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Theory of Ultrafast Spectroscopy
or
Feynman Diagrams Made Simple
Nonlinear-Spectroscopic Experiments:
Limiting Cases
Medium to be studied
Frequency 1
Domain
sig c (3) (sig ;1,2 ,3 )
2 or
cw monochro-
matic beams c (3) (1 , 2 , 3 )
3
t1
Time Domain
t2 t sig R(3) (tsig ; t1, t2 , t3 )
delta-function or
pulses R (3) ( 1 , 2 , 3 )
t3 where i ti 1 ti
Ultrashort
laser pulses
are an
intermediate time
case.
Ultrashort laser pulses are really short, so they appear to be time-
domain experiments waiting to happen.
But, unlike true d-function pulses, they have finite bandwidth.
So they can be resonant or nonresonant. This will be the key.
Ultrafast-Spectroscopy Experiments:
An Intermediate Case
Ultrashort pulses have large, but finite, bandwidth. So experiments
using them can be resonant or nonresonant.
b b
Resonant time domain
Nonresonant frequency domain
a a
In addition, ultrashort-pulse experiments can be b
―nearly resonant.‖ This involves much more complex
formulas. We won’t treat this case.
a
g
Also, ultrashort-pulse experiments
can be nonresonant for some input
b pulses and resonant for others.
We can treat this case.
a
Feynman Diagrams Made Simple
Quick quantum-mechanical derivation
Nonlinear-optical Feynman diagrams in
the frequency domain—cw experiments
Example and tricks
Nonlinear-optical Feynman diagrams in the time
domain—delta-function-pulse experiments
Example and tricks
Feynman diagrams for experiments with simultaneous time- and
frequency-domain character—ultrashort-pulse experiments
The ―ultrashort-pulse domain‖
Examples and tricks
A Sneak Preview of Feynman Diagrams
Each diagram corresponds to a term in a complex sum. We use such
diagrams because they’re easier to remember than the actual equation.
A Feynman diagram can be interpreted in the time or frequency domains.
Time domain Frequency domain
d g d g
t3 3
b t2 b 2 a, b, g, and d
t1 a a
1 a a
represent states.
delta-function pulse inputs cw beam inputs of
at times, t1, t2, and t3 frequencies, ω1 , ω2 , and ω3
In both domains, the particular ordering of the pulse times or beam
frequencies is referred to as a time-ordering.
Many time-orderings contribute to the total response/susceptibility.
Semiclassical Nonlinear-Optical
Perturbation Theory
Treat the medium quantum-mechanically and the light classically.
Assume negligible transfer of population due to the light.
Assume that collisions are very frequent, but very weak:
they yield exponential decay of any coherence
Use the density matrix to describe the system.
Effects that are not included in this approach:
saturation, population of other states by spontaneous emission,
photon statistics.
The density matrix b
If the state of a single two-level atom is:
a
t ca (t ) a cb (t ) b
ρaa(t) or ρbb(t) are the
population densities of
The density matrix, rij(t), is defined as: states a and b.
raa (t ) rab (t ) ca (t )ca (t ) ca (t )cb (t )
* *
r (t ) r (t ) * *
ba bb cb (t )ca (t ) cb (t )cb (t )
When laser beams with different k-vectors excite the atom,
rij(t) tends to have a spatially sinusoidal variation.
A grating is said to exist if ρaa(t) or ρbb(t) is spatially sinusoidal,
A coherence is said to exist if rab(t) or rba(t) is spatially sinusoidal.
The density matrix
For a many-atom system, the density matrix, rij(t), is defined as:
raa (t ) rab (t ) ca (t )ca (t )
*
ca (t )cb (t )
*
r (t ) r (t )
cb (t )ca (t ) cb (t )cb (t )
* *
ba bb
where the sums are over all atoms or molecules in the system.
Simplifying: ca (t ) 2 ca (t )cb (t )
*
cb (t )ca (t ) cb (t )
* 2
The diagonal elements (gratings) are always positive, while the off-
diagonal elements (coherences) can be negative or even complex.
So cancellations can occur in coherences.
Why do coherences decay?
A coherence is the sum over all the atoms in the medium.
Atom #1
The collisions
"dephase"
Atom #2
the emission,
causing
Atom #3 cancellation of
the total
emitted light,
typically
exponentially.
