Inspired by this question, How is the Schroedinger equation a wave equation?, I started playing around with the the Schrodinger equation.
Let's consider we're working in the energy basis, and so the time dependence of our wavefunctions will appear as $$\psi(\vec{r}, t) = e^{-iEt/\hbar}f(\vec{r}).$$
From the very simple differentiation of exponentials, we can write:
$$\partial_t \psi(\vec{r}, t) = -\frac{E}{\hbar}i\psi(\vec{r}, t).$$
If we now turn to the Schrodinger equation, we see a factor of $i \psi(\vec{r}, t)$ appears on the lefthand side;
$$\hbar\partial_t (i\psi(\vec{r}, t)) = -\frac{\hbar^2}{2m}\nabla^2\psi (\vec{r}, t).$$
Making our substitution, this yields:
$$ -\hbar\partial_t\bigg(\frac{\hbar}{E}\partial_t \psi\bigg) = -\frac{\hbar^2}{2m}\nabla^2\psi (\vec{r}, t).$$
Rearranging, this gives:
$$\partial_t^2 \psi = \frac{E}{2m}\nabla^2\psi,$$
which is just the wave equation, but with the relationship $v^2 = E/2m$, which would give us a velocity,
$$v = \sqrt{\frac{E}{2m}}.$$
My question is, is there any physical interpretation of this result?
Edit: Have fixed a rather significant sign error
Best Answer
You made a sign mistake right at the beginning; you should have $\psi({\bf r}, t) = e^{-i E t/\hbar} \psi({\bf r})$, where $\psi({\bf r})$ is an energy eigenstate wavefunction.
If I may simplify your argument: if $\psi$ is an eigenstate of the Hamiltonian, then
$$\begin{align*}\hat{H} \psi &= i \hbar \partial_t \psi \\ \hat{H}^2 \psi &= -\hbar^2 \partial_t^2 \psi \\ E \hat{H} \psi = -\frac{E \hbar^2}{2m} \nabla^2 \psi &= -\hbar^2 \partial_t^2 \psi \\ \nabla^2 \psi - \frac{2m}{E} \partial_t^2 \psi &= 0, \end{align*}$$ which is the wave equation with $v = \sqrt{E/(2m)}$. This almost makes sense: a nonrelativistic free particle has energy $E = \frac{1}{2} m v^2$, so we should have $v = \sqrt{2E/m} = 2 \sqrt{E/(2m)}$. Why are we only getting half the velocity that we should?
The answer is that we assumed that $\hat{H}$ took the nonrelativistic form $\hat{p}^2/(2m)$. What if we instead used the expression $\hat{H} = \hat{p} c$ for the energy of a photon? Then the third line above would become $$\hat{p}^2 c^2 \psi = -\hbar^2 \partial_t^2 \psi.$$ If we assume the usual representation $\hat{p} \to -i\hbar {\bf \nabla}$ in the position basis, then the $\hbar^2$'s cancel and we get $$\nabla^2 \psi - \frac{1}{c^2} \partial_t^2 \psi = 0,$$ which the the relativistic wave equation! If we were to allow for relativistic mass, then we'd use $\hat{H}^2 = \hat{p}^2 c^2 + m^2 c^4$, and we'd end up with the Klein-Gordon equation.
So to answer your question: the physical interpretation of your result is just the velocity of a classical nonrelativistic free particle of energy $E$, except off by a factor of two. But if you were to instead use the relativistic form of the energy, you'd get the Klein-Gordon equation, which is the exact quantum description of a massless scalar relativistic field rather than nonrelativistic particle.