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Lecture 26 – Quantum description of absorption
Ch 10
567-576; 555-559
Quantum mechanical description
The interaction leading to absorption of light is electromagnetic
in origin. The oscillating electromagnetic field associated with the
incoming photon generates a force on the charged particles in a
molecule (electrons); if the interaction results in a change in
electronic state, we say that a transition has occurred; this is
absorption
The only proper way to understand absorption of electromagnetic
radiation is through quantum mechanics. Although we have only
considered until now the time-independent Schrodinger equation,
you should recall that the time-dependent Schroedinger equation
is:
H 0  ( x, y , z, t )  
ih ( x, y, z, t )
2
t
Quantum mechanical description
If the Hamiltonian is independent of time, the wave function, i.e.
the solution to the time-dependent Schroedinger equation is:
( x, y, z, t )  n ( x, y, z )e 2En t / h
where En and n(x,y,z) are obtained by solving the timeindependent Schroedinger equation:
H 0 n ( x, y, z )  E n n ( x, y, z )
Let us now consider the system is perturbed because
electromagnetic radiation is applied. A light wave has an electric
field component:


V (t )  V0 cos 2t
 is the frequency of the radiation
Quantum mechanical description
This electric field interacts with an atom or molecule which has
an electric dipole moment m defined from the charges qi and their

position in space as follows:

m qr

i i
i
Classical electrodynamics has provided an expression for the
energy of interaction between the molecular dipole and the
electric field component of the light wave:
 
 
U   m V   m V cosq
where q is the angle between the electric dipole moment vector
and the electric field vector
Quantum mechanical description
This energy of interaction between the molecular dipole and the
light wave changes the Hamiltonian to a time-dependent
expression:
h2
2
H  H 0  U (t )
H0  
8 2 m
  U ( x, y , z )
reflects all time-independent energies and interactions and U(t)
reflects the interaction of the molecular dipole with a time

dependent electric field
V (t )  V0 cos 2t
Schroedinger’s equation is now:
H 0  ( x, y , z , t )  U (t )  ( x, y , z, t )  
ih ( x, y, z, t )
2
t
Quantum mechanical description
If there is no electromagnetic field V, then Schroedinger’s
equation is:
H 0  ( x, y , z, t )  
ih ( x, y, z, t )
2
t
The solution to this equation has the form:
 0 ( x, y, z, t )  n0 ( x, y, z )e 2En t / h
When the electric field is present, the Schroedinger’s equation has
solutions of the form:
( x, y, z, t )   a n n0 ( x, y, z, t )
n
Quantum mechanical description
 ( x, y , z, t )   ( x, y , z )e
0
0
n
2E n t / h
( x, y, z, t )   a n n0 ( x, y, z, t )
n
The physical interpretation is as follows. Suppose just before the
atom is exposed to light the atom is in its ground state described
by the ground state wave function
0
0 ( x, y , z, t )
When the light and the atom interact, energy is absorbed from
the light into the atom. The result of this absorption of energy is
that the atom may undergo transitions from its ground state into
higher energy states described by wave functions
0
( x, y, z, t )   a n n0 ( x, y, z, t )
n ( x, y, z, t )
n
is telling us the condition of the system after it has absorbed
radiation for a time t
Quantum mechanical description
( x, y, z, t )   a n n0 ( x, y, z, t )
 ( x, y , z, t )
0
0
n
The probability Pn that at time t the system has made a transition
 
0
0
0
n
Pn  a n (t )
2
If the interaction of the electric field with the electric dipole
moment is weak the probability amplitude has the general form:
t
i 2t ' ( E 0  E n )
2
a n (t ) 
U
(
t
'
)
exp(
)dt '
n
,
0
ih 0
h
U n ,0 (t )   nU (t )0 d  U n ,0 cos( 2vt)
U n,0   n* m V0 cosq0 d
Un,0 is the transition moment from state 0 to state n
Quantum mechanical description
00  n0
Pn  a n (t )
2
U n ,0 (t )   nU (t )0 d  U n ,0 cos( 2vt)
t
i 2t ' ( E 0  E n )
2
a n (t ) 
U
(
t
'
)
exp(
)dt '
n
,
0
ih 0
h
U n,0   n* m V0 cosq0 d
This is called transition dipole and is what allows us to
understand absorption of radiation and also calculate how
radiation is absorbed from atomic or molecular orbitals.
The probability amplitudes have the final form:
t
i 2t ' ( E 0  E n )
2
a n (t ) 
U
cos(
2

vt
'
)
exp(
)dt '
n ,0
ih 0
h
This integral is non-zero if and only if:
v
E0  En
h
Transition dipoles
The concept of transition dipole can be generalized, because the
dipole itself is a vector and not a scalar operator. The transition
dipole between two states m and n is defined as:
m n ,m   n* mm d
Because the transition dipole is a vector, it has direction in
addition to intensity. Most electronic transitions are polarized,
which means that the transition dipole is not the same in all
directions. The magnitude of the transition is defined as the
dipole strength:
2
Dn , m  m n , m
Its units are called Debyes, and are obviously the same units as
the square of a dipole (i.e. Cxm). 1 D=3.336x10-30Cxm
Transition dipoles
Example - Calculation of Transition Dipole Moments
Suppose the vibrational motion of a homonuclear diatomic
molecule is modeled as a simple harmonic oscillation. Assume the
system is initially in the ground state, described by the wave
function:
 
