How to show mathematically the equivalency between Ising Model and QUBO?

Question

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?

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.