The wave-function as a true ensemble

  • Context: Graduate 
  • Thread starter Thread starter bohm2
  • Start date Start date
  • Tags Tags
    Ensemble
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 1K views
Messages
828
Reaction score
55
"We argue that the ψ-ontic/epistemic distinction fails to properly identify ensemble interpretations and propose a more useful definition. We then show that all ψ-ensemble interpretations which reproduce quantum mechanics violate Statistical Independence."

https://arxiv.org/abs/2109.02676
 
Reply
  • Like
Likes   Reactions: Demystifier
Physics news on Phys.org
"Suppose you have a theory that isn’t ontic because two of the hidden variables map to the same wave-function. Well, then you can just declare the wave-function to be part of the hidden variables, so that the new hidden variables – now including the wave-function – will always map to only one wave-function. Such an easily malleable definition of ‘ontic’ is not what one wants to base theorems on."

This is an interesting point. If I have a quantum state ##\psi##, I can always define an observable $$\hat{O} = \lambda_\psi|\psi\rangle\langle\psi| + \lambda_{\not\psi}(\hat{I} - |\psi\rangle\langle\psi|)$$ We have a property ##\lambda_\psi## resolvable (in principle) by experiment, that ##\psi## predicts with certainty. All other pure states will predict ##\lambda_{\not\psi}## with certainty. Make this variable a real hidden variable and voila, your wavefunction uniquely characterises the physical state of the system.
 
Morbert said:
"All other pure states will predict ##\lambda_{\not\psi}## with certainty.
No, they won't. Only pure states orthogonal to ##\psi## will.
 
Reply
  • Like
Likes   Reactions: gentzen and vanhees71