The Interaction of Light and Matter: a and n
The interaction of light and matter is what makes life interesting.
Why is light absorbed or transmitted by a particular medium?
Light causes charged particles to vibrate. The particles then emit light,
which interferes with the incident light.
Prof. Rick Trebino
Georgia Tech
www.frog.gatech.edu
Destructive interference means absorption.
And ~ ±90º out-of-phase interference
changes the phase velocity of light.
The Forced Oscillator: Resonance
Complex Lorentzian: 1/(d - iG)
where d  w0 - w
Absorption coefficient, a,
and Refractive index, n.
Rayleigh Scattering and the blue sky.
Refractive
index
n–1
Natural
frequency of
medium
Absorption
coefficient
a
w0
w
Light
frequency
Forward vs. Rayleigh Scattering
Forward scattering from molecules yields the absorption coefficient
and refractive index. Rayleigh scattering (in all other directions) from
molecules yields the blue sky.
Rayleighscattered
beams
Input beam w
w0
Transmitted
beam
Forwardscattered beam
Rayleighscattered
beams
Forward
scattering is
important mainly
near resonance
(w = w0).
Rayleigh
scattering
occurs far from
resonance
(w << w0).
Adding Complex Amplitudes
When two waves add together with the same complex exponentials,
we add the complex amplitudes, E0 + E0'.
Constructive
interference:
1.0
Destructive
interference:
1.0
1.0
+
+
0.2
-0.2
=
1.2
Laser
+
-0.2i
=
=
0.8
time
Quadrature phase, ±90°,
interference:
1-0.2i
time
Absorption
time
Slower phase velocity
(when accumulated over
distance)
Light excites atoms, which then emit light
that interferes with the input light.
When light of frequency w excites an atom with resonant frequency w0:
Electric field
at atom
Electron
Emitted
field
Incident light
E (t )
xe (t )
E (t )
+
Emitted light
=
On resonance (w = w0)
Transmitted light
An excited atom vibrates at the frequency of the light that excited it
and emits energy as light of that frequency.
The crucial issue is the relative phase of the incident light and this
emitted light. For example, if these two waves are ~180° out of
phase, the beam will be attenuated. We call this absorption.
What do you think is happening here?
Resonance!
The Forced Oscillator and Resonance
When we apply a periodic force to a natural oscillator (such as a
pendulum, spring, swing, or atom), the result is a forced oscillator.
Examples:
Child on a swing being pushed
Periodically pushed pendulum
Bridge in wind or an earthquake
Electron in a light wave
Nucleus in a light wave
Tacoma Narrows Bridge oscillating and collapsing because
oscillatory winds blew at its
resonance frequency.
The forced oscillator is one of the most important problems in
physics. It is the concept of resonance.
Electric field
at nucleus Nucleus
The Forced Oscillator:
The amplitude and
Below
relative phase of the resonance
oscillator motion with w << w0
respect to the input
force depend on the
frequencies.
On
Let the oscillator’s resonant frequency be w0, and
the forcing frequency be w.
Let the forcing function be
a light electric field and the
oscillator a (positively
charged) nucleus in a
molecule.
Weak
vibration.
In phase.
resonance
w = w0
Strong
vibration.
90° out
of phase.
Above
resonance
w >> w0
Weak
vibration.
180° out
of phase.
The amplitude and
relative phase of
an electron’s
motion with
respect to incident
light also depend
on the frequencies.
The electron charge is
negative, so there’s a
180° phase shift in all
cases (compared to the
previous slide’s plots).
Electric field
at electron Electron
Below
resonance
w << w0
Weak
vibration.
180° out
of phase.
On
resonance
w = w0
Strong
vibration.
-90° out
of phase.
Above
resonance
w >> w0
Weak
vibration.
In phase.
The amplitude
and relative
phase of
emitted light
with respect to
the incident
light depend on
the frequencies.
Maxwell’s Equations
will yield emitted light
that’s -90° phaseshifted with respect
to the atom’s motion.
Electric field
at atom
Electron
Emitted
field
Below
resonance
w << w0
Weak
emission.
90° out
of phase.
On
resonance
w = w0
Strong
emission.
180° out
of phase.
Above
resonance
w >> w0
Weak
emission.
-90° out
of phase.
The Simple Harmonic Oscillator
Consider an electron with position Re{xe(t)}. If it’s attracted to a
nucleus (assuming it’s like a spring), there will be a restoring force.
Using Newton’s Second Law (F = mea, where me is the electron mass):
ma
Restoring
force, F
me d 2 xe / dt 2  - mew02 xe
The solution is:
xe (t )  xe 0 exp(-iw0t )
So the electron oscillates at its natural
frequency (w0).
The complex amplitude xe0 will depend on other factors.
This is a model of any type of oscillator.
xe (t )
0
Simple Harmonic Oscillator
The motion is
sinusoidal.
xe (t )  xe 0 exp(-iw0t )
xe (t )
Simple
harmonic
motion
time
But many forces act to slow down motion.
For example, atoms spontaneously decay to the ground state
after a random time (and so stop oscillating). So the average
motion of many electrons is a decaying exponential.
xe (t )
The
probability
of decay is
exp(-Gt).
Damped
simple
harmonic
motion
exp(-Gt)
time
Dephasing and the Damping Factor G
Also, the vibration of a medium is the sum of the vibrations of all the
atoms in the medium, and collisions cause cancellations in the sum.
Collisions
(1)
Atom #1 xe (t )
Collisions dephase
the vibrations,
reducing the
total medium
vibration, also
exponentially.
(2)
Atom #2 xe (t )
(3)
Atom #3 xe (t )
exp(-Gt)
Sum
G is often called the
dephasing rate.
time
The Damped Oscillator
The equation of motion must include a damping-force term,
proportional to the velocity:
ma
Damping
force, Fd
Restoring
force, F
me d 2 xe / dt 2  2me G dxe / dt  mew02 xe  0
The solution is:
where we’ll
assume that
the damping
is very weak:
G << w0.
xe (t )  xe 0 exp(-iw0t )exp(-2G t )
This solution is only
approximate, especially
in the limit of very strong
damping. But, in optics,
the damping is usually
very weak, so it will
suffice for our purposes.
xe (t )
Damped
simple
harmonic
motion
exp(-2Gt)
time
The Forced Oscillator
Consider an electron on a spring with position xe(t), now driven by
a light wave, E(t) = E0 exp(- iw t). The equation of motion becomes:
ma
Damping
force
Restoring
force
Lorentz force from
the light wave
me d 2 xe / dt 2  2me G dxe / dt  mew02 xe  eE0 exp(-iwt )
e is the electron charge
The solution is:
E (t )


