The following content is provided
under a Creative Commons license.
Your support will help MIT OpenCourseWare continue
to offer high quality educational resources for free.
To make a donation or view additional materials
from hundreds of MIT courses, visit MIT OpenCourseWare
at ocw.mit.edu.
PROFESSOR: We're going to continue
our discussion of correlation functions, and in particular,
g2 of 0.
If you think I'm focusing a lot on one quantity, g2 of 0,
that's correct.
But I have to tell you, I have been
involved in so many discussions with people
who looked at g2 of 0 with both Einstein condensates
when Bose-Einstein condensates in [INAUDIBLE] were discovered
here at MIT by [INAUDIBLE] and collaborators.
There was some mystery about, in essence, the g2 of 0 function.
So a lot of physics actually goes into it.
And I've learned a lot from my own research
and through discussions.
And in this unit, I want to summarize it.
So the confusion sometimes comes because people
use classical pictures, quantum pictures,
and they may not see the relation.
So let's continue the discussion.
I want to go through four different views
of the same physics that you will
recognize that they're interrelated.
The first one, just to repeat that,
is if you have light which is Gaussian intensity
fluctuations, light from random sources.
I think if you would look at the twinkle of starlight,
if you look at the light from a light bulb,
because of the central limit theorem,
you will find that the intensity that the electric field
has a Gaussian distribution, and the intensity
in exponential distribution.
And if you then ask, what is g2 of 0, you'll find it's 2.
Because for an exponential distribution,
the average of the intensity squared
is two times the intensity, the average of the-- what
did I say.
This average of the squared intensity
is two times the average of the intensity squared.
Before I forget it, one student asked me after class,
I'm discussing g2 of 0, but what about g2 of tau?
Well, my understanding is that almost all of the physics
is in g2 of 0.
And there is a simple number, 1 or 2, in those special cases.
What happens is, if you have completely independent sources,
if you completely independent statistics,
then the probability of finding a second particle,
a second photon, is independent of the first.
So it means g2 of tau would be unity.
This is the case for an uncoordinated system.
But if you have g2 of 0 which is higher, this means correlation.
This means some form of coherence.
And usually, the correlated value simply
decays to the uncorrelated value with a characteristic coherence
time.
And this characteristic coherence time
can be inhomogeneous, technical.
It can be limited by the Fourier transform of a light pulse,
the monochromaticity, the bend which
delta omega is 1 over tau.
So it has all of the usual suspects
for what limits the coherence time.
So therefore in general, we're not
learning something fundamental in looking at g2 of tau.
We're just learning the usual spiel about correlation times
and coherence times.
The second angle I gave you with the brief interference
is something which is deeply insightful.
It shows you the following.
If you have a plane wave, the intensity is constant.
If you have two plane waves which interfere,
you create density fluctuations, because the interference
pattern can be constructive and can be destructive.
So therefore, if the intensity fluctuates,
then i square average is no longer i average square.
And what happens is because of the nature of interference,
because the average of cosine square kx is-- well,
there are factors of 1/2 and 2 which just come out
of cosine and sine square function.
You get exactly g2 of 2.
But that should already tell you something,
which will be important for both quantum particles, and light,
and the classic limit.
The moment if you have light or particles in one single mode,
you can't have any form of interference.
Only two modes can interfere.
And you will always get a g2 function of 1.
But we'll come to that in a second.
Any questions at that point?
Yes.
AUDIENCE: I thought that when we talked about the g2 function
several weeks ago, you said that if we had a single mode
thermal light, it would have the g2 function of 2.
And that's therefore, all we were
using for lasers when we used them in experiments
is the fact that it's single mode, not the coherence
in the g2 function.
PROFESSOR: You have a good-- that's a good point, Cory.
Let's just try to connect the two pictures.
Slowly, slowly.
What did we discuss?
This question is great for me to repeat something.
When we looked at black body radiation,
I wrote down the partition function.
And I said I'm only considering a single mode.
And then, we have of course in thermal equilibrium
a Boltzmann type, or Bose-Einstein type,
distribution of finding certain numbers,
number of photons populating the single mode.
And that resulted in a g2 function of 2.
AUDIENCE: It's not really single mode though, right?
Because you're summing over many frequencies.
PROFESSOR: No, it was single frequency.
I have to think about it.
That's a really good question.
Right now, I would quickly say, in thermal equilibrium--
I'll give you partial answer, and when I think about it,
I'll give you full answer on Friday
at the beginning of the next class--
I think would be calculated where
the thermal fluctuations in the intensity.
And the intensity fluctuations, and those intensity
fluctuations.
Those intensity fluctuations result in a g2 function of 2.
That's correct.
Maybe what I want you to say here
is, if you have two modes which have all maybe
single occupation or constant occupation,
and they interfere, then they create a spatial interference
pattern if the phase between the two modes is random,
which gives rise to additional fluctuation.
So most likely, the two fluctuations are independent.
One is the fluctuations because of the interference
of two modes.
