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Specifically, why is the operator “sandwiched in” between $\Psi^*$ and $\Psi$? i.e. Why isn’t the formula just $$\langle \hat{Q} \rangle = \int \hat{Q}\cdot|\Psi|^2 dx = \int \hat{Q}\cdot\Psi \cdot \Psi^* dx$$ instead of $$\langle\hat{Q}\rangle = \int \Psi^* \cdot\hat{Q}\cdot\Psi dx~? $$

I get that operators are generally noncommutative in QM, which is why the expressions aren't equivalent (or, more specifically, they're only equivalent if $\hat{Q}$ is Hermitian), but I don't get why we use the second expression instead of the first, as the first is just applying the definition of expected value from statistics. Expected value isn't even a physics thing, it's a math/statistics thing with a very straightforward definition, so how does it make it any sense to go changing it?

Does it have something to do with the operators being complex-valued? That's the only thing I can think of, especially because, for observables, which always correspond to real numbers, the operators have to be Hermitian, and, if I understand correctly, in that case, the QM formula reduces to the usual statistics formula. Hence, it would seem the formulas are only different when the operator outputs a complex value. But, looking at the wikipedia article on complex random variables, the formula doesn't really seem any different than the usual one for real random variables:

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So what the heck is up with this formula? I know it's a postulate of QM, not derived from something else, but there still has to be some REASON for postulating it, and it can't just be empirical, since the formulas are only different for non-observables.

I did find this thread but it's mostly just people demonstrating that, for observables, the formulas are the same, which I already know. What I want to know is why, in QM, we start out with a different definition of expected value in the first place, regardless of whether it reduces to the usual definition in the special case of Hermitian operators.

The only other semi relevant thread I could find was this one, but that person was more wanting to know how it made sense to use expected value in QM at all, rather than about the difference in the two definitions.

Mikayla Eckel Cifrese
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    Your question is long and confusing, IMHO... What are you asking, exactly - in one sentence? To start: Operators are not complex -valued, the definition of the expected value in QM is a postulate and so on...And we keep it as a postulate because it agrees with experiment... – Tobias Fünke Mar 28 '23 at 07:56
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    Possible duplicate: https://physics.stackexchange.com/q/299286/2451 – Qmechanic Mar 28 '23 at 08:02
  • @TobiasFunke, how does it agree with experiment though? For observables, which, by definition, are the only things we can measure, the QM formula is the same as the statistics formula. But it's different for nonobservables. I don't understand why the QM definition of expected value of nonobservables is different from the statistics definition of expected value. – Mikayla Eckel Cifrese Mar 28 '23 at 08:21
  • nonobservables == not observable in experiments... But see also this and the answers therein – Tobias Fünke Mar 28 '23 at 08:52
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    As an experimentalist I should probably point out that the only actual physical "observables" that exist are energy, momentum, angular momentum and charges. What the theory calls "observables" are linear functions of the spectra of these quantities. We can always form such linear functions, both on paper and experimentally (with lenses, gratings, magnetic fields etc.), but the linearity is guaranteed by the independence of individual quanta (i.e. repetitions of the same experiment). In other words... it's an artifact of how we define "experiment" and observable and not fundamental. – FlatterMann Mar 28 '23 at 09:17

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