U1L6j.txt

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So Alice and Bob share this state.

Alice gets row.

Measures the observables in that row.

So Alice and Bob are sharing this state.
When Alice gets a second row, she
measures sigma z on the second qubit, sigma
z on the first qubit.
And she, I should say, red if eigenvalue of minus 1
and green if eigenvalue of plus 1.

So she measures sigma z on the first qubit, sigma
z on the second qubit, and then if the first one was a minus 1
and the second one was a 1, she would
mark this square red, this square green,
and this square red.
Now, we know because all three of these matrices
multiply to i that the observables must follow i to 1.

So that means that there are two greens in Alice's rows
or even number of reds.

Because the reds were minus 1, and their product is 1.
But Bob's columns all multiply to minus i,
which means you get an odd number of red
in each column when Bob measures these.

So Bob gets odd number of reds.

And the last thing is that Alice and Bob
are sharing the state so Alice measures on her qubits.

And I guess this would be AB, AB, AB, AB.
And Bob measures on his qubits.
But when they measure the same observable,
because this is in the 1/2 0 0 plus 1 1 state,
they get the same result.
So when both of them measure sigma z, they
get the same result. When both of them
measure sigma x, they get the same result.
Actually, when both of them measure sigma y,
they get opposite results, but you know this is opposite.
This is opposite.
And we're taking the parity.
So they get the same plus or minus 1
for this one, even though each of their qubits
they get opposite.

Rather, if Bob measured a plus 1, plus 1 for both qubits,
Alice would measure minus 1, minus 1 for both qubits.
But then they are looking at the eigenvalue plus 1 times plus 1
equals minus 1 times minus 1.

And any questions?

Yeah?
So under what circumstances would they lose then?
They never lose if they--
Oh, well, what's the [INAUDIBLE] for?
Oh, that's a classical strategy.
Oh, I see.

Classic rule strategy.
Remember, the classical strategy,
they had to agree on a square.
And you cannot color all the squares so that every row has
an even number of reds and every column has an odd number
of reds.
And that's why they cannot win with probability more than 8/9.
But a quantum strategy, because you have these observables
such that any product is identity here
and any product is minus the identity in the columns,
means they can always win.

Any other questions?
OK.
So the last thing I wanted to do was
say what the simultaneous eigenvalues
are in sigma x tensor sigma in this column.
So sigma x tensor sigma x is equal to 1, 1, 1, 1.
Sigma z tensor sigma z equals 1, minus 1, minus 1, 1.

OK.

And there are four eigenspaces.
And first, if you know sigma x tensor sigma x,
and you know the eigenvalue for sigma z tensor sigma z,
you know the eigenvalue for sigma y tensor
sigma y, because this times this times this is minus 1.
So we don't need to worry about sigma
y tensor sigma y and the eigenvectors are 0,0,
plus 1, 1.

So that's in the plus 1 eigenspace for this.
And that's the plus 1 eigen space for this.

And that's plus, plus.

And 1 over root 2, 0, 0, minus 1, 1.
Well, that's still in the plus eigenspace for this
because 0, 0 and 1, 1--
that's 0, 0, 1, 1-- are both in the plus eigenspace.
But this vector is just 1, 0, 0, minus 1.
And you can see that it's taken to its negative by this matrix.
So that's in the minus plus eigenspace.

And then 1 over root 2, 0, 0, 0, 1, minus 1, 0.
And that's in the minus, minus eigenspace.
And 1 over root 2, 0, 1, plus 1, 0, and that's
in the plus, minus eigenspace.

So that if Alice measured this row, and Bob measured,
let's say, this row, this is Bob column three.

Alice column-- I want her to do column two--

eigenspaces are 0, 0, 0, 1, 1, 0, 1, 1.

So if Alice and Bob shared this thing, Alice would and Bob got,
so Alice is measuring this row, and Bob is measuring this row.
Bob might see 0,0 plus 1, 1.
And Alice might see 1, 1.

But they will agree on this eigen.
And 0, 0, plus 1, 1 is in the plus 1 eigenspace of sigma z
tensor sigma z.
1, 1 is in the plus one eigenspace of sigma
z tensor sigma z.