M1L7d.txt

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# Captions for 8.422x module
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I introduced to you, and I just want
to remind you of that because we need that, I introduced
to you this famous phenomenon of Hong-Ou-Mandel interference,
which of the following situation.
If you a 50/50 beam split and you have two identical photons,
you will never get single photons out.
The photons are bozons.
They want to be together and so you will always get two photons
but you don't know on which side of the beam splitter.
Now, this is important because we can now
use the Hong-Ou-Mandel interferences
effect to entangle atoms.
So the element we will take from this beam splitter
is the following.
We will have atoms emit photons, but then we create entanglement
probabilistically.
When these photons have passed through beam splitters and they
tell you everything about it, and both detectors
make a click.
If both detectors make a click, you know for sure
that the two photons at the input of the beam splitter
were not identical.
Because if they were identical, only one detector
receives light.
So we will actually use this HOM interference to project out
photon states where the two photons are not identical
because they have different polarization.
Any questions?
Yes?
[INAUDIBLE]

The phase factors out.
[INAUDIBLE]
Frequency, very important.
Polarization, important.
Timing is important.
A single photon has no phase.

It's a global phase, and if you have many single photons,
your many photon state is a product state
of those single photon states, and all the individual phases,
if you want to hold onto this concept for a second,
just become one big multiplicative phase.
[INAUDIBLE]
Yes, Nancy?
[INAUDIBLE]

Oh, everything matters experimentally.
This is not easy experiment but I
think the beam splitter is probably not-- I mean,
beam splitter is a piece of wonderfully polished optics
and in Germany and elsewhere people
have really learned how to do exquisite optics.
I don't think you're limited by optics right now.
What I would think is difficult, if you
have two photons on a beam splitter,
they have to be in the same spatial mode.
So if you have two fibers and the fibers are not
fully aligned, or you have another mode which
is an admixture, or if you have a lens which
is distorting your Gaussian mode you mix other modes in.
So all this creates non-distinguishability.
But I think optics, the quality of optics,
is probably the least of your concerns.

Actually for today, because I'm out of town next week,
I want you to have something to think about it,
I prepared, I already pre-wrote some of the slides
so I can go a little bit faster.
Give me feedback.
If you think I'm going too fast I will not do it again.
But I felt I can probably have a reasonable speed
by making notations to what I already prepared.
So the situation is the following.
We want to entangle two atoms.
We have two atoms which are both in the excited
state in this scatter light.

And they can go to two different ground states.
Think about two different type of [INAUDIBLE] states.
I call them U and D, up and down.
And when, because of selection rules,
one state can be reached by one polarization,
the other state can be reached with the other polarization.
And I call this polarization H and V, horizontal and vertical.
In reality, this may be circular polarization,
so you may want to transform from linear to circular
after the fact.
But for just the discussion right now,
I assume that horizontal polarization leads
to one state, vertical polarization
leads to the other state.
So therefore, if you have two atoms-- and Chris Monroe
in Michigan did the experiment where he really
had two different vacuum chambers with two
different ions, they both emitted photons,
and then the photons came together on a beam splitter.
So that you should sort of have in mind.
Two ions here and there, they both emit a photon.
And after they have force emitted a photon,
each ion is in a superposition before you
make any measurements.
It can be in up with a horizontal photon
or it can be in down with a vertical photon.
So that is the state of the ion.

So we have two of those.

And then you want to detect photons.
And the goal now is, by using beam splitters and all
those tricks, that the detection of photons
and the outcome of a measurement will project this product
state of two atoms with a photon into a Bell state
for the atoms.
So a measurement on photons can take
two atoms which were completely remote,
had nothing to do with each other,
and suddenly they're in a Bell state.
And this is done by the probabilistic measurement.
So let's develop it.
Our initial state is UH plus DV.
This is system one.
And we have the direct product with our second ion trap
apparatus which has done the same.
The ion has emitted a photon and we don't know anything about it
at this point.
And now it is important if you want to have entanglement,
we can not go and measure the polarization
because this would project an individual ion into a state
and would not have entangled it.
So what we are doing is, we are allowing the two photons--
one photon here, another photon here--
they come together at a beam splitter.
Actually, before I do it, let me just perform the product here.
So what we get is, well, it's just multiplying it out.
UU for the atoms, HH for the photons.
UDHV plus DUVH plus DDVV.

