Q1L2a.txt

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We have described quantum mechanics
using four postulates.
However, for quantum error correction we need more.
We need classical statistics.
Consider this question.
With probability 1/2 you were given either psi or phi.
How do you describe this state?
It is not a single quantum state, but rather
a statistical combination of two possible states
with equal probability.
You may recall that in quantum mechanics
this is described by a density matrix.
Here it is the sum of the two outer product
vectors of the quantum state.
We call this rho.
To further appreciate the need for density matrices
in describing stochastic combinations of quantum states,
consider the following scenario.
We have a two-part state-- psi AB--
and this state we may draw as this quantum circuit which
originates from a single place.
The two qubit state will be this particular superposition,
and have parts A and B. A will be the first label
and B will be the second label in each of the kets.
The question is, suppose B measures B's part of the state.
What describes A's part of this bipartite quantum state?
This state will be a statistical mixture,
because what A obtains differs depending
on what B's measurement result. If B measures a 0,
then A has a 0.
If B measures a 1, A has a 1.
And each one of these happens with a different amplitude,
so we get one or the other of these two states
with the two different amplitudes.
Let us call this statistical mixture Mixture 1.
And now, imagine another separate scenario.
In this scenario, B measures, but in a different basis,
rather than in this computational 0 and 1 basis
as shown above.
We have the same bipartite state as an input
to this quantum circuit.
B, before measurement, let us say,
performs a Hadamard transform.
The question, again, is what is A's state?
A will see something that we may interpret as being different,
depending on what B's measurement result is.
The state before B's measurement is written here.
It is a superposition where the second ket has had a Hadamard
operation applied to it.
And therefore we can read off from this
what A's state is going to be.
A will see when B measures a 0 a superposition of root 3 over 4
|0> and root 1 over 4 |1>.
If B measures a 1 instead, A's state
will be different, with a minus sign instead of a plus sign.
Now, this is another statistical mixture.
And the mixture is of these two states-- one with a plus
and one with a minus-- and they are superpositions in contrast
to Mixture 1.
Let us call this Mixture 2.
And, certainly, it looks very different from Mixture 1.

The question is, how are these two statistical mixtures
different?
Or might they be the same mixture in some sense?

To answer this question, we will need
to turn to the definition of the density matrix.
For the statistical mixtures we have written here
using this circle plus as an or sign, the density
matrix is defined as the sum over the outer product of each
of the elements in this statistical mixture,
as given here.
We call rho the density matrix.

Let us explore some sample density matrices.
Recall that the state 0 is a two component vector 1 0,
and the state 1 is 0 1.
The density matrices corresponding to these
are given by the outer products, so 0
is 1 0 0 0, and 1 is 0 0 0 1.
The cross term 0 1 has the 1 in the upper right hand corner.
And this is easy to understand by labeling the rows
and columns appropriately, with 0 on the left and on the top.

You may also explicitly multiply the column and row vector
to obtain this matrix form.

Using these matrices, we may now write out
the density matrices which correspond to the mixtures
that we have studied.
The first statistical mixture-- let
us call it rho 1-- turns out to be the density matrix given
by this sum.
3/4 of 1 and 1/4 of 1 gives you 1/4 times 3 0 0 1.
The second statistical mixture is
where we need the cross terms because the states are
superpositions.
These cross terms give rise to off-diagonal elements
in the density matrix.
Summing, as before, we find, though,
something very interesting.
The result is the same density matrix as for Mixture 1.
We conclude from this that the two are actually
the same state.
More about this can be understood
by returning to the formalism of density matrices.
Let me begin by defining a density matrix.
A density matrix is a matrix rho which
has two particular properties.
First, the trace of rho must equal 1.
Second, the density matrix rho must be positive.
In other words, for all possible states that one might choose,
the expectation value of rho must be real and greater than
or equal to 0.
How do I construct a density matrix from states?
First, let me make this claim, that a density matrix may
be constructed from a probabilistic combination
of pure states.
P, here, is a probability.
Psi are states which for now let us take to be orthogonal.
We may prove that this statistical combination
is a density matrix by checking the two constraints.
First, we note that its trace is equal to 1,
since Ps are probabilities.
And second, it is clear that the matrix is positive.
So we can construct density matrices this way.
In the reverse direction, given a density matrix,
how can we interpret it?
Well, I claim that any density matrix
rho can be expressed as a stochastic combination
of pure states.
So that is exactly in the form we
have just seen, using a stochastic combination
of pure states.
Let us prove this claim as follows.
A density matrix is positive, and thus it
has a spectral decomposition into eigenvalues
and eigenvectors.
And this spectral decomposition gives us exactly what we want.
We may see this straightforwardly
by writing out the spectral decomposition with lambda sub k
being the eigenvalue for state k.
Note that the trace is 1, and thus the eigenvalues
must sum to 1, meaning they are a probability distribution.
They are also real-valued, and rho
is positive by construction.
This decomposition-- known as an unraveling--
allows us to consider some useful definitions.
Rho is known as being pure when its decomposition results
in a single pure state alone.
Otherwise, rho is known as being a mixed state,
meaning that it is a stochastic combination
of multiple pure states.
OK, given these two constructions
let's consider a quick question.
How about a stochastic combination of density
matrices, given Ps as probabilities and rhos
as density matrices.
Is this combination a legal density matrix?
This is a simple thing to work out.
The answer is yes.
So try it yourself and see if you can prove this to be true.
The concept of unraveling a density matrix
into a stochastic combination of pure states
was considered by von Neumann and is very interesting.
This is an important lemma about unravelings.
Let rho be unraveled as a stochastic combination
with probabilities P and states psi.
It turns out that this unraveling is not unique,
in general, for mixed states.
For a mixed state the density matrix
may also be unraveled with probabilities P and states phi,
under the following condition.
That is, that the square root-- in a sense--
of this decomposition-- square root of P and state psi--
is related by a unitary transform
to the other decomposition, with square root of q and states phi.
The physical origin of this infinity of possibilities
of unravelings may be interpreted
as arising from the fact that it doesn't matter which
basis B measures in, in the original scenario we started
out our discussion with today.

Let us now turn to another definition.
We will say that a purification of a density matrix
rho sub A is a state psi sub AB-- a pure state-- such
that rho sub A is obtained by taking the partial trace over B
of the density matrix given by the pure state.
This partial trace operation removes the degrees
of freedom associated with B, and can
be viewed in terms of the quantum circuit
that we started out today with.
Namely, that we have a bipartite system,
and the B part of the system is measured,
leaving rho sub A as the final state.
This concept of having infinite unravelings,
and the interpretation of density matrices
in purified forms will be key ideas
in understanding the intuition behind quantum error correction.