Recall the duality functional $\mathfrak{H}(\vec x,\vec y)=\prod_{i=0}^{N}q^{y_i(x_i+i)}$ which, by agreement, is zero unless $y_0=0$. Here $\vec x\in \mathbb{X}^N$, $\vec y\in \mathbb{Y}^{N}$. By $\mathfrak{H}_k$ denote the restriction of $\mathfrak{H}$ to $\mathbb{X}^N\times \mathbb{Y}^N_k$.

Let $(q,\mu,\nu)$ be the parameters of the $q$-Hahn PushTASEP satisfying . For any $k\ge 1$ there exists $\mu_0>0$ (depending on $k$) such that for all $0<\mu<\mu_0$ we have \[thm:push\_duality\] $$\label{push_duality_formulation_equation}
        \mu^k \mathsf{P}^{\textnormal{PushTASEP}}_{q,\mu,\nu} \mathfrak{H}_k
        \bigl(\mathsf{P}^{\textnormal{Boson}}_{q,q/\mu,\nu}\bigr)^{T}
        =
        \nu^k
        \mathfrak{H}_k
        \bigl(\mathsf{P}^{\textnormal{Boson}}_{q,q/\nu,\nu}\bigr)^{T}.$$ Here “$T$” means transpose, that is, the Boson operators act in the $\vec y$ variables while the $q$-Hahn PushTASEP transition operator acts on the $\vec x$’s.

The condition that $\mu$ is sufficiently small guarantees the convergence of the infinite series coming from the action of $\mathsf{P}^{\textnormal{PushTASEP}}_{q,\mu,\nu}$. This convergence issue is the reason that only finitely many of the $q$-moments of the $q$-Hahn PushTASEP exist and are given by the contour integrals (, see also ).

Note also that unlike in the $q$-Hahn TASEP duality (), in the Boson operators do not necessarily have nonnegative matrix elements. This duality is [*not*]{} a Markov duality due to this lack of positivity as well as the factors $\mu^k$ and $\nu^k$. Despite this, we will see that it still provides meaningful information about how the 