Sum:
Grating and coherence decay: T1 and T2
A grating or coherence decays as excited states decay back to ground.
A coherence can also cancel out if each atom has different phase.
The time-scales for these decays to occur are:
Grating [raa(t) or rbb(t)]: T1 ―relaxation time‖
Coherence [rab(t) or rba(t)]: T2 ―dephasing time‖
Collisions dephase; so, except in dilute gases, T2 << T1.
The measurement of these times is the goal of much of nonlinear
spectroscopy!
Nonlinear-Optical Perturbation Theory
The Liouville equation for the density matrix is:
dr
i V , r (in the interaction picture)
dt
which can be formally integrated:
t
r (t ) r (t0 ) 1/ i
V (t '), r (t ') dt '
t0
which can be solved iteratively: r (t )
n 0
r ( n ) (t )
t t1 tn-1
r (t ) 1/ i
dt1 dt2 ... dtn V (t1 ), V (t2 ), ...V (tn ), r (t0 ) ...
(n) n
t0 t0 t0
Note that t0 tn tn-1 ... t1 t i.e., a “time ordering.”
Perturbation Theory (continued)
Expand the commutators in the integrand:
V (t1 ), V (t2 ), ... V (tn ), r (t0 ) ...
Consider, for example, n = 2:
V (t1 ), V (t2 ), r (t0 ) V (t1 ),V (t2 ) r (t0 ) r (t0 )V (t2 )
V (t1 ) V (t2 ) r (t0 ) V (t1 ) r (t0 ) V (t2 )
V (t2 ) r (t0 ) V (t1 ) r (t0 ) V (t2 ) V (t1 )
Thus, r
(n)
contains 2n terms.
Perturbation Theory (continued)
Now, V is the perturbation potential energy due to the light
and is of the form, E, where E is the total light electric field.
But V is in the interaction picture, so we have:
VI (t ) U *(t ) [ E (t )] U (t ) where: U (t ) exp( iH 0 t / )
[Note that U(t) U*(t’) = U(t-t’) ]
So a typical term looks like:
VI (t1 ) r I (t 0 ) VI (t2 ) U *(t1 ) [ E (t1 )] U (t1 )
U * (t0 ) r (0) (t0 ) U (t0 )
U *(t2 ) [ E (t2 )] U (t2 )
r (t ) is also in the interaction picture: r I (t ) U *(t ) r (t ) U (t )
Dividing out these U(t)’s yields:
U (t t1 ) [ E (t1 )] U (t 1 t0 ) r (0) (t0 )
U (t0 t2 ) [ E (t2 )] U (t2 t )
Notice that time propagates from to to t along two different paths.
Perturbation Theory (continued)
So a typical term (in second order) is:
t t1
(1/ i)2
dt1 dt2 U (t t1 ) [ E (t1 )] U (t1 t 0 ) r (0) (t0 )
t0 t0
U (t0 t2 ) [ E (t2 )] U (t2 t )
But, in nth order, the E-field is typically the sum of n input light fields:
E (t ) E1 (t ) exp(i1t ) E2 (t ) exp(i 2t ) ... En (t ) exp(i nt ) c.c.
As a result, each of the above type of terms expands into many terms.
Allowing each field to occur only once yields n! as many more.
Thus, in nth order, there are 2nn! terms!
How do we remember all these terms?
Use diagrams!