But when I say the single mode has a g2 function of 1,
I meant actually a single mode with constant amplitude.
I'll double check and give you a more complete answer on Friday.
Other questions?
OK.
The third view is classical versus quantum statistics.
And I think this really shows that we need at least
some quantumness to find correlations.
Let's assume we have n particles.
And we have N possible states.
And if it's a classical system, let's
assume we want to find one particle in a certain state.
We have a big box and want to-- we have a big box,
have a small sub-volume, and say what
is the probability of finding one particle here?
So this probability to find one particle
is P1 is small n over big N. The probability to find
two particles.
If you have non-interacting classical particles,
well, it's simply P1 squared.
Independent classical particles, they
don't care what they are doing.
You grab into a sub-volume, which
is maybe phase space cell.
And therefore, your quantum state,
you have probability P of finding one.
But since each particle moves around independently,
the probability to find two particles is just P squared.
It's a little bit like if you toss two coins, one
with your left and one with your right,
and you ask what is the probability
to find a head or tail, and you simply
multiply the probabilities.
But now, following reasonings which Bose and Einstein
introduced, you want to use counting statistics, which
takes into account the indistinguishability
of particles.
So if we go from the classical distribution
to the distribution of indistinguishable particles,
then the classical probability P 1 is reduced by n factorial,
because we are not-- let me put it this way.
The number-- I don't want to go through
a mathematical argument.
If you grab one particle, it can be the probability
to find one particle is represented
by a microconfiguration classically.
And each configuration when you permute particles
counts as independent.
But if you have quantum indistinguishability,
you're not counting permutations as
an independent configuration.
And therefore, you have a reduction by n factorial.
If you look at the probability of finding two particles,
the reduction is-- this is just the counting statistics
n factorial by 2.
So therefore, you will find that classically, that quantum
mechanically for bosons, is two times.
The probability to find two bosons in one quantum state
is two times the probability squared to find one boson.
So this is just counting statistics.
This can be applied to conditions
in a Bose-Einstein in a Bose gas.
When you have inelastic collisions, spin relaxation,
dipolar collisions, two body collisions,
two-body collisions have a rate gamma 2,
which is proportional to the probability of finding
two particles at the same time at the same location.
And that brings in the g2 function.
Whereas three-body collisions, gamma 3,
reflects the third order correlation
function g3, which is defined in an analogous way.
So if you now compare at the same density
the two-body rate coefficient, like the same density
means we are looking at the rate of two-body collisions.
And we compare between two-body collision
in a thermal cloud, which is a g2 factor of two,
a thermal cloud of bosons.
Whereas a Bose-Einstein condensate
is a constant occupation of particles in one mode
and has therefore a g2 factor of 1, exactly as [INAUDIBLE]
of the Bose-Einstein condensate is
for particles for meta waves, but the laser is for light.
Whereas, if you look at three-body collisions,
those who do experiments with Bose-Einstein condensates
will know that usually the lifetime
of Bose-Einstein condensates is limited
by three-body collisions.
Well, I didn't derive it, but actually
yes, we have the Gaussian statistics for randomness.
For Gaussian statistics, the n-body collision function
has a factor of n factorial.
So therefore, three-body collisions
scale with 3 factorial, and two-body collisions with 2
factorial.
So in other words, you have a thermal cloud,
you have a Bose-Einstein condensate at the same density.
But what matters for two-body collision
is not the average density.
It is the average of n square.
And the fluctuations in the thermal cloud
because of the g2 function are two times enhanced
compared to Bose-Einstein condensate.
And therefore, you find that at the same density,
you have more loss.
You have a higher rate of two and three-body collisions.
Some of this was not-- became better understood
soon after Bose-Einstein condensates were realized.
The Boulder group studied three-body collisions.
And for two-body collisions, myself
in the post-doc clarified the situation
that even the mean field energy, which everybody had measured
before, requires two particles to interact with short range
interaction is therefore proportional to g2.
And therefore, when people had determined the mean field
energy of a condensate without really knowing
they had already determined the g2 function
of Bose-Einstein condensates.
OK, so this is pretty much the counting statistic
which you do in undergraduate class
when you derive the three different statistics--
Boltzmann statistics, Bose-Einstein statistics, Fermi
statistics.
And it gives you the result of that is also for bosons,
you have the factor of 2.
For fermions, of course, you get 0.
The average between bosons 2 and fermions 0 is 1.
And this is the classical statistics.
Any questions?
Then let me finish this discussion
with a quantum mechanical description.
If you describe the detection or collision of two
particles in two modes.
So you have mode.
A mode operator, annihilation operator a1 you and a2.
But if you are asking what happens
if I measure two particles, you annihilate two particles
with an operator a1 and a2.
But then, because of the indistinguishability
of particles, you have to consider
an exchange term, which is a2a1.
So whenever you detect two particles
in two different modes, your signal
is proportional to something related to a1a2 plus a2a1.
