Page 318 - Electrical Properties of Materials

300 Lasers

A lossless resonant circuit has a well-defined resonant frequency. However,
in the presence of losses the resonance broadens. In what form will the voltage
decay in a lossy resonant circuit? If the losses are relatively small, then circuit
theory provides the equation

U(t)= U 0 exp(–t/τ) cos 2πν 0 t. (12.21)
What is the corresponding frequency spectrum? If the oscillations decay,
then they can no longer be built up from a single frequency. The range of
necessary frequencies, that is the spectrum, is given by the Fourier transform
∞

f (ν)= U(t)exp(i2πνt)dt. (12.22)
0
Restricting ourselves to the region in the vicinity of ν 0 and after proper
normalization, we obtain
g(ν) is known as a Lorentzian (1/2)πτ
lineshape. g(ν)= 2 2 , (12.23)
2π[(ν – ν 0 ) +(1/2(πτ)) ]
If we work out now the frequency range between the half-power points, we
obtain
1
ν = , (12.24)
2πτ
which is the same as eqn (12.19), provided we identify the decay constant of
the circuit with the spontaneous lifetime of the quantum-mechanical state. So
again, a simple argument based on the uncertainty relationship agrees with that
based on a quite different set of assumptions.
In a practical case spontaneous emission is not the only reason why a state
has finite lifetime. Interaction with acoustic waves could be another reason
(electron–phonon collision in quantum-mechanical parlance) or collisions with
other atoms. The latter becomes important when lots of atoms are present in a
gas, leading to so-called pressure broadening.
All those mentioned so far belong to the category of homogeneous broad-
ening, where homogeneous means that conditions are the same everywhere in
the material. When conditions differ (say strain varies in a solid) then we talk
of inhomogeneous broadening.
The best example of inhomogeneous broadening is the so-called Doppler
broadening, owing to the fact that an atom moving with velocity, v, will emit
at a frequency,
v

ν = ν 0 1+ . (12.25)
c
In thermal equilibrium the atomic gas has a Maxwellian velocity distribu-
tion, hence the corresponding broadening may be calculated. The result (see
Exercise 12.8) for the normalized lineshape is
2
g(ν)= C 1 exp[–C 2 (ν – ν 0 ) ], (12.26)
where
M is the atomic mass.
c M 1/2 M c 2
C 1 = and C 2 = . (12.27)
v 0 2πk B T 2 k B T v 0