# Hamiltonian of the unused qubits

When a problem is embedded only a subset of the physical qubits is required. A number of qubits are left unused. What exactly is the initial and final Hamiltonian for these unused qubits? How is their initial state prepared?

1 comment
• Hi Michele,

Given the following representation of the Hamiltonian, where σ are Pauli matrices operating on a qubit q, and h and J are the qubit biases and coupling strengths:

Q1: What is the initial Hamiltonian State for those “unused” qubits?

The initial (s = 0) Hamiltonian for the unused qubits with indices i is given above with A(s = 0) = 1 and B(s = 0) = 0. The initial state of the unused qubits is the ground state of that initial Hamiltonian, which is a uniform superposition state of 0 and 1.

Q2: What is the final Hamiltonian State for those “unused” qubits?

The final Hamiltonian state for the unused qubits and couplers, both h and J are set to 0 and follows the usual envelopes of the energy scaling functions, A(s) and B(s).

Q3: How is the initial Hamiltonian state prepared?

The initial state is prepared by setting up the Initial Hamiltonian shown above for all qubits, wherein there is a gap much larger than temperature to the first excited state and waiting for the system to cool to the (trivial) ground state. This initial ground state is a uniform superposition of 0 and 1.

Since Ji,j = 0, the unused qubits play no role in the dynamics of the used qubits during annealing as the Hamiltonian above can be separated into two distinct parts, one for the used qubits and one for the unused qubits. Even in the case that Ji,j is not perfectly zero, the effect on the evolution of the used qubits is negligible. If one wishes to follow this path of wondering about the unused qubits, then analytic or numerical simulations of the above Hamiltonian would be a good first step.

I hope this helps. Please let us know if you have any further questions.

Best Regards,

Tanjid