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Quantum optics
Eyal Freiberg
Question no.1
Derive the condition for lasing:
The equations for the two levels are:
dN1
dN 2

 BW ( N 2  N1 )  AN 2
dt
dt
Where now the radiation density
The thermal black body density
W
has two independent components.
WT21 and the external density WE
(the Latter did not exist in Einstein's treatment)
W  WT21  WE
Let’s write the eq. for N1
.
in terms of N1
We obtain: N 1  ( N  N 1 ) A  ( N  2 N 1 ) BW
and N
knowing that N  N1  N 2
Question no.1
The General solution of this equation is in the form:
N1  e  t  
And if we plug it into the rate eq. we obtain:
 e  t  ( A  2BW )  ( A  2BW )e  t  ( A  BW )  0

Is determined from the initial condition

N1 (0)  N
thus:
( A  BW ) N
BWN
;   A  2 BW ;  
A  2 BW
A  2 BW
The final solution is: N 1 (t ) 
BWN
AN
(1  e ( A 2 BW )t ) 
A  2 BW
A  2 BW
Also, the population of N 2 is immediately given as N 2 (t )  N  N1 (t )
Question no.1
Steady state implies that there is no change in population.
Then we can deduce from the rate equation:
BWT
N2
1


N1
A
N1  N 2
1
N2
We know that in equilibrium
N1
e
N2
w1 2
k BT
Therefore:
A
WT 
B (e
w1 2
k BT
 1)
12

 2c 2
3
comparing to plank’s law
WT
1
(e
w1 2
k BT
 1)
Question no.1
12
We see that A 
B
2 2
 c
3
The condition for lasing is the so-called population inversion, N 2  N1
It is clear that this cannot be achieved with a two level system as the maximum value
can be reached is
N
2
when
W 
Question no.2
The name stands for Light Amplification by Stimulated Emission of Radiation
We can think of it as the Bose condensate for light: one photon that bounces
back and forth in a cavity with two highly reflecting mirror, stimulates
two photons into the same state as the original photon. So we get a huge
amount of coherent radiation very quickly in this way.
If one of the mirrors can also transmit then this output gives us the laser light.
Its properties are:
high intensity, coherence and directionality.
Question no.2
The modes allowed are those whose wavelength satisfies:
The corresponding frequency is
n 
2c


nc
L
n  2L
N
E   E i  E 0 ie ii t
So we need to calculate
i 1
Suppose that we have N different waves in a cavity. We then have
e
i
The intensity is
inct
L

1 e
iNct
1 e
ict
L
L
sin 2 ( Nct
2L
I
sin 2 (ct
)
2L
The peaks occur when
t  2 Ln
)
c
Question no.2
The width of each pulse is given by
2L
cN
(distance from the central maximum at t = 0 to the first Minimum ).
The extra phases come from the fact that the atoms in the walls of
the cavity emit photons at random times. The intensity is now
I  E *  E   jk E 0 e
2
 i j
e ik  E 0 ( N   j  k e
2
i (k  j )
 E 0 ( N  2 j  k cos( k   j ))
2
And thus:
2
I  E0 N
The average of cosine is zero since the phases are random.
)
Question no.2
The pulses can be made short by making sure that the extra phase
of as many modes as possible are the same (known as mode locking)
and this would also lead to an increase in intensity which would then
be proportional to
N2
Interactions of Light with Matter II
In the semi-classical model atoms are quantized, but light is not.
as an additional part of the Hamiltonian. The evolution of the system is
obtained by solving the Schrodinger equation with the total Hamiltonian.
This is frequently impossible to solve analytically and we have to resort to
approximations or numeric. The effect of oscillating field is usually taken
as a perturbation of the basic non-interacting atomic Hamiltonian. This
leads to the time-dependent perturbation theory where the most useful
result is Fermi's golden rule. This tells us the probability to obtain a
transition from one level to another under a time-dependent perturbation.
Interactions of Light with Matter II
1 and 2 represent the two atomic levels. The
E1
and
E2 are the corresponding
energies of the two states. They are the eigenvalues of the atomic Hamiltonian with
1 and
2
being the eigenvectors.
Now, when this atom interacts with a field the Hamiltonian contains the transition
elements for jumping from 1 to 2 and vice versa. We have the creation
and annihilation operators 2 1
and
1 2
The Schrodinger equation is
( E1 1 1  E 2 2 2  e
it
1 2  e
 it
2 1 )  (t )  i
  (t )
t
Interactions of Light with Matter II
By substituting in the wave-function
(t )  c1 (t )e iE1t 1  c2 (t )e iE1t 1
We obtain two equations:





c 2  i c1 e i (1 2  ) t

c1  i c2 e i (1 2  ) t
This is a system of coupled equations which we solve for c2 by differentiating
the second equation and substituting in the first equation into it.
Interactions of Light with Matter II
We obtain:
..
.
c2  i(12   ) c 2 
Trial solution
2

