Laser Amplification
Fundamentally, an electro-magnetic wave travelling through a medium may gain or lose
energy depending on the state of excitation of that medium
Consider a small segment of medium, with
refractive index  and cross-sectional area A,
between z and z + z. The incident intensity is
I(z) integrated over all frequencies. It is also
assumed that the medium is composed of simple,
two-level atoms of the type previously discussed.
The population densities in the lower and upper
states are nm and nn respectively. We will also
assume propogation in one direction and
therefore ignore non-directional spontaneous
emission.
The change in total power in passing through the
segment of medium is:
(I(z)  I(z  z))A  
I(z)
zA
z
This may also be expressed in terms of the radiation density U and Einstein B coefficients:

I(z)
zA  BU(, z)h (n m  n n )zA
z
(note that the radiation density is due to the incident power, not black body radiation which
we may assume is negligible)
Now, the intensity is integrated over all frequencies. So, the intensity at a particular
frequency, , is I(z,  )  I(z )f ( ) where f ( ) is the spectral line profile function, defined
such that:

 f ( )d  1 .
0
The intensity can therefore be related to the radiation density by I(z )f ( ) 
c

U (, z ) .
We can therefore write the rate of loss of intensity as: (substituting for U)
BI(z)f ( )h (n1  n 2 )
I(z)

z
c
This function is clearly of the form
dy
 ky which has as its solution y  y 0 e  kx
dx
Hence, we can identify an absorption coefficient, K:
k ( ) 
Bf ( )h (n1  n 2 )
c
Taking into account the degeneracies of the states, and the relationship between the A
and B coefficients, this relationship may also be written:
k ( ) 
c2A
8 
2
2




gn 
gm
gm
 nm 
n n  f ( )  ( )   n m 
n n  f ( )
gm 
gn
gn



This is known as the Füchbauer–Ladeberg formula.
Clearly, for a normal sample of atoms in thermal equlibruim, the lower state will have
overwhelmingly the larger population. k() will therefore be positive, and absorption of the
beam will occur. However, if the population of the upper state can be made such that k()
is negative, amplification can occur. In this case, k() is referred to as the small-signal gain
coefficient.
Example, The lifetime of the upper level of the 632.8 nm transition in the HeNe laser is
1107 s. Calculate the degree of population inversion required to give a gain coefficient of
0.07 m1.
Assumptions and simplifications:
1. Assume the degeneracies of the upper and lower states are equal.
2. Assume the refractive index is 1.
3. Ignore any broadening of the spectral line except the natural width due to the lifetime.
4. A common simplification in such problems is to assume the line profile function is given
by f ( )  0,    0   2 and f ( )  1 ,    0   2 .
The lifetime is 1107 s, hence the Einstein A coefficient is 1107 s1. The wavelength is
632.8 nn, so the frequency is 0.5 THz. Given the relationship between natural lifetime and
bandwidth, f ( )  6.28  10 7 Hz 1 at the wavelength of the transition.
Using the Füchbauer–Ladeberg formula derived above, the required density of population
inversion is 7.821011 m3 (c.f., density of atoms in a HeNe laser running at a pressure of
10 mmHg is 1022 m3) i.e., there must be 7.821011 m3 more atoms in their upper state
than in the lower.
Threshold Condition for Laser Action
What conditions must be satisfied in the laser medium for laser action to occur?
Considering again a simple two-level system;
1. To get amplification at all, a population inversion (i.e., Nn > Nm)
2. A strong stimulating field is necessary, to ensure that stimulated emission is stronger
than spontaneous.
8h 3
1
 h
I.e., U ( ) 
, the stimulating energy density >> black body mode
3
c
e kT  1
density.
(1) is difficult. E.g., for h = 5 eV, and temperature = 1000 K (kT = 0.086 eV)
Nn
5
 e 0.086  15
.  10  25 !!
Nm
In any case, a population inversion is a non-equilibrium situation, and in fact would require
a negative temperature!
(2) can always be achieved, with a powerful enough lamp!
Assuming the above conditions can be satisfied, it is also necessary to achieve a
threshold condition for laser action.
Consider a laser cavity containing a medium of length L with a gain of k per unit length
and mirrors with reflectivities R1 and R2. All other sources of losses (e.g., diffraction) are
ignored. Then, the gain in energy in the beam, after one round-trip in the laser cavity is:
G  R1 R2 e 2 kl  e 2( kL  ) , defining R1 R2  e 2 (in this, of course, other sources of loss could
be included)
Now, to get amplification, we require G  1 i.e., kL   . Now, k is a function of wavelength,

so we will only amplify those frequencies such that k ( ) L  L .

gm 
Using k ( )  ( )   nm 
n  f ( ) (note – ““ because here we are dealing with gain
gn n 

coefficients rather than absorption coefficients)
the threshold condition becomes:
 gm



nn  nm  
 gn
 ( ) f ( ) L
If no atoms are excited, the absorption coefficient at the centre frequency of the transition,
0 is k 0  (0 ) f (0 )n0 where n0 is the total number density of atoms. The threshold
condition simplifies further to:
 gm
 n 0

, giving a fractional population inversion, n p  k L

nn  nm  
0
 gn
 k0L
e.g., for pink ruby,
( r1  100%, r2  90% )
k 0  0.28
cm1. Take a 10 cm rod with silvered ends,
Then, e 2  r1r2    0.05
 n  175
. %
(Note – gm = gn = 4 for the laser transition in ruby)
Bandwidth of a Laser
Any oscillator posesses a property known as the quality factor, Q. This may be defined in
many equivalent ways, but for our purposes, we will define it in terms of the laser
frequency and bandwidth:
Q


Again, for any oscillator, the rate at which energy is lost has the form:
W (t )  W0 e

2  0t
Q
At what rate is energy lost in the laser cavity?
Suppose a group of photons (P0 at time zero) is injected into a laser cavity, and suffers
losses described by the quantity  defined earlier.
Then, the number of photons remaining in the cavity at time t is:
P(t )  P0 e


 t
where  is the round-trip time of a photon in the cavity.
Hence, the characteristic time for a photon remaining in the cavity is t p 


Comparing with the earlier expression involving Q, we can see that:
Q
2 0

 

or, Q  2 0    2 0 t p
 

Hence, from our original definition of Q, the bandwidth of the laser is simply:
 
1
2t p
*This result is quite astonishing! You might think that the bandwidth of a laser line should
depend on the laser medium. It doesn’t – it depends on the laser construction. You might
as well regard the laser medium as a coherent ensemble of atoms, molecules or whatever
behaving in consert rather than individual objects.
Examples:
1, Suppose we have a passive (unpumped) laser cavity 0.5 m long, with 2% losses per
round trip.
0.98  e     0.02 Hence, t p 
2L
 0.17 s
c
The bandwidth is then simply:
 
1
 0.94 MHz
2t p
2, In an ideal laser, all losses are replenished, hence t p   and   0 . In practice,
t p  0.16 ms and   1 kHz  Q  5  1015 – a possible time and frequency standard.