U1L6b.txt

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I'm going to present Bell's argument first.
Originally, it's a statement about marginal probabilities
of a joint probability distribution.
But I find it makes much more intuitive sense
if you talk about it as a game that two players are
cooperatively trying to play.

And so what is this game?

And this is called the CHSH game after Clauser, Horne, Shimony,
and someone else.
Sorry.
I should have looked it up.
And so here's the game with Alice and Bob.
So Alice and Bob have migrated from cryptography
to quantum information theory, if you want
to know where they came from.

So Alice and Bob are two of the prototypical participants
in cryptosystems.
So they're playing a game.
And they want to cooperate to win with as high probability as
possible.
So you have a referee who doesn't actually
make any decisions.
He just flips coins and does what the instructions tell him.
So sends a bit to Alice and Bob.
So we'll have him send A to Alice and B to Bob.
And then Alice and Bob, who are in separate rooms and cannot
communicate, get these two bits and output two more bits,
xy two bits.
And I should say the referee chooses the bits completely
at random.

So great.
So now, we need to tell you when Alice and Bob win
and when they lose because, otherwise, that would be--
I mean, yeah.
So we need to know when people win
and lose to determine the game.
So Alice and Bob win if x plus y equals A times B.
Or I should say x plus y mod 2.
So that's a x exclusive r y.

Wait, no.

Yeah, no, that's right.

OK.
So that should be pretty clear, but I
want to put up a little chart to make it well.

So A can be 0 or 1.
B can either be 0 or 1.
And Alice and Bob win if x plus y equals 0.
x plus y equals 0.
x plus y equals 0.
And x plus y equals 1.

Now, let's pretend that Alice and Bob have
to do this deterministically.
So Alice has a little chart.

If x equals 0, output 0.
If x equals 1, output 1, or something like that.

And there is a general theory in games like this,
Alice and Bob cannot do any better with a probabilistic
strategy than a deterministic strategy.
And we'll get to that later.
So here, well, here, this is x0 plus y0, x0 being the bit
that Alice outputs if A is 0.
This is x0 plus y1.
This is x1 plus y0.
And this is x1 plus y1.
And these are all mod 2.
Yeah?
Oh, that was my question.
Isn't it possible to circle mod 2?
Yes.
I am being lazy.

OK.
So this is what Alice and Bob want.
Now, can anyone tell me why they can't
win with probability 100%?

Following this y--

Yeah?
So if you sum everything up, you get 0.
That's right.
So you sum everything up.
So that's 2x0 plus 2x1 plus 2y0 plus 2y1 is equal to 1 mod 2.
That doesn't work because, you know,
there's 2x1s in this chart, two x0s, 2y1s, and 2y0s.

Now, I want to say Alice and Bob can only
win with probability 75%.

Why is this?
Well, so there's no choice of x0, x1, y0,
and y1 that makes this perfect.
So let's suppose that Alice and Bob have
agreed in advance to some combination of x0, x1, y0,
and y1.
The referee comes along and sends them
two bits, which randomly select one of these four boxes.
Well, we know one of these four boxes
has to disagree with what Alice and Bob want
because they can't all work.
So at least one has to be wrong.
And if one is wrong, and the referee
picks that one, which he does 1/4 of the time,
then Alice and Bob lose.
So that's 75%.

Now, I want to argue why a random strategy can't
do any better.

So for a random strategy, we have a probability
that they win is equal to some probability of, let's
let r be the string of random bits that Alice.
So a random strategy you can always
think of this as Alice and Bob get some set of random bits.
And we don't care whether they're correlated or not.
And then for these random bits, they
need to pick a deterministic outcome of x0, x1, y0, and y1.
So the probability that Alice and Bob
win with a random strategy is equal to sum
times the probability of string r
random bits times the probability
that Alison and Bob win given r.

So this is essentially an average
of how often they win given any particular string
of random bits.
So if this was greater than 3/4, there
must be some string of random bits
which gives Alice and Bob a deterministic strategy
of winning with probability at least 3/4.
And we just argued that was impossible.

So that's the classical part.
So now, we want to say that if Alice and Bob have
an EPR pair of qubits, 0, 1, minus 1, 0, then
they can win with probability larger than 75%.

But the problem with 1/2 0, 1, minus 1,
0 is that Alice and Bob get opposite answers every time.
And I find it much harder to think
about getting opposite answers than getting the same answer.
So I want to find a way that Alice and Bob can