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Absorption and Emission of Radiation:
Time Dependent Perturbation Theory Treatment
Want Hamiltonian for Charged Particle in E & M Field
Force on Charged Particle:
1


F  e E  V  B 
c




Force (generalized form in Lagrangian mechanics)
jth component:


U d  U 
U is the potential
Fj  

qj are coordinates
 q j dt   q 
j 

Example:
U d  U 
Fx  
 
 x dt   Vx 
since:
dx 
Vx 
x
dt
Copyright – Michael D. Fayer, 2007
Use the two equations for F to find U,
the potential of a charged particle in an E & M field
Once we have U, we can write:
H  H  H
0
Where H' is the time dependent perturbation
Use time dependent perturbation theory.
Copyright – Michael D. Fayer, 2007
Using the Standard Definitions from Maxwell’s Eqs.
B   A
1
E
A  
ct
A  vector potential
  scalar potential
Force on Charged Particle:
1


F  e E  V  B 
c



Then:



1 A 1
F  e    
 V  A 
c t c




Components of F, Fx, etc…
(  ) x 

V  A

x
d
dx
  Ay  Ax
 Vy 

 y
  x

  Ax  Az 

V

 z
 x 
 z

Copyright – Michael D. Fayer, 2007

V  A

x
  Ay  Ax
 Vy 

 y
  x
Adding and Subtracting Vx
V    A
x

  Ax  Az 

V

 z
 x 
 z

 Ax
x
 Ay
 Ax
 Ax
 Ax
 Az
 Ax
Vx
 Vy
 Vz
 Vy
 Vz
 Vx
x
y
z
x
x
x
Total time derivative of Ax is
d Ax  Ax   Ax
 Ax
 Ax 

  Vx
 Vy
 Vz
dt
t  x
 y
 z 
Due to explicit variation of
Ax with time.
Due to motion of particle Changing position at which Ax is evaluated.
Copyright – Michael D. Fayer, 2007
Then:
 Ax dAx
 Ax
 Ax
 Ax

 V y
 Vz
 Vx
t
dt
y
z
x
Using This

V  A

x

dA  Ax

VA  x 
x
dt
t


Since:
 Ay
 Az
 Ax

 A A V
V A V

 Vx
 Vy
 Vz
x
x x
x
x
x


0
Copyright – Michael D. Fayer, 2007
Substituting these pieces into equation for Fx
   1  Ax 1

Fx  e  

 (V    A) x 
c
  x c t

cancels
But
V    A


  
1
 1 d Ax 
Fx  e  
  A V  



x
c
c
d
t




Then
Ax 
x
dAx  Ax


VA 

x
dt
t

A V
 Vx
because


 Vx


substitute

A V 

AxVx  AyV y  AzVz 

 Vx
V y
V x
Ax
 Ax
 Vx
 Ay

V x
V x
V x
1 +
0
+
0
A not function of V &
Vy Vz independent of Vx
+
0
Copyright – Michael D. Fayer, 2007
Therefore
  

1
 1 d  
Fx  e  
  A V  
A V 


c
 c dt   Vx

  x


  
 d   
 
e
e
Fx   
e  c A  V   
e  c A  V   


 dt   Vx 
 
  x
The general definition of Fx is:
Since  is independent of V, can add in.
Goes away when taking 
V x


d 
Fx  
U 
U
x
dt   Vx 
U  potential
Therefore:
e
U  e   A V
c
Copyright – Michael D. Fayer, 2007
Legrangian:
L=T–U
T  Kinetic Energy
For charged particle in E&M Field:
L  T  e 
e
A V
c

1
T  m x2  y2  z2
2

The ith component of the momentum is give by
Pi 
Therefore,
L
 qi
Px  mx 
where:
dqi
 Vi
dt
q x  x  Vx
qi 
e
Ax
c
Copyright – Michael D. Fayer, 2007
The classical Hamiltonian:
H  Px x  Py y  Pz z  L
Therefore

 
 

e
e
e
H   mx 2  Ax x    my 2  yAy    mz 2  zAz 
c
c
c

 
 





1
e
 m x 2  y 2  z 2  xAx  yAy  zAz  e 
c
2
This yields:


