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In ordinary non-relativistic quantum mechanics, the eigenvalues of the Hamiltonian, which represent the allowed energies of the system, are often quantized. For example, the energy levels of harmonic oscillators, those of the hydrogen atom, and so on. However, the eigenvalues of position and momentum, are always continuous (never quantized). Why?

Qmechanic
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    This is not true in general. Consider for example the momentum operator on the ring. And the hydrogen also contains a continuous part of the spectrum by the way. Other than that, from a mathematical point of view, I don't see any other reason than: Because these operators are defined the way they are. – Tobias Fünke May 06 '23 at 14:25
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    Does this answer your question? What entities in quantum mechanics are known to be "not quantized"? – Quillo May 06 '23 at 14:59
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    Hint. If you have a commutator of finite dimensional matrices, its trace must be zero. – Connor Behan May 06 '23 at 15:34
  • The discreteness of a measurement is guaranteed by the definition of "physical measurement": it has to provide a unique physical value that persists for eternity in the change of a physical system (e.g. the printout on some paper). There is no difference between classical mechanics and quantum mechanics in that regard. What is different in quantum mechanics that all energy and momentum exchanges have to be accompanied by a quantized change in angular momentum. This does, however, not automatically quantize energy and momentum. – FlatterMann May 06 '23 at 15:48
  • The "comparison" of momentum/position versus energy is not justified. The comparison is not reasonable. But we can ask separately: (1) Why Energy can often be quantized? (2) Why momentum-position, in general, are dealt continuously? - The answer to (1): It is a property of the Hamiltonian, representing the total energy under certain boundaries (we said “often”!) resulting in quantized eigenvalues- The answer to (2): Because of the Heisenberg uncertainty, you cannot have momentum and position fixed discrete, the non-zero commutation relation will not allow quantizing their eigenvalues. – al-Hwarizmi May 06 '23 at 16:59
  • @ConnorBehan This is not a sufficient argument for the fact that (say on $L^2(\mathbb R)$ both operators have a continuous spectrum, tho. It only tells you that we have to work with infinite-dimensional Hilbert spaces. The Wintner-Wielandt theorem then gives that at least one of both operators must be unbounded, which itself still is not enough to conclude anything about the spectrum. – Tobias Fünke May 06 '23 at 17:22
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    @Mauricio particle in a box doesn't even have conserved momentum, thanks to the walls breaking translation invariance. – JEB May 07 '23 at 00:37

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