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Observables in QM are postulated to be self-adjoint operators. Those have to obey $\hat A \vphantom{A}^+ \! = \hat A$, including the equality of their domains. If we work on a finite interval $(a, b)$, an example of such an observable is the momentum operator: $$ \hat p: {\rm D}(\hat p) \to L^2 \big( (a,b) \big) \\[5pt] {\rm D}(\hat p) = \big\{\; f \in W^{1,2} \big( (a,b) \big) \; \big| \; f(a+) = f(b-) \;\big\} \\[5pt] \hat p f = -{\rm i} f' $$

We can easily inspect that $\hat p$ with this domain is indeed self-adjoint using integration by parts: $$ \big( \hat p f, \; g \big)_{L^2} = {\rm i} \big( f', \; g \big)_{L^2} = \big[ fg \big]_a^b - {\rm i} \big( f, \; g' \big)_{L^2} = \big[ fg \big]_a^b + \big( f, \; -{\rm i}g' \big)_{L^2} $$ Here, $g$ has to be from $W^{1,2}$ in order to have a derivative and the necessary and sufficient condition for $[fg]_a^b$ to be zero is $g(a+) = g(b-)$, hence ${\rm D}(\hat p^+) = {\rm D}(\hat p)$ and $\hat p$ is self-adjoint.

However, this doesn't work on infinite intervals. In $L^2(\mathbb{R})$, functions either don't have a limit at infity, or it's zero. If we require that $f(-\infty) \to 0, \;\; f(+\infty) \to 0$, it is sufficient for $g$ to be only bounded at infinity and we get ${\rm D}(\hat p^+) \subsetneq {\rm D}(\hat p)$. On the other hand, if we require that $f$ is bounded at infinity, we get that $g$ has to vanish at infinity, therefore ${\rm D}(\hat p^+) \!\supsetneq {\rm D}(\hat p)$.

How do I achieve ${\rm D}(\hat p^+) = {\rm D}(\hat p)$ on $L^2(\mathbb{R})$? What is the domain of the momentum operator on $\mathbb{R}$?

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  • $\begingroup$ What makes think it is not a Sobolev space also for $\mathbb R$. $\endgroup$ – DanielC Dec 13 '20 at 19:30
  • $\begingroup$ @DanielC I thought it couldn't be $W^{1,2}$ because I thought that functions from this space don't have to have a limit at $\pm\infty$. Since I discovered that this is not the case, the domain really is just $W^{1,2}(\mathbb{R})$. $\endgroup$ – m93a Dec 13 '20 at 19:52
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As it turns out, this actually does work on the infinite interval. The important observation the question is missing is that all functions $f \in W^{1,2}(\mathbb{R})$ are guaranteed to vanish at infinity – see this proof by Valter Moretti. This means that all the “different” domains that I considered were actually the same set: $$ \big\{\, f \in W^{1,2}(\mathbb{R}) \;\big|\; f(-\infty) = f(+\infty) = 0 \,\big\} = \big\{\, f \in W^{1,2}(\mathbb{R}) \;\big|\; f \text{ is bounded at } \infty \,\big\} = W^{1,2}(\mathbb{R}) $$

This means that the problem with ${\rm D}(\hat p)$ and ${\rm D}(\hat p^+)$ not being equal for different conditions is solved and the one true domain for the self-adjoint momentum operator is: $$ {\rm D}(\hat p) = W^{1,2}(\mathbb{R}) $$

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