In my studies I came across Polymer Quantum Mechanics. Last line on p. 4 states:

These operators are weakly continuous in the parameters $\lambda$, $\mu$ and this ensures that the self-adjoint operators $\hat{x}$, $\hat{p}$ exist in $\mathcal{H}_{Sch}$.

By "these operators", it means the translation operators in the position and momentum space.

Could someone please provide a proof of this? I posted this here instead of math stackexchange because I would like to get a physical insight for this as well.

  • $\begingroup$ This doesn't make that much sense because $X$ and $P$ do not act on all of $L^2(\Bbb R)$ anyway. It is clear that $D(X)$ and $D(P)$ are nonempty since $0\in $ both of them. So I don't know what it means for them to "exist in $\mathcal H_{Sch}$" and why this needs to be proved. $\endgroup$ – Ryan Unger Apr 11 '17 at 0:23

Asking for a proof of this is a bit too much, especially on this site. The slides allude to the Stone theorem on one-parameter unitary groups $U_t$ of operators in Hilbert spaces. Stone proved that if the group is strongly continuous in $t$ then it is generated by a self-adjoint operator, i.e. $U_t=e^{itA}$ for some self-adjoint $A$ (densely but not everywhere defined in general). Von Neumann later proved that it is enough for $U_t$ to be weakly measurable in $t$, in particular weakly continuous suffices. There is a more general result that for one-parameter semigroups in Banach spaces weak continuity always implies strong continuity, the Yosida theorem. Proofs can be found in most functional analysis texts, Reid and Simon, Fourier analysis, Self-adjointness is physicist friendly, but still not a cakewalk. Perhaps, Stone's original paper will be a shorter study, only 7 pages.

The idea is to construct the infinitesimal generator as the limit (in a certain weak sense) $\lim_{t\to0}\frac1t(U_t-I)$, which intuitively should give $iA$, and then use the properties of unitary groups and continuity to show that $A$ must be self-adjoint. Taking $U_t$ as the group of position shifts $U_t\psi(x):=\psi(x+t)$ gives us $\hat{x}$ as $A$, and similarly the group of momentum shifts $V_t$ gives us $\hat{p}$.

Later the linked slides also use the Stone–von Neumann theorem, which is stronger. It states that if groups of shifts satisfy the kind of relations that position and momentum shifts $U_t$ and $V_t$ satisfy, then their generators satisfy the canonical commutation relations. And if they are also jointly irreducible (no invariant subspaces), then they are unitarily isomorphic to the position and momentum shifts.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.