# How to keep the initial Hamiltonian in D Wave

So the D wave machine claims to start in an eigenstate of the Pauli X term in the transverse Ising model Hamiltonian, $$-\frac{A(s)}{2}\sum_{i} \sigma^{x}_{i}$$ and then $$A(s)$$ is turned off slowly while the desired Hamiltonian is turned on in the form $$\frac{B(s)}{2} \left( \sum_{i}h_i \sigma^{z}_{i} + \sum_{\langle i j \rangle} J_{ij} \sigma^{z}_{i} \sigma^{z}_{j} \right)$$ and $$B(s)$$ is increased, with $$s$$ running from 0 to 1. My question is how to use the interface to, say, simply sample the initial state (no annealing whatsoever), where I should get a random sampling of 0s and 1s for each qubit. Moreover, how can I stop the annealing at any point in between where $$A(s)$$ is still finite?

In order to sample only A(s), e.g. the transverse field Hamiltonian, you simply need to program every qubit and coupler to have a coefficient of 0. This then means that there are no non-zero terms in the $$B(s)$$ part of the summation. You can actually do this by submitting an empty dictionary of linear and quadratic terms to the D-Wave backend because the software assumes that all coefficients not specified are 0. Or you can specify the dictionaries to contain all zero coefficients for the hardware qubits and couplers you are interested in (or the whole chip).
You can not stop the anneal at an arbitrary point - you need to end the anneal at $$s=1$$, which corresponds to a readout of the states of the qubits. What you can do is quench the anneal, meaning you transition the anneal schedule up to $$s=1$$ as fast as you can (the restriction on how fast this can be done is hardware specific). This does not leave the state as it was when you started the quench though, there is an effect that the quench has on the annealing process. I worked on a couple of papers that attempted to design some Ising / QUBO models which would allow us to have a reduced impact from that quenching step, in order to ideally better observe what was occurring mid-anneal. Paper1, Paper2 if you are interested.
• Thanks, this is helpful. Is it clear that at the end of the annealing, when $s=1$, with $h$ and $J$ both zero from the start, the final state is an eigenstate of $X$? Since the coupling in front of the transverse field has been brought to zero at the end it seems the final Hamiltonian is zero. Jun 4 at 19:43