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Understanding the Commutator

  1. Jun 7, 2004 #1
    The commutator plays a central roll in quantum mechanics. I guess it is hard to study any aspect of quantum mechanics without running into a commutator.
    I understand you can accept the fact that commutators of compatible measurables equal zero, while those of incompatible measurements equal something like -ih. and after accepting these relationships you can derive a lot of other properties of your system.
    I have seen explanations of the commutators in quantum mechanics as the quantum-mechanical version of the poisson brackets. But this explanation is not intuitive enough for me. ( I think better in terms of pictures rather than symbols on paper).
    I have also read the approach by Sakuray, which almost led me to understand the issue but not quite.
    I'll appreciate any insights on this topic.
  2. jcsd
  3. Jun 7, 2004 #2


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    Blame it all on Heisenberg. He's the one who first thought of using matrices to describe atomic-scale happenings, and once you have matrices as your operators, the fact that matrix multiplication is noncommutative has to be dealt with. Here is a brief website that just scratches the surface:


    But you may already know that much.
  4. Jun 7, 2004 #3
    Thank you Janitor,
    But I am looking for a deeper meaning. I am not happy with "this is the way it is" or the argument that you get the commutators because matrices don't commute. As a matter of fact, the matrices of compatible operators do commute.
    I would like to understand "why" (sorry Dr. Feynman) incompatible operators don't commute and "why" you get what you get when you calculate the commutator.
    I have tried to get an answer to this by looking at the behavior of Pauli matrices, but that didn't make it much more clear for me.
    I can intuitively understand how two rotations in different planes don't commute. If the measurable operators represented rotations, then I could understand why you get a non-zero commutator. But as far as I know these operators don't represent rotations in physical space or in Hilbert space. Right?
  5. Jun 7, 2004 #4


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    Take a coffee cup, stand it on a flat surface. Let the z axis be perpendicular to the bottom of the cup, the x axis goes through the handle and lies in the plane of the handle, perpendicular to the z axis. The y axis is, of course perpendicular to both x and z axes. Rotate the cup 90 degrees about the x axis, then 90 degrees abou the y axis. Then go back to the initial configuration, do 90 degrees about the y axis, then 90 degrees about the x axis. Each set of rotations leads to a different final configuration; rotations do not commute. The mathematics must reflect the physical reality, and so rotation operators do not commute.

    Indeed, non-commutivity can be hard to understand at a gut, physical level. But, usually a study of group theory helps to gain confidence and understanding -- most groups involve non commuting elements. Many professional physicists scratched their heads upon their initial encounters with non commuting operators.

    Another simple example is the operator x d/dx vs. d/dx x applied to a function of x.

    Hang in there, you will get it sooner or later.
    Good luck and regards,
    R. Atkinson
  6. Jun 7, 2004 #5
    Thanks for your post. I do understand why rotations (as long as they are about different axes) do not commute.
    What I don't understand is why the operators for incompatible observables do not commute. I will "hang in there" but I'll make an attempt to understand these things before I get used to them and just accept them because "they work"
    I understand my approach to learning may be today somewhat unorthodox and even anoying. But i'll keep trying. Let's see if I get somewhere or if someone points me in the right direction.
    Somewhere in the web I found this explanation, which I think is not bad, but I am looking for a better explanation:
    AB-BA is not generally zero. In programming talk, it is actually very similar to how ++C and C++ differ. Ie, it depends if the increment comes before/after the value is returned"
    I think this explanation would be perfect if one operator just returned the value (in this case the eigenvalue) and the other just "moved the state" without returning any value.
  7. Jun 7, 2004 #6


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  8. Jun 7, 2004 #7
    No Janitor,
    I don't think that thread can help me at this point. I am looking for something simpler and more intuitive. Thanks for trying to help though.
  9. Jun 9, 2004 #8


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    Maybe someone would be so kind as to comment on anticommutators as well. The only place within quantum theory that I have come across them is in connection with the creation and annihilation operators for fermions.

    As Zee puts it in one of his books, for a spin 0 field (spinless boson), if you try to use the anticommutator instead of the (correct) commutator, "We get a nonvanishing piece of junk. A disaster if we quantize the scalar field as anticommuting! A spin 0 field has to be commuting. Thus relativity and quantum physics join hands to force the spin statistics connection... We would also get into trouble if we quantize the Dirac field with commutation instead of anticommutation rules."
  10. Jun 9, 2004 #9
    I think this line is really the key alexepascual. reilly has an excellent example of non-commutation of rotations and then references the position (x) and momentum (d/dx) operators in one dimension (to whithin factors).

    I think that while your desire to really understand why some observables are non-compatible is justified, some very bizarre things happen in the quantum prescription. Things like position and momentum are transformed to operators which can act on a wavefunction.

    Now for two operators/properties to be simultaneously observable, you need their commutator to be 0 (that's the math). But what does this mean? Since the commutator [a,b] = ab-ab, it means that acting on your wavefunction in either ordering yields the same result.

    So for the position/momentum example, we should think about what it means to make a measurement of 1 of those two properties. We know from Uncertainty principle that the product of the 2 uncertainties must be greater than h-bar. I think that this means a measurement of position of the object affects the wavefunction.

    Ultimately, the ordering is important, and the intrinsic relationship between the QM position and momentum operators is the cause of their incompatiblility.

    Although I have strayed from my original point of clarification, I think that reilly pointing out the x and d/dx is really the key to understanding how these QM properties are related.
  11. Jun 9, 2004 #10


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    Just to complement what bkfizz02 is saying a little bit, I have read that a measurement leaves a system in a state, and (ideally, at least!- don't know about in the real world) any subsequent measurements of the system will not perturb the state if they are from the set of ones compatible with the first one. Of course, there might be objections to this from the camp that says that what I just wrote is B.S., because QM only applies to ensembles of systems, not to a single system.
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