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Now (this is the unfinished part, I don't have a clear sense of how to tell the story from here) we can bring in algebras of operators on Hilbert spaces, the Stone-von Neumann theorem about representations of the canonical commutation relations $[X, P] = i \hbar$, and Stone's theorem on one-parameter unitary groups, and the Skolem-Noether theorem for the algebra of bounded operators on a Hilbert space, which together are also quite suggestive, although the leap to Hilbert spaces and the self-adjoint / unitary constraints haven't been motivated (for that matter we haven't at all motivated the decision to work over $\mathbb{C}$ as opposed to $\mathbb{R}$). This still does not get us to $C^{\ast}$-algebras but IMO this is a feature and not a bug: $C^{\ast}$-algebras can only talk directly about bounded operators and the position and momentum operators aren't bounded! Also there's no reason an arbitrary $C^{\ast}$-algebra should satisfy Noether's theorem; this line of reasoning specifically pulls us towards the special ones that do, which IMO is also a feature and not a bug.

Now (this is the unfinished part, I don't have a clear sense of how to tell the story from here) we can bring in algebras of operators on Hilbert spaces, the Stone-von Neumann theorem about representations of the canonical commutation relations $[X, P] = i \hbar$, Stone's theorem on one-parameter unitary groups, and the Skolem-Noether theorem for the algebra of bounded operators on a Hilbert space, which together are also quite suggestive, although the leap to Hilbert spaces and the self-adjoint / unitary constraints haven't been motivated (for that matter we haven't at all motivated the decision to work over $\mathbb{C}$ as opposed to $\mathbb{R}$). This still does not get us to $C^{\ast}$-algebras but IMO this is a feature and not a bug: $C^{\ast}$-algebras can only talk directly about bounded operators and the position and momentum operators aren't bounded! Also there's no reason an arbitrary $C^{\ast}$-algebra should satisfy Noether's theorem; this line of reasoning specifically pulls us towards the special ones that do, which IMO is also a feature and not a bug.

This suggests a potential route to an answer to the title question about why addition of observables in quantum mechanics must be commutative: it's because addition in a Lie algebra must be commutative. I like this idea because in the Lie algebra case it's also unclear what the "physical" meaning of addition is: it doesn't correspond in a straightforward way to an operation on one-parameter groups if the corresponding generators don't bracket to zero, which is exactly the issue we see with observables.

Now (this is the unfinished part, I don't have a clear sense of how to tell the story from here) we can bring in algebras of operators on Hilbert spaces, the Stone-von Neumann theorem about representations of the canonical commutation relations $[X, P] = i \hbar$, and Stone's theorem on one-parameter unitary groups, which together are also quite suggestive, although the leap to Hilbert spaces and the self-adjoint / unitary constraints haven't been motivated (for that matter we haven't at all motivated the decision to work over $\mathbb{C}$ as opposed to $\mathbb{R}$). This still does not get us to $C^{\ast}$-algebras but IMO this is a feature and not a bug: $C^{\ast}$-algebras can only talk directly about bounded operators and the position and momentum operators aren't bounded! Also there's no reason an arbitrary $C^{\ast}$-algebra should satisfy Noether's theorem; this line of reasoning specifically pulls us towards the special ones that do, which IMO is also a feature and not a bug.

This suggests a potential route to an answer to the title question about why addition of observables in quantum mechanics must be commutative: it's because addition in a Lie algebra must be commutative (I see that Gianmarco Bramanti has also suggested this). I like this idea because in the Lie algebra case it's also unclear what the "physical" meaning of addition is: it doesn't correspond in a straightforward way to an operation on one-parameter groups if the corresponding generators don't bracket to zero, which is exactly the issue we see with observables.

Now (this is the unfinished part, I don't have a clear sense of how to tell the story from here) we can bring in algebras of operators on Hilbert spaces, the Stone-von Neumann theorem about representations of the canonical commutation relations $[X, P] = i \hbar$, and Stone's theorem on one-parameter unitary groups, and the Skolem-Noether theorem for the algebra of bounded operators on a Hilbert space, which together are also quite suggestive, although the leap to Hilbert spaces and the self-adjoint / unitary constraints haven't been motivated (for that matter we haven't at all motivated the decision to work over $\mathbb{C}$ as opposed to $\mathbb{R}$). This still does not get us to $C^{\ast}$-algebras but IMO this is a feature and not a bug: $C^{\ast}$-algebras can only talk directly about bounded operators and the position and momentum operators aren't bounded! Also there's no reason an arbitrary $C^{\ast}$-algebra should satisfy Noether's theorem; this line of reasoning specifically pulls us towards the special ones that do, which IMO is also a feature and not a bug.

Now (this is the unfinished part, I don't have a clear sense of how to tell the story from here) we can bring in algebras of operators on Hilbert spaces, the Stone-von Neumann theorem about representations of the canonical commutation relations $[X, P] = i \hbar$, and Stone's theorem on one-parameter unitary groups, which together are also quite suggestive, although the leap to Hilbert spaces and the self-adjoint / unitary constraints haven't been motivated (for that matter we haven't at all motivated the decision to work over $\mathbb{C}$ as opposed to $\mathbb{R}$). This still does not get us to $C^{\ast}$-algebras but IMO this is a feature and not a bug: $C^{\ast}$-algebras can only talk directly about bounded operators and the position and momentum operators aren't bounded!

Now (this is the unfinished part, I don't have a clear sense of how to tell the story from here) we can bring in algebras of operators on Hilbert spaces, the Stone-von Neumann theorem about representations of the canonical commutation relations $[X, P] = i \hbar$, and Stone's theorem on one-parameter unitary groups, which together are also quite suggestive, although the leap to Hilbert spaces and the self-adjoint / unitary constraints haven't been motivated (for that matter we haven't at all motivated the decision to work over $\mathbb{C}$ as opposed to $\mathbb{R}$). This still does not get us to $C^{\ast}$-algebras but IMO this is a feature and not a bug: $C^{\ast}$-algebras can only talk directly about bounded operators and the position and momentum operators aren't bounded! Also there's no reason an arbitrary $C^{\ast}$-algebra should satisfy Noether's theorem; this line of reasoning specifically pulls us towards the special ones that do, which IMO is also a feature and not a bug.