M1L4b.txt

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# Captions for 8.422x module
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Let's start with classical squeezing.
For squeezing, we need a harmonic oscillator.
Means we have parabolic potential,
we have potential v of x.
And then we study the motion of-- that
should be x square-- the motion of a particle in that.
But maybe before I even get started with any equation,
let me explain what the effect of squeezing will be about.
If you have a harmonic oscillator,
you have actually the motion of a pendulum
has two quadrature components-- the cosine motion
and the sine motion, and they are 90 degrees out of phase.
And what happens now is if you parametrically
drive the harmonic oscillator, you
modulate the harmonic oscillator potential at 2 omega.
I showed you mathematically-- it's very, very easy to show--
that depending on the phase of the drive,
you will actually exponentially amplify the sine motion
and exponentially damp the cosine motion
or, if you change, vice versa.
So by driving this system, you can amplify one quadrature
component and exponentially die out
the other quadrature component.
And that is called classical squeezing.
So let's do the math.
It's very simple.
Our equation of motion has the two solutions
I've just mentioned.
It has a solution with cosine omega 0
and one with sine omega 0 t.
And we have two coefficients.
The cosine is called c.
The sine coefficient is called s.
So-- I have to call it c0 because I want to call that c
and s.
And so what we have here is we have
the two quadrature components of the motion
in a harmonic oscillator.
And graphically, we need that for the electromagnetic field
as well.
When we have our two axes, like, you know,
the complex plane for the cosine probabilities,
I call one the s plane.
The s axis one is the c axis.
If you have only cosine motion, the s component is 0
and the harmonic oscillator will just oscillate here.
If you have only a sine component,
you stay on the x-axis.

And now if you have an equal amount of a cosine and sine,
then you can describe the trajectory to go in a circle.
OK, this is just the undriven harmonic oscillator,
which is-- I don't want to dwell on it any longer.
But what we are doing now is we are adding
a small parametric drive.
Parametric drive means we modulate the spring constant
or we replace the original harmonic potential,
which was this, by an extra modulation term.
So we have a small parameter epsilon.
And as I pointed out, the moderation
is at twice the resonance frequency.
So now, we want to solve the equation of motion
for the harmonic oscillator using this eddy potential.
And the way how we want to solve it
is, well, we assume epsilon is very small.
So if the pendulum is swinging with cosine omega t,
it will take a while for the epsilon term,
for a smaller term, to change the motion.
So therefore, we assume that we can actually
go back and use our original solution
and assume that over short term, the epsilon term is not
doing anything.
So for a short time, it looks like a harmonic oscillator
with a sine omega 0 and cosine omega 0 t oscillation.
But over any longer period of time,
the small term will have an effect.
And therefore, the coefficients c of t, c and s,
are no longer constant, but change as a function of time.
OK, so we want to solve now the equation of motion.
That means we use now this here as our ansatz,
and we calculate the second derivative.
We assume that the coefficients c and s are changing slowly.
So therefore, the second derivative of c and s
can be neglected.
By taking the derivative of the second derivative
of the cosine term and the sine term,
of course we simply get minus omega 0 square x of t.
And now, we have the second order derivatives.
Since we neglect the second order derivative of c and s,
the other terms we get when we take the second derivative
is first derivative of c times first derivative
of cosine, first derivatives of s times first
derivative of sine.
So we get two more terms, which are minus omega 0
c dot times sine omega 0 t plus omega 0
s dot times cosine omega 0 t.
So this is the second derivative of our ansatz for x.
And this has to be equal to the force provided
by the potential.
So, taking the potential, we need know the derivative
of the potential.
For the potential, we use, of course, this line.
The first part is the unperturbed harmonic
oscillator, which gives us simply omega 0 square times x.
And the second term, due to the parametric drive, is 2
sine omega 0 t.
And now for x, we use our ansatz for x, which
is the slowly changing amplitude c time cosine omega 0 t
plus s times sine omega 0 t.
Those two terms cancel out.
So now, we have portraits of trig function, sine 2 omega
times cosine omega.
Well, you know, if you take the product of two trig functions,
it becomes a trig function of the sum or the difference
of the argument.
So if you take sine 2 omega 0 times cosine 2 omega 0
and we use trigonometric identities,
we get an oscillation at 3 omega 0, which is 2 plus 1,
and one at a difference, which is omega 0.
So let me right down the terms which are of interest to us,
namely the ones at omega 0.
So let me factor out epsilon omega 0 squared over 2.
Then, we have the terms c times sine omega 0 t
plus s times cosine.
And then we have terms at 3 omega 0,
which we are going to neglect.
So now, we look at-- we compare the two sides of the creations.
We have sine omega 0 term.
We have cosine omega 0 term and the two sides of the equations
are only consistent if the two coefficients of the sine
and the sine term are the same.
So therefore, we obtain our two equations-- one for c dot, one
for s dot.
And these are first order differential equations.
The solution is clear in exponential,
but one has a plus sign, one has a minus sign.
So the c component, the c quadrature component,
is exponentially amplified with this time constant
whereas the sine component is exponentially deamplified.
And this finishes the mathematical discussion
of classical squeezing.
We find that s of t and c of t are exponential functions.
In one case, it's exponentially increasing.
In the other case, it is exponentially decreasing.
And that means that, well, if we go to our diagram
here-- and let's assume we had an arbitrary superposition
of cosine and sine amplitude.
So we have an-- this is cosine, this is sine.
We had sort of cosine oscillation and sine
oscillation, which means that as a phaser,
the system was moving on an ellipse.
If the sine component is exponentially deamplified
and the cosine component is exponentially amplified,
that means whatever we start with is squashed horizontally,
is squashed vertically, and is amplified horizontally
that in the end, it will become a narrow strip.
So this is classical squeezing.

