Timeline for Why is addition of observables in quantum mechanics commutative?
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8 events
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| Mar 10, 2021 at 12:05 | comment | added | saolof | So this answer is a misinterpretation of classical probabilities and does not even touch QM. What is actually different in QM is that conditional probabilities when measuring probabilistic variables, depend on the order in which one makes the measurement (but non-conditional probabilities are independent of the order) | |
| Mar 10, 2021 at 11:57 | comment | added | saolof | Right. Quantum mechanics produces the exact same result as classical mechanics if all operators and density matrices are diagonal, so this answer would get the same failure for a correlated classical system. The squaring rule is not compatible with the addition rule. X + Y for commuting variables classically means measuring both and adding the result, and that's what you need to use if you multiply them. The error that this answer makes is using an equivalence for expectation values and viewing it as an equality of observables | |
| Aug 20, 2020 at 22:41 | comment | added | Vectornaut | Agreed, @TobyBartels: this addition procedure is only associative on the level of expectation values. I don't see anything wrong with that, though! Having many physically different ways to measure the same observable is very typical. Your observation reminds me of the composition rule Hatcher uses to define the fundamental group, where you traverse each loop at double speed. It's non-associative in the same way, and it too becomes associative at the level of equivalence classes. | |
| Aug 16, 2020 at 4:39 | comment | added | Toby Bartels | A decade later, let me note another way that this attempt to define $X+Y$ goes pear-shaped: it's not associative. $(X+Y)+Z$ measures $X$ or $Y$ each with probability $1/4$ and $Z$ with probabilty $1/2$, while $X+(Y+Z)$ swaps the probabilities on $X$ and $Z$. | |
| Jan 3, 2012 at 17:58 | comment | added | Toby Bartels | On the contrary: in quantum mechanics as it exists, measuring $Z$ and squaring the result does measure $Z^2$, but flipping a coin (then making an appropriate measurement and doubling the result) does not measure $X + Y$. | |
| Jul 28, 2011 at 23:05 | comment | added | Vectornaut | @Steven Stadnicki: To me, given an observable $Z$ and a function $f \colon \mathbb{R} \to \mathbb{R}$, it seems totally natural to define $f(Z)$ as the observable you measure by measuring $Z$ and then applying $f$ to the result. What do you mean by an "invariant meaning," and why doesn't this definition give an "invariant meaning" to $f(Z)$ for any $Z$ and $f$? | |
| Jul 28, 2010 at 17:12 | comment | added | Steven Stadnicki | 'Given an observable Z, it makes sense to define Z<sup>2</sup> to be the observable Measure Z and square the result' - I think this is where that argument falls over, and not just on a quantum but even a classical level. I expect to be able to define the observable X+Y readily simply because linearity is a 'plausible' thing to have, but I see no reason to believe that I'll be able to ascribe any sort of invariant meaning to Z<sup>2</sup> in general. | |
| Jul 28, 2010 at 15:20 | history | answered | Vectornaut | CC BY-SA 2.5 |