In quantum annealing/adiabatic quantum computation, tunneling is utilized as a quantum effect to settle at the ground state. Yet, I do not quite understand the tunneling effect in the context of adiabatic quantum computation, where the system starts at the ground state of an Hamiltonian and by evolving slow enough, it remains (with high probability) at the ground state of the time dependent Hamiltonian and ends up at the ground state of the problem Hamiltonian.

Particularly, in this picture in wikipedia,

I now have the following question:

1. Is this cost configuration in this diagram the energy spectrum of the final Hamiltonian, so cost = energy of the the final Hamiltonian?

2. Where is the initial point(s), in this diagram at the start of the computation? Is it a single point at, for example a local minima, or a distribution of points spread out in the cost configuration (this diagram)? Naively, since I think the initial point = energy of the ground state of initial Hamiltonian, it is just one point in the diagram. Yet, i don't think it lies on the curve as the energy of the ground state of initial Hamiltonian can be anything?

2. ) This requires some explanation. The general Hamiltonian in the case of quantum annealing (which may be thought of as a special case of adiabatic QC) is of the form, $$H = J\sum_{<i,j>} S^z_i S^z_j + h \sum_i S^x_i.$$
$S^x$ and $S^z$ are the usual Pauli matrices. At start of the computation $h$ is set to be very high, so the Hamiltonian effectively consists only of the second term. Now the eigenstate of the Hamiltonain is the uniform superposition state. This state canot obiviosuly be represented as a single point on the above figure. As the computation proceeds $h$ is slowly decreased to zero, so that after the computation finishes we can get the ground state of the first term of the Hamiltonian (by adiabatic theorm, this will happen if we go about it slow enough). The final state is an eigenstate of the first term and the initial state is an eigenstate of the second term. As these two terms donot commute their eigenstates both cannot be shown as single points in the figure.