e / me
xe (t )   2
 E0 exp(-iwt )
2
 w0 - w - 2iw G 
E (t )
xe (t )
Now the electron oscillates at the incident light-wave frequency w,
not the electron’s natural frequency w0.
It has a complex amplitude that depends on w, w0, and G.
Checking Our Solution
E (t )
me d 2 xe / dt 2  2me G dxe / dt  mew02 xe  eE0 exp(-iwt )
Substitute the solution for
xe(t) to see if it works:


e / me
xe (t )   2
 E (t )
2
 w0 - w - 2iwG 
Immediately cancel the
factor of E(t) in all terms.
where E (t )  E0 exp(-iwt )



e / me
e / me
2
me  -iw   2

2
m
G
i
w
 2 2


e 
2
 w0 - w - 2iw G 
 w0 - w - 2iw G 
Multiply both sides
2
2
by w0 - w - 2iw G :


e / me
mew  2
e
2
 w0 - w - 2iw G 
2
0
me  -iw   e / me   2meG  -iw  e / me   mew02  e / me   e w02 - w 2 - 2iw G 
2
Checking Our Solution (cont’d)
me  -iw   e / me   2meG  -iw  e / me   mew02  e / me   e w02 - w 2 - 2iw G 
2
Canceling the electron masses:
e  -w 2   2eG  -iw   ew02  e w02 - w 2 - 2iw G 
Multiplying everything out and canceling common terms:
-ew 2 - 2iewG  ew02  ew02 - ew 2 - 2iew G
QED
So the solution for the electron motion is:


e / me
xe (t )   2
 E (t )
2
 w0 - w - 2iwG 
Notice that, if damping
weren’t present, the
amplitude would go to ∞
when w  w0.
Damped-Forced-Oscillator Solution for
Light-Driven Atoms
The forced-oscillator response is sinusoidal, with a relative phase
that depends on the frequencies involved:
Here, e < 0. Also, w, w0 >> G.




