# Hamilton operator in absence of causal order?

I hope, this question isn't too broad or vague.

In a recent paper, Ognyan Oreshkov et al. worked out a theory of quantum correlations in absence of any causal order, dropping the assumptions of a space-time: See their paper here.

Now I wonder (please refer to the linked paper), would it even be possible to define a Hamiltionian in absence of a space-time to refer to? And if so, how would one do that?

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Very interesting. I never really liked the idea of casualty anyway. Is the fact that events could be seen as happening in different orders from different observers in special relativity relevant? How does this formulation address the holistic-ness of states? –  namehere Dec 5 '12 at 13:39
I'm not entirely sure. At the very least they mention general relativity in their paper in two contexts: Assuming only local correctness of QM is equivalent to assume only local flatness of space in GR. And in the sublementary informations to the paper (see the very bottom of the linked page for a pdf of that), they mention that one term of the generic solution to a two-observer-problem, which, however, breaks unity of probability, is equivalent to a certain kind of time-like loops as described in one of their references. So if GR gets a mention, SR surely is related as well. –  kram1032 Dec 5 '12 at 13:47
I'll be sure to look all this up when I have more time. However, as the the formulation is completely local, the validity of a Hamiltonian does become less clear. –  namehere Dec 5 '12 at 14:00

## 1 Answer

The Hamiltonian is the generator of time-translations. If you eliminate time from the description no true Hamiltonian can exist.

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