So what you want to do is, we want
to send this beam-- the photons, of course,
the atoms stay put-- onto a beam splitter.
Now, I can do it more rigorously by keeping all the terms
until the end of the calculation,
but I hope it's rather obvious that we want to have it--
if you want to send it on a beam splitter
and then we want to detect that both detectors make click.
That requires the photons to be distinguishable,
so therefore those two terms will not
contribute because the photons have the same polarization.
So we are now using a projective measurement.
We want to find out what is the wave function of the atoms.
Measurement of one photon in the modes a
and b, which are the output modes of our beam splitter.
So it needs one more line.
The output of the beam splitter is--
so I'm only focusing now on the two terms which can give rise
to two clicks at the two different detectors--
so I will factor out UD.
And now we have the two photons hitting the beam splitter,
and each photon goes into a 50-50 superposition.
So I take now this horizontal photon,
and since we don't mess around with the polarization
it's still a horizontal photon, but the photon
can be now in mode b or in mode a, in one of the output
modes of the beam splitter, and I label that as 0 1 plus 1 0.
Now, for the second photon, it had a vertical polarization
so it's a vertical photon.
The second photon is coming onto the beam splitter
in the other input part because we have taken the two
photons from two different experiments,
from two different experimental set ups
and now we combine them.
So that will also be now in a superposition of 0 1 and 1 0.
But you know that one mode transforms with a plus sign
and the other mode transforms at the beam splitter with a minus
sign.
So that's how you should look at those terms.
And of course, from this here, from this term,
we get something very analogous, 0 1 plus 1 0.
And for the other spatial mode, the minus sign
from the beam splitter, 1 0.
Now we are done.
Now now we are detecting photons.
So we want to look at this wave function and ask, what happens,
if in part a-- the output part-- we find one horizontal photon,
and in b-- it must be vertical because the photons have
to be distinguishable-- so where do we have one photon in mode
a in horizontal?
This is here.
And here it is this product where
we have, with a minus sign, where we have-- no, sorry,
with the plus sign.
Here we have one photon in the vertical
so this is with a plus sign.
But we have a second possibility.
One photon in mode a horizontal happens here with a minus sign.
And one photon in mode b vertical happens here.
So in other words, from the term which gives rise
to these measurement process, we have
a plus UD here, and because of the minus sign, a minus DU.
So therefore, if it would detect the photon
with the polarization at this point, which we don't do,
this polarization gives rise to UD minus DU.
We have a second possibility to have a click at each detector,
and this is when we reverse the two polarizations.
But in this case, you can just stare at the wave function.
You get the same atomic state.
The sign doesn't matter.
So you have two measurement, two possibilities, in polarization
but they have the same outcome.
They generate this state.
So the result of all that is I've
shown you that the two atoms are left now in--
and now I put it back in the correct normalization-- in one
of the Bell states.
So whenever you detect the two single photons in coincidence
on both counters, at that moment you
know what you have is an entangled state,
a Bell state of atoms.

Now, the big issue here is efficiency.
Your photon detector are not highly efficient.
And I was only looking at combinations where
we have this double detection.
Most components of the wave function
will give the Hong-Ou-Mandel effect
the two photons going one way.
So therefore, you have to prepare your system,
have photons emitted, and in most of the cases
you will not get the state you want
so you have to keep on trying.
And usually those experiments are
heavily limited by the very, very dismal success probability
of creating the state you want.
And when you want to extend it, not just to two atoms
entangled, to three atoms and four atoms entangled,
then you get a small number to the power n
for your efficiency.
So this probabilistic preparation
doesn't scale up but it is a very simple scale,
it's a very powerful scale, and it
has been used to create teleportation of atomic states