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What does does [H1,H2] = 0 mean if each H is a hamiltonian of a quantum system

I'm trying to get through a research paper on theoretical quantum biology and I just want to make sure I'm interpreting this relation nearly correctly. Is this equivalent to saying that the two Hamiltonians are orthogonal? Or that the eigenstates of the Hamiltonian are orthogonal?

Here is an excerpt from the paper for context.

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I don't have enough time to learn all of the math right now as this is partially for a class presentation for the undergrad physics seminar course i'm in, but I want to get the gist of the theory.

Here's a link to the paper for anyone interested. http://arxiv.org/abs/1408.5798

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    $\begingroup$ It means that the Hamiltonians commute (or, in the case of the paper, don't commute). Two Hamiltonians that don't commute cannot, in particular, have the same eigenstates. If $A, B$ are two operators then $[A, B]$ denotes their commutator $AB - BA$. $\endgroup$ – Qiaochu Yuan May 2 '15 at 23:54
  • $\begingroup$ @QiaochuYuan Sorry, was busy typing as you answered. Should I delete my answer? $\endgroup$ – user12802 May 3 '15 at 0:04
  • $\begingroup$ @QiaochuYuan So does writing [A,B] = 0 say that the two Hamiltonians do not commute, which says that the order in which you apply them to a state matters, whereas if [A,B] =/ 0 the Hamiltonians do commute and the order does not matter? $\endgroup$ – lthermin May 3 '15 at 0:25
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    $\begingroup$ No, it is precisely the opposite. $[A,B] = AB - BA$ so $[A,B] = 0$ if and only if $AB = BA$. $\endgroup$ – JHance May 3 '15 at 0:26
  • $\begingroup$ @JHance Awesome thanks for clearing that up. $\endgroup$ – lthermin May 3 '15 at 0:37
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[A,B] is the commutator. It means A and B commute when AB - BA = 0. Otherwise they do not commute. To make things ultra simple, two observables cannot be in the same (eigen)state if they don't commute. Thus Heisenberg's Uncertainty Principle where position and momentum operators do not commute, thus position and momentum cannot be simultaneously known. You might think of lack of commutability as incompatibility.

ADDENDUM IF you want a great way to learn about this an QM in general, consider these quite accessible videos by James Binney, head of theoretical physics at Oxford. One of them is titled explicitly mentioning commutators but you may have to back up a bit to get the full flavor:

https://www.youtube.com/playlist?list=PLE73AA240E8655D16

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  • $\begingroup$ Thanks for the link! I may try to go through those videos over the Summer before my QM course in the Fall. $\endgroup$ – lthermin May 3 '15 at 0:27
  • $\begingroup$ Just to clarify when you say "two observables cannot be in the same (eigen)state if they don't commute" do you mean "two observables cannot be in the same (eigen)state if the Hamiltonians don't commute"? $\endgroup$ – lthermin May 3 '15 at 0:36
  • $\begingroup$ @LIEVBIRMAN Yes, that's correct. $\endgroup$ – user12802 May 3 '15 at 1:04

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