Question about Position and Momentum operators

Question

This is the question I'm struggling with:

Given

$\left<p|q\right> = \frac{e^{-ipq/h}}{\sqrt{2\pi h}}$

and that

$\hat{H} = \frac{\hat{P^2}}{2m} + V(\hat{Q})$

show that:

$\left<p|\hat{H}(\hat{P}, \hat{Q})|q\right> = \frac{e^{-ipq/h}}{\sqrt{2\pi h}} \cdot H(p,q)$

My working is as follows:

$\left<p|\hat{H}(\hat{P}, \hat{Q})|p\right> = H(p,q)$

$\left<p|\hat{H}(\hat{P}, \hat{Q})|q\right> = \left<p|\hat{H}(\hat{P}, \hat{Q})|p\right> \left<p|q\right>$

$= \frac{e^{-ipq/h}}{\sqrt{2\pi h}} \cdot H(p,q)$

but this is wrong. Any help would be very much appreciated.

Answer 1

\begin{align*} \langle p|H|q\rangle &= \left\langle p\left|\frac{P^2}{2m}+V(Q)\right|q\right\rangle \\ &= \left\langle p\left|\frac{P^2}{2m}\right|q\right\rangle + \langle p|V(Q)|q\rangle \\ &= \frac{1}{2m}\langle p|P^2|q\rangle +\langle p|V(Q)|q\rangle \\ &= \frac{p^2}{2m}\langle p|q\rangle +V(q)\langle p|q\rangle \\ &= H(q,p)\cdot \langle p|q\rangle \end{align*} where we have used : $$P|p\rangle=p|p\rangle$$ $$Q|q\rangle =q|q\rangle $$