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It is well known that one of the operators satisfying the Canonical Commutation Relation $[x,p]=i$ must be unbounded. In most cases I have seen, either both are unbounded or only $p$ is (e.g. Particle on $S^1$ has bounded position variable $\theta\in[0,2\pi)$ which trivially implies the operator is also bounded). Are there cases where the momentum operator is bounded and the position variable is unbounded?

And in that case are there Hamiltonian operators which are bounded in an infinite dimensional Hilbert space (so that it's domain must be the full space). Again here, there are countless examples in the finite dimensional case and also other bounded operators in infinite dimensional spaces like time evolution operator, S matrices, density matrices etc.

So, what are some examples of bounded momentum/Hamiltonian operators in infinite dimensional Hilbert spaces? This question can be regarded as a special case/near duplicate of this question.

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What about a projector onto a trace-one pure state? Like a projector onto a wavepacket.

Take $$|\psi\rangle=\int \psi(x)|x\rangle,$$ whose normalization condition from $\langle y|x\rangle=\delta(y-x)$ must be $\int |\psi(x)|^2=1$. Then the operator $$|\psi\rangle\langle \psi|$$ is bounded (it has trace one, after all) and is defined over the entire Hilbert space. In fact, it can give nonzero results when acting on all states $|x\rangle$ if $|\psi(x)|$ has support everywhere, which could indeed be true if it is a Gaussian function.

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  • $\begingroup$ As far as I understood, the question asks for bounded Hamiltonians and not generic operators. $\endgroup$ Commented Sep 29, 2023 at 13:25
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    $\begingroup$ @TobiasFünke good point. What stops me from saying my Hamiltonian is $E|\psi\rangle\langle \psi|$? I don't have any physical system in mind but it should qualify in general. Is the point to look for a Hamiltonian with a standard form of kinetic and potential energy terms? $\endgroup$ Commented Sep 29, 2023 at 16:37
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    $\begingroup$ Yes, sure, I agree. I don't know what exactly OP has in mind, though. $\endgroup$ Commented Sep 29, 2023 at 16:42

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