2
c2  e it
c2  0
  i(12   )  
2
leads to
2

2
0
which has two roots:
1, 2 
Therefore
But
12  
2
2

 (
2

  12 2
)
4
c2 (t )  Aei1t  Be i2t
c2 (0)  0
thus
A  B
the solution is therefore
c2
2
 4 A sin 2 (t )
Interactions of Light with Matter II
If on –resonance we have
For
c2
2
1
c2
2
 sin 2 (
we require
t

And so we obtain the time for a flop to be
Given that
  10 24 J
we find

t

)

2
t

2
t  1.6510 10 sec
Field Quantization
From the Schrodinger equation it follows that
ˆ
 (t )  e iHt   (0)
(note that this is only true if the Hamiltonian is time independent. Therefore
ˆ
 (t )  e iH (t t0 )   (t 0 )
and so
 iHˆ ( t t0 ) 
ˆ
U (t , t 0 )  e
is the time-evolution operator.
The usual expectation value of an observable is given by
O(t )  (t ) Oˆ (t )
Field Quantization
But this can be rewritten using our time-evolution operator as
O(t )   (0) e
iHˆ ( t t0 ) 
iHˆ ( t t0 ) 
ˆ
Oe
 (0)
and so
ˆ
ˆ
O(t )  e iH (t t0 )  Oˆ e iH (t t0 ) 
is the time dependent operator in the Heisenberg picture.
Let's now derive the evolution equation for the above operator. Differentiate the equation
obtaining
i

d ˆ
O(t , t 0 )   Hˆ Oˆ (t )  Oˆ (t ) Hˆ  Oˆ (t ), Hˆ
dt

Interaction of Light With Matter III
If the fluorescent light from a coherently - excited atom is collected and
the mean intensity
I
and the mean (intensity squared)
I2
is measured, you will find for a light beam from a single atom that
I2
< I
2
This has been measured (by Kimble, Mandel and others)
and is called photon anti-bunching.
Interaction of Light With Matter III
This is surprising because it says that r.m.s (root mean square) intensity
is smaller than the average intensity.
It is not true for thermal radiation nor for coherent states photon statistics.
This relationship is impossible for a classical random process.
Therefore photon anti-bunching is a signature of a purely quantum effect and
quantum state of radiation.
If spontaneous emission is sufficiently large and
the coherent state amplitude small,
this will lead to observation of anti-bunching.
This is therefore ultimately an effect of quantum vacuum.
Interaction of Light With Matter III
The variance of the photon number distribution is
.
Using commutation relations, this can be written as
This can be written as
The second order intensity correlation function (for zero delay time) is defined as
Interaction of Light With Matter III
This quantity is basically the probability of detecting two simultaneous photons,
normalized by the probability of detecting two photons at once for a random photon source.
Here and after we assume stationary counting statistics.
Then we have
Then we see that sub-Poisson photon statistics, one definition of photon antibunching,
is given by g(2)(0) < 1.
We can equivalently express anti-bunching by Q < 0 where the Mandel Q Parameter is defined as
Interaction of Light With Matter III
If the field had a classical stochastic process underlying it,
say a positive definite probability distribution for photon number,
the variance would have to be greater than or equal to the mean.
Sub-Poissonian fields violate this,
and hence are nonclassical in the sense that there can be
no underlying positive definite probability distribution for photon number
(or intensity).
Interaction III: 6th Question
A coherent state is the minimum uncertainty state in position
and momentum of a harmonic oscillator.
This is a natural Hamiltonian from the physical perspective as it
says that when a photon is lost in the field it is absorbed by the
atom as vice versa.
Going to interaction picture the Schrodinger equation reduces to

 (  a    a)   i


t
Interaction of Light With Matter III
We assume that there are n photons in the field.
Then due to energy conservation only the following superposition
  c1 e, n  c2 g , n  1
is possible
The Schrodinger equation becomes
 (c1 n  1 g , n  1  c2 n  1 e, n )  i(c1 e, n  c2 g , n  1 )
Multiplying
g
and
e
we obtain
 n  1c1  ic2
 n  1c2  ic1
Interaction of Light With Matter III
By taking the derivative of the second equation and substituting in the first:


2
c1   n  1 c1  0
The solution is
Therefore
But at t = 0,
c1 (t )  A sin(  n  1t )  B cos( n  1t )
n   n  1
c1 (0)  0
, so that
c1 (t )  sin(  n  1t )
The probability is therefore
p1 (t )  c1 (t )  sin 2 ( n  1t )
2
Interaction of Light With Matter III
If the field is in the coherent state
 e

2
2

n
n!
n
n
then the amplitude for the ground state at time t is
c1 (t )  e

2
2

n
n
n!
sin(  n  1t )
Thus the probability is
p1 (t )  e

2

n

n
n!
2
sin(  n  1t )
Interaction of Light With Matter III
If
 0
and
, hence
, then
p1 (t )  1
t 
4
2
sec
p1 (t )  sin( t )
implies
2
sin( t )  1
2
thanks