1
H  m x2  y2  z2  e
2
Copyright – Michael D. Fayer, 2007
Want H in terms of momentum Px, Py, Pz
since we know how to go from classical momentum to QM operators.
Multiply by m/m
H
1
2m

 mx    my    mz 
2
2
2
  e
Using:
Pi  mqi 
e
Ai
c
Classical Hamiltonian is:
1 
e  
e  
e 
H
 px  Ax    p y  Ay    pz  Az 
2m 
c  
c  
c 
2
2
2

  e

Copyright – Michael D. Fayer, 2007
QM Hamiltonian
px   i
Make substitution:

x
Then the term:
2
e 

p

 x c Ax   


2
 2 e2
e 
e

2

A

i
A

i
A
x
x
x
 x2 c2
c x
c
x
The operator operates on a function, y.
Using the product rule:
Same. Pick up
factor of two
e 
e Ax
e
y ( x )
i
A y ( x)  i
y ( x )  i Ax
c x x
c x
c
x
2
e 

p

 x c Ax   


2
 2 e2
e  Ax
e

2

A

i

2
i
A
x
c  x
c x x
 x2 c2
Copyright – Michael D. Fayer, 2007
The total Hamiltonian in three dimensions is:

1  2 2 e2 2 i e
i e
H



A



A

2
A



  e
2
2m 
c
c
c

This is general for a charged particle in
any combination of electric and magnetic fields
For light  E & M Field
=0
(no scalar potential)
And since
1 
 A 2
0
c t
Then:
(Lorentz Gauge Condition)
 A 0
Weak field approximation:
2
A is negligible
Copyright – Michael D. Fayer, 2007
Therefore, for a weak light source
H
1  2 2 2i e




A



2m 
c

kinetic energy of particle
For many particles interacting through a potential V,
add potential term to Hamiltonian.
Combine potential energy term with kinetic energy term to get normal
many particle Hamiltonian.
H  
2
0
j
2m j
 2j  V
Time independent
The remaining piece is time dependent portion due to light.
H  
j
e
i Aj  j
m jc
The total Hamiltonian is
H  H  H
0
Use H0 + H' in time dependent perturbation calculation.
Copyright – Michael D. Fayer, 2007
E & M Field  Plane wave propagating in z direction (x-polarized light)
unit vectors
i x
jy
kz
A  iAx

z 

0
A

A
cos
2

t

with
x
x


c  


vector potential
To see this is E & M plane wave, use Maxwell's equations
E
1 
2  0
z

Ai
Ax sin 2    t  
c t
c
c

B x A j
2  0
z

Ax sin 2    t  
c
c

 x A
i
j
k

x

y

z
Ax
0
0
Equal amplitude B and E fields,
perpendicular to each other,
propagating along z.
Copyright – Michael D. Fayer, 2007
To use time dependent perturbation theory we need:
 0m H '  0n   0m  
j
e
A x j P x j  0n
m jc
Dipole Approximation:
Most cases of interest, wavelength of light much larger
than size of atom or molecule
  2 x 103 Å
atom, molecule
part of a cycle
of light
1  10Å
molecule
Take Ax constant spatially  two particles in different parts of molecule will
experience the same Ax at given instant of time.
Copyright – Michael D. Fayer, 2007
Dipole Approximation:
Pull Ax out of bracket since it is constant spatially
e
1
 0m H '  0n   Ax 
 0m P x j  0n
c
j mj
i

 xj
e
1

Ax 
 0m
 0n
c
 xj
j mj
doesn’t operate on time dependent part of ket, pull
time dependent phase factors out of bracket.
 0m (q, t ) H   0n (q, t )  i
0
Need to evaluate y m
e
Ax e i ( Em  En ) t /
c

 xj
1

0
y
(
q
)
j m m  x y n0 (q)
j
j
y n0
Can express in terms of xj rather then

 xj
Copyright – Michael D. Fayer, 2007
First for one particle y 's can write following equations:
(1)
d 2y m0* 2m
 2  Em  V ( x )y m0*  0
2
dx
(2)
d 2y n0 2m
0

