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Given the time dependent equation: $$\partial_t\,\hat{\psi}(p,t) = \dfrac{p^2}{2\,m}\,\hat{\psi}(p,t) + \hat{\psi}(p,t)\star{\hat{V}(p)}$$

and forcing through some kind of separation: $\hat{\psi}(p,t) = P(p)\,T(t)$

I end up getting: $$\partial_t\,T(t)\,P(p)= \dfrac{p^2}{2\,m}\,P(p)\,T(t) + (P(p)\star{\hat{V}(p)})\cdot T(t) \\[12pt] \dfrac{\partial_t\,T(t)}{T(t)} = \dfrac{p^2/(2\,m)\,P(p)+ P(p)\star \hat{V}(p)}{P(p)}$$

thus the awkward equation $$C\,P(p) = p^2/(2\,m)\,P(p)+ P(p)\star \hat{V}(p)$$

Awkward because there is nothing to be solved in the equation. Something's wrong?

Also what's the meaning of $C$. Some kind of energy?

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    $\begingroup$ If you include factors of $\hbar$ and solve $\partial_t T=CT$ you'll find the units of $C$ to agree with energy $\endgroup$ May 10 at 17:36

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Here, $\hat{V}(p)$ is an operator that should be acting on $\hat{\psi}(t,p)$ and therefore that should be acting on $P(p)$. If $\hat{V}(p)$ simply multiplies the wavefunction by some function of $p$ and so does $p^2/2m$ then $P(p)$ can be anything and is not constrained by these equations.

For something like the simple harmonic oscillator, for example, the potential energy is proportional to $x^2$, and we recall that $x=id/dp$ in momentum space, so $V$ would be an operator and not a scalar (this is indeed true in general).

Note the factor of $i$ that should be present.

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