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?

  • 1
    $\begingroup$ 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

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.