first move particles from location 1 to location 0, then from location 2 to location 1 (note that the distribution of the particles taken from location 2 is no affected by what happened at location 1), and so on. No particles are moved from the location $0$, and no particles are added to the location $N$. This implies that $\mathsf{P}^{\textnormal{Boson}}_{q,\alpha,{\boldsymbol\upnu}}$ preserves each of the spaces $\mathbb{Y}^{N}_k$.

We put no restrictions on the parameter $\alpha$ in $\mathsf{P}^{\textnormal{Boson}}_{q,\alpha,{\boldsymbol\upnu}}$. In particular, we do not require it to be a Markov transition operator: it is allowed to have negative matrix elements. However, the rows in $\mathsf{P}^{\textnormal{Boson}}_{q,\alpha,{\boldsymbol\upnu}}$ still sum to one, and for ${\boldsymbol\upnu}\le \alpha<1$ the operator $\mathsf{P}^{\textnormal{Boson}}_{q,\alpha,{\boldsymbol\upnu}}$ defines a discrete time Markov process on $\mathbb{Y}^{N}$ called the $q$-Hahn Boson system.

The $q$-Hahn TASEP and $q$-Hahn Boson operators with the same parameters $\alpha={\boldsymbol\upmu}$ are dual to each other via the duality functional $\mathfrak{H}\colon \mathbb{X}^{N}\times \mathbb{Y}^{N}\to\mathbb{R}$ defined as $$\label{H_duality_functional}
    \mathfrak{H}(\vec x,\vec y):=
    \begin{cases}
        \displaystyle\prod_{i=1}^{N}q^{y_i(x_i+i)},&y_0=0;
        \\[10pt]
        0,&y_0>0.
    \end{cases}$$

\[thm:TASEP\_duality\] We have $$\mathsf{P}^{\textnormal{TASEP}}_{q,{\boldsymbol\upmu},{\boldsymbol\upnu}}\mathfrak{H}
        =
        \mathfrak{H}\bigl(\mathsf{P}^{\textnormal{Boson}}_{q,{\boldsymbol\upmu},{\boldsymbol\upnu}}\bigr)^T,$$ where “$T$” stands for transpose of an operator. In other words, $\mathsf{P}^{\textnormal{TASEP}}_{q,{\boldsymbol\upmu},{\boldsymbol\upnu}}$ acts in the variables $\vec x$, and $\mathsf{P}^{\textnormal{Boson}}_{q,{\boldsymbol\upmu},{\boldsymbol\upnu}}$ acts in the variables $\vec y$, and their actions on $\mathfrak{H}(\vec x,\vec y)$ 