Decay of the First Derivative of the Quantum Wave Function I understand that the Hilbert space of all physical solutions of the Schrodinger equation have the property where
$$
\lim_{x\to\infty}\Psi=0
$$
For one of my assignments, I wanted to use
$$
\lim_{x\to\infty}\cfrac{\partial\Psi}{\partial x}=0.
$$
To make sure that this is indeed a property of $\Psi$, I asked if this was indeed a property of a physical solution to the Schrodinger equation. After all, I figured that the Hilbert space is invariant under differentiation. However, I have found that simply because a function is square integratable, it is not guaranteed to have a first derivative that is as well. For example, I know the function
$$
f(x)=\frac{\sin(x^{2})}{x}
$$
which decays to 0 and is normalizable, but
$$
\frac{df}{dx}=2\cos(x^2)-\frac{\sin(x^2)}{x^2}
$$
which no longer decays to $0$ as $x$ goes to $\infty$ and whose first derivative is not square integratable.
Now, my question is does there exist a physical solution of the Schrodinger equation whose first derivative does not decay to zero?
 A: *

*An element $\psi$ of the  Hilbert space $L^2(\mathbb{R})$ of square-integrable functions is in general neither differentiable nor does (as your example shows) differentiability of some $\psi \in L^2(\mathbb{R})$ guarantee that $\psi^\prime$ is still an element of the Hilbert space $L^2(\mathbb{R})$. This reflects the fact that a meaningful definition of an unbounded linear operator $A$ in a Hilbert space $\mathcal{H}$ always requires the specification of its domain $D(A) \subset \mathcal{H}$.


*In a physical context, differentiation of wave functions is related to the definition of the momentum operator $P$ as a selfadjoint linear operator ($P^\dagger =P$),
$$D(P)=\{\phi \in L^2(\mathbb{R}) | \, \phi^\prime \! \in L^2(\mathbb{R}) \, \, {\rm and}\, \,  \phi \, \, {\rm absolutely \, continuous}\},$$ $$ (P\phi)(x) = -i \phi^\prime(x)\,  \forall \, \phi \in D(P). $$ The domain $D(P)$ is a dense subspace of $L^2(\mathbb{R})$, meaning that any element $\psi \in  L^2(\mathbb{R})$ can be written as the limit $\lim\limits_{n \to \infty} \phi_n = \psi$ (with respect to the $L^2$-norm) of a sequence $\{\phi_n\}_{n=1}^\infty$ of elements $\phi_n \in D(P)$. The requirement "$\phi$ absolutely continuous" is needed for partial integration, ensuring $D(P^\dagger) = D(P)$ (for details, see e.g. W. Thirring, Quantum Mathematical Physics). The essential point with regard to your question is that $\phi \in D(P)$ implies $\phi(x) \to 0$ for $x\to \pm \infty$, whereas $\phi \in D(P^2)$ guarantees $\phi^\prime(x) \to 0$ in the same limits. (See also the answer given by Valter Moretti to a similar question: Is there a function which is square integrable and doesn't tend to zero at infinity but it belongs in the domain of the momentum operator?)


*Let me also remark that you should have stated more precisely what you mean by  "a physical solution of the Schrödinger equation". (a) If you simply ask for an arbitrary normalized wave function $\chi \in L^2(\mathbb{R})$ with $\langle \chi |\chi\rangle= \int_{\mathbb{R}} dx \ |\chi(x)|^2 =1 $ describing a pure state, there is no reason why the derivative of $\chi$ should even exist. Nevertheless, the vectors $\psi_t=U_t \chi$ with a one-parameter group of unitary operators $U_t$ describe a perfectly well defined time evolution with the initial condition $\psi_{t=0}=\chi$. (b) On the other hand, if you write down the time dependent Schrödinger equation of, say, a free particle, $i \partial_t \psi_t (x) = -\frac{1}{2m}\partial_x^2 \psi_t(x)$, you might require $\psi_t \in D(P^2)$ for the right-hand-side to be defined properly. (c) If it is the so-called time-independent Schrödinger equation $-\frac{1}{2m} \psi^{\prime \prime}(x)= E \psi(x)$ (better: eigenvalue equation of the Hamilton operator $H= -\frac{1}{2m} \partial_x^2$) you have in mind, its solutions $\sim e^{\pm \sqrt{2mE}x}$ are not even elements of of $L^2(\mathbb{R})$. (d) Finally, you might consider a Hamilton operator $H= P^2/2m +V(X)$ admitting a point spectrum ${E_1, E_2, \ldots}$ (the corresponding eigenfunctions being normalizable) and a continuous spectrum, where the associated eigenfunctions are again non-normalizable (i.e. $\notin L^2(\mathbb{R})$).