This extra exchange term gives you an extra factor of 2
for bosons, which is exactly what appears
in the g2 function.
And, of course, if you have an exchange term for fermions,
you get 0.
And this, of course, leads to the antibunching,
or the g2 value of 0 for fermions.
However, and this is obvious, if you have a single mode,
then the only operator to detect two particles is a1 times a1.
And this has no exchange term.
And this is the situation of the laser
and the Bose-Einstein condensate.
And that means they have a g2 function of 1.
So Cory, this is another argument
why single mode occupation has a g2 function of 1.
But I have to reconcile it with your question.
Right now, I think there may be difference
between a canonical ensemble and grand canonical ensemble.
In a canonical ensemble, we have [INAUDIBLE]
the total particle number is fixed.
But in a grand canonical it fluctuates.
And maybe this leads to additional fluctuations
for the case of a single mode chaotic light.
But I will have to think about it, your question.
But here is the same argument why
if you occupy only a single mode-- if you look,
for instance, at a Bose-Einstein condensate,
you do not have the second exchange term.
And the missing exchange term just
propagates through the equations and gives you
the g2 factor, the g2 function of 1.
OK so let me sort of wrap it up.
I've given you different angles at the g2 function.
Some are simply classical intensity distribution.
Some is simply interference.
Another aspect is Poissonian counting statistics.
My understanding and my interpretation
is they look as different as they can possibly look,
but they reflect the same physics.
Because interference of to light only happens because light
consists of photons.
And photons are bosons.
So the classical interference and
the quantum mechanical counting of bosons
lead not for random reasons, but lead for profound reasons
to the same result.
And chaotic light, which seems to be determined
by just random fluctuation, well,
the random fluctuations if you have single mode
light, a thermal distribution of single mode photons,
comes from random phases.
And these random phases lead to random interference.
And so we are back to interference.
Finally, let me give you my view on the measurements of g2.
When you look through the literature,
you find the famous Henry [INAUDIBLE] experiment.
In one of your homework assignments,
you looked at a seminal experiment
where people dropped atoms out of cold atom clouds
and measured the g2 function for cold atoms.
Sometimes it's confusing when you directly
compare the two experiments.
But let me try to give you a common description,
or the common denominator, between all these experiments.
You can say that all experiments to measure the two particle
correlation function is about comparing
the probability of finding two particles with a probability
of finding one particle.
And well, you restrict your measurement, your detection
of particles, to either one quantum
state, one mode of the electromagnetic field, one
coherence length, or coherence volume, of light.
Or if you use the semi-classical argument for the description
of particles, you take one phase space cell.
This is one phase space cell, one mode, one coherence volume,
one quantum state, is sort of the different definitions
of what a quantum state is, or what a wave packet
is with Heisenberg limited uncertainty.
You cannot define a particle in phase space to within better
than the phase space volume of a quantum state,
which is h square.
Or is it h bar?
I don't think there is a bar-- h square.
And so this is how you can relate wave packets
to quantum states to phase space volume.
It's pretty much mathematically I could say in different basis
sets, in different basis sets, it is one quantum state.
But you can use wave packets.
You can use time dependent description.
But what I said here-- one quantum state,
one mode, one coherence volume-- this
is as well as you can define a particle or photon
to be by fundamental reasons by Heisenberg's uncertainty
relation.
AUDIENCE: What do you mean by one coherence volume?
PROFESSOR: What?
AUDIENCE: What do we mean by one coherence volume?
PROFESSOR: Well, if you have a laser beam which propagates,
and it is TEM00, you would say the transverse coherence volume
is the size of the laser beam but then
the longitudinal coherence volume
is given by the coherence length.
Just envision you have a laser beam here.
And the coherence volume where all of the photons are coherent
is the area of the laser beam times the coherence length.
And whenever you find a photon in this kind of volume,
then it is in a Heisenberg Uncertainty
limited or Fourier-limited state.
It is in, if it's a wave packet and pulse laser,
it's in a time dependent quantum state.
But it is as coherent.
It is fully coherent.
So you have to figure out for your system what
is this minimum-- what is the coherence volume, what defines
the fundamental phase space cell of your system?
And now you do the following.
You are asking, what is the probability
to find two particles?
And what is the probability to find one particle?
So if p is the probability to find one particle,
then you have three options.
p square, 2 p square or 0 to find two particles.
One is the classical case of distinguishable particle.
This is the case of bosons, bosonic atoms or photons.
Collins Therefore includes the Hanbury Brown-Twiss experiment.
And 0 is fermions.
And the question is now different experiments,
how do you define, with spatial resolution
or temporal resolution, the coherence volume?
And that can involve transverse coordination,
temporal resolution, and whatever
you can use to define the phase space
cell or the coherence links, the coherence volume of a laser
beam.
Let me give you one example, which I think illustrates it.
And this is the example of an atom cloud.
If you have an atom cloud, all the particles
which are in a volume of a thermal de Broglie wavelengths
are coherent.