e / me
1
xe (t )   2
 E (t )  -  2
 E (t )
2
2
 w0 - w - 2iw G 
 w0 - w - 2iw G 
When w << w0:
xe (t )  -
1
w02
E (t )  - E (t )
1
E (t )  -iE (t )
When w = w0: xe (t )  -2iw G
When w >> w0:
xe (t )  -
1
-w
2
E (t )  E (t )
The electron vibrates 180° out
of phase with the light wave.
Vibration magnitude is small.
The electron vibrates -90° out
of phase with the light wave.
Vibration magnitude is large.
The electron vibrates in phase
with the light wave.
Vibration magnitude is small.
The amplitude and
relative phase of
an electron’s
motion with
respect to incident
light depend on the
frequencies.
Recall that the
atom’s resonant
frequency is w0,
and the light
frequency is w.
Electric field
Electron
at atom
Below
resonance
w << w0
Weak
vibration.
180° out
of phase.
On
resonance
w = w0
Strong
vibration.
-90° out
of phase.
Above
resonance
w >> w0
Weak
vibration.
In phase.
The atom’s response is approximately
a Complex Lorentzian.
Consider: xe (t ) 
e / me
e / me

w02 - w 2 - 2iw G (w0  w )(w0 - w ) - 2iw G
Assuming w ≈ w0, w0 + w ≈ 2w,
so this becomes:
e / me

2w (w0 - w ) - 2iw G

e / me
1
2w (w0 - w ) - i G
In terms of the variable d  w0 - w, the function 1/(d – iG), is called a
Complex Lorentzian. Its real and imaginary parts are:
1
1 d  iG
d
iG

 2
 2
2
d - iG d - iG d  iG d  G d  G 2
Complex Lorentzian
Real
Imaginary
1
1 d  iG
d
G

 2
 i 2
2
d - iG
d - iG d  iG
d G
d  G2
1/G
G
Imaginary
(even)
component
Real (odd)
component
0
d
Okay, so we’ve determined what the
light wave does to the atom.
Now, what does the atom do to the
light wave?
At every plane in the medium, light emitted
from excited atoms interferes with the
incident
light
beam.
The resulting light then goes
on to excite atoms in the next
plane, where the same thing
happens, etc., etc.
What will be the light’s field a
distance z into the medium?
Medium
E (0,t)
E (L,t)
z
0
L
So consider the total electric field:
z
Incident light
E (z,t) = E incident(z,t) + Eemitted(z,t)
+
Emitted light
=
Transmitted light
Maxwell's Equations will allow us to
solve for the total field, E (z,t).
The incident field E (z = 0, t) will be
the initial condition.
The Inhomogeneous Wave Equation
The induced polarization, P , is due to the medium:
2
2
2E
 2P
E 11 2E
E
P
-- 22 22  0 22
22
tt
zz cc00 tt
where:
PP((tt)) NNeexRe{
e (t )xe (t )}
x
P (t )
where N = the electron number density.



e
1
P   0 Ne 

 E  0 c E
 2wm e  0   (w0 - w - iG) 
2 P
2 E 1 2 E
 0 2   0  0 c 2  2 c 2
t
t
c0  t
which defines c,
the susceptibility.
where c0 = the speed of
light in vacuum
Substituting this into the inhomogeneous wave equation:
2 E 1 2 E 1 2 E
- 2 2  2c 2
2
z
c0  t
c0  t
2 E 1  c 2 E

- 2
0
2
2
z
c0  t