E

V
(
x
)
y


n
n  0
2
2
dx
complex conjugate of
Schrödinger equation
0
Left multiply (1) by xy n
Left multiply (2) by xy m0*
(1)
(2)
xy n0
d 2y m0* 2m
2m
0 0*

x
y
y
E

V ( x ) xy n0y m0*  0
n m
m
2
2
2
dx
xy m0*
d 2y n0 2m
2m
0* 0

x
y
y
E

V ( x ) xy m0*y n0  0
m
n n
2
2
2
dx
Subtract
xy n0
2 0
d 2y m0*
d
y n 2m 0* 0
0*

x
y
 2 y m xy n ( Em  En )  0
m
2
2
dx
dx
Copyright – Michael D. Fayer, 2007
xy n0
2 0
d 2y m0*
d
y n 2m 0* 0
0*

x
y
 2 y m xy n ( Em  En )  0
m
2
2
dx
dx
Transpose
2m
2
( En  Em )y m0* xy n0  xy n0
2 0
d 2y m0*
d
yn
0*
 xy m
d x2
d x2
Integrate
2m
2

2 0
 0 d 2y m0*
0* d y n 
 En  Em   y xy dx    xy n dx 2  xy m dx 2  dx

 


0*
m
0
n
0
0
This is what we want. y m x y n
Need to show that it is equal to
y m0

 xj
y n0
Copyright – Michael D. Fayer, 2007
2m
2


2 0
 0 d 2y m0*
0* d y n 
 En  Em   y xy dx    xy n 2  xy m 2  dx
dx
dx 

 
0*
m
0
n
Integrate right hand side by parts:
u  xy n0
d 2y m0*
dv 
dx
2
dx

 udv  uv





 vdu
and collecting terms

Because wavefunctions vanish at infinity, this is 0, so we have:




0*
0
 d
d
0 dy m
0* dy n 
  d x xy n d x  d x xy m d x  dx






using product rule
0
dy m0*
dy n0 dy m0*
dy m0* dy n0
0* dy n
 y
x
y m
x
dx
dx dx
dx
dx dx
0
n
2nd and 4th terms cancel

0
 0 dy m0*
0* dy n 
   y n
y m
 dx
d
x
d
x

 
Integrating this by parts  equals second term
Copyright – Michael D. Fayer, 2007
Therefore, finally, we have:
d
m
0
0
ym
y n   2  Em  En  y m0 x y n0
dx
This can be generalized to more then one particle by summing over xj.
Then:
 0m (q, t ) H   0n (q, t )   i
1
Ax ( Em  En ) xmn e i ( Em  En ) t /
c
with:
xmn  y m0 e  x j y n0
j
x-component of "transition dipole."
Copyright – Michael D. Fayer, 2007
Absorption & Emission Transition Probabilities
dCm
i

dt
 Cn  (q, t ) H   0n (q, t )
0
m
Equations of motion
of coefficients
n
Time Dependent Perturbation Theory:
Take system to be in state  0n (q, t ) at t = 0
Cn=1
Cmn=0
Short time  Cmn  0
Using result for E&M plane wave:
dCm
1
i E E t /
  2 Ax  Em  En  xmn e  m n 
dt
c
xmn  y m0 e  x j y n0
No longer coupled equations.
Transition dipole bracket.
j
Copyright – Michael D. Fayer, 2007
For light of frequency :
Ax  Ax0 cos(2 t )


1 0 i 2 t
Ax e
 e  i 2 t
2
vector potential

note sign difference
Therefore:

dCm
1
i  Em  En  h  t /
i  Em  En  h  t /
0

A
x
E

E
e

e


x mn
m
n
dt
2c 2

Copyright – Michael D. Fayer, 2007
Multiplying through by dt, integrating and
choosing constant of integration such that Cm = 0 at t = 0
note sign differences
i  Em  En  h  t /
i  Em  En  h  t /

i
e

1
e
 1
0
Cm 
Ax xmn  Em  En  


2c
E

E

h

E

E

h





 m

n
m
n
Rotating Wave Approximation
Consider Absorption Em > En
Em
This term large, keep.
Drop first term.
En
(Em – En – h)  0
as h  ( Em  En )  denominator goes to 0
Copyright – Michael D. Fayer, 2007
For Absorption – Second Term Large  Drop First Term
Then,
0
*
Probability of finding system in  m as a function of frequency,   Cm Cm
Using the trig identities:
e