The momentum uncertainty of particles in a thermal cloud
is 1 over the thermal-- is the thermal momentum.
And according to Heisenberg, the position uncertainty
related with this momentum spread
is just the thermal de Broglie wavelengths.
So you can say that all the atoms in a cubic de Broglie
wavelengths are coherent.
All the atoms are in one semi-classical quantum state.
So therefore, this is sort of atoms in one single mode.
So the Hanbury Brown-Twiss experiment with atoms,
or the measurement of the g2 function,
could be defined as follows.
You have an atom cloud.
And if you had an electron microscope
or some high resolution device, you grab into your cloud
and ask, what is the probability for one particle.
What is the probability for two particles?
And you will find that p2 is 2 times p1 squared.
If your volume is too big, you lose the factor of 2,
because you average over uncorrelated volumes.
Now, in your homework, you were looking at the question,
how can I really grab into a cloud
and just pick out atoms out of one phase space
cell of effective size, lambda de Broglie cubed?
And the way how it was done is that you take an atom cloud
and you drop it and expand it.
When the cloud expands, there is a mapping from momentum space
into position space.
Then you use some form of pinhole,
which provides transverse coordination.
You use a detection laser, which gives you temporal resolution.
And well, this was part of your homework,
but I just wanted to give you the bird's view on it.
By controlling the transverse correlation
and the temporal resolution of the detection,
you create a situation that what you count only atoms which
originated from one phase space cell in your cloud.
So in other words, let's say experiment
where without electron microscope,
without submicron spatial resolution,
you can literally grab into a cloud,
capture a volume of the thermal de Broglie wavelengths,
open your hand, and figure out how
is the probability for two particles related
to the probability of finding one particle.
Any questions?
All right.
OK, good.
So it's time for new chapter.
So in this chapter, we want to look
at interactions between light and atoms
using Feynman diagrams.
And I know this part of the course is a little bit formal.
We are using exact solutions for the time evolution
operator in quantum physics.
But I'm not doing it to teach you sophisticated formalism.
Well, it's interesting to learn it anyway.
But it is this picture of really writing down
the exact solution of the time evolution operator, which
helped me to understand much better what virtual states are,
what certain virtual photons are.
So there are things we talk about it all the time.
And the question is, how do you define a virtual state?
What is the virtual emission of a photon?
And the only way how I can really explain it to you
is by showing you the equation and say look,
the virtual photon is just a term in this equation.
So that's the goal of this chapter.
I don't want to overemphasize the mathematical rigor,
but I want to really show you what
it means to have virtual photons and what
exactly virtual states are.
So let me first motivate that by reminding you of two diagrams.
In physics, we always like to draw something--
a few lines, a few doodles.
And when we have a two level system,
we have the two processes which are emission and absorption.
Well, so this is easy as long as the light is in resonance.
But if the light is not in resonance,
we may ask what about this process?
And since somehow the weekly line of the photon ends here,
I may even put in a dashed line and say,
this is a virtual state.
What does it mean?
Or we can even ask, is it possible--
I'm just playing with straight lines which are quantum state
and wiggly lines-- is it possible
that this process happens.
That would actually mean that an atom in the ground state,
just out of the blue, emits a photon.
Can that happen?
And I can again draw a dashed line
and say, here's a virtual state.
Or, we can say we start in the excited state.
The atom is in the excited state.
But can the atom-- its a two level system,
so we don't include higher states-- but can a two level
atom in the excited state absorb another photon?
So let me just change the color of these photons
to red, because I want to-- and this would remain black.
All right.
Good.
So question is, are those processes possible?
Yes or no?
AUDIENCE: Yes.
PROFESSOR: Yes, they are.
And they have experimental, observable consequences.
However, they look funny because something
seems to be strange with energy conservation.
And what we will see is that in the end,
at the end of the day, that means
if you let the system evolve for a long time, after a long time,
energy has to be conserved.
And therefore, we will always need a second photon
to conserve energy.
So those weirder processes, where
photons are emitted by ground states out
of the blue or excited states absorb other photons
without-- also, there is no higher lying state.
That looks a little bit strange.
So what you will see is-- and I want
to show you that mathematically--
if an atom emits a photon out of the ground state,
you would say, where does the energy come from.
But quantum mechanic allows us to violate energy conservation
by an amount delta e for a time which is h bar over delta e.
So for short movement, energy can be violated.
But then, you need the second photon
to reconcile energy conservation.
You can violate energy only for short times.
And if you would say, what does it really
mean to violate energy conservation for short times?
Well, I want to show you the equation,
which exactly explains what it means.
Next question, just to see.
If you have an excited state, and you just
say it's possible to absorb another atom,
you are in the excited state, and you absorb another photon.
And at least you said, yes this is possible.
My question is, when you are in the excited state
and absorb a photon, in which quantum state is the atom
after absorbing a photon?
AUDIENCE: Ground state.
PROFESSOR: Pardon?
AUDIENCE: Ground state.
