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DarMM
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Firstly I will say note they are only that equivalent in QM, in QFT the Schrodinger picture can require additional renormalizations, so they aren't unitarily equivalent as the Schrodinger picture will have a slightly different Hamiltonian due to the new counter terms.akvadrako said:How does this require more than the well-known equivalence between the Heisenberg and Schrodinger pictures, given the algebra stuff is the Heisenberg observables?
Deutsch's algebra perspective moves beyond the equivalence of the Schrodinger and Heisenberg picture and has nothing to do with it really. That's just moving time evolution between the state and observables. He is staying that the quantum state could ultimately be eliminated from the theory as it has no ontic existence. When a measuring device and a particle interact, that is purely an interaction between the device's algebra-stuff and the particle's algebra-stuff, the state doesn't exist. It's only useful to convey constraints on measuring the algebra (i.e. extracting a real number from an algebra element) when you don't know the device algebra details, i.e. the state is epistemic.
This does make Deutsch's view local as the evolution of the algebra is provably locally, most generally in algebraic QFT where regional algebra are even called "local algebras". However it is typically viewed that the state is nonlocal, nobody questions the locality of the observable algebra. Another way of putting it is that the dynamics are thought to be local, but the states are not. Deustch avoids this by dropping the ontic status of the quantum state and thus all his ontic elements are local.
However the state not being ontic, but just an epistemic constraint, and the algebra being is very different from MWI or indeed any interpretation of QM, it's not just the Heisenberg picture. It's also a very undeveloped view.
If you retain the state as ontic, then Deutsch's proof is just a (interesting) Heisenberg picture demonstration of no-signalling, not locality.
The algebra doesn't have the spacetime as an element, it's a sheaf over the spacetime (and always a sheaf over a spacetime, can't be without it). The state is then nonlocal as it doesn't factorise across the algebras of regions.akvadrako said:You're saying that because when it's defined on an algebra without spacetime it's alocal by definition and when defined on an algebra with spacetime ##\mathcal{M}^4## as a primitive element that it contains global properties and becomes non-local.
Before we continue and I think this might be core to the whole thing, I would add:akvadrako said:This isn't the definition of *local I'm using. A theory is local if it possibly can be reformulated in terms of separate regions, so that actions on separated regions don't effect each other.
"can be reformulated in terms of ontic elements in separate regions, so that actions on separated regions don't effect each other".
If the restrictions to regions that don't effect each other are necessarily epistemic, then any locality demonstrations would only be non-signalling demonstrations, agreed?