Here's an argument that might support the statement that such a non-smooth wavefunction is not physical:
You cannot add a finite number of smooth functions to get a non-smooth function. By fourier expansion theorem, any function can be expressed as a sum of plane waves (which are smooth with respect to spatial dimensions). Hence you must need an infinite number of smooth functions to get a non-smooth function. Now here's the problem. The different plane waves are momentum eigenfunctions and the non-smooth function is a superposition of these momentum eigenfunctions. If you now try to calculate the expectation value of the momentum, due to the fact that the momentum eigenvalues associated with the momentum eigenfunctions are unbounded from above, the expected value of the momentum could blow up (go to infinity). This is specifically what would make the wavefunction non-physical.
$$\psi(x) = \int \widetilde{\psi}(k) e^{ikx} \mathrm{dk}$$
$$E(k) = \int |\widetilde{\psi}(k)|^2 k \ \mathrm{dk}$$
But for a general smooth function, how do I know whether the fourier coefficients associated with larger and larger momentum eigenmodes taper off fast enough for the expected value of the momentum to converge?
Best Answer
There is no physical requirement for a wave function to be smooth, or even continuous. At least if we accept the common interpretation that a wavefunction is nothing else than a "probability amplitude". I.e. it represents, when multiplied with its complex conjugate, a probability density. Now a probability density can be discontinuous or even undefined on sets of Lebesgue measure zero. As a matter of fact, the continuous and smooth wavefunction $$c(x)=e^{-x^2}$$ and the discontinuous wavefunction \begin{equation}d(x)=\left\{\begin{aligned}&e^{-x^2}&x\neq 0\\&5&x=0 \end{aligned}\right.\end{equation} give the same probability distribution. In other words, for the purpose of quantum mechanics they are perfectly indistinguishable, and in fact they belong to the same equivalence class of $L^2(\mathbb{R}^d)$ functions. In again different words, the Hilbert space vector $\psi$ on $L^2(\mathbb{R}^d)$ defined by $c(x)$ and $d(x)$ is the same, and therefore it corresponds to the same quantum state.
For the reason above already, continuity is in my opinion meaningless when we deal with wavefunctions. Of course it is always possible (and convenient) to embed, let's say, functions of rapid decrease $\mathscr{S}(\mathbb{R}^d)$ in $L^2(\mathbb{R}^d)$, i.e. considering, by an abuse of notation, $f\in \mathscr{S}(\mathbb{R}^d)\subset L^2(\mathbb{R}^d)$; however we are implicitly considering the equivalence class of almost everywhere equal functions $[f]\in L^2(\mathbb{R}^d)$ to which $f$ belongs, in place of $f$ itself. So let's suppose now that we modifiy OP's assertion to be:
Now this is a better defined assertion from a mathematical point of view, and it means that we consider physically meaningful only wavefunctions such that there exist a continuous and differentiable representative in its equivalence class. However, also this requirement is not physical. In fact, there are potentials $V(x)$ such that the Hamiltonian $H=-\Delta/2m +V(x)$ is self-adjoint, and $$e^{-iHt}\Bigl[\mathscr{C}^1_{L^2}(\mathbb{R}^d)\Bigr]\nsubseteq \mathscr{C}^1_{L^2}(\mathbb{R}^d)\; .$$ In other words, there are quantum evolutions for which some "smooth" (in the sense just above) wavefunctions become non-smooth as time passes. One relevant concrete example would be the Coulomb potential $V(x)=\pm \frac{1}{\lvert x\rvert}$.