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THOMAS VIDICK: So so far we've seen two different kinds

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of quantum states-- pure states and mixed states.

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And we've been thinking of mixed states as probability distributions.

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The mixed state is simply a mixture over pure states.

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So a natural question is what does this mixture represent?

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So we say that the mixed state is in state psi with probability p.

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Where does that probability come from?

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Where does this uncertainty arise from?

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And so what we're going to see in this lecture

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is that, in fact, every mixed state arises as the reduced density matrix

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of a pure state on a larger system.

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So the fact that we are in a mixed state, a state that is not

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pure-- so a mixed state that has rank larger than 1--

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reflects the fact that this mixed state must be part of a larger system.

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That larger system is in a pure state, but we

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don't have access to the whole system.

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And this is why we end up with a density matrix that has rank larger than 1,

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so a mixed state that is not pure.

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So let's see how these purifications work.

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So we've already seen before that density matrices can

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arise by taking the reduced density.

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And the sort of prototypical example is--

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if you start with an entangled state like this psi

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here, which is initialized in an EPR pair.

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So let's quickly compute the reduced density on A.

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We know how to do that.

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We have to choose a basis B. For the B system

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here, it's appropriate to choose the standard basis for B.

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So let's choose this.

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To this basis we can associate POVM elements

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that represent the partial measurement of the B system

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in the basis B. So this would be M0, which

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is the identity on A tensored rank 1 projection on 0 for B and M1,

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which is the identity on A tensored the rank 1 projection on 1 for B.

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We can choose as Kraus operators the same

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as the M's, because they're projectors.

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This lets us compute the reduced densities.

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If you do the computation, you'll see that you'll

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get the outcome 0 with probability 1/2.

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And in which case, you get the state rho 0 on A and B, which is simply 0 0

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or you could get a 1 with probability 1/2.

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The state will be rho 1.

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And so this tells us that the reduced density on A

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is the totally mixed state-- half of the identity

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matrix, half of the projection on 0 plus the projection on 1.

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So here we have a pure state, a global state.

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It's entangled.

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I take the reduced density on system A and I obtain a mixed state.

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And the question is whether this process is reversible or more generally,

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if whether given the density matrix, we can always

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write it as the reduced density on the system of a pure state of a larger

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system.

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And the answer is yes, and let's see how that works.

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So let's take an arbitrary density matrix rho

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A. We know that we can write it as a mixture

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with here some coefficients p_i that are non-negative and sum to 1.

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These are the eigenvalues of the reduced density matrix

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and of psi i, which are the eigenvectors.

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So you can always compute this decomposition using the singular value

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decomposition.

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So the way we interpreted this is by saying that, all right,

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this is a mixture.

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And what it means is that with probability p_i the state on A

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is in the pure state psi i.

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So where does this probability arise from?

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Who gets to decide what state we're actually on?

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So now what I claim is that I can introduce a new state

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on A and a new system B, where B is going to be the system that

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controls which i we get.

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So it's natural to introduce this state, psi A B, which

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is going to be the sum over all i.

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Let me put root p_i psi i on A tensored I on B. Where B is a new system,

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let's have this index A_i range up to the dimension of the system B.

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So B-- this is a new system.

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It has the same dimension as the A system.

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And it has a basis that I called i.

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So this is certainly a pure state.

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It's a valid state.

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It's normalized, because the norm of psi will

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be the sum of the p_i's, which is 1 because rho A was

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the normalized density matrix.

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And now let's check that it achieves what I claim it does.

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So we have to compute the reduced density of psi A B on A.

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How do we do that?

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We choose a basis for the B system.

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We measure.

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We see what happens.

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So, of course, what basis are we going to choose?

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The one we just introduced, so let's use that basis for B.

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Then if we compute the probability that we get a certain outcome i--

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we're getting used to this now.

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It's just going to be the squared norm of this vector square root p_i psi

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i, which is p_i and the corresponding projected state rho

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i will be psi i on the A system tensored i on the B system.

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And so this tells us that the overall reduced density rho A on the A system,

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obtained by taking the partial trace operation,

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will be the sum over i of the probability that we get i.

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This is p_i tensored with the reduced density on A

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that's associated with rho i.

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So that's just a rank 1 projection on psi i.

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And you can verify that, as claimed, the results that I obtain here

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is exactly the state that I started from.

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And I didn't make any assumption about that state.

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This was just an arbitrary density matrix on A. So

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any such state that can be obtained the way that I've obtained it,

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or in any way that is such that it's reduced density on the system A--

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this is called a purification of the reduced density.