Get:
C Cm 
*
m
1
c
2

 1 e  ix  1  2(1  cos x )  4sin 2 x / 2
ix
2
0 2
x
A
 ( E  En  h )t 
sin 2  m

2
2
2

xmn  Em  En 
2
(
E

E

h

)
 m n

Em  En  E energy difference between two eigenkets of H0
E  E  h
amount radiation field is off resonance.
Copyright – Michael D. Fayer, 2007
Em  En  E
E  E  h
energy difference between two eigenkets of H0
amount radiation field is off resonance.
Plot of Cm*Cm vs E:
2
Q t 2
(2)
Maximum at E=0
0
4/t
E
Cm
2
max
t2
Q
(2 )2
Q
1
c2
2
0 2
x
A
xmn
2
 Em  En 
2
Maximum probability  t 2 – square of time light is applied.
Probability only significant for width ~4/t
1 ps  67 cm-1
Determined by uncertainty principle. For square pulse: t = 0.886
1 ps  30 cm-1 from uncertainty relation
Copyright – Michael D. Fayer, 2007
The shape is a square of zeroth order spherical Bessel function.
t increases 
Height of central lobe increases, width decreases.
Most probability in central lobe
10 ns pulse 
Width ~0.03cm-1, virtually all probability
t 

Cm* Cm  Dirac delta function (E=0) ;
h = (Em - En)
Total Probability  Area under curve



1
Q sin 2 ( Et /2 )
*
*
 CmCm d   h  CmCm d E  h   E 2 d E
Q t

h2

Qt
4 2

sin 2 x
 x 2 d x
Probability linearly proportional to
time light is applied.
Copyright – Michael D. Fayer, 2007
Since virtually all probability at E = 0,
0 2
x
evaluate A
(in Q) at frequency ( Em  En )/   mn
Therefore:
C Cm 
*
m
2
 2 mn
c2
2
A  mn  xmn t
0
x
2
related to intensity of light
see below
2
transition dipole bracket
Probability increases linearly in t.
Can’t let Cm* Cm get too big if time dependent perturbation theory used.
Limit by excited state lifetime.
Must use other methods for high power, “non-linear” experiments (Chapter 14).
Copyright – Michael D. Fayer, 2007
Have result in terms of vector potential, Ax0 .
Want in terms of intensity, I.
Poynting vector:
S
c
ExB
4
For plane wave:
c 4 2 2 0 2 2
Sk
Ax sin 2 ( t  z / c )
4 c 2
Intensity  time average magnitude of Poynting vector
Average sin2 term over t from 0 to 2  1/2
Therefore:
and
Ix 
 2
2c
0 2
x
A
2
2
C Cm  2 I x xmn t
c
*
m
Linear in intensity.
Linear in time.
Copyright – Michael D. Fayer, 2007
Can have light with polarizations x, y, or z. i. e. Ix, Iy, Iz
Then:
Cm* Cm 
2
2
2
2 
t
I
x

I
y

I
z
2
x
mn
y
mn
z
mn

c 
xmn, ymn, and zmn are the transition dipole brackets for
light polarized along x, y, and z, respectively
Copyright – Michael D. Fayer, 2007
Another definition of “strength” of radiation fields
radiation density:
average
 ( mn ) 
1 2
E ( mn )
4
2
2
2 2 mn
0
E ( mn ) 
A ( mn )
2
c
2
Then
A  mn  
2
0
2c 2

2
mn
 ( mn )
Isotropic radiation
A  mn   A  mn   A  mn 
2
0
x
0
y
2
0
z
2
2
1 0
 A  mn 
3
For isotropic radiation


