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M1L4e.txt
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M1L4e.txt
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#
# File: content-mit-8422-1x-captions/M1L4e.txt
#
# Captions for 8.422x module
#
# This file has 176 caption lines.
#
# Do not add or delete any lines. If there is text missing at the end, please add it to the last line.
#
#----------------------------------------
The definition of the displacement operator
is given here.
The displacement by a complex number alpha,
is done by putting alpha and alpha-star
into an exponential function.
And in many quantum mechanic courses
you show very easily, the elementary properties
that if the displacement operator is
used to transform the annihilation operator,
it just does that.
And if you take the complex conjugate of it--
so in other words, what that means is it's
called the displacement operator.
Just take that as a definition, but you immediately
see why it's called the displacement operator.
When we do the unitary transformation
of the annihilation operator, we get the annihilation operator
displaced by a complex number.
So the action, the transformation
of the annihilation operator, is the annihilation operator
itself minus a c number.
So therefore, we say that the annihilation operator
has been displaced.
So this is the action of the displacement operator
on an operator, on the annihilation operator.
The question is now, how does a displacement operator
act on quantum states?
And the simplest quantum state we want to test out
is the vacuum state.
And well, not surprisingly, the displacement operator
displacing the vacuum state by alpha,
is creating the coherent state alpha.
This can be proven in one line.
We take our displaced vacuum, and we act on it
with the annihilation operator.
If we act with the annihilation operate on something,
and we get the same thing back times an eigenvalue
we know it's a coherent state because this was a definition
of coherent states.
So therefore, in order to show that this is a coherent state,
we want to show that it's an eigenstate of the annihilation
operator.
So this is what we want to do.
Well, the proof is very easy.
By multiplying this expression with unity, which is DD dega,
we have this.
And now we can use the result for the transformation
of operators, namely that this is simply the annihilation
operator plus alpha.
And if the annihilation operator acts on the vacuum state,
we get zero.
If alpha acts on the vacuum state, we get alpha times zero.
So therefore, what we obtain is that.
So therefore, when that annihilation operator
acts on the state, we get alpha times the state,
and therefore, the state is a coherent state
with eigenvalue alpha.
So therefore, in a graphical way,
if you have a vacuum state, the displacement operator D-alpha
takes a vacuum state and creates a coherent state alpha.
So if we want to have three states with a finite value,
well, we just discussed the electric field.
When we relate it to the harmonic oscillator,
we want squeezed states which are not
centered at the origin, which have a finite value of x or p.
We can now create them by first squeezing the vacuum,
and then displacing the state.
[INAUDIBLE]
What is a physical realization of the displacement operator?
The physical representation of the displacement operator,
we do that on Monday, is the following.
If you pass an arbitrary state through a beam splitter,
but it's a beam splitter which has very, very
high transmission, and then from the-- Let me just tha
show that.
If you have a state, this is trans beam splitter, which
has a very high transmission.
T is approximately one, then the state passes through.
But then from the other side of the beam splitter,
you come with a very strong coherent state.
You have a coherent state which is
characterized by a large complex number, beta.
And then there is a deflection coefficient,
r, which is very small.
It sort of reflects the coherent state with an amplitude r-beta.
And if you mix together the transmitted state and r-beta--
I will show that you explicitly, by doing a quantum
treatment of the beam splitter, what
you get is the initial state is pretty much transmitted
without attenuation, but the reflected part
of the strong coherent state-- you compensate for the small r
by a large beta.
That's actually an exact displacement of r-beta.
So you think it's--
I mean, it's actually great.
The beam splitter is a wonderful device.
You think you have a displacement operator
formulated with a's and a-degas.
It looks like something abstract,
but you can go to the lab, simply get one beam splitter,
take a laser beam, and whatever you send through the beam
splitter gets displaced-- gets acted upon
by the displacement operator.
Yes?
So we showed that the displacement operator-- when
you acted on the vacuum state will displace the vacuum state
to a state alpha.
Does it still hold if you acted on another coherent state,
or in this case a squeeze state like that?
Yes.
I haven't shown it, but it's really--
it displaces-- when we use this representation
with quasiprobabilities, it's simply
does a displacement in the plane.
But now to be honest, when I say it
does a displacement on the plane,
it reminds me that we have different ways
of defining quasiprobabilities, the w, the p, and the q
representation.
I know we use it all the time, that we displace things
in the plane, but I'm wondering if the displacement operator
does an exact displacement of all the representations,
or only of the q representation.
That's something I don't know for sure.
Yeah.
I was thinking of it also like [INAUDIBLE] displaced
all types of light like thermal light,
or any sort of representation of light
that you put in is the same displacement
operator [INAUDIBLE] or is it still
needs just like the vacuum in coherent states?
See, the fact is the coherent states,
I've shown you that it's a vacuum state.
I know that's the next thing to show,
the displacement operator.
If you have a displacement by alpha,
followed by a displacement by beta,
it is equal to displacement by alpha plus beta.
So displacement operator forms a group,
and if you do two displacements, they
equal into one displacement, which is
the sum of two complex numbers.
So what I'm just saying-- but if you do the first displacement,
you can get an arbitrary coherent state.
So therefore, the displacement operator
is exactly displacing a coherent state
by the argument of displacement operator.
And now, if you take an arbitrarily quantum state
and expand it into coherent states,
coherent states are not only complete,
they're even over complete.
All you've done is you've done a displacement.
Now, the over completeness of course,
means you have to think about it because you
can represent states in more than one way
by coherent states.
But if you have your representation,
you just displace it, and this is the result
of the displacement operator.
So since q representation is simply
you take the statistical operator
and look for the diagonal elements in alpha,
and if you displace alpha, the q representation
has been moved around.
So I'm sure that for the q representation, for the q
quasiprobabilities, the displacement-- whatever
we have, the displacement operator shifts it around
in this plane.
For the w and p representation, not sure.
Maybe there's an expert in the audience who
knows more about it than I do.