PROFESSOR: In the ground state.
Yes.
There is no other state.
And the operator-- and I want to show you
that-- when photons are exchanged,
always transforms the ground to the excited state,
because a dipole operator connects
the ground to the excited state.
So therefore-- but we'll see that all in the time evolution
operator in the formal solution that this dash line is actually
the ground state.
And while it's not obvious, what is this dash line?
AUDIENCE: The excited state.
PROFESSOR: It is the excited state.
The atom in the ground state emits a photon.
Therefore, the atomic system has now lower energy,
because a photon has been emitted.
But the character of this state is now the excited state.
So we will later see that this means that in a perturbation
analysis, we are violating energy.
We are violating energy, because the real excited
state is omega photon plus omega atom, the resonant energy
for the atom.
The real excited state is omega photon plus omega atom higher.
So you will actually find that in a perturbation analysis,
this term has an energy denominator, which
is of resonant by exactly this separation.
And this term, because the real count state is here,
the virtual state here which is the count state is here,
has actually the same energy denominator as that state.
And for this state, of course, this
is something you've seen of resonant light scattering.
The energy [? denominator ?] is there.
So, OK.
So I'll explain to you that we involve virtual states.
So when we have virtual states, we
want to clarify what is their energy.
But we discussed that already in the discussion.
And what is the wave function, ground or excited state.
Yes.
So just to repeat it, when we draw those diagrams,
those diagrams show the energy of the atom,
but taking into account the energy of the photon.
In other words, if you have a ground state,
and we absorb a photon, the photon has disappeared,
and the energy of the atom is now
the ground state and the photon energy.
And this is a dashed line.
Similarly, when an atom emits a photon in the ground state,
the photon is emitted.
Therefore, the atomic energy is lower than the ground state
by the photon energy.
And we draw the dashed line here.
This is what we mean when we say that is
the energy of the virtual state.
And this is what we draw in those diagrams.
By the way, yesterday we had the CUA seminar.
And the speaker was actually talking about trapping atoms
with quantum fluctuations.
And he conceded explicitly Casimir forces,
forces of the vacuum.
And diagrammatically, forces of the vacuum
come, because the ground state atom emits a virtual photon
and reabsorbs.
This is actually, I will mention it later,
the same diagram is actually also
the diagram which leads to the Lamb shift.
These are all sort of when you have atoms in the lowest state,
and they interact with the electromagnetic field.
Well, if they interact with the vacuum,
they cannot absorb photon.
All we can do is emit a photon, and this leads to this diagram.
But just to tell you that you have to be careful,
yesterday's speakers actually used this process.
But he drew this diagram.
He drew it in the opposite way, which
I think confused some people in the audience.
The correct way is to draw it like this.
The virtual state for vacuum fluctuations
for the Lamb shift for the Casimir force
is below the current state.
OK, so we have-- let me just redraw those three diagrams.
I want to now introduce the time evolution.
So I want to use those.
Seems I'm running out of space.
So we want to introduce the time axis
for-- so we had this diagram.
We had this diagram.
And we had-- this is a virtual state.
OK.
So those three diagrams correspond
to the following situation.
If time evolves from the bottom to the top,
we can now draw the first process
when atom in the ground state first
absorbs and then emits a photon.
And let's label the second photon in the purple color.
So an atom starts out in the ground state.
Then it reaches the time where it
interacts the electromagnetic field.
There is a photon which propagates
from earlier times to this time, t equals t prime.
At the time t prime, the photon disappears.
And the atom, which was in the ground state,
goes to the excited state.
It may propagate in the excited state for a while.
And then it emits the photon and is back into the ground state.
So this is the temporal diagram for this process.
For the next situation, we also start
with an atom in the ground state.
But now we have the situation that a photon
is emitted by the ground state.
As a result, the atom switches to the excited state.
Strong violation of energy, but possible for short times.
Then there is a real photon, which is now absorbed,
absorbed by an atom in the excited state.
And that takes us back down to the ground state.
And finally, in the third scenario,
we have an atom in the excited state.
There is a real photon which is absorbed
that switches the atom into the ground state.
And then the atom in the ground state emits a photon.
And as a result, it is back in the excited state.
So let me just make a note that when a ground state
atom absorbs a photon, this is the co-rotating term
in the quantum description, which a lot of you have seen.
The opposite process of emitting a virtual photon-- I
shouldn't say virtual photon.
It's a photon.
What makes it virtual we will see later
when we have the more accurate mathematical formulation.
So this is a counter-rotating term.
And we have used the rule that every photon, which
with a real or virtual-- every photon which is interchanged
with the atom absorbs or emitted--
changes the atomic state from ground to excited.
So an atom goes from the ground to the excited state
either by photon absorption or by photon emission.
Both is possible.
So whenever photon appears or disappears,
it changes-- let's say we start in the ground state.
We go to the excited state.
And this is possible by photon absorption--
this is a co-rotating term-- or emission.
