# Initial Hamiltonian in Quantum Annealing

Background: The quantum annealing process performed by the D-Wave machines is described as follows: We have a problem Hamiltonian, $H_{P}$, whose ground state encodes the solution to a problem of interest, and a 'disordering' Hamiltonian, $H'$, which does not commute with $H_{P}$. The energy of the D-Wave system can be described by the Hamiltonian $H = H_{P} + \Gamma H'$, where $\Gamma$ changes from a large value to $0$ throughout the computation, so that $H=H_{P}$ at our computation's completion. http://vanilla47.com/PDFs/Quantum/1/An%20Introduction%20to%20Quantum%20Annealing.pdf

Question: What state is the system set up in (so what is $H(0)$)? Is it an easy to encode Hamiltonian, such as the one used in adiabatic quantum computing (normally labelled $H_{0}$)? Or my other guess would be that $\Gamma$ is initially $0$, so $H=H_{P}$ initially, and then $\Gamma$ is increased to enable 'quantum tunneling' through barriers separating local minima (to help the system get into its ground state), and then $\Gamma$ is slowly decreased to $0$ again.

In the D-Wave setup (and most other setups), the initial Hamiltonian $H'$ is a local magnetic field on each spin (in the D-Wave case, in a basis dual to the basis of the Ising glass to be solved), and $\Gamma$ is large. Since the ground state of this Hamiltonian is simple, one expects the system to easily thermalize into this state.
• Thanks for your answer Norbert. But does your description fit the quantum annealing model of $H=H_{P} + \Gamma H'$ with $\Gamma$ large initially? For no matter how large $\Gamma$ is initially, $H \neq H'$ initially in this model. Your description seems to fit the adiabatic model where one evolves the Hamiltonian like $H = (1-t/T)H_{0} + (t/T)H_{P}$ better. Jan 5, 2017 at 1:39