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It follows from momentum space definition of position operator as $x\; =\; i\backslash hbar\backslash frac\{\backslash partial\; \}\{\backslash partial\; p\}$, that by using Schrodinger's equation and taking $m\_0=\backslash frac\; 1\; \{m\backslash omega^2\}$, we get:

$\backslash begin\{align\}$

\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 $x\; \backslash rightarrow\; p$, $m\backslash rightarrow\backslash frac\; 1\; \{m\backslash omega^2\}$, $\backslash omega\; \backslash rightarrow\; \backslash omega$ and $|n\backslash rangle\; \backslash rightarrow\; (-i)^n|n\backslash rangle$ in any equation expressed in terms of those variables gives another valid equation.

Finding $x(t)$ from given $v(x)$ curve in configuration space, $x\backslash left(t=t\_0+\backslash int\_\{x=x\_0\}^x\; \backslash frac\; \{dx\}\{v(x)\}\backslash right)=x$.