Reply To: Consistency of the Everett ‘world branching’

Miroljub Dugic

Dear Dieter [if I may]

Thanks a lot for your comments, which I fully agree and helped me to realize that my first post hadn’t been clear. Let me briefly draw the ‘DIS’ to your attention.

For the sake of the argument, let me illustrate DIS (decomposition into subsystems) via the standard [non-relativistic] treatment of the hydrogen atom (HA). HA is defined as a pair “electron+proton”, e+p, which is a pair of ‘elementary’ (structureless) particles. The pair e+p is one DIS for HA. However, as we all know very well, the standard quantum theory of HA regards the alternative DIS of the atomic “center of mass + internal (relative)” virtual systems that can be presented as CM+R. This CM+R is another DIS for the hydrogen atom and for this DIS, the CM and the R systems represent the ‘elementary particles’.

If the atom were an isolated system—a model Universe—then our main argument starts with Entanglement Relativity (ER) for the atom. Actually, if (as it is typically the case) we assume a tensor-product pure state for the CM+R, the pair e+p *must* be entangled; please notice: this ER argument is kinematical and no interaction (such as the realistic Coulomb interaction) is required for the e+p pair. Already at this point (i.e. even without any regard of decoherence), I can illustrate a challenge posed for the original Everett world branching: Let me assume that the tensor-product state for the CM+R structure (DIS) is a result of ‘world branching’. Then, and this is the point, due to ER, the e+p atomic structure has not branched. As long as we consider the two atomic structures (DISs), e+p and CM+R, on the equal footing, the Everett world branching cannot be physically justified—only one of these structures, either e+p or CM+R, is allowed to branch. Therefore a need to choose one and only one fundamental DIS (atomic structure) that may be allowed to branch.

Is there any realistic, sufficiently ‘macroscopic’ decoherence model in support of this argument? Yes, there is, the QBM model mentioned in my previous post.

I should better stop here with the hope that this time I have better performed.

Best regards,

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