130_notes.dvi

29.13Examples

29.13.1The 2P to 1S Decay Rate in Hydrogen

29.14Derivations and Computations

29.14.1Energy in Field for a Given Vector Potential

We have the vector potential
A~(~r,t)≡ 2 A~ 0 cos(~k·~r−ωt).

First find the fields.

E~ = −^1

c

∂A~

∂t

= 2

ω

c

A~ 0 sin(~k·~r−ωt)

B~ = 2∇×~ A~= 2~k×A~ 0 sin(~k·~r−ωt)

Note that, for an EM wave, the vector potential is transverse tothe wave vector. The energy density
in the field is

U=

1

8 π

(

E^2 +B^2

)

=

1

8 π

4

(

ω^2

c^2

+k^2

)

A^20 sin^2 (~k·~r−ωt) =

2 ω^2

2 πc^2

A^20 sin^2 (~k·~r−ωt)

Averaging the sine square gives one half, so, the energy in a volumeV is

Energy=

ω^2 A^20 V

2 πc^2

29.14.2General Phase Space Formula

If there areNparticles in the final state, we must consider the number of statesavailable for each
one. Our phase space calculation for photons was correct even for particles with masses.

d^3 n=

V d^3 p

(2π ̄h)^3

Using Fermi’s Golden Rule as a basis, we include the general phase space formula into our formula
for transition rates.

Γi→f=

∫ ∏N

k=1

(

V d^3 pk

(2π ̄h)^3

)

|Mfi|^2 δ

(

Ei−Ef−

∑

k

Ek

)

δ^3

(

~pi−~pf−

∑k

~pk

)

In our case, for example, of an atom decaying by the emission of onephoton, we have two particles
in the final state and the delta function of momentum conservationwill do one of the 3D integrals
getting us back to the same result. We have not bothered to deal with the free particle wave function
of the recoiling atom, which will give the factor of V^1 to cancel theV in the phase space for the
atom.