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Quantum Harmonic Oscillator
It follows from momentum space definition of position operator as $$x = i\hbar\frac{\partial }{\partial p}$$, that by using Schrodinger's equation and taking $$m_0=\frac 1 {m\omega^2} $$, we get:

$$\begin{align} \frac 1 {2m} p^2 \psi(p) -\frac 1 2 m \omega^2 \hbar^2\frac{\partial^2 }{\partial p^2}\psi(p) = E \psi(p)\\ \frac 1 2 \frac{1}{m_0} (-i\hbar \nabla_p)^2\psi(p)+\frac 1 2 m_0 \omega^2 p^2 \psi(p)=E\psi(p) \end{align}$$

Since momentum basis Schrodinger equation is same as that of position basis Schrodinger equation, the real-valued momentum wavefunction solution is same as momentum wavefunction corresponding to spatial wavefunction up-to an arbitrary phase which can be found to be $$(-i)^n$$ for $$E_n$$ energy eigenfunction. Hence the change $$x \rightarrow p$$,  $$m\rightarrow\frac 1 {m\omega^2}$$, $$\omega \rightarrow \omega$$ and $$|n\rangle \rightarrow (-i)^n|n\rangle$$ in any equation expressed in terms of those variables gives another valid equation.

Configuration space
Finding $$x(t)$$ from given $$v(x)$$ curve in configuration space, $$x\left(t=t_0+\int_{x=x_0}^x \frac {dx}{v(x)}\right)=x$$.