# The sum of two observables: Can its interpretation depend on more than just those two observables?

In the context of quantum theory, suppose we have two models $$M_1$$ and $$M_2$$ formulated on the same Hilbert space. Suppose that the operator $$A$$ is an observable in both models, with the "same" physical interpretation in both models. Suppose also that the operator $$B$$ is an observable in both models, with the "same" physical interpretation in both models.

Question: Can $$A+B$$ be an observable in both models with a physical interpretation in $$M_1$$ that differs from its interpretation in $$M_2$$?

My gut says yes, $$A+B$$ can have different interpretations in the two models, but I haven't come up with a satisfying example.

The question assumes that we're using a systematic approach (not ad-hoc) for associating physical observables to self-adjoint operators, like what is used in quantum field theory. I considered adding the quantum-field-theory tag for that reason, but I decided against that because there could be other systematic approaches, too.

Notes:

• This question was inspired by the question What is the physical meaning of the sum of two non-commuting observables?, but this new question is posed differently to solicit a specific kind of answer, namely answers that compare two specific models, rather than merely addressing generalities about measurement.

• In the phrase "same" physical interpretation, I put "same" in scare-quotes because one could question whether or not that's a meaningful concept, but I'm hoping to avoid that issue here.

• I'm not asking whether or not $$A+B$$ is necessarily an observable. That might also be an interesting question, but here I'm only considering models in which $$A+B$$ is taken to be an observable.

• I'm also not asking whether or not we have any hope of measuring $$A+B$$ in practice. I suspect there are cases where measuring $$A+B$$ would hopelessly complicated even if measuring $$A$$ and $$B$$ individually is simple, and that thought motivated me to post a related question on Math SE (https://math.stackexchange.com/q/3334439). For the present question, measurability in principle is sufficient. I admit that the distinction between practice and principle might itself be a non-trivial issue (cf chapter 7 in Omnes [1994], The Interpretation of Quantum Mechanics, Princeton University Press: "Some observables cannot be measured, even as a matter of principle"), but I'm hoping the present question can be answered without getting into that.

• Full disclosure: This is a sincere question, and I'd really like to know the answer, but it's also an experiment. This is arguably a duplicate of physics.stackexchange.com/q/498675 (to which I linked in my question), but it's posed differently to solicit a different kind of answer. I've read some Meta posts about duplicates, but I'm still not sure how this question will be received, so I'm trying it as an experiment. – Chiral Anomaly Oct 3 '19 at 14:23
• If you are wanting to "experiment", why not first ask about your ideas of doing this on meta first? – BioPhysicist Oct 3 '19 at 14:43
• @AaronStevens Yeah, I went back and forth about whether or not that would be the best approach. My motive for posting this question is to get answers to the physics question itself. I did consider posting on Meta first, but I struggled to come up with a way of expressing the Meta question without saying "Here's what I'd like to ask... how would this be received?" Pre-posting a physics question on Meta didn't seem appropriate, so I just posted the real Q here and acknowledged that I'm aware of the potential for being a duplicate. I can delete the question and go the Meta route if that's better. – Chiral Anomaly Oct 3 '19 at 14:57
• I don't know what would be better. I think this is fine tbh :) – BioPhysicist Oct 3 '19 at 15:29