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Quantum teleportation is not
quite like the science fiction
idea of teleporting yourself
from the spaceship down
to the surface of the planet,
but it shares a core idea.
The idea of quantum
teleportation
is to transfer a quantum state
from one place to another
without transferring the
specific carrier of that state.
So without moving the object
itself from the spaceship
to the surface of
the planet, can we
put a similar object or entity
on the surface of the planet
into the same state
as our original object
in the spaceship?
Of course, we know from
the no-cloning theorem
that we can't, in
general, replicate
the object in the spaceship
in an arbitrary quantum state.
So we may have to destroy
the one was on the spaceship,
or at least destroy
its quantum state.
But in return, we'll be putting
the corresponding object
on the surface of the
planet into the same state
as the original one
on the spaceship.
This is actually
quite hard to do, even
for something as
simple as a photon.
It is, however, an interesting
concept to think about,
if only to understand if
it can be done quantum
mechanically, even in principle.
Apparently, it can be
done, at least in theory.
So let's look at how
we might attempt that.
The idea of quantum
teleportation
is to transfer a quantum state
from one place to another
without actually transferring
the specific carrier
of that state.
So for example, we
might have photon 1
that's in an unknown
superposition
of horizontal and
vertical polarization.
We want the output photon
to be in the same state
but without sending
photon 1 there.
We might even actually
destroy photon 1,
possibly absorbing it.
But we know, from the
no-cloning theorem,
that we cannot clone photon 1
to produce another output photon
in the same arbitrary
superposition as photon 1.
We also know that
simply measuring
photon 1-- for example, with
a polarizing beam splitter
together with photo
detectors-- will not reliably
tell us the full quantum
state of photon 1.
We end up statistically
collapsing the state
and throwing away
quantum information
about the original quantum
state of the photon.
The key to quantum teleportation
is to share entanglement.
And we can do that by sharing
an EPR pair of photons.
So here's our EPR source.
And EPR stands for Einstein,
Podolsky, and Rosen.
And those photons, this
pair that we create here,
are going to be in a Bell
state, an entangled state.
For example, we
could presume, here,
that the EPR photon pair is
in this particular Bell state.
So note, we've got
labels here, 2 and 3,
corresponding to
photons 2 and 3 here.
And we've chosen a
particular entangled state
with horizontal and
vertical polarizations
in here called Psi minus 2 3.
The input photon is in
an unknown superposition
that we could write this way.
So it's a superposition
of horizontal and vertical
polarizations with
some coefficients here.
But we don't know
what those are.
And we're not going
to know what those
are throughout this
entire process.
The state of all three
photons, therefore,
can be written like this.
So here's the state, the
unknown state, of photon 1.
And here's the state,
actually a known state,
of photons 2 and 3.
We can be sure that
we are generating them
in a state like
this, for example.
A core trick in
teleportation is to note
that this state can be rewritten
as a linear superposition
of Bell states of
photons 1 and 2.
So we can easily check that this
corresponds to the version here
written out in terms
of Bell states.
But again, to emphasize,
we're using the Bell states
of photons 1 and 2,
not photons 2 and 3.
Note that we've
managed, therefore,
to write this state in
terms of the Bell states
of photons 1 and 2,
which are these states.
If we now make a measurement in
Alice's Bell state measurement
box-- so here's Alice's
Bell state measurement box,
and Bob is over here-- of the
Bell state of photons 1 and 2,
we collapse the state into
just one of the four Bell state
terms.
For example, suppose Alice
measures this particular Bell
state for photons 1 and 2.
We can know this
answer classically.
This is a measurement.
It gives a classical result.
And hence we know that the
overall system of three photons
would now be in the
state just the part
corresponding to that
particular Bell state.
Because Alice can tell Bob
the result of her measurement
by communication over an
ordinary classical channel--
for example, a
telephone line-- Bob
now knows that photon
3 is in this state,
though he doesn't know what
these coefficients are.
What do we mean by that?
He knows that whatever those
coefficients were, of the input
photon, this photon
three is in this state,
because the system has been
collapsed into that state.
Now this state is not the
same as the original state
of photon 1.
That, by definition,
was this state,
with a horizontal
polarization associated
with this coefficient and
a vertical polarization
associated with that one.
That's not what we have here.
We have a vertical
polarization associated
with this coefficient and
a horizontal polarization
associated with that one
for this output photon 3.
So as it stands at
the moment, photon 3
is not in the same
state as photon 1,
although it does have the same
coefficients involved with it
that are still unknown.
Well, we can sort this out.
Bob could simply
rotate the polarization
90 degrees clockwise,
turning vertical polarization
into horizontal and horizontal
into minus vertical.
That's what happens if you
do that 90 degree rotation.
So we would change vertical
to horizontal and horizontal
to minus vertical by
rotating by 90 degrees.
It's just a change of
axes in the system.
And then we could insert what's
called a half wave plate.
That's a piece of
optics that will
delay one polarization--
the vertical polarization,
in this case-- by 180 degrees.
And that would turn the cv
coefficient to minus cv,
undoing this minus sign here
that was a bit of a nuisance.
So Bob is implementing a
controlled unitary transform
here.
And his choice of
that unitary transform
is determined by the
classical information
that Alice sent over to him of
the result of her Bell state
measurement of the state
of these two photons.
Now she doesn't actually
know the state of photon 1,
but she knows that
this pair of photons
was in a particular Bell state.
By this controlled
unitary transformation,
Bob has changed this
state for photon 3
into this state, which is
the one he really wanted.
And he's done this, basically,
by putting in some polarizing
components inside
this path in here.
And he's chosen the
settings of those components
based on the information he
got over the classical channel.
So photon 3 is now in
exactly the same state
as photon 1 was without
either Alice or Bob
actually even knowing
what that state was.
They have not actually
measured the state of photon 1.
But somehow, they've
managed to take that state
and put it onto the
state of photon 3.
So they didn't
actually have to know
ch and cv, these coefficients
of the horizontal
and vertical
polarization of photon 1.
And yet they've managed
to make photon 3 have
exactly these same coefficients.
For other results that Alice
might have got from Bell state
measurement, Bob implements
other polarization
manipulations over here.
And this presents no
fundamental problem.
He could, for example,
use electrically
controlled phase shifters here
that can operate quite quickly.
In general, Bob is
therefore implementing
a specific unitary
transformation
on photon 3, a
combination, here,
of phase delays and
polarization rotations,
that simply depends on the
result he gets from Alice
of what she says her Bell state
measurement had given her.
Hence, for any
result from Alice,
Bob can put photon 3 into
exactly the same state
as photon 1 originally
had, thus completing
the teleportation of the quantum
mechanical state of photon 1
onto photon 3.
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