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U2L1f.txt
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#
# File: content-mit-8370x-subtitles/U2L1f.txt
#
# Captions for course module
#
# This file has 151 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 this is a circuit for teleportation.
Now, what I want to do is take this circuit for teleportation
and use it to show you that teleportation works.
And this involves doing a bunch of, I guess,
operations on circuits, which leave what the circuit does
unchanged, but make it clearer, or at least [INAUDIBLE]
But first, what did I want to do?
First, I wanted to divide this into Alice and Bob.
So Alice has the first two parts.
So the first operation, if we take plus and 0
and apply a control know, we get 1 over root 2 0, 0 plus 1, 1.
So that means that the circuit is equivalent to 0 plus 0.
I'm going to put a control knot here,
and then a control knot here.
And then, I have [INAUDIBLE] and then a measurement,
and a classically controlled--
well, two measurements-- and a classically
controlled x, which I guess I should be drawing in green,
and a classically controlled z.
So this does the same thing as this circuit.
So if you can prove that this one teleports,
you can prove this one teleports.
OK, now, the second change I want to make in the circuits
is moving these controls to the left of these measurements.
So what does that look like?
OK, we're starting with our unknown qubit.
Yeah, our new and computed qubit was supposed to be zeta.
And I seem to have removed it in the first picture.
So let's fix that.
Zeta and plus and a 0.
And the. first thing is a control knot here.
The second thing is a controlled knot here.
And then, there's that Hadamard up there.
And then, there's control z, control x, and a measurement
here, and a measurement here.
Let's see.
OK, good, now, I really should have--
OK, so what did we do?
What we're saying is it doesn't matter
if we do this measurement, and then do the control
classically, or if we just do a quantum control knot,
and then do the measurement.
So why is this true?
So we have a unitary--
well wait, we have a qubit here, say.
And we have another qubit here, alpha 0 plus beta 1.
And we want to say that a control
knot before a measurement is the same as a classical measurement
followed by a control knot.
Well, alpha 0-- so this is actually
not the most general situation.
But I'll just show you what happens in this situation.
Y-- and we want to do a c knot on this--
is equal to alpha 0--
and the c knot does nothing, If alpha is 0--
plus beta 1 sigma x phi.
So doing a control knot on this gives you
alpha 0 phi plus beta 1 sigma x phi, measuring 0.
If you get 0 as a result of the measurement,
you have alpha phi.
And if you get 1 as a result of the measurement,
you have beta sigma x times phi.
Now, if you measured first--
this should be alpha squared, I'm sorry, and beta squared.
Now, if you'd measured first, with alpha squared probability,
you would have gotten 0.
And with beta squared probability,
you would have gotten 1.
And then, when you apply the controlled classical sigma
x to phi, with probability alpha squared, you get phi.
And with probability beta squared, you get sigma x phi.
So in this case, these two results are the same.
And in fact, the two results are always the same.
And you can show it with a little more computation
in quantum computing.
Yeah, you have a question?
What specifically do you mean by they're always the same?
Oh, well, if you have a measurement and a basis,
well, you have a controlled unitary followed
by a measurement in this basis, it's
the same as doing the measurement
and then doing the classically controlled unitary.
The results here, the probability
of, the probability skew outcomes for the basement,
and the quantum states at the end of the measurement
will be identical.
OK, so now, our original circuit was equal to this.
And now, there's only one more step
we need to take to get to the circuit,
well, actually two steps.
A controlled z is equal to a controlled z up.
So they both give unitary 1, 1, 1, minus 1.
So this is just saying, change this phase of the entire system
if both the top and bottom qubits are 1.
Because you do nothing if the top qubit is 0.
And if the top qubit is 1, you change the phase
only if the bottom qubit is 1.
Because sigma z was what?
Sigma z was this matrix, 1 minus 1.
But this matrix is symmetric.
And [? the force ?] qubit and qubit,
so it's the same as this thing.
And a control z is equal to a control x, Hadamard, Hadamard.
Because z equals H sigma x H. So if you put a control on them,
you still get the same identity.
So what we're going to do is we're going to take this
and we're going to turn that over.
And let's do it up here.
Zeta plus 0--
OK, there's an H here.
And there is another control knot here,
and then a control knot up this way with an H
on the left and an H on the right.
And yeah, OK, good.
And the last transformation we need to do is just--
remember, if you apply to H's, they give the identity.
So this is completely equivalent to our teleportation circuit.
Of course now, we have control knot gates
between Alice and Bob, which weren't
present in the original.