It is said that the Ising Model using spin variables $s ∈ \{−1, 1\}$ $$H(s)=\sum_{i}h_is_i+\sum_{i<j}J_{ij}s_is_j,$$ and a Quadratic Unconstrained Binary Optimization (QUBO) problem with binary variables $x ∈ \{0, 1\}$ $$Obj(x,Q)=x^{T} ·Q·x,$$ are equivalent under a simple change of basis. How one could mathematically show this?


1 Answer 1


The relation between "Ising" and binary variables is following $$ x_i = \frac{1 + s_i}{2}, $$ where $s_i$ is a spin and $x_i$ is a binary variable. Clearly setting $s_i = -1$ leads to $x_i = 0$ and if $s_i = 1$ we get $x_i$ = 1. So, this simple linear transform changes spins to binary variables and conversely.

Quadratic terms in QUBO objective functions are equivalent to two-body interactions in Ising Hamiltonian ($Z_i \otimes Z_j$ terms in the Hamiltonian) and linear terms are equivalent to independent $Z_i$ terms in the Hamiltonian.

You can find more about the equivalence between spins glasses (Ising Hamiltonians) and QUBO task here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.