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Commutation of operators

  1. Nov 30, 2013 #1
    What do we mean by the two operators are commutative or non commutative? I wanted to understand the physical significance of the commutative property of the operators. We are doing the introduction to quantum mechanics and there are many things that are really confusing. Any help will be appreciated. Thanks.
     
  2. jcsd
  3. Nov 30, 2013 #2

    bhobba

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    Non-commutivity means, roughly, they do not have exactly the same eigenvectors.

    The physical interpretation of operators is their eigenvalues are the possible outcomes of observations, and the eigenvectors are the state the system will be in if that eigenvalue occurs.

    If observables commute it means, roughly, they have common eigenvectors, hence you will get 'compatible' outcomes if observed. In fact it can mean they are really the same observable in disguise - since you may get the other merely by relabeling the outcome with a different value - but the exact sense that is true requires a bit of experience with the formalism.

    If they do not commute, in a sense they are incompatible, and it can be shown they can not have simultaneous values - this is the essence of Heisenberg's famous uncertainty relations.

    Just so you understand what's going on a bit better I will tell you a more advanced view. Don't worry if it doesn't gel right now - come back to it later as you progress in your understanding.

    Imagine we have a system and some observational apparatus that has n possible outcomes associated with values yi. This immediately suggests a vector and to bring this out I will write it as Ʃ yi |bi>. Now we have a problem - the |bi> are freely chosen - they are simply man made things that follow from a theorem on vector spaces - fundamental physics can not depend on that. To get around it QM replaces the |bi> by |bi><bi| to give the operator Ʃ yi |bi><bi| - which is basis independent. This is the first axiom of the treatment in Ballentine - QM - A Modern Development (a very well respected book on QM that develops it in a fairly careful and rigorous way from just two axioms - not good to start with since its graduate level - but good to work towards which is why I am mentioning it), and heuristically why its reasonable.

    Next we have this wonderful theorem called Gleason's theorem which, basically, follows from the above axiom:
    http://kof.physto.se/theses/helena-master.pdf [Broken]

    This is the second axioms in Ballentine's treatment.

    This means a state is simply a mathematical requirement to allow us to calculate expected values in QM. It may or may not be real - there is no way to tell. But its very similar to the role probabilities play in probability theory, and most would not say they are real.

    Further in that vein, nowadays its often thought of as just a novel version of probability theory - there basically being just two reasonable models applicable to physical systems. Check out:
    http://arxiv.org/abs/quant-ph/0101012
    http://arxiv.org/abs/0911.0695

    Again - the above is just for future reference. You will appreciate them more when you have gone a bit deeper. For now simply keep in mind what I said at the start and plough on.

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  4. Nov 30, 2013 #3
    Thanks a lot bill. That was very helpful.
     
  5. Dec 4, 2013 #4
    Bill,

    Is it correct in saying that |bi><bi| is a projection operator? I'm thinking that the projection operator would give us an eigenvalue projected in the 'direction' of some given eigenstate (mathematically speaking) since it is the inner product, but I feel like I am missing something, or I am not on the right track of thinking.

    Also, what do you mean when you say the operator Ʃ yi |bi><bi| is basis independent?
     
  6. Dec 5, 2013 #5
    It means it doesn't matter what order you do them in. If you apply A first then B you will get the same answer as when you apply B first then A.
    ABψ = BAψ
    This means in terms of measuring two properties A and B that you will get the same results (eigenvalues, a and b) whichever order you make the measurement in.
     
  7. Dec 5, 2013 #6

    bhobba

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    It is - but it's a projection onto a one dimensional subspace. The general form of a projection operator is Ʃ |bi><bi| where the |bi> are orthonormal.

    A projection operator only has 1 and 0 as eigenvalues. The exact properties it has are easily deduced from the trace form of the Born rule.

    Its trivial - the eigenvalues of operators do not depend on your basis. A vector is Ʃ yi |bi>, and the yi change with your basis. Mapping the outcomes of observations, yi, to vectors means you also need to know the basis to determine the yi. Mapping it to projection operators avoids that. Generally you would expect fundamental physics to not depend on your choice of basis - which are entirely arbitrary man made things.

    Its a more sophisticated version of the one Victor Stenger used to justify the principle of superposition:
    http://www.colorado.edu/philosophy/vstenger/Nothing/SuperPos.htm [Broken]

    I use that the physics should not depend on the basis and Gleason's Theorem, which also requires that assumption. This would seem to be one of the rock bottom fundamental foundations of QM. Just my view of course - others may not agree. Not with the math - that's beyond doubt - but what you think the foundational 'basis' is.

    The above is the slickest justification of QM I know - my actual favorite however is the seminal paper by Hardy:
    http://arxiv.org/pdf/quant-ph/0101012.pdf

    But that does require a bit of an understanding of it in the first place. Mine however directly translates to the 2 axioms in Ballentine, so can be used from the start.

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  8. Dec 13, 2013 #7
    I'm currently in the middle of finals, but those resources look very interesting. Can't wait to check them out and thank you for your help!
     
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