Consider two input beams and this second-order term, noting that
time propagates from t0 to t along two paths:
t t1
(1/ i)2 dt1 dt2 U (t t1 ) E1 (t 1 ) exp(i1t1 ) U (t1 t 0 ) r (0) (t0 )
U (t0 t2 ) E2 (t2 ) exp(i2t2 ) U (t2 t )
t0 t0 *
time
t r (2) (t )
t1 U (t t1 ) U (t2 t )
E1 (t1 ) exp(i1t1 ) E2 (t2 ) exp(i 2t2 )
*
t2
t0 U (t1 t0 ) U (t0 t2 )
r (t0 )
(0)
Perturbation Theory (cont’d) c
1 2
Now expand in terms of the atomic eigenstates:
b
For our second-order term, for example: a
t t1
(1/ i)2 dt1 dt2 U (t t1 ) E1 (t1 ) exp(i1t1 ) U (t1 t0 ) r (0) (t0 )
t0 t0
U (t0 t2 ) E2 (t2 ) exp(i2t2 ) U (t2 t )
*
we find:
t t1
(1/ i)2 dt dt
a , b ,g t
1 2 exp ia (t t1 ) ba E1 (t1 ) exp(i1t1 )
exp ib (t1 t0 ) rbb (t0 ) exp ib (t0 t2 )
0 t0 (0)
e gb E2 (t2 ) exp(i2t2 ) exp ig (t2 t )
*
Computation of the number of terms now is an exercise left to the student…
Doing the integrals… [ Set t0 = 0 ]
t t1
dt1 dt2 exp ia (t t1 ) ba E1 (t1 ) exp(i1t1 )
0 0
exp ib t1 rbb (0) exp ib t2
(0)
gb E2 (t2 ) exp(i2t2 ) exp ig (t2 t )
*
Dipole moment matrix elements at the ith beam polarization
rbb (0) (1)2 gb ba exp i(a g )t
(0) (2) (1)
t
dt1 E1 (t1 ) exp i( b a 1 )t1
t1
0
dt2 E2 (t2 ) exp i(g b 2 )t2
*
0
Now, to go further, we’ll consider limiting cases.
The Frequency
E1 (t1 ) E1 ; E2 (t2 ) E2
Domain: cw beams
rbb (0) (1)2 gb ba E1E2 exp i(a g )t
(0) (2) (1)
t
dt1 exp i( b a 1 )t1
t1
0
dt2 exp i(g b 2 )t2
0
t1
exp{i(g b 2 )t2 } exp[i(g b 2 )t1 ] 1
i(g b 2 ) 0
i(g b 2 )
t
dt1 exp i( b a 1 )t1 exp[i(g b 2 )t1 ]
exp i(g a 1 2 )t 1
0
t
exp i(g a 1 2 )t1
0 i (g a 1 2 )
c
Including dephasing
1 2
Before we evaluate these integrals, b
a
we must include dephasing.
Every time a transition frequency ab a b (e a e b ) /
occurs, we must subtract off the dephasing rate i /(T2 ) ab
for that transition.
This is the usual method for adding width to a transition. Thus:
ab (e a e b ) / i /(T2 ) ab
This addition comes from a complex analysis that takes into
account collisions.
Now we can do the integrals in the various cases.
The Frequency Domain
Evaluating a single second-order term for monochromatic fields yields:
g
(-1) r (0)
bb
(1)
ba
(0 )
ag
( 2)
gb
c (1 2 )
(2)
1 2
bg 2 ag (1 2 )
i
b
a
where ab (e a e b ) / i /(T2 ) ab
and where:
The (-1) occurs in terms with an odd number of V(t)’s to the right of rbb (t0 )
(0)
The factor of r bb is the population density of the initial state.
(0)
The factors of ab are dipole-moment matrix elements between the
(k )
states a and b for the polarization of beam k.
The denominators contain the line shape--the dynamical information.
Drawing Feynman Diagrams in the cw Limit
1. Draw two vertical line segments.
2. Draw a rightward-pointing diagonal arrow for each input field. Upward-
pointing arrows correspond to absorbed photons, and downward-
pointing photons correspond to emitted photons. Choose an ordering for
these “interactions,” and also choose which side each should appear on.
Label each interaction with a light frequency.
3. Write in states (a, b, c, d, e) at the base and just above each interaction.
d
ω4 Every possible diagram of this form
ε
ω3 corresponds to a term in the expression for
c(n), where n is the number of interactions.
