Wave function of a particle in a gravitational field 
Suppose we have a particle with mass $m$ and energy $E$ in a gravitational field $V(z)=-mgz$. How can I find the wave function $\psi(z)$? 

It should have an integral form on $dp$. Any help would be appreciated.
What I've tried
One way to solve the problem is use of change of variable
$$
x~:=~\left(\frac{\hbar^2}{2m^2g}\right)^{2/3}\frac{2m}{\hbar^2}(mgz-E)
$$
we can reduce Schroedinger equation to 
$$
\frac{d^2\phi}{dx^2}-x\phi(x)~=~0
$$
This is a standard equation, its solution is given by
$$\phi(x)~=~B~\text{Ai}(x)$$
where $\text{Ai}$ is the Airy function. But my solution should be (not exactly) like this:
$$
\psi(z)= N\int_{-\infty}^\infty dp \exp\left[\left(\frac{E}{mg}+z\right)p-\frac{p^3}{6m^2g} \right]
$$
 A: The basic idea for this is to use the momentum space version of the Schroedinger equation:
$$
\hat{p}\to p,\quad\hat{x}\to i\hbar\frac{\partial}{\partial p}
$$
and then solve the system1,
$$
\left[\frac{p^2}{2m}+img\hbar\frac{d}{dp}\right]\phi=E\phi
$$
which should be solvable (e.g., complex exponentials). You can then Fourier transform to physical space to get
$$
\psi(x)=\frac{1}{\sqrt{2\pi\hbar}}\int dp\,e^{ipx/\hbar}\phi(p)
$$
This should match to what your professor expects.


 More formally/generalized, you should get$$
\frac{p^2}{2m}\phi\left(p\right)+\int dp'\,U\left(p-p'\right)\phi\left(p'\right)=E\phi\left(p\right)$$
where
$$U(p)=\frac{1}{\sqrt{2\pi\hbar}}\int dx\,e^{-ipx/\hbar}U(x)$$
but this will reduce to the above anyway
A: $$
\left[\frac{p^2}{2m}+V(i\hbar\frac{d}{dp})\right]\phi(p)=E\phi(p)
$$
$$
\left[\frac{p^2}{2m}+(-mg)(i\hbar\frac{d}{dp})\right]\phi(p)=E\phi(p)
$$
$$
\frac{1}{i\hbar mg}(\frac{p^2}{2m}-E)\phi(p)=\frac{\phi(p)}{dp}
$$
When integrate we have:
$$
\frac{i}{\hbar mg}(Ep-\frac{p^3}{6m})=Ln\frac{\phi(p)}{\phi(p_{o})}
$$
$$
\phi(p)=\phi(p_{0})e^{\frac{E}{mg}p-\frac{p^3}{6m^2g}}
$$
$$
\psi(z)=\int dp e^{ipz/\hbar} \phi(p)
$$
$$
\psi(z)=\phi(p_{0})\int_{-\infty}^\infty dp e^{i/\hbar \left[ (\frac{E}{mg}+z)p-\frac{p^3}{6m^2g}\right]}
$$
