Why is the Laplace operator used in the Schrödinger equation? Why is the Laplacian necessary in the time-dependent Schrödinger equation in a position basis for a non-relativistic particle?
$$i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + \hat{V} \Psi?$$
where the second derivative of the wave function $\Psi$ is the Laplacian. 
What does the Laplacian operator tell us physically about the wave function?
What's the physical meaning behind the Laplacian?
I heard there's also a potential function defined in the Schrödinger equation. Are those just the initial conditions or constraints on the equation, and not related to the Laplacian? 
 A: The Schrödinger equation is 
$$i\hbar\partial_t \Psi(\mathbf{r},t) = \hat{H}\Psi(\mathbf{r},t),$$
reflecting the fact that the integral of motion associated with translation in time (expressed by operator $i\hbar\partial_t$) is the energy of the system, i.e. its Hamiltonian, $\hat{H}$. Then the canonical quantization procedure (closely related to the correspondence principle) demands that the energy operator has the same form as in the classical mechanics, except for replacing all the quantities with their respective operators. Thus, in the non-relativistic case
$$\hat{H}=\hat{K} + \hat{V},$$
where $\hat{K}, \hat{V}$ are respectively the kinetic and the potential energies of the system. In absence of magnetic field
$$\hat{K} =\frac{\hat{\mathbf{p}}^2}{2m},$$
where $\hat{\mathbf{p}}=-i\hbar\nabla$ is the momentum operator. Therefore
$$\hat{K} =\frac{\hat{\mathbf{p}}^2}{2m} = -\frac{\hbar^2}{2m}\nabla^2.$$
A slightly different way of thinking about it is in terms of energy momentum relationship for a free particle: $E=\frac{\mathbf{p}^2}{2m}$ - it leads to the same conclusion, but including the potential term requires a bit more explaining. This second way of thinking serves to make intuitive transition to the relativistic case, by adopting relativistic energy-momentum relationship:
$$E^2 = \mathbf{p}^2c^2 + m^2c^4.$$
Replacing here $E\rightarrow i\hbar\partial_t$ and $\mathbf{p} \rightarrow -i\hbar\nabla$ results in so-called Klein-Gordon equation, which also contains the Laplace operator... but proves problematic when describing massive particles with positive charge density. This is why one has to do away with the Laplacian and write Dirac equation, which contains only the first order derivatives.
