Is it guaranteed that wavefunction is well behaved everywhere? I don't really know much about Quantum mechanics, but would like to know one simple fact.
The state function $\Psi(r, t)$ whose magnitude gives the probability density of the position of the particle and the magnitude of its ($\Psi(r, t)$) fourier transform gives probability density of its momentum. Is there any rule that these state functions are smooth (possess infinite order derivatives everywhere) (derivatives of all orders exist)?
 A: The only general requirement on the state function for a single, spinless, quantum particle (quanton) in a physically realistic state is that the state function be square integrable, i.e., the integral of its absolute value squared over all space be finite. Non-square integrable state functions are used for many purposes, but they are all idealizations that do not, individually, represent realistic states. If the state function is also to belong to the domain of definition of the Hamiltonian, then, in non-relativistic QM, the state function must be spatially differentiable to second order as well. State functions which are square integrable but not second order differentiable do not satisfy the Schroedinger equation. But their time evolution is still determined by continuity considerations since the second order differentiable state functions are everywhere dense in the state space, i.e., Hilbert space.  
A: Some of the conditions for wavefunctions $\Psi(x)$, for all elements $x$ of a subset of $\mathbb{R^{d}}$ (in the hyperphysics link, they use $x \in \mathbb{R}$).
1.- Must be a solution of the Schrodinger equation.
2.- Must be normalizable.
3.- Must be a continuous function of $x$.
4.- The slope of the function in $x$ must be continuous, that is, $\displaystyle \frac{\partial \Psi(x)}{\partial x}$ must be continuous.
The property of being square-integrable is included in the condition 2.
