^2*ρ_0 (ω)=(ω^3 |p|^2)/(3ϵ_0 πc^3 ℏ)$


Equation 9.54 relates A (spontaneous emission rate) coefficient to B coefficients (stimulated emission rates) between states b and a

$A=(ω_0^3 ℏ)/(π^2 c^3 )*B_{b a}$


$B_{b a}=π/(3ϵ_0 ℏ^2 ) |p|^2$   with $p=e<ψ_b |r| ψ_a>$

The allowed transition is $(b→a)$

$|210> →|100>$

Normally because $Y_{l m}$  states are orthogonal since in the initial state $l=1$ and in the final state l=0, p should be zero, but since it is its modulus that counts we only take the scalar product between the radial parts of the wave functions

$R_{21}=1/√(3)*1/(2a)^{3/2} *r/a*exp⁡(-r/2a)$   and $R_{10}=2/a^{3/2} *exp⁡(-r/a)$

so that

$p=e 1/(√(6) a^3 ) ∫_0^∞ r/a*exp⁡(-r/2a)*exp⁡(-r/a)*r*r^2 d r =$

$= e/(√(6) a^4 )*(256a^5)/81=(256e*a)/(81√6)$ 

where $r^2 d r$ comes from the element of volume $d V=r^2 d r*dΩ$ is spherical coordinates

so that

$B_{b a}=π/(3ϵ_0 ℏ^2 ) |p|^2=π/(3ϵ_0 ℏ^2 )*((256e*a)/(81√6))^2$

$A=(ω_0^3 ℏ)/(π^2 c^3 )*B_{b a}=(ω_0^3)/(3ϵ_0 c^3 πℏ) ((256e*a)/(81√6))^2$     and $τ=1/A$