THOMAS VIDICK: So so far we've seen two different kinds
of quantum states-- pure states and mixed states.
And we've been thinking of mixed states as probability distributions.
The mixed state is simply a mixture over pure states.
So a natural question is what does this mixture represent?
So we say that the mixed state is in state psi with probability p.
Where does that probability come from?
Where does this uncertainty arise from?
And so what we're going to see in this lecture
is that, in fact, every mixed state arises as the reduced density matrix
of a pure state on a larger system.
So the fact that we are in a mixed state, a state that is not
pure-- so a mixed state that has rank larger than 1--
reflects the fact that this mixed state must be part of a larger system.
That larger system is in a pure state, but we
don't have access to the whole system.
And this is why we end up with a density matrix that has rank larger than 1,
so a mixed state that is not pure.
So let's see how these purifications work.
So we've already seen before that density matrices can
arise by taking the reduced density.
And the sort of prototypical example is--
if you start with an entangled state like this psi
here, which is initialized in an EPR pair.
So let's quickly compute the reduced density on A.
We know how to do that.
We have to choose a basis B. For the B system
here, it's appropriate to choose the standard basis for B.
So let's choose this.
To this basis we can associate POVM elements
that represent the partial measurement of the B system
in the basis B. So this would be M0, which
is the identity on A tensored rank 1 projection on 0 for B and M1,
which is the identity on A tensored the rank 1 projection on 1 for B.
We can choose as Kraus operators the same
as the M's, because they're projectors.
This lets us compute the reduced densities.
If you do the computation, you'll see that you'll
get the outcome 0 with probability 1/2.
And in which case, you get the state rho 0 on A and B, which is simply 0 0
or you could get a 1 with probability 1/2.
The state will be rho 1.
And so this tells us that the reduced density on A
is the totally mixed state-- half of the identity
matrix, half of the projection on 0 plus the projection on 1.
So here we have a pure state, a global state.
It's entangled.
I take the reduced density on system A and I obtain a mixed state.
And the question is whether this process is reversible or more generally,
if whether given the density matrix, we can always
write it as the reduced density on the system of a pure state of a larger
system.
And the answer is yes, and let's see how that works.
So let's take an arbitrary density matrix rho
A. We know that we can write it as a mixture
with here some coefficients p_i that are non-negative and sum to 1.
These are the eigenvalues of the reduced density matrix
and of psi i, which are the eigenvectors.
So you can always compute this decomposition using the singular value
decomposition.
So the way we interpreted this is by saying that, all right,
this is a mixture.
And what it means is that with probability p_i the state on A
is in the pure state psi i.
So where does this probability arise from?
Who gets to decide what state we're actually on?
So now what I claim is that I can introduce a new state
on A and a new system B, where B is going to be the system that
controls which i we get.
So it's natural to introduce this state, psi A B, which
is going to be the sum over all i.
Let me put root p_i psi i on A tensored I on B. Where B is a new system,
let's have this index A_i range up to the dimension of the system B.
So B-- this is a new system.
It has the same dimension as the A system.
And it has a basis that I called i.
So this is certainly a pure state.
It's a valid state.
It's normalized, because the norm of psi will
be the sum of the p_i's, which is 1 because rho A was
the normalized density matrix.
And now let's check that it achieves what I claim it does.
So we have to compute the reduced density of psi A B on A.
How do we do that?
We choose a basis for the B system.
We measure.
We see what happens.
So, of course, what basis are we going to choose?
The one we just introduced, so let's use that basis for B.
Then if we compute the probability that we get a certain outcome i--
we're getting used to this now.
It's just going to be the squared norm of this vector square root p_i psi
i, which is p_i and the corresponding projected state rho
i will be psi i on the A system tensored i on the B system.
And so this tells us that the overall reduced density rho A on the A system,
obtained by taking the partial trace operation,
will be the sum over i of the probability that we get i.
This is p_i tensored with the reduced density on A
that's associated with rho i.
So that's just a rank 1 projection on psi i.
And you can verify that, as claimed, the results that I obtain here
is exactly the state that I started from.
And I didn't make any assumption about that state.
This was just an arbitrary density matrix on A. So
any such state that can be obtained the way that I've obtained it,
or in any way that is such that it's reduced density on the system A--
this is called a purification of the reduced density.