January 26, 2004 16:26 WSPC/Book Trim Size for 9in x 6in b ook2
Atoms in a Classical Light Field 23
Writing the imaginary part of the dielectric function of the atom as
(ω)=4πχ
(ω),usingχ(ω) from Eq. (2.23) and employing the Dirac
identity, Eq. (1.69), we obtain
(ω)=ω
2
pl
π
2
n
f
nl
nl
[δ(ω −
nl
) − δ(ω −
ln
)] , (2.34)
with ω
2
pl
=4πn
0
e
2
/m
0
.Since|l is the occupied initial state and |n are
the final states, we see that the first term in Eq. (2.34) describes light
absorption. Energy conservation requires
n
= ω +
l
, (2.35)
i.e., an optical transition from the lower state |l to the energetically higher
state |n takes place if the energy difference
nl
is equal to the energy ω
of a light quantum, called a photon. In other words, a photon is absorbed
and the atom is excited from the initial state |l to the final state |n.This
interpretation of our result is the correct one, but to be fully appreciated it
actually requires also the quantum mechanical treatment of the light field.
The second term on the RHS of Eq. (2.34) describes negative absorption
causing amplification of the light field, i.e., optical gain. This is the basis of
laser action. In order to produce optical gain, the system has to be prepared
in a state |l which has a higher energy than the final state |n, because the
energy conservation expressed by the delta function in the second term on
the RHS of (2.34) requires
l
= ω +
n
. (2.36)
If the energy of a light quantum equals the energy difference
ln
, stimulated
emission occurs. In order to obtain stimulated emission in a real system,
one has to invert the system so that it is initially in an excited state rather
than in the ground state.
2.3 Optical Stark Shift
Until now we have only calculated and discussed the linear response of an
atom to a weak light field. For the case of two atomic levels interacting
with the light field, we will now determine the response at arbitrary field
intensities. Calling these two levels n =1, 2 with
2
>
1
, (2.37)