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?
Best Answer
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?