Timeline for Why are diagonal Hamiltonians considered classical?
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| Mar 25, 2021 at 6:14 | comment | added | Adam Zalcman | Yes, I had clarified language of the statement that related commutation with local bases, turned it into a theorem to make it clear what it says and added a proof. Just now, also added a sentence to address (i). Regarding (ii), TC is mentioned in remark 2. Thank you for the link to the paper. It'll take me some time to go through it. Is there a specific fragment you could refer me to for the special character of two-body interactions? | |
| Mar 25, 2021 at 6:07 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Mar 24, 2021 at 22:50 | comment | added | Norbert Schuch | True. My point was just that (i) it is important to formulate clearly what the "local" in local basis means, and (ii) I still stand that two-body interactions are special (see arxiv.org/abs/quant-ph/0308021). That's why I felt that the TC was the only clear counterexample (note: I just see that you significantly rewrote it ... let me see). | |
| Mar 24, 2021 at 22:38 | comment | added | Adam Zalcman | At the end of the day, the criterion is equivalent to the existence of a local basis of energy eigenstates. And it's formulated in terms that make it easier to apply directly to the situation described by the OP. | |
| Mar 24, 2021 at 22:37 | comment | added | Adam Zalcman | Agreed. Implicit in the problem is a choice of partitioning of the system into subsystems which corresponds to a choice of single-body Hilbert spaces whose tensor product yields the full Hilbert space. In the situation you describe, you choose a 4-dimensional single-body spaces which is fine. If you account for this in the criterion in the theorem above things work out as expected, i.e. it confirms that your Hamiltonian is classical. | |
| Mar 24, 2021 at 19:35 | comment | added | Norbert Schuch | By "single two-body term" I mean the 2nd example you give. This is classical in the Bell basis. If I have a system - e.g. on a ladder - which is diagonal in the Bell basis on the rungs I would say this is classical. Say, a Toric Code on the ladder: This is classical, just on 4-level systems, not 2-level systems. Genuine non-classicality would mean that it is impossible to find a local basis (local $\ne$ single site!) in which the problem is diagonal. | |
| Mar 24, 2021 at 19:15 | comment | added | Adam Zalcman | Regarding all two-body Hamiltonians having a ground space spanned by products: Perhaps I'm misunderstanding, but this appears to be false. The $H=-\sigma_1^X\sigma_2^X-\sigma_1^Z\sigma_2^Z$ example above is a counterexample. Its ground space is not spanned by products because it is a 1-dimensional space spanned by a Bell state. (Thanks for terminology remark. Removed "poor".) | |
| Mar 24, 2021 at 19:10 | comment | added | Adam Zalcman | Regarding single two-body term being classical: You are right and this is consistent with the above characterization. If $H=A_0\otimes B_0$ then all $A_0$ operators commute vacuously among themselves, as do the $B_0$ operators among themselves. | |
| Mar 24, 2021 at 19:08 | comment | added | Adam Zalcman | @NorbertSchuch I did not mean to imply that we can judge classical character of a Hamiltonian from a single term, but I concede the text was not very clear or rigorous. Made it so now. I think the above characterization of classical Hamiltonians is interesting, useful and correct. In particular, it explains why we need a transverse field to make the Ising model exhibit quantum features. | |
| Mar 24, 2021 at 18:59 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Mar 24, 2021 at 10:35 | comment | added | Norbert Schuch | ... and illustrate this with the toric code. As for the terminology, it might be poor, but it is used by reputable people ;) | |
| Mar 24, 2021 at 10:35 | comment | added | Norbert Schuch | I still feel that the "Commuting tensor factor" part contains more misleading (and probably wrong) parts than not. A single two-body term can always be describes as classical. The question is whether there is one local (not necessarily single-site) basis which makes all Hamiltonian term diagonal. You can't judge this from a single term. In fact, all systems described by two-body Hamiltonian have ground states which are "almost" classical (in the sense that their ground space is spanned by products.) I feel it would be safer off to just say that commuting does not imply classical ... | |
| Mar 24, 2021 at 5:51 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Mar 24, 2021 at 5:12 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Mar 24, 2021 at 4:20 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Mar 24, 2021 at 1:53 | comment | added | Adam Zalcman | @NorbertSchuch Good catch! Thanks! Fixed both errors. | |
| Mar 24, 2021 at 1:53 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Mar 23, 2021 at 23:21 | comment | added | Norbert Schuch | For "classical", being diagonal in any product basis is sufficient. Also, commuting does not imply classical, see e.g. the toric code. | |
| Mar 23, 2021 at 21:25 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Mar 23, 2021 at 21:02 | history | answered | Adam Zalcman | CC BY-SA 4.0 |