a
g
ω1 This diagram corresponds to:
ω2 0 2 1 3 4
β β
Drawing Feynman Diagrams in the cw Limit
Diagram Include a factor of –1 if there are an odd
number of interactions on the right:
β
ω1 (1)
g
ω2 Include a factor of the initial population
density of the state at the base of the
a a diagram:
raa
(0)
Drawing Feynman Diagrams in the cw Limit
Piece of diagram At each interaction, we write down a dipole-
moment matrix element:
a
ω1
ba
(1) ―(1)‖ means ―for the
polarization of beam 1‖
β
After each interaction (reading upward), we
Diagram write a resonant denominator of the form:
a
1 2
1
ag
ω1
g
2
1
bg
β β ω2
where ab (e a – e b ) / – i /(T2 ) ab
Interpreting Feynman Diagrams in the cw Limit
The contribution to c(n) is the
product of all factors shown below:
ω4
d Resonant
ed
1
ε denominator: 1 2 3 4
Resonant denominator: eg
ω3 1
1 2 3
a
Resonant denominator: ag
1
ω1 1 2
g
Resonant denominator: (bg 2 )
1
ω 2
Matrix elements: ba ae ed dg gb
(1) (3) (0) (4) (2)
β β
The population density of the state at the base: r bb
(0)
Two interactions on the right (a factor of –1 for each): (-1)2
Example: Linear Optics—The Absorption
Coefficient and Refractive Index vs. Frequency
Linear optical problems involve only one photon: β
ω1
raa ab ba
(0) (1) (1)
β c (1)
a
ba 1
i
ω1
raa) ab) 2
(0 (1
a a (e b - e a ) / 1 i /(T2 )ab
Resonance frequency Dephasing time
Light
frequency
This is just the well-known complex Lorentzian line shape, whose
even (imaginary) component is the absorption coefficient and whose odd
(real) component is the refractive index.
How do you know which diagrams to include?
First consider the process, and include only the g
most resonant, and hence strongest, terms.
1 2
For example, consider difference-frequency
generation, 0 1 2 β
a
Maximally resonant denominators:
ω2 β a
(ba 1 2 ) 1
ω1
g (ga 1 )1 (bg 2 )1 g
ω1 ω2
a a β β
a Anti-resonant denominators:
b ω2
ω1 (ab 1 2 ) 1
g (gb 2 )1 (ag 1 )1 g
ω2 ω1
b b a a
Example: Higher-Order Wave Mixing
g
12-wave mixing
51 1 42 2
Signal
frequency:
1 2 41 32
β
a
g
ω1
A 12-Wave-Mixing β
Feynman Diagram ω2 All denomi-
g nators are
maximally
ω1
g resonant.
β
ω1
g
ω2 a ω1
1 2 g
ω2
ω2 β
β g a
a ω1 ω1
g
β β
ω2
Unfortunately, there are about 10,000 more such diagrams to consider…
Drawing Feynman Diagrams in the Time Domain
Now, suppose that the input light is a sum of delta-function pulses.
The relevant variables are
now the pulse relative
1 2
delays, 1 and 2 time
t1 t2 t
We can now write Feynman diagrams for this class of processes, but we
must label the interactions with times, rather than frequencies.
e t4 i ti 1 ti
d
t3
g Every possible diagram of this form
β corresponds to a term in the expression
t2
t1 for the response, R(n), where n is the
number of interactions.
a a
The Time Domain: E1 (t1 ) E1d (t1 1 )
d-function pulses E2 (t2 ) E2d (t2 2 )
The integrals are now even easier.
rbb (0) (1)2 gb ba E1E2 exp i(a g )t
(0) (2) (1)
t
dt1 d (t1 1 ) exp i( b a )t1
t1
0
dt2 d (t2 2 ) exp i(g b )t2
exp i(b a )1
0
exp[i (g b ) 2 ]
rbb (0)(1)2 gb ba E1E2 exp[iga t ]exp[iab1 ]exp[ibg 2 ]
(0) (2) (1)
Note that the result is a product of propagators.
Interpreting Time-Domain Feynman diagrams
As before, include a factor of –1 if there are an odd number of
interactions on the right:
( 1) 2
e t4
As before, include a factor of the initial population d
density of the state at the base of the diagram. t3
g
raa(0)
β
t2
Also, as before, at each interaction, we write down t1
a dipole-moment matrix element: a a
ab
(1)
etc.
Instead of resonant denominators, we write
simple exponential propagators:
exp(-i ba 1 ) exp(-i bg 2 ) exp(-idg 3 ) exp(-ide 4 )
Example: The Linear Response
As before, linear optical problems involve β
only one photon and use the same diagram:
a
Ri(1) raa ab ba exp(iba1 )
(0) (1) (1)
β
t1 where 1 t t1 . Dropping the subscript, 1:
(1) 2 e b ea
raa ab
(0)
exp i exp /(T2 )ab
aa
Resonance frequency Dephasing time
This is just the well-known fact that the molecules oscillate at their own
frequency, emitting ―free-induction decay,‖ and dephase exponentially.
The Fourier transform of this response is the complex Lorentzian line
shape, whose even component is the absorption coefficient and whose
odd component is the refractive index.
