# 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?

• "After all, I figured that the Hilbert space is invariant under differentiation" -what do you mean? Jan 27 at 13:55
• @TobiasFünke by that, I meant that I figured that the image of P on the space would be the space, but clearly that’s not true, via the answer below Jan 27 at 16:27

1. 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 $$D(A) \subset \mathcal{H}$$.
2. 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?)
3. 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. $$\notin L^2(\mathbb{R})$$).
• Could you comment on the first and second equation in the question? IIRC, absolutely continuity implies (at least in the 1D case) that $\psi \in \mathcal D(P)$ actually obeys the first equation. And hence, I guess, that if the derivative is a.c. again, the second equation should hold, too. (?). That being said, we have that if $\psi \in \mathcal D(H)$ then $\psi \in \mathcal D(P^2)$ and hence $\psi \in \mathcal D(P)$, no? So this leaves me with the impression that if $\psi$ is in the domain of $H$, it actually should satisfy both the first and second equation...? Jan 27 at 8:41
• @Tobias Fünke $D(P^2)= \{\psi \in L^2(\mathbb{R}) | \psi, \psi^\prime \, {\rm a.c.}, \, \psi^\prime \in D(P) \} \subset D(P)$. "$\psi$ absolutely continuous" can be characterized by the property that $\psi(x) = \psi(0)+\int\limits_0^x dx^\prime g(x^\prime)$ ($g$ integrabel), i.e. $\psi^\prime = g$ a.e. (almost everywhere). The condition "absolutely" cannot be relaxed, as there are strictly monotic functions with $\psi^\prime =0$ a.e. Jan 27 at 9:25
• @Joshua G-F As $D(P^2)$ is a proper subset of $D(P)$, you can find vectors with $\psi \in D(P)$ but $\psi \ne D(P^2)$. $\psi \in D(P)$ implies $\psi(x) \to 0$ for $x \to \pm \infty$, but not necessarily $\psi^\prime (x) \to 0$ in these limits. For the PDE $-i \partial_t \psi_t(x)= H \psi_t(x)$ with $H=P^2/2m+V(x)$ to make sense, you have to require $\psi_t \in D(H)= D(P^2) \cap D(V)$, so $\partial_x\psi_t(x) \to 0$ for $x \to \pm \infty$. Jan 27 at 17:37
• @Joshua G-F This should not be confused with my last remark on the solutions of the eigenvalue equation $H \phi = E \phi$. The bound state solutions (related to the point spectrum of $H$) are necessarily elements of $D(H)$. On the other hand, the non-normalizable scattering wave-functions $\phi_E$ (related to the continuous spectrum $\sigma_c(H)$) are not even elements of the Hilbert space. However elements of the Hilbert space can be constructed as superpositions $\int_{\sigma_c(H)} dE \, c(E) \, \phi_E$ with suitable weight functions $c(E)$. Jan 27 at 17:50