What I've just said reflects that when
we derived the dipole approximation,
that the essential term of the dipole operator is of diagonal.
The dipole operator is an operator
between calm and excited state.
Therefore, it has those two matrix elements.
Whereas electric field operator is the equation
and annihilation operator a and a dagger.
Or which is very elegant if I use for the two level
system a spin one half description.
And I use sigma plus and sigma minus,
raising and lowering operator.
Sigma plus takes the ground state to the excited state.
Sigma minus takes the exciting state to the ground state.
Then in the fundamental atom light interaction,
we have those four terms.
And now you see that sigma plus takes the atom
from the calm to the excited state.
But sigma plus appears both with photon absorption
and with photon emission.
So therefore, the atom can-- this
is what the quantum mechanical operator
tells us-- go from the ground to the excited state
either by photon absorption or by photon emission.
Any questions?
Yes.
AUDIENCE: So if we were to look beyond the dipole
approximation, would we see transitions
that leave the [INAUDIBLE] state?
PROFESSOR: Yes or no.
What we need here is a bilinear interaction,
which has an raising and lowering operator for the atom.
And here it has a raising and lowering atom
operator for the photon number.
And it has the co-rotating term where you raise here
and you lower there.
You lower here and you raise there.
But it has also terms where you raise the atomic excitation,
and you raise the photon number by one,
which of course violates energy.
But the message I want to give you is this
is possible for short times.
And this is exactly what the quantum mechanical equations
tell us.
So the short answer to your question
is, as long as we describe the system
by the dipole interaction, or what we have even beyond--
I'm not using a perturbation argument.
I'm writing down the operator.
And this operator can now effect the system
to all possible orders.
So even the non-perturbative strong coupling limit
will always have products of atomic raising operators,
lowering operators with photon raising and lowering operators.
The little bit longer answer is, some of what I'm telling you
may be gauge dependence.
If you work in the-- if you use, not the dipole description,
but you use the p dot a Hamiltonian,
this Hamiltonian has a e square term.
The e square term allows the atoms
to exchange two photons at the same time.
So in other words, in the a square interaction,
we may have a vertex point in time
where the atom interact with the photon field
where two photons are exchanged.
And then the rules are different.
So in that sense-- but this is quite often
when we give a quantum mechanical pictures, which
has terms, which nicely shows the time evolution.
And you think wow, yeah, this is what happens.
One photon at a time, the atom goes
ground, excited, excited, ground state.
And you have your picture.
And it is fully consistent with the exact quantum
mechanical result.
You may be able to get the same result out
of a different picture.
That point will actually occur again
when we talk, for instance, when we
talk about density operators.
A density operator is an average of our quantum states.
But you can get the same density operator
by averaging quantum states in different basis sets.
So if you get one picture out of one specific basis set,
it gives you the correct intuition,
it's a correct description, but it is not
the only possible description.
[? Collin ?].
AUDIENCE: Maybe the thing, I don't know.
AUDIENCE: I think what you're trying
to get at is that picture works as long as you can limit
yourself to a two level system.
Because that's correct on a whole.
Or is it like strong coupling?
But as long as you can identify the two level system,
or I think the dipole versus particle approximations
[INAUDIBLE] an atom may have transition probabilities
to others, like from an [INAUDIBLE] state.
But--
AUDIENCE: Actually, I was just wondering
if it was possible to have it like a transition matrix
element to [INAUDIBLE] coupling the ground state to the ground
state.
PROFESSOR: Yes.
The e square operator does that, because
in the other gauge, which is pa, pa.
p is in dipole operator. p is the momentum operator
for the atoms.
And the p operator will switch the atom from ground
to excited state.
a, the vector potential, involves a and a daggers.
So the p dot a term does pretty much the same
as the d dot e term, because it's
a bilinear product of an operator which
has an effect on the atoms and has an effect on the photons.
And it means you always switch the quantum state in your atom
and create or annihilate the photon.
But the a squared term-- it's just a square,
it doesn't have a p in front of it--
will allow an atom in without changing the state to create
two photons, emit two photons.
And then there's probably a cross term
which means to absorb and emit a photon.
You will actually, in one of your next homework assignments,
you will actually look at the a square term.
And I think it's a very educational problem.
So you really see that you have two
very different descriptions, but the results fully agree.
But this is already an expert discussion.
Why don't we for now just hang onto the dipole approximation.
We have d dot e, and we just learn
what is inside the dipole approximation,
and what is an exact quantum mechanical description using
the dipole approximation.
But and then we can come back to the discussion,
is everything we describe really real.
And while the answer is, what is real
in quantum physics is the final result,
the intermediate results, you have
to say that this is only one possibility
to go to the final result.
OK, so we want to-- let me just put in one more page.
So what we want to do now is we want to do a calculation.
It's a perturbative calculation, but we
can take it to all orders.
So therefore it's exact and general.
We want to do a perturbative calculation of transition
amplitudes.
And what I am following here the discussion in atom photon
interaction on those pages.