Example: The Excite-Probe Experiment
Excitation pulse Observe change in probe-
t1
t2 t sig pulse energy vs. delay,
1 t2 t1
Probe pulse
The observed signal vs. delay
is complex, with three Coherence spike
components:
Photon echo Excited-
(PFID) State decay
Delay, τ1
The Excite-Probe Experiment (cont’d) Signal
pulse
The excite-probe experiment is a third-order process, g
with the excitation pulse providing two photons: Probe t2
pulse
t1 g t2 g t2 g β
t1 g t1 Excite
Pulse(s)
β t2 β β t1 t1 β a
t1 β t1 β β t1
t2 β t1 Only state a
is populated
a a a a a a aa initially.
t 2 t1 t1 t1 t 2 t1 t1 t1 t 2 t1 t1 t 2
Coherence
spike
Photon echo
Excited-
(PFID)
state decay
Delay, τ1 t2 t1
The intermediate domain: ultrashort pulses
Ultrashort pulses have finite bandwidth and finite pulse length. Can we
define Feynman diagrams for nonlinear-optical experiments with them?
Yes!
tj
All the integrals
are of the form: exp i ab
i ti E j (ti k ) dti
t0
For ultrashort pulses, two important cases yield simple results.
Case 1. Resonant excitation: r 1/ ab i
Set E(t) = d (t ) time domain
Case 2. Nonresonant excitation: r 1/ ab i
Set E(t) = constant frequency domain
The Ultrashort-Pulse E1 (t1 ) E1
Domain E2 (t2 ) E2d (t2 2 )
rbb (0) (1)2 gb ba E1E2 exp i(a g )t
(0) (2) (1)
t
dt1 exp i( b a 1 )t1
t1
0
dt2 d (t2 2 ) exp i(g b )t2
0
exp[i (g b ) 2 ]
t exp i( b a 1 ) t 1
exp i( b a 1 ) t1
0 i ( b a 1 )
The Ultrashort Pulse Domain (cont’d)
Purely resonant ultrashort-pulse experiments are pure time-domain
experiments, and we use the time-domain Feynman diagrams.
Purely nonresonant ultrashort-pulse experiments are pure frequency-
domain experiments, and we use the frequency-domain Feynman
diagrams.
What about experiments that are resonant at some steps and
nonresonant at others?
δ
g t4
ω3 Resonant steps, so
Nonresonant steps,
ε label with times
So label with t2
β
frequencies
ω1
a a
But can we define rules for interpreting this “time-
frequency” hybrid diagram that make sense? Yes!
Doing the integrals, we see that time-domain steps yield the same factors,
But frequency-domain (nonresonant) steps yield slightly different
denominators (we must take into account the existing coherence from the
previous time-domain step).
Also any resonant pulse must be simultaneous with all nonresonant pulses
prior to it since the last resonant step. Here, for example, pulses 1 and 2
must be coincident in time (same for 3 and 4).
δ t4 Propagator: exp(igd 4 )
ω3 g Resonant denominator: be 3
1
ge
ε
t2 Propagator: exp(i be 2 )
β
1
1
ω1 Resonant denominator: ba
a a
Matrix elements: ab bg gd de ea
(1) (2) (0) (4) (2)
The population density of the state at the base: raa
(0)
β
Example: Femtosecond CARS
δ
1 2 3 0
g
In fsec transient CARS, a two-photon
Raman resonance is excited, and its decay
is probed vs. delay, all by pulses. a
The decay of the coherence, ga, is to be measured by varying a delay.
t3 δ Propagator: exp(ida 3 )
g Propagator: exp(iga 2 )
t2
β
ω1 Resonant denominator: (ba 1 )1
a a
Signal
Plotting the signal intensity vs. 2 t3 t2
yields the dephasing time, T2ag:
τ2
Conclusions
Nonlinear-optical Feynman diagrams better allow us to
remember the large number of terms in the complex
perturbation-theory expansion.
We can define Feynman diagrams for several cases:
general input fields
cw input fields (frequency domain)
delta-function input fields (time domain)
ultrashort-pulse input fields: time or frequency-domain or both
An understanding of nonlinear-optical Feynman diagrams
can make almost any nonlinear-optical or nonlinear-
spectroscopic problem (and even linear ones!) relatively easy!
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