9.1
(1)
So, by setting , we have:
where
(2) We apply an adiabatic change to .
On the other hand,
Hence, from the Schroedinger equation , we have:
From the definition of , we have:
So,
--- (2-1)
By differentiating the both sides of ,
So, from (2-1), we have:
(3) Since , we have:
(4)
where
The eigenvalues of are
The eigenvector with is given by:
Hence, the correction is
9.2
The change should be slower than the spontaneous photon emission time of the qubit.
9.3
Fix the direction of the 1st spin as , and decide the remaining spins with the following induction:
9.4
(1) We usually choose as a non-interacting system such as in (9.18) or (9.12). So the time evolution under is local.
(2) We need at least one gate for one interaction. So the number of gates is .
9.5
The variance of sample average scales as .
So, the deviation scales as
Since the Hamiltonian has terms, the total variance of the estimated Energy (as a sample average) scales as . So, to keep the total variance below , we have:
9.6
In the limit of , we ignore other terms. Then the ground state is apparently the one where all spins have the same direction. Flipping one spin, we have the minimum gap .
As the initial state (the ground state of is a superposition of all states, it has an overlap with the ground state of . So, with the adiabatic process (in the limit of ), it achieves the state where all spins have the same direction, that is, the state of qubit 1 is copied to the N intermediate qubits.
9.7
In this case, "Success in experiments" means "Success at least one of experiments". So the success probability is calculated as "1 - probability of fails in all experiments." Hence, the success probability is:
If , we need to achieve .
Practically, is defined as where is the number of experiments to achieve the required success probability.