2
2
2
2
C Cm  2 xmn  ymn  zmn  ( mn )t
3
*
m
Copyright – Michael D. Fayer, 2007
Einstein “B Coefficients” for absorption and stimulated emission
Probability of transition taking place in unit time (absorption)
for isotropic radiation
Bnm  ( mn ) 
2
2

mn  ( mn )
2
3
where
mn  xmn  ymn  zmn
2
2
2
2
transition dipole bracket
Copyright – Michael D. Fayer, 2007
For emission (induced, stimulated) everything is the same except
keep first exponential term in expression for probability amplitude.
Em stimulated emission
need radiation field
En
Previously, initial state called n.
Now initial state m, final state n.
Em > En.
i  En  Em  h  t /
i  E m  E m  h  t /

i
e

1
e
 1
0
Cn 
Ax xmn  En  Em  


2c
E

E

h

E

E

h





 n

m
n
m
Rotating wave approximation. Keep this term.
Bmn  ( mn )  Bnm  ( mn )
Einstein B coefficient for absorption
equals B coefficient for stimulated emission.
Restrictions on treatment
1.
Left out spontaneous emission
2.
Treatment only for weak fields
3.
Only for dipole transition
4.
Treatment applies only for Cm* Cm 1
If transition dipole brackets all zero
5.
mn  0
Higher order terms lost when we took vector potential
constant spatially over molecule:
Lose  Magnetic dipole transition
Electric Quadrupole
Magnetic Quadrupole
Electric Octapole
etc…
Only important if dipole term vanishes.
Copyright – Michael D. Fayer, 2007
Einstein “A coefficient” – Spontaneous Emission
Bmn
induced
emission
, Bnm
Einstein B Coefficients
absorption
Want: Amn spontaneous emission coefficient
Nm = number of systems (molecules) in state of energy Em (upper state)
Nn = number of systems (molecules) in state of energy En (lower state)
At temp T, Boltzmann law gives:
 Em / k BT
Nm e
  En / kBT  e  h mn / kBT
Nn e
Copyright – Michael D. Fayer, 2007
At equilibrium:
rate of downward transitions = rate of upward transitions
N m  Amn  Bmn  ( mn )  N n Bnm  ( mn )
spontaneous emission
Using
stimulated emission
absorption
N m e  Em / kBT
  En / kBT  e  h mn / kBT
Nn e
e  h mn / kBT 
Bnm  ( mn )
Am n  Bm n  ( mn )
Solving for  ( mn )
 ( mn ) 
Am  n e  h mn / kBT
 Bm  n e
 h mn / k BT
 Bnm
Copyright – Michael D. Fayer, 2007
Bnm  Bmn
Am n
Bm n
Then:
 ( mn ) 
e
h mn / k BT
1
Take “sample” to be black body, reasonable approximation.
Planck’s derivation (first QM problem)
3
8 h mn
1
 ( mn ) 
h mn / k BT
c3
e
1
Gives
Am  n
3
8 h mn

Bm n
3
c
Am  n
3
32 3 mn
2

mn
3
3c
Spontaneous emission – no light necessary,
I = 0,  3 dependence.
Copyright – Michael D. Fayer, 2007
Spontaneous Emission:
 3 dependence
No spontaneous emission - NMR
  108 Hz
Optical spontaneous emission
  1015 Hz
Typical optical spontaneous emission time, 10 ns (10-8 s).
  NMR

  optical
3
  108 
21
   15   10
  10 
3
NMR spontaneous emission time – 1013 s (>105 years).
Actually longer, magnetic dipole transition much weaker than
optical electric dipole transition.
Copyright – Michael D. Fayer, 2007
Quantum Treatment of Spontaneous Emission (Briefly)
Radiation Field  Photons
State of field n
same as Harmonic Oscillator kets
Number operator
aa n  n n
number of photons in field
Absorption:
a n  n n1
annihilation operator
Removes photon – probability proportional to bracket squared
 n  intensity
need photons for absorption
Copyright – Michael D. Fayer, 2007
Emission
one more photon in field
a n  n  1 n  1
creation operator
Probability  n + 1
when n very large n >> 1, n  Intensity
However, for
n=0
a 0  1 1
Still can have emission from excited state in absence of radiation field.
QM E-field operator:


E k  i (  k /2 oV )1/ 2  k a k exp(  i k t  ik r )  a k exp( i k t  ik r )
Even when no photons, E-field not zero. Vacuum state. All frequencies have
E-fields. “Fluctuations of vacuum state.”
Fourier component at E   induces spontaneous emission.
Copyright – Michael D. Fayer, 2007