I am given the following action functional $$S=\int dt \left[p_1\dot q_1+p_2\dot q_2-\frac{p_1p_2}{m}-\sum_{n=0}^\infty\frac{\partial^nV(q_1)}{\partial q_1^n}q_2^n\right]$$
where $p_i$ is the momentum conjugate to the coordinate $q_i$. I am then asked to quantise this action as a two dimensional system and show that $\psi:=\psi(q_1,p_2,t)=\left\langle q_1,p_2\middle|\psi(t)\right\rangle$ follows time evolution $$\frac{\partial\psi}{\partial t}=-\frac{p_2}{m}\frac{\partial\psi}{\partial q_1}+\sum_{n=0}^\infty(-\hbar^2)^n\frac{\partial^{2n+1}V(q_1)}{\partial q_1^{2n+1}}\frac{\partial^{2n+1}\psi}{\partial p_2^{2n+1}}$$
My attempt:
I first find the classical Hamiltonian $$H=p_1\dot q_1+p_2\dot q_2-L=\frac{p_1p_2}{m}+\sum_{n=0}^\infty\frac{\partial^nV(q_1)}{\partial q_1^n}q_2^n$$
I was thinking of then using $$\frac{\partial\psi}{\partial t}=-\left\{\left\{\psi,H\right\}\right\}$$ where $\{\{\ \}\}$ is the Moyal bracket, but it doesn't really get me anywhere.
UPDATE My bad. In this question I gave the wrong expression for the action, which in fact was $$S=\int dt \left[p_1\dot q_1+p_2\dot q_2-\frac{p_1p_2}{m}-\sum_{n=0}^\infty\frac{1}{(2n+1)!2^n}\frac{\partial^{2n+1}V(q_1)}{\partial q_1^{2n+1}}q_2^{2n+1}\right]$$
I appreciate Cosmas for taking the time to answer the question. However, it turns out that the answer I was looking for follows quite simply from the conventional quantisation of the action above.