1. Mar 8, 2016

### mupsi

Hi everyone,

For adibatic quantum computation one prepares the groundstate of a particular Hamiltonian and than adiabatically evolves the system to a problem Hamiltonian Hp which groundstate encodes the desired solution to a satisfiability problem. If the evolution takes place slowly the system will be in a groundstate Hp with a high probability. My question: why can't I start with Hp to begin with, cool down the system such that the wacefunction relaxes into the goundstate?

2. Mar 8, 2016

### Staff: Mentor

How do you cool the system if your Hamiltonian requires that you don't have decoherent interactions with the environment?

3. Mar 8, 2016

### mupsi

Why? I could couple the Hamiltonian to a bath for a short time, cool it down and seperate the system from the environment. I don't see any problem with that. Decoherence just causes the system to have a Boltzmann-Distribution

4. Mar 8, 2016

### Staff: Mentor

I think as soon as you do that coupling, your Hamiltonian does something, but not solving the problem. Which means you are doing what you described in post 1: cool to the ground state of one Hamiltonian, then go to a different one.

5. Mar 8, 2016

### mupsi

I agree, but when you seperate the system from its environment you get the original hamiltonian again and the wave function should be in the groundstate.

6. Mar 9, 2016

7. Mar 9, 2016

### mupsi

but that approach is fundamentally different. I think that the argument is time complexity. In general it takes more time for the system to relax into the groundstate (as a function of the number of qubits) than if you perform an adiabatic evolution. The computation time is related to the number of qubits via the inverse of the spectral gap between groundstate and the first excited state (which itself is a function of # qubits) and the spectral gap might (in some cases) only increases polynomially depeding on the computational problem whereas the relaxation time increases exponentially (this is speculative). A further assumption is that the groundstate of the initial hamiltonian is easy to prepare. I could be wrong though.