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M1L5r.txt
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M1L5r.txt
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#
# File: content-mit-8422-1x-captions/M1L5r.txt
#
# Captions for 8.422x module
#
# This file has 133 caption lines.
#
# Do not add or delete any lines. If there is text missing at the end, please add it to the last line.
#
#----------------------------------------
So that's what the beam splitter does to the mode operators.
Now we want to talk about what does
the beam splitter do to quantum states,
to single-photon states.
So now we want to transform the single-photon states.
It's a little bit like, well, we know
what to do in the Heisenberg picture to the operator,
but now we want to see what happens to the wave functions.
And I want to be-- say it very slowly because you
have to-- it shouldn't lead to any confusion later.
When we have single photons, we can
have the single photon in mode b or we can have it in mode a.
These are the two possibilities.
And right now, I label it like this,
where the first number is the occupation number of mode a
and the second number is the occupation in mode b.
You have to be little bit careful.
I'm just saying that keep your alert level high.
Later, I will use a qubit representation
where the qubit 1 means the photon is
in mode a and the qubit 0 means the photon is in mode b.
So when we talked about logical states
of our two-level system-- the photon can be here
or the photon can be there-- 0 means the photon is there.
It doesn't mean that we have zero photons.
But sometimes I have to talk about the photons,
and that's what I'm doing right now.
And 0 means no photon in this mode.
But I will remind you of that as we go along.
So we have the two quantum states 1, 0 and 0, 1.
And let's just see what the beam splitter is
doing to one photon in mode a.
Sorry, that was b.
So let's use this convention.
So just to elaborate, what I just said is this convention 1,
0 is the direct product of the Hilbert space for mode
b with one photon with the Hilbert space of mode a,
and we have zero photons there.
And sometimes you want to denote this state
by putting a comma in between.
That's all the same.
How do we transform the quantum state
with a beam splitter operator?
Well, we could now apply it to the quantum state.
But we just learned how to apply to operators.
So let's just use what we already know because we
can write the state like this.
And now we want to insert the unity operator, b dega b,
and now we can know-- we can use our knowledge of the operator.
We know how this transforms.
We had this above-- the transformation of the operators
b and b dega.
And when we apply b to nothing, to no photons,
we get no photons.
So now by taking from the page above,
the transformation of the operator b dega,
it's a linear combination of a dega and b dega.
We find, well, what we expect.
The photon can now be in the same mode
or it can appear in the other mode.
And the coefficients from the transformation
are cosine theta minus sine theta.
And if we would send the other state through the beam
splitter, we find the cosine theta, sine theta component.
So since the total probability to have a photon
is cosine square plus sine square is unity,
we find what we expected.
But it's nice to see it, namely that b conserves the photon
number.
Of course, this should have been obvious from the outset
because we used a Hermitian operator, a unitary time
evolution, and that is energy-conserving.
What happens if we use a state which
has one photon in each mode and we act on it with the beam
splitter?
Well, it gives a superposition of-- the two photons
can now be in either of the two modes
or they can be distributed to one in each mode.
And the coefficients are-- because we
have two transformations-- products of sine and cosine.
Here is cosine square plus sine square theta--
here is a plus sign-- Square root 2 plus square root 2.
So that means if we allow one photon to be in each mode,
it leads us to actually out of the Hilbert
space of single-photon states because we
have a certain probability now that we
have two photons in each mode.
So we don't want that.
So if we want to deal with only one photon,
we should restrict our attention, our formalism,
to the states which have no more than one photon.
And now if we act on those states with beam splitters,
we don't get out of this subspace of Hilbert space.
However, we also want to omit this state because for-- well,
it's more elegant to do it.
But also, if we have a single-photon state
and we have a detector which has not 100% efficiency
and we do a measurement, well, we
don't know if we're at the state 0, 1
but we just didn't detect it or if we're at the state 0,
0, whereas if you deal with two states
where you always have a photon, exactly one photon,
and you read out your system and you detect nothing,
well, you discard the measurement and you perform--
and you only take the measurements
which have detected one photon.
So you have the ability to deal with finite efficiencies
and losses in a very straightforward way.
Yes?
The operation of [INAUDIBLE] on 1, 1
doesn't look so unitary to me.
Should be a negative sine square root.
It's close to square root of minus sine [INAUDIBLE]
Yes, and it is in my lecture notes-- minus.
Also in the coefficient of 1, 1.
And co square root of minus--
Because otherwise that's just 1.
Yeah.
Otherwise--
Yes.
And it's right in my lecture notes.
Sorry.
Sometimes when you're talking, explain it, build it up.
Yes.
Thank you.