Deriving Schrödinger equation from Ehrenfest Theorem

Question

Im reading the following article https://arxiv.org/abs/1105.4014. There they derive the Schrödinger equation from Ehrenfest‘s Theorem. By starting with the following equations: \begin{equation} \left\langle\frac{\mathrm{d}}{\mathrm{d} t}\psi\right|\hat{p}|\psi\rangle+\langle\psi|\hat{p}\left|\frac{\mathrm{d}}{\mathrm{d} t}\psi\right\rangle=\left\langle-V'(\hat{x})\right\rangle \end{equation} \begin{equation} \left\langle\frac{\mathrm{d}}{\mathrm{d} t}\psi\right|\hat{x}|\psi\rangle+\langle\psi|\hat{x}\left|\frac{\mathrm{d}}{\mathrm{d} t}\psi\right\rangle=\frac{1}{m}\langle \hat{p}\rangle. \end{equation} Then they use the canonical commutator relation and remove the mean. My question is now why can the averaging dropped out? They justify this by saying that it must apply to all initial conditions, but for me it‘s not clear why than the averaging can be dropped out.

Answer 1

This step is a mathematical one; in general, if $$ \int f = \int g, $$ you can't conclude that $f=g$. But if $$ \int f \psi = \int g \psi $$ for a large class of functions $\psi,$ then you sometimes can conclude that $f=g$. For example, to argue that $f(x)=g(x)$ for some $x$, you can let $\psi$ be a sharply peaked function at $x$ (so nonzero near $x$, zero elsewhere, think of an approximation to the dirac delta).

Your situation is a little more difficult, here instead we want to conclude from $$ \int \psi^* \hat{A} \, \psi = \int \psi^* \hat{B} \, \psi $$ for some operators $\hat{A}$ and $\hat{B}$ and a large class of wavefunctions $\psi$, that $\hat{A} = \hat{B}$. Here the argument to convince yourself of this is more complicated, and depends on the operators $\hat{A}$ and $\hat{B}$, but I think this is roughly what is intended.