U1L6d.txt

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So I want to say, the first column here,
the output is going to be 0.
And the right column here, the output is going to be 1.
So let's label this diagram.
So this is Alice, Alice, Alice, Alice, Bob, Bob, Bob, Bob.
And here, this is, I guess, 0.

I want A equals 0, A equals 1, B equals 0, and B equals 1.
And the outputs are going to be 0, 0, 0, 0.
So this is x.
This is y.
This is x.
This is y.
x equals 1, y equals 1 x equals 1, y equals 1.
So is this fairly clear?

So Alice gets a bit from the referee.
And if the bit is 0, she chooses this basis.
And if the bit is 1, she chooses that basis--
ditto from Bob.
And if Bob gets this state, he outputs 0.
And if he gets this state, he outputs 1.

So suppose Alice gets A equals 0 and Bob gets B equals 0.

Alice gets observable sigma z.
And now, I see that I never should
have used x and y as the bits, because we
have sigma z, which is--
and Bob-- I shouldn't say gets, I should say measures.

Bob measures the observable 1 over root 2 sigma
x plus sigma z.

So I suppose Alice measures 0.
Then Bob has qubit 0.
He measures it at with sigma x plus sigma z over 2.
So what is the probability that he gets the eigenvector plus 1
for this?
Well, one way to do this is compute the eigenvector plus 1
and take the inner product of it with 0 and square it.
But there's an easier way.
Ah, this should be a root 2, sorry.

So the expectation of Bob's value
is sigma x plus sigma z over root 2 times 0, 0.

OK, we can compute that.

So how do we compute that?
Well, it's just 1, 0--
that 0-- 1, 1, 1, minus 1.
So that's sigma x plus sigma z, because sigma x was 1, 1 here,
and sigma z was 1 minus 1.
We must put a 1 over root 2 in front of that.
Times-- OK, this should have been a row vector,
and this should be a column vector.

So this times this times that is 1.
So this is 1 over root 2.
But this is equal to the probability
that Bob gets 0 minus the probability
that Bob gets 1, because it's the expectation
of this observable.
And this observable has two eigenvalues--
1 and minus 1.
And the 1 eigenvalue corresponds to Bob getting a 0,
or getting y equals 0.
And the minus 1 eigenvalue corresponds
to probability that Bob gets y equals 1.

And so we want two numbers, whose difference
is 1 over root 2 and whose sum is 1,
because probabilities sum to 1.
So this is 1/2 plus 1 over root 2.
And this is 1/2 minus 1 over root 2.
So this is the probability that Bob wins if Alice measures a 0.

So what is that probability?
It's 0.854 roughly.
And let's go back and look--

and it's also equal to cosine of pi over 8 squared.

So let's go back and look at this picture again.
That was the probability that Alice and Bob won if they
got A equals 0, B equals 0.
Well, the probability that we win
if they get B equals 0 and A equals 1 is also--
yeah?
Question?
Shouldn't that be 1 over 2 root 2?
Yes, 1 over 2 root 2 and 1 over 1/2 minus 1 over 2 root 2.
Thank you very much.

But this is, indeed, 0.854.
I'm sorry.
I have a tendency to leave out 2's and get minus signs wrong,
and all those other things.
So let's go back and look at this diagram some more.
That was the probability that Alice and Bob win
if they use these two axes--
or these two-- yeah--
these two points on the Bloch sphere,
which differ by an angle of pi over 4.
But if they use these two points on the Bloch sphere, which also
differ by pi over 4, they also win with that probability,
because the only thing that matters
when telling whether two axes on the Bloch sphere agree
is the angle between the points--
and ditto for B equals 1 and A equals 1.
Well, how about B equals 1 and A equals 0?
Well, we know that--

let's see, if A equals 1 and B equals 1, well, Bob gets 0
and Alice gets 1.

So that's this point and this point.
And we want to know how often these occur together.
Well, they occur together with 1 minus the probability
that these two get the same answer, because this answer is
always the--
I mean, Bob here, Bob's y is 0, and here, y is 1.
So x and y agree for these exactly when
x and y disagree for these.
But x and y disagree exactly when--
but Alice will get this answer and Bob
will get this answer with probability also 0.854.
So that means that because if Bob and Alice got
0, 1; 0, 0; and 1, 0, they wanted their bits to agree.
And if they chose A equals 1 and B equals 1,
they want their bits to disagree.
And that's exactly-- and they disagree
with exactly the same probability
that these two things would agree.
I don't think I said that very well,
but hopefully, you understand what I mean.
So A equals 1 and B equals--
so x equals 0 and B equals 1 are the vice versa.
x plus y equals 1 with probability 0.854.
So this is what we wanted to happen.

So Alice and Bob can win this game
with probably 0.854, which is bigger than 0.75.
There's two more things I want to say.

If this angle was theta--

well, let's draw another thing.
If this is theta, these two observables
agree with probability cos squared theta over 2.

So that's cos squared pi over 8.
But if you've measured these two observable simultaneously,
the probability that they would be agreed
would be cos squared pi over 4, which is 1/2.
And we know that these two observables--
if Alice measures 0 and Bob measures plus,
they agree with probability of 1/2.
And these two observables never agree.
And that's cosine squared pi over 2.

So this is the general formula for if Alice and Bob have
an entangled state and they measure something-- two
observables theta apart, it's cosine squared theta over 2.
And I'm not going to prove that.
Yeah?
So this means that this particular strategy
is the optimal one?

That's the next thing I was going to say.
You can prove that this particular strategy
is the optimal one.
They cannot do better than 0.854 at this game.
And that is-- well, I'm not going
to prove it in this lecture, because it's
a little more complicated than everything I've done so far.
But Tsirelson, I think, is the one who proved it.

So now, this gives a game which Alice and Bob can
win with higher probability if they have an entangled