# What is the domain of momentum operator on $\mathbb{R}$?

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 $$\mathbb{R}$$?

• What makes think it is not a Sobolev space also for $\mathbb R$. – DanielC Dec 13 '20 at 19:30
• @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 $W^{1,2}(\mathbb{R})$. – m93a Dec 13 '20 at 19:52

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: $${\rm D}(\hat p) = W^{1,2}(\mathbb{R})$$