I am attempting to solve the Schroedinger equation for a particle with potential energy: $$V(x) = -Fx$$ The associated Hamiltonian is: $$ H = \frac{p^2}{2m} - Fq $$ In the Schroedinger representation, $p$ and $q$ are represented by the operators: $$q = x \qquad p = -i\hslash \frac{\partial}{\partial x} $$ Plugging in the standard $H\psi = E\psi$ yields: $$ \psi '' = \left ( \frac{2m}{\hslash} (E-Fx)\right )\psi $$ Which I cannot solve.
However, if I instead use the momentum representation $$ p = x \qquad q = i\hslash \frac \partial {\partial x} $$ the Hamiltonian becomes $$ H = \frac{x^2}{2m} - F\frac \partial {\partial x} $$ Which means the Schroedinger equation becomes $$ F \psi ' = \left ( \frac{x^2}{2m} - E \right ) \psi $$ Which is a first-order equation.
Would solving this yield a valid wave function, or did I miss something on the way?
Also, how do I solve the fist case (the second order equation)?