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U1L4c.txt
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U1L4c.txt
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#
# File: content-mit-8370x-subtitles/U1L4c.txt
#
# Captions for course module
#
# This file has 101 caption lines.
#
# Do not add or delete any lines. If there is text missing at the end, please add it to the last line.
#
#----------------------------------------
OK, so the first thing I want to do
before I get to the no-cloning theorem
is talk about tensor products.
So suppose we have two qubits.
What is the state space?
Well, let's think.
What is a basis for the state space?
How many distinguishable states can you find?
Anybody?
Can somebody give me-- yeah?
0, 0; 0, 1, 1; 0, 1, 1.
Yeah.
0, 0; 0, 1; 1, 0; and 1, 1.
So how can you distinguish them?
Well, if you wanted to distinguish these two,
you can measure the second bit.
If you wanted to distinguish these two,
you can measure the first bit.
And, in fact, there are four distinguishable states.
Which means that if you have two qubits,
your state space has dimension four and n qubits.
Can someone tell me what the size of the state
space for n qubits is?
2 to the n.
And I probably don't need to write down all the basis
states.
And I should say something, that this
is one way of expressing this state space.
We can also write it as [? 0, ?] 0,
where this applies to the first qubit
and this applies to the second.
Or 0 tensor, 0.
So this is 1.
And if we name our systems AB, we
can make it clear which qubit belongs
to which system by putting, say, AB here or AB or AB.
OK, so that's the no-cloning theorem--
that's the tensor products.
And let's do a little bit more tensor products
before we get to the no-cloning theorem.
So how many degrees of freedom for a single qubit?
Well, there were three, remember?
Because you had a block sphere and a qubit
it could be anywhere on the block sphere.
So how many degrees of freedom for two qubits?
Well, we have alpha 0, 0 plus beta 0, 1
plus gamma 1, 0 plus delta 1, 1.
So that's 2, 4, 6, 8.
But then we have to subtract some things.
But some alpha I squared is equal to 1.
So that gives you 1 fewer degree of freedom.
And-- wait, 3?
No, this is 2.
I mean, it's on the surface of the block sphere
so it's just 2 degrees of freedom.
So I mean-- yeah.
And two qubits.
And we also have a global phase, which is irrelevant.
So that's 6.
And n-- well, we really have 2 to the n
minus two because, again, we have
the sum of the squares is 1 and a global phase.
So what does this say?
Well, this says that there are 4 degrees of freedom
if we take the product of 2 qubits, each of which
is in its own state, and there are 6 degrees of freedom
if we look at the general state space of two qubits.
So that means there are some states of 2 qubits which cannot
be factored into 2 things.
So a separable state--
each qubit is in a definite state.
And alpha 0 plus beta 1--
let's call this qubit--
A times gamma 0 plus delta 1--
let's call this qubit B--
is equal to alpha gamma 0, 0 plus alpha delta 0,
1 plus beta gamma 1, 0 plus beta delta 1, 1.
So this is a qubit in a separable state
where you can say these two states each exist.
So otherwise the qubit is an entangled state, e.g.
1 over root 2 0, 1 minus 1, 0.
And here you cannot separate this state into one state
for the first qubit and one state [? for ?] the second