8/13/2019 Quantum.46

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SPONTANEOUS EMISSION   31

We have used the constraint imposed by the delta function when we replaced(EB − EA)2 by ω2

k. We now see why this is called the Dipole Approximation:

the transition probability depends upon the dipole moment (squared, as in

classical radiation theory). The factor   ε

∗λ(k) ·D implies that the probability

for photon emission is a maximum when   ε∗λ(k) is parallel to  D and thus  k

perpendicular to  D, again in agreement with classical radiation theory fordipole radiation. We also see that energy is conserved because the energyEB + ωk   in the final state is equal to  EA, the energy in the initial state,thanks to the  δ-function. Look over the calculation to see where the + signcomes from and verify that photon absorption would involve   aλ(k) andgive δ(EB − EA − ωk) (note the − sign in front of  ωk) automatically and thetransition probability would again conserve energy. As mentioned before,  D

is calculable for specific cases but one can sometimes see from symmetry thatD has to be zero. An example is radiation in a H-atom when it transitionsfrom a 2s   state to a 1s   state. The   s   states have   l = 0 and are thereforespherically symmetric, ex is odd, therefore the transition dipole moment iszero. Another way to see this is by looking at the integration over space tobe done to calculate  D. The wave functions are even functions but  x is oddso the integral is zero.

Experimentally we ask for the probability that a photon is emitted in aspecific region of phase space  d 3k. We know from statistical physics that

such a phase space volume is given by

Vd 3k

(2π )3  =

V k2d |k|d k

(2π )3  =

V ω2k d ωk d k

(2π)3  (1.119)

The product of the transition probability wfi  and the volume of phase spacein which we find the photon is the desired quantity

dwfi  =4π2

V   ωk | ε

∗λ(k) ·D |2

 V ω2kd ωkd k

(2π )3  δ(EB − EA + ωk) (1.120)

We see that V  cancels as it should (so we can let V  → ∞). We can integrateover the photon energy   ωk   because it is constrained by the   δ-function,enforcing energy conservation. If the experiment does not observe thepolarization of the photon we must sum over λ. We get

dwfi

d k

=ω3k

2π

λ=1,2

| ε∗λ(k) ·D |2 (1.121)

We perform the sum over  λ  using the three vectors   ε1(k), ε2(k),k/ωk   thatform a triplet of (real) unit vectors. We can write

|D|2 = D2x +D2

y  +D2z  = (ε1 ·D)2 + (ε2 ·D)2 +

k ·D

ωk

2

(1.122)