I am trying to derive an equation which is a standard result in physics (the momentum space Schrödinger equation).
(Background: The wavefunction is a complex valued function of position coordinates and time $\psi:\mathbb{R^3}\times\mathbb{R^+}\rightarrow\mathbb{C}$. It is square integrable ($L^2$) and is generally assumed to behave such that as ${r\rightarrow\pm\infty}$ then $\psi$ and $\nabla \psi$ fall off to zero sufficiently fast. Starting from the fact the position-space ($\psi$) and momentum space ($\phi$) wave functions are fourier transforms of each other):- \phi(\mathbf{p},t) = \int \psi(\mathbf{r'},t) e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}d^3r' (Integral runs over all space) Taking the partial time derivative \partial_t\phi(\mathbf{p},t) =\partial_t\left(\int \psi(\mathbf{r'},t) e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}d^3r' \right) = \int \partial_t\psi(\mathbf{r'},t) e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}d^3r' Because $t$ is a parameter while $p$ and $r$ are dynamical variables, i.e have no explicit $t$ dependence. Now using a postulated differential equation of $\psi$ $i\hbar \partial_t\psi(\mathbf{r},t)=-\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r},t) + V(\mathbf{r})\psi(\mathbf{r},t) $ I get i\hbar \partial_t\phi(\mathbf{p},t)= \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}\left[ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r'},t) + V(\mathbf{r'})\psi(\mathbf{r'},t)\right]d^3r'
First Question: I moved my exponential term to the left, so that the operator $\hat{H} =-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbb{r}) $ acts only on $\psi(\mathbb{r},t)$ and not on \psi(\mathbb{r},t)e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}. What is the justification for this exchange mathematically?
Evaluating the first integral by using integration by parts and identifying the resulting integral as the fourier transform I get -\frac{\hbar^2}{2m}\int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}\nabla^2\psi(\mathbf{r'},t)d^3r' = \frac{p^2}{2m}\phi(\mathbf{p},t)
Second Question: I am having trouble understanding the second integral perhaps for the same reason as my first question: after fourier transforming $\psi$ \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}V(\mathbf{r'})\psi(\mathbf{r'},t)d^3r'= \int \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}V(\mathbf{r'})e^{\frac{i}{\hbar}\mathbf{p'}\cdot\mathbf{r'}}\phi(\mathbf{p'},t)d^3r'd^3p' I thought I had done it but when I looked it up, there was a caveat that $V(\mathbb{r})$ must be analytic and the following must be true:$\color{blue}{e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}V(\mathbf{r'}) = V(i\hbar\nabla_p)e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}}$
Under this assumption I have the correct equation in momentum representation. $i\hbar \partial_t \phi(\mathbf{p},t) = \frac{p^2}{2m}\phi(\mathbf{p},t)+V(i\hbar\nabla_p)\phi(\mathbf{p},t)$
I am not getting where the blue thing came from. Can someone explain what conditions hold for $\hat{f}(x)\hat{g}(y)=\hat{g}(y)\hat{f}(x)$ particularly in this context.
Third Question: Any other less clumsy way to derive this?