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This question I want to deal with the basics of modelling a physical theory: Let's say we start with observing in the world (be it little bubbles in the water, a particle moving, a pattern in the traffic jam, whatever), and we want to build a theory to describe what we see, that is capable of making predictions about future observations.

The first step in building a physical theory is translating observed objects into mathematical objects: For example translating the perceived position of a particle into an element of $\mathcal{R}^3$.

The second step is searching a set of assumptions about those mathematical objects, formulated in the language of mathematics, that is fully capable of describing the observations.

While the 2nd step of course involves formlating assumptions, my question would be wether the first steps does as well contain formulating "hidden" assumptions:

At the moment I identify an observed object from my surrounding nature with a mathematical object, is this already considered an assumption?

My question stems from the case of quantum mechanics, where observables are identified with operators acting on a Hilbert space $\mathcal{H}$ over the field of complex numbers. Together with the demand that all propabilities have to add up to 1, this naturally leads to an unitary time evolution for those operators. I have the feeling that choosing another representation for observables (representation here is not meant as a term from representation-theory) might have lead to different "natural consequences", and hence the question.

To give my own thoughts on this: Of course the choice of the mathematical objects affects the assumptions that afterwards should make statements about those mathematical objects, This makes it an "assumption".

But if for any other choice of mathematical object (for example, representing the position of a particle not by 3 real values, but instead by one complex one, and one real one), translating all the assumptions from one choice to the other choice would result in the same predictions - then all choices would lead to the same predictions, and it weren't important what mathematical object to choose to model an observation.

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  • $\begingroup$ One comment that is probably not useful is that, whenever designing a mathematical model or theory, defining one object mathematically rapidly induces other such definitions on other objects (for example, saying observables = operators leads to wavefunctions = vector space elements). As a result, it’s tough to pin down a single mathematical definition of that type in a model, and it is more proper to define some collective definition of the math involved in the subject (Schrodinger picture = linear algebra in infinite dimensions). $\endgroup$ – aghostinthefigures Jul 1 '18 at 15:03
  • $\begingroup$ Some assumptions are explicit, some are implicit, but it's not always easy to notice the implicit ones, or they may be so obvious in the context that nobody bothers to mention them explicitly. $\endgroup$ – PM 2Ring Jul 1 '18 at 23:55
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At the moment I identify an observed object from my surrounding nature with a mathematical object, is this already considered an assumption.

Yes.

When you take into account that the ordering of operators is not set in stone, then we introduce a new pathway for "natural consequences"

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The first step in building a physical theory is translating observed objects into mathematical objects: For example translating the perceived position of a particle into an element of R3

My question stems from the case of quantum mechanics, where observables are identified with operators acting on a Hilbert space H over the field of complex numbers. Together with the demand that all propabilities have to add up to 1, this naturally leads to an unitary time evolution for those operators.

(Emphasis added)

I think this is framed in a way that’s not consistent with how it historically happened..

The development of quantum mechanics involved decades of false starts and alternative formulations. These were considered on their own, via thought experiments and calculations, but also in concert with experiment: theories suggested experiments, and experiments confronted theories.

It wasn’t a process where there was a “choose identification” step and a “demand constraint” step that then inevitably and without further choice “leads” somewhere. Many early theoretical construct “led” to inconsistency, and were pruned off.

I have the feeling that choosing another representation for observables (representation here is not meant as a term from representation-theory) might have lead to different "natural consequences",

That’s probably true. But lots and lots of representations were examined. They just mostly didn’t work.

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  • $\begingroup$ I guess your last sentence would be an argument to answer the question by "yes". $\endgroup$ – Quantumwhisp Jul 1 '18 at 15:32

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