And the idea is the following.
We have the Hamiltonian for the atom.
And then we have a Hamiltonian which
describes the transverse field.
And the interaction of the transverse field
with the atoms.
What we want to use is we want to use the interaction
picture, which is often the nicest picture
to describe the effect of the interaction between two
systems.
In the interaction picture, you are transforming
from you Schrodinger type e function
to wave function psi tilde.
And those wave function psi tilde include already
the dynamics of the Hamiltonian H naught.
So that means that all operators in the normal Schrodinger
picture become now operators in the interaction picture
by canonical transformation, which
involves the dynamics due to the unperturbed Hamiltonian.
And so that's just a reminder what
the interaction picture is.
And we are now interested what happens in the interaction
picture to an initial wave function psi of ti.
This wave function already includes the time evolution
due to H naught.
So the only change now comes because we have interactions
with the electromagnetic field.
So this allows us now to focus only
on the relevant interactions, the interactions we
want to understand.
And the formalism which is used in atom photon interaction
focuses on the time evolution operator,
which I'm sure all of you have seen.
So the time evolution of our system,
how the wave function involves from the initial
to the final time, is described by the time evolution operator.
I will come back to the wave function later,
but in order to derive the time evolution of the wave function,
it's at least convenient to focus first
on the time evolution operator.
So in other words, I want to show
you a formal solution for the time evolution operator.
And once we know the time evolution operator,
we apply it to the wave function.
And then we are talking again about the wave function.
OK, since the derivation involves many equations
and they are all printed in the full beauty in atom photon
interaction, I decided to use pre-written slides here.
Also, most of it is fairly elementary.
The equations are complicated, but the concepts behind them
are very, very simple.
It's actually the beauty of using
a kind of a time evolution operator.
It's a little bit mathematically formal,
but it allows us to express what is really
going on in the solution in very, very simple terms.
So the goal is to do a perturbative expansion
for the time evolution operator.
If we have no interaction potential v, the interaction
picture, nothing happens in the interaction picture,
and therefore, the time evolution operator
is a unity matrix.
But if you have a coupling term v, there is time evolution.
And the time evolution can be treated
in lowest or in higher orders in the perturbation.
And we want to use now perturbation expansion, where
we have correction terms, which is
the first order, second order, and nth order
correction for time evolution operator.
So what we simply want to do is we
want to do an iterative solution.
We want to find an iterative solution for the time evolution
operator.
The next two lines is just a reminder.
What is the differential equation for the time evolution
operator?
Well, we are really just talking about Schrodinger's equation.
Schrodinger's equation for the wave function
turns into differential equation for the time evolution
operator.
Schrodinger's equation in the interaction picture
says that the time derivative of the wave
function in the interaction picture
has nothing to do H naught.
H naught has already been absorbed
in the definition of the wave function.
So the time evolution in the interaction picture
only comes from the interaction term.
And now, if you write the wave function at time tf
as the time evolved wave function from the initial time
to the final time, then you take the derivative here.
You actually take a derivative of the time evolution operator.
And therefore, in one step, from Schrodinger's equation,
you find a differential equation for the time evolution
operator.
In other words, this is now the operator equation
for the time evolution operator.
And it's nothing else than a 100% rewrite of the Schrodinger
equation.
Interrupt me if you have questions.
This part should be-- we'll go through that
for solely pedagogical reasons.
And that means you should really understand it.
So this equation can now be formally solved
in the following way.
Take this expression for the time evolution operator
and insert it in this equation, and you find an identity.
Of course, you haven't really solved it,
because you have the time evolution operator expressed
by the time evolution operator.
This looks like a circular conclusion,
which is just nonsense.
But if you inspect it more closely,
you observe that the time evolution operator here
is expressed by the time evolution operator multiplied
with v. And if you now think in an iterative solution
that we want to express.
We want to expand the time evolution operator
in the different orders in the interaction parameter v. Then
you find actually that if you are only
interested in the first order of the interaction operator,
you can use the [INAUDIBLE] order here,
because you get one more power of v.
So that's how you get the first order.
If you want to know the second order of the time evolution
operator, you can plug-in the first order
on the right hand side.
So therefore, you get an iterative solution.
The first order solution is one.
The [INAUDIBLE] order solution is the unity operator.
The first order solution is by just putting the unity operator
here.
The second order solution is by taking the first order
solution, plugging it in in here.
And then you get two integrals over the operator
v. So in that sense, we have formally
solved the time evolution of the system in all orders.
So this is what we get out of it.
Yes, the equations are getting longer.
But the structure is fairly obvious.
So you saw that the-- let me just flashback.
The first order term had a temporal integral
over v. The second order term has two temporal integrals
over v. And there's a time order in between tau 1 and tau 2.
And the nth order term now involves n temporal integrals
where the times are ordered in such a way.
Now we want to go back to the Schrodinger equation.
So we want to use this already fairly complicated expression
for the time evolution operator and apply it
to the initial wave function.
