Actually, this is used this all the time; it is just another way of describing the $q=1$ case! A symmetric way to state the situation for curves is that, for any locally free sheaf of $\mathcal O_C$-modules $\mathcal F$ on $C$, there is a natural perfect pairing $H^0(C,\mathcal F) \times H^1(C, \Omega_C\otimes \mathcal F^{\vee}) \to k$ (where $k$ is the ground field). Replacing $\mathcal F$ by $\Omega_C \otimes\mathcal F^{\vee}$ shows the symmetric role played by the $H^0$ and $H^1$, and shows that the $q=0$ and $q=1$ statements from Wikipedia are equivalent (in the case of a curve).
In fact even more canonically, the pairing is $H^0(C,\mathcal F) \times H^1(C,\Omega_C \otimes \mathcal F^{\vee}) \to H^1(C,\Omega^1),$ the pairing being induced by cup product from $H^0 \times H^1$ to $H^1$, together with (the tensor product with $\Omega^1$ of) the natural evaluation map $\mathcal F\otimes \mathcal F^{\vee} \to \mathcal O_C$. This is then combined with the fact that there is a canonical isomorphism (the trace map, or --- in more topological terms --- the fundamental class) $H^1(\Omega^1_C) \cong k$. (So the fact that $H^1(\Omega_C^1)$ --- which was noted in another answer --- is not just an incidental consequence of Serre duality; it is one of the linchpins.)