Question
What is the physical reason (i.e. without any maths) that the derivative of a wavefunction (except with infinite potentials) has to be continuous?
Other info
I know that in the classical wave case it is because otherwise an infinitesimal mass would feel an infinite acceleration, but since since the wavefunction is not 'physical' I don't think a direct analogy can be made.
For the time-dependent Schrödinger equation there might be one. The spatial derivative of the wave function is connected to a "flow of probability" associated with the squared absolute value of the wave function which gives a probability density.
You can view this probability density as a fluid with mass conservation (probability conservation). A discontinuous derivative of the wave function would cause a discontinuous current. And in case of an ordinary fluid this would be rather strange physically, since you are forced to uphold the conservation laws.
There is no need, nor mathematically nor physically, for the wavefunction or its (spatial) derivative to be continuous.
In fact, the space of wavefunctions is usually considered to be an Hilbert space (and there are very poignant physical and mathematical motivations, the first is that any algebra of physical observable that satisfies reasonable assumptions is represented isomorphically as operators on some Hilbert space).
The aforementioned Hilbert space is usually of infinite dimension, and to (irreductibly) represent the finite dimensional canonical commutation relations (a very natural physical requirement) it must be isomorphic to $L^2(\mathbb{R}^d)$. And the functions of $L^2(\mathbb{R}^d)$ can easily be non-continuous and non-differentiable.
With respect to time instead, there is a continuity requirement of the map of wavefunctions $\psi(t):\mathbb{R}\times \mathscr{H}\to\mathscr{H}$ that associates to an initial wavefunction its time evolution. That requirement is, roughly speaking, related to a suitable existence and uniqueness theorem for the solution to the Schrödinger equation, i.e. of the quantum dynamics that is described by the equation $$i\partial_t \psi(t)=H\psi(t)\; .$$ As a matter of fact, since $H$ is required to be a (real-numerical-value) observable, then it should be a self-adjoint operator. That is very nice, since the self-adjoint operators are in 1-1 correspondence with strongly continuous unitary groups of operators; and the group $e^{-itH}$ corresponding to $H$ solves in a unique fashion the Schrödinger equation: i.e. $$\psi(t)=e^{-itH}\psi_0\; ,\; \psi_0\in D(H)$$ is the unique solution in $\mathscr{H}$ of the Cauchy problem $$\left\{\begin{aligned}&i\partial_t \psi(t)=H\psi(t)\\&\psi(0)=\psi_0\end{aligned}\right .\; ;$$ and can be extended to a non-differentiable solution when $\psi_0$ is a generic vector of $\mathscr{H}$. This unique solution is also always continuous in time, i.e. $\psi(\cdot)\in C^0(\mathbb{R},\mathscr{H})$; however as I said it is only differentiable on $D(H)$, that is usually a densely defined Hilbert subspace of $\mathscr{H}$.
Comment (about the referred book). Mathematically speaking, there is a weaker notion of derivative that make the point raised by the authors incorrect. It is possible to define derivatives in a distributional sense, and this does not require continuous functions (it is possible to define a derivative for distributions). Also, exploiting the properties of the Fourier transform there is a quite natural notion of functions of $L^2$ that admit square-integrable derivatives, and they form Hilbert spaces that are called the Sobolev spaces. Again no smoothness at all is required, and this framework is ideal for quantum mechanics. In addition, the existence of solutions to the elliptic equation they cite (the time-independent Schrödinger equation) can be studied (and it is quite customary) on these function spaces with "weak" derivatives.
I'd say that a good physical reason for the first derivative being continuous is for the probability current to be continuous too, since it's constructed from $\psi(x,t)$ and $ \nabla \psi(x,t)$.
You need second derivative to exist because that is the kinetic energy. And for that, the first derivative must be continuous!
This is the physical reason.
