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If you have two operators, one the true Hamiltonian $H$ and one we call an effective Hamiltonian $H_{eff}$ and say they agree on every eigenvector with eigenvalue up to $E_{eff}.$ Above that, they can disagree but if either has an eigenvector with eigenvalue $E<E_{eff}$ then it is an eigenvector of the other operator too and with the same eigenvalue.

Then for all those eigenvectors with eigenvalues below that limit they act the same, and the unitary evolution they generate acts the same on the span of those vectors.

If we don't have any evidence that either is correct (or even know which is correct) then isn't it the case that we have experimentally only explored the space spanned by the eigenvectors with eigenvalues less than $E_{eff}$?

Now there can be vectors with finite expectation where they disagree. And even vectors that have expectation value less than $E_{eff}$ where they disagree. But those still aren't accessible if we can't make interactions with that high of an energy. If $E_{eff}$ is $10^{10^{10^{10^{10^{10}}}}}J$ then we can't see it.

And sure our reduced domain where the two operators agree isn't the whole space so the eigenvectors aren't complete for the whole space. But again, the larger space isn't experimentally accessible.

Our theories are about what we test. So since one of them could be bounded and we can't tell the difference between them what besides convenience do unbounded operators give us?

Edit for answers that bring up canonical commutation operators.

Isn't $\hat x$ already unbounded? I agree that if you like unbounded operators then you like unbounded operators. That seems a bit circular though. Imagine I'm talking to an experimentalist and trying to convince them to learn about unbounded operators but they tell me it is just Math and not Physics, what can I say?

I agree that people like the $\hat x$ operator because then they can talk about all kinds of correspondence with classical mechanics. But assume someone already learned basic quantum mechanics and is happy and is happy with any specific operator that corresponds to a real experimental setup or observation.

But they aren't going to see an energy eigenvector with an energy that is $10^{10^{10^{10^{10^{10}}}}}J.$ And they aren't going to use $\hat x$ they are going to use an operator corresponding to a screen that can scintillate in a small but finite region or a Geiger counter that can fit in a small but finite region or other similar realistic real world devices. If they say unbounded operators are too complicated and irrelevant anyway what is the best response?

How can we explain a physical relevance to real observations and real experiments that can be done? Feel free to assume the same person objects to surface charge and line charge.

Potentially useful approximations but not real Physics.

Additional edit for answers that bring up canonical commutation operators.

In case it is unclear what it means to not be excited by the canonical commutation operators. Assume we have a Hilbert Space and an evolution equation and some operators that match things that can actually be done in the lab and if that means we don't have $\hat p_x$ or $\hat x$ that is fine as long as we have the operators for the experiments we can actually do. So the basis is the that we have a set of vectors in a Hilbert Space and we know how they evolve and we know how they act with a family of operators rich enough to include all the things we can actually do in the lab.

That's the direct description. If you insist on some connection to the space and operators you are used to you might be able think of it as an experimentalist working with something like the quotient with the subspace of things that can't be distinguished or resolved by their equipment.

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  • $\begingroup$ The premise is wrong. Suppose one of the two operators is bounded by $E_{eff}$ and compact, then its eigenvectors form a basis of the Hilbert space. If the other one is unbounded with non-compact resolvent then its eigenvectors cannot form a basis of the space. Therefore your premises cannot be realized in that (possible) situation. In addition, I strongly suspect that the evolution generated by two different operators is different even when acting on a subspace of common definition (where they are equal). $\endgroup$ – yuggib Aug 19 '15 at 17:12
  • $\begingroup$ I may elaborate further tomorrow or tonight, now sadly I can't... $\endgroup$ – yuggib Aug 19 '15 at 17:19
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tl;dr: The canonical commutation relations require unbounded operators.

Your truncation procedure seems ill-defined because it destroys the Hilbert space structure:

  1. Option one: You truncate to the span of finitely many vectors. The canonical commutation relations are not representable on finite-dimensional Hilbert spaces, this space is useless (because the action of $x$ and $p$ on vectors inside it will not give vectors that lie inside it, for one)

  2. Option two: You truncate to the span of infinitely many vectors. You have to take the closure of the span to get a Hilbert space (since we require completeness). Can you tell what vectors will lie in that closure? Is the restriction of $x$ and $p$ an endomorphism or does it lead out of the space, making it useless again? Additionally, the Stone-von Neumann theorem guarantees that any "good" restriction will be equivalent to the usual unbounded representation of the canonical commutation relations.

Only the second option has a chance at succeding, but it's chances are slim. $x$ and $p$ are unbounded operators unless the particle is confined, there's no way around this. Let's say the particle is free, so your energy truncation is essentially a momentum truncation. But the Fourier transform of an interval $[-p_\text{max},p_\text{max}]$ is an unbounded discrete lattice, $x$ is not unbounded here. This example shows that truncating one operator doesn't really get rid of unbounded operators, and introduces undesirable side effects like discretizing the spectrum of the position operator here - can you guarantee that the granularity the position values here get is compatible with your experimental precision?

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  • $\begingroup$ It doesn't even matter that much if one is bounded and the other isn't if they agree up to some huge energy I don't see how you can experimentally distinguish them. And position and momentum aren't sacred, experimental results are what is important. A theory isn't useless if it predicts laboratory results. For instance the nonrelativistic Schrödinger equation disagrees for large energies with the version with higher order relativistic corrections but there are lots of experiments that can't distinguish the two. A Hamiltonian we use now could start being wrong at very very high energies. $\endgroup$ – Timaeus Aug 19 '15 at 17:45
  • $\begingroup$ A theorist might see this as an argument against ignoring parts of the spectra of operators. On the other hand, an experimentalist (and even a logical positivist) could take this as an indication that $x$ and $p$ are pretty bad operators to be working with. $\endgroup$ – user10851 Aug 20 '15 at 23:19
  • $\begingroup$ @ACuriousMind There is an alternative view on Option 1 (what the OP proposes) and I think it is referred to as the "low energy approximation". Instead of simply truncating the Hilbert space to some (in)finite subspace and ignoring the discarded part, it sets up a proper projection approach and follows the consequences where interactions and observables are concerned. It works fine as far as the processes described involve/are confined to mainly the target subspace. $\endgroup$ – udrv Aug 22 '15 at 2:00

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