0   
 
1/ 4
e x
2
/2

2
m
h
The system is irradiated at the fundamental frequency:
 
E1  E0
h
Will a transition occur to the n=1 state?
The answer rests on the value of the transition dipole moment
Transition dipoles
If we define the dipole moment as qx, the transition dipole
moment has the form:
 
   e
U   qx dx
 

1/ 4
*
1
1, 0
x 2 / 2
0
0

 
1   
 
1/ 4
the integral has the form:
 1 / 2 xex
2
/2

U1,0 
2 x
 x e dx  0
2

(the function is even so the transition dipole is not zero). Because
the transition dipole moment is non-zero, the transition has nonzero intensity if DE=h
Transition dipoles
Will a transition from n=0 to n=2 if we irradiate at (E2-E0)/h? The
transition moment is now:

U 2, 0 
  2 x 0dx 


 2x
3

 x e x dx  0
2

the function is odd so the transition dipole is zero
If we irradiate a harmonic oscillator exactly at   DE / h
we will observe a transition from n=0 to n=1 but not from n=0 to
n=2. This is an example of a selection rule. For a harmonic
oscillator, resonant irradiation will only induce transitions for
which
Dn  1.
Stimulated and Spontaneous Emission
Thus far we have discussed the response of a two energy level
system to a resonant field, where the system is initially in the
ground state:
0
 t  0  0
Suppose instead that the system is initially in the excited state:
t  0  10
A transition to the ground state from the excited state will occur
when the system is irradiated with a resonant field. Because the
transition is from a higher energy state to a lower energy state,
energy is emitted instead of absorbed. This event is called
stimulated emission.
Stimulated and Spontaneous Emission
t  0  10
In fact, the probability per unit time that a stimulated emission
will occur has exactly the same value as the corresponding
absorption problem given above. In addition to stimulated
emission, spontaneous emission may also occur, whereby in the
absence of a resonant field the system spontaneously emits energy
and drops from the excited state into the ground state.
Two Level system
We can learn more about absorption by considering in greater
detail a simple example of the absorption of energy by a
quantized system consisting of only two energy levels:
The ground state corresponds to n=0. If the system is exposed to a
time dependent electric field the wave function of the system is:
00   00e i 2 tE01 / h
 t   a0 t  00e i 2tE0 / h  a1 (t ) 10e i 2tE1 / h
Two Level system
 t   a0 t  00e i 2tE0 / h  a1 (t ) 10e i 2tE1 / h
00   00e i 2 tE01 / h
If the energy of interaction
U (t )  m  V (t )   m V0 cos q cos 2t
between the system’s electric dipole moment m and the electric
field V(t) is not large, the system changes slowly, that is to say,
the transition from the ground state into the excited state occurs
slowly and the probability amplitudes have the form:
 i E  E1 t  
 2 
a1 (t )     U1, 0 cos 2t exp  0
dt 
h
 ih  0


t
t
 ei 2t   e i 2t  
 2 
 iDEt   
 exp 
    U1, 0 
dt
2
 ih  0
 h 



 DE
 
 exp  i
 t  
 2 
 

 h
  U1, 0 

 DE

 ih 
 



 h



a0 (t )  1
Two Level system
00   00e i 2 tE01 / h
 t   a0 t  00e i 2tE0 / h  a1 (t ) 10e i 2tE1 / h
U (t )  m  V (t )   m V0 cos q cos 2t
Using the fact that:
eix  1  2ieix / 2 sin x / 2
sin (DE  h )t / 2h
 2 
a1 (t )   U1,0ei ( DE  h ) / 2h 
(DE  h ) / 2h
 ih 
The probability of finding a particle in the excited state is:
2
 2  2 sin DE  h t / 2h 
   U1,0 
DE  h t / 2h2
 h 
2
P1  a1
2
Two Level system
2
 2  2 sin DE  h t / 2h 
   U1,0 
DE  h t / 2h2
 h 
2
P1  a1
2
There are two interesting features of this probability. First of all,
it oscillates sinusoidally as a function of time, reaching the larger
maximum probability as the splitting between the energy levels
DE becomes closer to h, the frequency of the radiation times
Planck’s constant. Second, the probability of a transition to the
excited state approaches 100% only as the difference between DE
and h approach zero. The probability of observing a transition
to the excited state increases as the frequency of the radiation
approaches resonance i.e. DE=h.
Two Level system
2
 2  2 sin DE  h t / 2h 
   U1,0 
DE  h t / 2h2
 h 
2
P1  a1
2
The transition probability is sharply peaked around DE-h=0
Transition Probability
Probability
1.2
1
0.8
0.6
0.4
0.2
-5
-4
-3
-2
-1
0
1
2
3
4
5
0
Frequency Match (MHz.)
As DE approaches h, the transition probability approaches:
2
 2  2 sin DE  h t / 2h  2  2
P1    U1,0 
   U1,0  t
2
h
DE  h  / 2h
 
 h 
2
2
The probability per unit time that a transition will occur to the
excited state is, on resonance: P  2 
2
2

 U1, 0
t  h 
1
DE  hv