Time-independent form of the Schrodinger equation states $$\hat H\psi=E\psi$$ For a Hamiltonian in form of $$\hat H=\frac{\hat p^2}{2m}$$ Which indicates a free particle, In the position space is routine and starts with plugging in the momentum operator in position space as $$\hat p=-i\hbar\frac{\partial}{\partial x}$$ And we can obtain eigenvalues and eigenfunctions as $$E=\frac{\hbar^2k^2}{2m}$$ $$\psi^+(x)=e^{ikx}\space\space,\space\space\psi^-(x)=e^{-ikx}$$ $$\psi(x)=A\psi^++B\psi^-$$ I also know we can derive the wavefunction in the momentum space with a Fourier transform. But I want to solve the SE in the momentum space. So $$\hat H\tilde\psi(p)=E\tilde\psi(p)$$ $$\frac{p^2}{2m}\tilde\psi(p)=E\tilde\psi(p)$$ $$\tilde\psi(p)(\frac{p^2}{2m}-E)=0$$ One answer is the same as the previous method $$E=\frac{\hbar^2k^2}{2m}$$ But here $\tilde\psi(p)$ can be any function of $p$. But we know it should be the same as the result of the Fourier transform on $\psi$.

How can we obtain $\tilde\psi(p)$ with this method?

  • $\begingroup$ @AlfredCentauri No, only the eigenfunctions. But how can we obtain eigenfunctions of momentum for a free particle without using FT on psi? $\endgroup$ – Alireza Dec 12 '17 at 18:21
  • $\begingroup$ @AlfredCentauri I will state my problem in another way. We can solve the SE in position space, obtain $\psi$ and use FT to get $\tilde\psi$. This is done in every textbook. But I want to solve the SE in momentum space and obtain $\tilde\psi$ then use a FT to get $\psi$. $\endgroup$ – Alireza Dec 12 '17 at 18:44

But here $\tilde{\psi(p)}$ can be any function of $p$.

In $p$ space, $p$ is the variable, not a constant and so, in general

$$p\,f(p) \ne P\,f(p)$$

where $P$ is a constant. Only for the case that $f(p) \propto \delta(p - P)$ can we write, e.g.,

$$p\, \delta(p - P) = P\, \delta(p - P)$$

Can you take it from here?

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  • $\begingroup$ Thanks, But can I get a little more insight and explanation on this? $\endgroup$ – Alireza Dec 12 '17 at 19:03
  • $\begingroup$ @Alireza, what is the momentum operator in the $p$ basis? $\endgroup$ – Alfred Centauri Dec 12 '17 at 19:08
  • $\begingroup$ It is simply $p$. As in position space the position operator is $x$. $\endgroup$ – Alireza Dec 12 '17 at 19:11
  • $\begingroup$ @Alireza, now let $|P\rangle$ be an eigenket of momentum such that $\hat{p}|P\rangle = P\cdot |P\rangle$. Let $\psi_P(p) = \langle p|P\rangle$ be the $p$ space representation of $|P\rangle$. Then it must be that $p\cdot\psi_P(p) = P\cdot\psi_P(p)$. What is the only 'function' of $p$ that satisfies this? $\endgroup$ – Alfred Centauri Dec 12 '17 at 19:19

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