Q2L6e.txt

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What quantum circuits prepare stabilizer states
and how do stabilizers respond to measurement?
First, let us describe the preparation
of stabilizer states.
For example, consider this stabilizer.
It is a state 0 0 plus 1 1 divided by root 2.
We know how to prepare this state
with a simple circuit that utilizes a Hadamard
and a controlled NOT gate.
It is convenient to describe this state entirely
in terms of how the stabilizer evolves from S1 through S3.
S1 is just IZ, ZI, the stabilizer
for 2 qubits in the 0 state.
S2 is the stabilizer given by IZ and XI.
This is because a Hadamard acts on the first qubit
to change the stabilizer from Z to X.
For S3, it is convenient to review
the action of the controlled NOT gate
on input stabilizer states, here labeled as Control and Target.
The main thing to note is that a CNOT propagates an X
from the Control to an XX, and a Z on the Target to become a ZZ.
Thus it is straightforward to see
that S3 is the stabilizer ZZ, XX, exactly
the stabilizer we desired.
This idea of using the evolution of stabilizers
to describe the action of a circuit
can also be generalized to measurement.
We focus on the measurement of operators, which
we have defined previously.
Consider the measurement of m, a Pauli operator.
Recall that this measurement is done by adding a control
qubit in a superposition which performs a controlled m
operation.
A second Hadamard is applied to the control
qubit, which is then measured.
The measurement result A corresponds
to the eigenvalue of the eigenstate psi sub a
resulting from the measurement.
We employ here this convenient shorthand notation
for an operator measurement circuit.
How does such a measurement affect the stabilizer?
Suppose the stabilizer initially is g1 through gn,
and we measure a Pauli operator m.
Note that m squared is Identity and the eigenvalues of m
are plus and minus 1.
Without loss of generality, we may assume that m anticommutes
with g1, but commutes with all of the other generators.
You will explore this in the exercises.
Of course, S may already be an eigenstate of m,
and thus it is possible that m commutes
with all of the generators.
Measurement thus changes the stabilizer
to add m in place of g1 with a plus or minus 1
to the a-th power.
We know this must be the form of the output stabilizer,
because the output is an eigenstate of m with eigenvalue
plus or minus 1.
So let us see how this works in some examples.
First, consider the measurement of the operator ZZ on the state
00.
This initial state has stabilizers IZ and ZI.
What is the stabilizer of the output state?
Well, actually it should be obvious
that S prime is IZ, ZI, and that is because the measurement
operator ZZ commutes with ZI and IZ,
thus the stabilizer is left unchanged.
Consider a second example, just like the first, but now
measuring XX instead of ZZ.
The initial state has stabilizer ZI and IZ,
which is equivalent to the stabilizer IZ, ZZ.
Now S is in the form where the first generator anticommutes
with m, but the remaining generator commutes with m,
Thus the output stabilizer is minus 1 to the A, XX and ZZ.
This illustrates the general rule we gave earlier.
Stabilizer measurement also provides a mechanism
for preparing an arbitrary stabilizer state,
namely given a stabilizer generated by g1 through gn,
just measure g1 through gn.
Let's look at how this works with our example.
Consider the output stabilizer S prime.
This depends on the measurement result A.
If we get minus XX and ZZ, but 1 plus XX and ZZ,
then we can fix this by applying ZI,
a unitary fix operator U. Generally there
will always be a way to fix up the signs to get the desired
plus 1 eigenstate.
Later we will see how to find the fix up operator systematically.