And we want to calculate matrix elements
between the initial wave function and some other phi
f, some other wave function in a given basis.
So we are also going back.
I don't want to go through too many steps.
Here, we're also going back to the original Schrodinger
picture.
So I've taken all the tildes off.
But you know the step form the Schrodinger
picture to the interaction picture
was only taking out the Hamiltonian H naught.
And the Hamiltonian H naught for eigenstates phi i and phi f,
is simply a time evolution, e to the e ei with a certain time.
So all what this transformation to the interaction picture
gave us, it eliminated, at least for a short while,
all of those phase factors, which are simply
the evolution of the eigenfunction
in the Schrodinger picture.
So let me look at a typical term which we have right now.
And you can't expect something simpler
than this, because this is the general solution.
It's an exact solution to all orders.
It's a summation over terms to all orders.
And eventually the summation of an infinite number of diagrams
is an exact solution.
So what we've got here is we have an integral over n times.
And the n times are time ordered.
This is the formal solution for the time evolution operator.
And that means that we integrate overall times,
but what it is under the integral sign
is a product of the operator v at times
tau 1, tau 2, tau 3, tau 4.
Since we got back to the Schrodinger picture, what
happens is between times tau 1, and tau 2,
the particle simply evolves with the phase factor
given by the Hamiltonian H naught.
So the picture is the following.
The time evolution operator in the interaction picture
is a product over the operator v tilde at different times.
If you simply plug it back into the above equation--
and you have to look at it for a little while
to see that everything works out correctly.
But what that means is then that if you go to the Schrodinger
picture and write down the operator in the Schrodinger
in the basis of unperturbed eigenfunction,
that you have products of the operator v at times
t1, tau 2, tau 3, they're time ordered.
And because we have left the interaction picture,
you also get the propagation with the phase factors
given by the eigenenergies of the unperturbed Hamiltonian H
naught.
So therefore, you get just an infinite number
of terms which have all the following structure,
that the particle propagates in its eigenstate.
It reaches a vertex where the interaction switches
the particle from one state to the next.
Then in the next state, which may now be the excited state,
we have propagation in the excited state.
Then the excited state is, again, exchanging a photon,
has an interaction term, and is switched to-- well,
for the two level system, it has to go back to the ground state.
In a more general situation, it goes to another quantum state.
So therefore, what we have exactly derived
is that the propagation, or the time evolution,
of the wave function is nothing else then
many, many of those factors.
And each of those factors can be represented
by a diagram like this.
So let me just write down what this diagram here
means mathematically.
It means that initially, we have a photon of energy epsilon,
of polarization epsilon, wave factor k, and a certain energy.
And we have particles in state a.
That means that we get from the initial time to the time tau 1.
And tau 1 is the time of the first vertex
of the first photon exchange, that the wave function evolves.
And it evolves with an energy of the unperturbed Hamiltonian,
which is the atomic energy.
And we have one photon h bar omega.
At the time tau 1-- and this is when
the diagram on the left hand side
has a vertex-- we are now bringing in the interaction
operator, which acts on the atom.
It takes the atom from state A to state B.
And it changes the state of the photon.
In this case, a photon with a certain wave factor
and polarization simply disappears.
So this is now the time tau 1.
The next Vertex is reached at time tau 2.
And between the time tau 2 and tau 1,
the system evolves according to the Hamiltonian H naught.
And the energy is now, because we have no photon,
is simply the energy of the atom in state B.
Then we reach the next vertex.
We start with a state B and no photon.
The interaction switches, as now I
assumed we have more than two levels.
Two are state C. And we have now a photon, which
may be different, k prime epsilon prime.
Since we are now, for this given example, I've selected n
equals 2, something which is second order
in the perturbation.
But that means now that between the time tau
2 and the final tie-in for the time propagation, that's
now everything is done.
Things propagate.
And the unperturbed Hamiltonian H naught
gives us now a phase factor, which
reflects the energy of the atomic state,
and the energy of the photon in mode k
prime epsilon prime with energy omega h per omega prime.
So in other words, I hope you see that at the end of this--
I mean, this is why people have Feynman diagrams.
It's a complicated, involved mathematical formulation
with multiple integrations.
But at the end of the day, what we have derived
is that the most general solutions
to all orders in the probation is nothing else than a sum
and integral over diagrams like this.
So when I depict it here, the second order diagram, of course
this has an exact mathematical meaning.
And what we have to do is when we solve
a quantum mechanical equation, we
have to now allow those interactions, which
happen at time tau 1 and tau 2, to happen at arbitrary times.
So therefore, we have to sum over amplitudes
by integrating over times tau 1 and tau 2.
So on Friday-- I have to stop now.
I think there's a seminar right now.
On Friday, I will show you that with this description,
we have actually captured everything I explained to you
earlier-- those virtual states, the emission
of virtual or real photons.
So everything that which was maybe
qualitative at the beginning of this section
has now a precise, mathematical meaning.
But that's what we do on Friday.