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? There are some counterexample for functions that are square-integrable but doesn't tend to zero at infinity. However these counterexamples are not member of the domain of the momentum operator.


The short answer is: No, there is not such a function.

Indeed, it is false that $f\in L^2(\mathbb R, dx)$ vanishes at infinity as is well known (there are also some answers in PSE concerning this issue) but it is also true that, if $D(P)$ is the domain of the momentum operator $P$ over the real line,

$\qquad\qquad\qquad\qquad\qquad\qquad$$f \in D(P)\quad$ implies that $\quad f(x) \to 0\quad$ as $\quad x\to \pm\infty$.

Let us prove this fact.

First of all, notice that one of the equivalent ways to define $D(P)$ in order to have a properly selfadjoint momentum operator in $L^2(\mathbb R , dx)$ is $$D(P) := \left\{\: \left.f \in L^2(\mathbb R, dx)\:\right|\: \exists \: f' \mbox{in weak sense and } f' \in L^2(\mathbb R, dx)\right\}\:,$$ and then, where $f'$ is the weak derivative of $f$, the momentum operator is defined as the selfadjoint operator $$Pf = -i\hbar f'\:.$$ So, let us assume $f\in D(P)$. Since $[s,s']$ has finite Lebesgue measure $f\in L^2([s,s'], dx)$ implies $f\in L^1([s,s'], dx)$, so $F(s) := \int_{s'}^s f'(x)dx$ exists. It is obvious that it is also a continuous function in view of the properties of the integral. From known theorems of real analysis we also known that $$f(s')-f(s) = \int_s^{s'} f'(x)dx \quad \mbox{almost everywhere}\:.\tag{1}$$ In particular we can fix $f$ to be continuous everywhere since, modifying $f$ over a zero-measure set, $f(x)= f(s)+ F(x)$.
Now we can take advantage of Chaucy-Schwartz inequality in (1): $$|f(s')-f(s)| \leq \int_s^{s'} |f'(x)| |1|dx \leq \sqrt{\int_s^{s'} |f'(x)|^2 dx}\sqrt{\int_{s}^{s'}|1|^2 dx} \leq ||f'||_{L^2} \sqrt{|s-s'|}\:.$$ Notice that $||f'||_{L^2} <+\infty$ by hypothesis. The estimate $$|f(s)-f(s')| \leq ||f'||_{L^2} \sqrt{|s-s'|},$$ which is valid everywhere with our choice of $f$, implies that $f$ is uniformly continuous over the whole $\mathbb R$.

To conclude I prove that

PROPOSITION. If $f: \mathbb R \to \mathbb C$ is uniformly continuous and $f\in L^p(\mathbb R, dx)$, for some $p\in \mathbb R$ (in particular $p=2$) then $f(x) \to 0$ both for $x\to +\infty$ and $x\to -\infty$.

PROOF. Suppose that it is false that $f(x) \to 0$ for $x\to +\infty$ (the other case is analogous). Hence, there is $M>0$ and a sequence $x_n \to +\infty$ as $n\to +\infty$ such that $|f(x_n)| >M$. As a consequence, I can extract a subsequence satisfying $f(x_{n_k})>M$ for every $k$ or $f(x_{n_k})< -M$ for every $k$. I suppose valid the former since the latter can be treated analogously. Since $x_{n_k} \to +\infty$ as $k\to +\infty$, I can extract another subsequnce $x_{n_{k_h}} \to +\infty$ as $h\to +\infty$ such that $x_{n_{k_{h+1}}}- x_{n_{k_h}}>1$ and, as said $f(x_{n_{k_h}})>M$.

For the sake of simplicity I henceforth define $s_h := x_{n_{k_h}}$.

Now observe that, by uniform continuity, if $\epsilon = M/2$, there is $\delta>0$ such that $$|f(s)-f(s_h)|< M/2 \quad \mbox{if $|s-s_h|<\delta$ for every $h\in \mathbb N$.}$$
Hence $$-M/2 <f(s)- f(s_h)< M/2$$ so that, in particular $$M/2 < f(s_h) -M/2 < f(s)\quad \mbox{if $|s-s_h|<\delta$.} $$ In summary, taking $\delta < 1/2$ if necessary, we have an infinite class of pairwise disjoint intervals $I_h = [s_h-\delta,s_h+\delta]$ with identical length $2\delta>0$ where $f(s) > M/2 >0$. Therefore $$\int_{\mathbb R} |f(x)|^p dx \geq \sum_{h\in \mathbb N} \int_{I_h} |f(x)|^p dx \geq \sum_{h\in \mathbb N} 2\delta M^p/2^p= +\infty\:.$$ This is impossible since $f\in L^p(\mathbb R, dx)$ and thus the said sequences do not exist and $f(x) \to 0$ for $x\to \pm \infty$. QED

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    $\begingroup$ Does your conclusion stay the same for momentum operators in more than one dimensions? $\endgroup$ – higgsss Oct 31 '18 at 13:22
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    $\begingroup$ I never investigated what happens in more than one dimension. Several steps above are not valid (first of all the considered functions are not necessarily continuous). However it does not automatically imply that the statement is false. My feeling is that it is false however for $n>1$, but I did not try to prove it. $\endgroup$ – Valter Moretti Oct 31 '18 at 13:26
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    $\begingroup$ The assumption of n dimensions is relevant iff one thinks of compacts subsets of R^n, but otherwise L^2 (R^n) is isometrically isomorphic to the tensor product L^2 ((0,infty)) times L^2 (S^(n-1)). The momentum operator acts only in the first space, therefore the analysis of self-adjointess and maximal domain is reduced to 1D, thing which has been solved for many, many years. $\endgroup$ – DanielC Oct 31 '18 at 18:53
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    $\begingroup$ It is not completely true. Details matter. For instance, a function in the first Sobolev-Hilbert space $H^1$ is continuous in one dimension. So functions in the domain of $P$ in 1D are continuous. Instead they are not in nD in general. Continuity plays a crucial role in my proof above. $\endgroup$ – Valter Moretti Oct 31 '18 at 18:58
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    $\begingroup$ It is also wrong that the momentum operator acts only on the radial factor. You have $n$ components of the momentum operator. Each component is a different selfadjoint operator and it sees the angular directions. In 3D these components form a spherical tensor. $\endgroup$ – Valter Moretti Oct 31 '18 at 19:03

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