# Stationary Schrödinger Equation in Momentum space

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?

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

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.