You may want to ask, why did I neglect the three omega 0 term?
Well, I have to.
Otherwise, I don't have a solution
because I have to be consistent with my approximations.
So, what I did here is I had an equation
where I have a clear vision that the solution has slowly varying
c and s coefficient.
And then I simply use that.
I take the second order derivative
and I have only Fourier components
with omega 0, the sine and cosine.
Now, I've made an approximation here.
For the derivative of the potential,
the first line is exact.
But in order to match the approximation
I've done on the other side, I can only
focus on the two Fourier components resonant with omega
0, which I have here.
So in other words, the 3 omega 0 term
would lead to additional accelerations, which I have not
included in the treatment.
So it's consistent with the-- that's
consistent with the assumption that we
have resonant oscillations with the slowly changing amplitude.
But there will be a small admixture at 3 omega 0.
But it will be small.
Any questions about that?
Let me then show you an animation of that.
The basic period.
Then, we will parametrically amplify one quadrature
component and we will unamplify the other one.
So now, I'm gonna start doing that.
And then you notice that its motion turns into an ellipse.
We've amplified this quadrature component
but we've unamplified that one.
And that's squeezing.

Feel free to try it at home.
Actually, you may start to think about this demonstration.
What he has shown was when you have a circular pendulum which
goes in a circle or an ellipse and you
start pulling on the rope with a certain phase
that one quadrature component will be deamplified,
the other one will be amplified.
And as a result of that, no matter what
the circular or the elliptical motion was,
after driving it for a while, it will only
swing in one direction.
And this is the direction we've amplified.

There's one thing which should give you pause.
I have discussed, actually, a single harmonic oscillator.
What Dave Bridger demonstrated was actually
two harmonic oscillators.
The harmonic oscillator has an x motion and a y motion.
However, you can say this was just
sort of a trick for the demonstration
because when you have a circular motion
initially, you have the sine omega 0 and the cosine omega 0
component present simultaneously.
And you can see what happens to the sine
and the cosine component in one experiment.
So in that sense, he did two experiments at once.
He showed what happens when you have initially a sine component
and what happens when you initially