U1L5b.txt

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There's another way of doing measurement.
Suppose we measure in phases plus minus.
And we'll take the same state for our example.
Well, one way of doing it is just
rewriting this state and the basis plus-minus
and doing exactly the same thing.
Or at least rewriting the first qubit in the basis plus-minus
and doing exactly the same thing.
There's another way of doing it.

1 over root 2 0 plus 1--
so this is plus--

times 1 over root 3 0 0 plus 0 1 plus 1 1.

And now we just take the inner product
of this guy with this guy.

And this is something you're not used
to doing in linear algebra.
In fact, you know, we're taking, you know,
the dimensions in a product of a two-dimensional vector
with a four-dimensional vector, which is something
you would never do in 18:06.
But what you do is you can lose the rules.

x A x, y AB is equal to delta x--
what's-- I need to--
let's put a tilde on this x.
delta-x x-tilde, where this is a Kronecker delta function,
which is 1 if x equals x-tilde, times y sub B.
And I suppose you should call this a partial inner product,
because later, we will get to partial traces and partial--
maybe partial transposes, and they work exactly the same way
as this partial linear product, although I
haven't heard it called partial linear product.
I haven't-- I don't remember any real name for it.
But anyway, if we use this thing, we take 0 0.
So first, I guess we get a 1 over root 6 here.
And 0 0 gives you with 0 B plus.

Here we have a 1 B. And here we have a 1--

another 1 B. 1 over root 6.

Let's see, this is going to be 0 sub B plus 2 1.

Now, it feels like I should rewrite this in the basis plus
to make sure that it works.
To prove to you that it works the same way,
but I don't think I'm going to do that.
Because I didn't-- and if you took 0 minus 1 B,
minus 1 root 2 0 A. This would give you, well,
the same 1 over root 6.
And the 1 would be--

am I doing this right?
0 B plus 1 B minus 1 B. Yeah.
Is equal to 1 over root 6 0 A.
Well, I think we should figure out the probabilities.
So the probability of seeing this
is just the length of this vector.
Can anybody tell me what the length of this vector is?

5 or a 6.
Or actually the square of the length of the vector.
And the probability of this, probability is 1/6.

This is-- so if we take--
measure this plus with 5 6 probability,
we get the first vectors in the state plus,
and the second vector is in this state.
With probability 1/6, we get the first vector
is in the state minus, and the second vector
is in the state 0.

And if you want to check that this works exactly the same way
as if you wrote this out in the--
wrote the first qubit in the 0 plus 1 basis
and the plus-minus basis and did this procedure,
I'll let you do that at home.