Consider Schrödinger's *time-independent* equation
$$
-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=E\psi.
$$
In typical examples, the potential $V(x)$ has discontinuities, called *potential jumps*.


Outside these discontinuities of the potential, the wave function is required to be twice differentiable in order to solve Schrödinger's equation.


In order to control what happens at the discontinuities of $V$ the following assumption seems to be standard (see, for instance, Keith Hannabus' *An Introduction to Quantum Theory*):



> 
> **Assumption**: The wave function and its derivative are continuous at a potential jump.
> 
> 
> 


**Questions**:


1) Why is it necessary for a (physically meaningful) solution to fulfill this condition?


2) Why is it, on the other hand, okay to abandon twofold differentiability?


Edit: One thing that just became clear to me is that the above assumption garanties for a well-defined probability/particle current.



Justification for the matching condition for the wave function at potential jumps. Why is it both restrictive enough and sufficiently general?
To answer your first question:


Actually the assumption is *not* that the wave function and its derivative are continuous. That follows from the Schrödinger equation once you make the assumption that the probability amplitude $\langle \psi|\psi\rangle$ remains finite. That is the physical assumption. This is discussed in Chapter 1 of the first volume of *Quantum mechanics* by Cohen-Tannoudji, Diu and Laloe, for example. (Google books only has the second volume in English, it seems.)


More generally, you may have potentials which are distributional, in which case the wave function may still be continuous, but not even once-differentiable.


To answer your second question:


Once you deduce that the wave function is continuous, the equation itself tells you that the wave function cannot be twice differentiable, since the second derivative is given in terms of the potential, and this is not continuous.


