130_notes.dvi

vector~k). Using classical E&M to compute the energy in a field (See Section 29.14.1) represented
by a vector potentialA~(~r,t) = 2A~ 0 cos(~k·~r−ωt), we find that the energy inside a volumeVis

Energy=

ω^2

2 πc^2

V|A 0 |^2 =N ̄hω.

We may then turn this around and writeA~in terms of the number of photonsN.

|A 0 |^2 = N ̄hω

2 πc^2

ω^2 V

=

2 π ̄hc^2 N

ωV

A~(~r,t) =

[

2 π ̄hc^2 N

ωV

]^12

ˆǫ

(

2 cos(~k·~r−ωt)

)

A~(~r,t) =

[

2 π ̄hc^2 N

ωV

]^12

ˆǫ

(

ei(

~k·~r−ωt)

+e−i(

~k·~r−ωt))

We have introduced the unit vector ˆǫto give the direction (or polarization) of the vector potential.
We now have a perturbation that may induce radiative transitions. There are terms with both
negative and positiveωso that we expect to seeboth stimulated emission of quanta and
absorption of quantain the the presence of a time dependent EM field.

But what about decays of atoms with no applied field? Here we need togo beyond our classical
E&M calculation and quantize the field. Since the terms in the perturbation above emit or absorb
a photon, and the photon has energy ̄hω, letsassume the number of photons in the field is
thenof a harmonic oscillator. It has the right steps in energy. Essentially, we are postulating
that the vacuum contains an infinite number of harmonic oscillators,one for each wave vector (or
frequency...) of light.

We now want to go from a classical harmonic oscillator to a quantum oscillator, in which the ground
state energy is not zero, and the hence the perturbing field is never really zero. We do this by
changingNtoN+ 1in the term that creates a photonin analogy to the raising operatorA†
in the HO. With this change, ourperturbation becomes

A~(~r,t) =

[

2 π ̄hc^2

ωV

]^12

ˆǫ

(√

Nei(

~k·~r−ωt)

+

√

N+ 1e−i(

~k·~r−ωt))

Remember that one exponential corresponds to the emission of a photon and the other corresponds
to the the absorption of a photon. We viewA~as an operator which either creates or absorbs a
photon, raising or lowering the harmonic oscillator in the vacuum.

Now there is aperturbation even with no applied field(N= 0).

VN=0=VN=0eiωt=

e

mc

A~·~p= e

mc

[

2 π ̄hc^2

ωV

]^12

e−i(

~k·~r−ωt)

ˆǫ·~p

We can plug this right into our expression for the decay rate (removing theeiωt into the delta
function as was done when we considered a general sinusoidal time dependent perturbation). Of
course we have this for all frequencies, not just the one we have been assuming without justification.
Also note thatour perturbation still depends on the volumewe assume. This factor will be
canceled when we correctly compute the density of final states.