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Position-momentum commutation relation

  1. Aug 7, 2014 #1
    Hi,

    what is the physics experiment that leads to the position-momentum commutation
    relation

    xpx - px x = i hbar

    What does it mean to multiply the position and momentum operators of a particle?
    What is the corresponding physical quantity?
     
  2. jcsd
  3. Aug 7, 2014 #2

    bhobba

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    Gold Member

    Interestingly its not really an experimental thing as far as I know - others more into experimental stuff may correct me on that though.

    It follows quite easily from the position and momentum operator. Believe it or not they follows from symmetry considerations via the principle of relativity so isnt really a separate axiom, but a result of a well accepted principle.

    Multiplying them together is simply applying an operator then another - in and of itself its doesn't mean anything.

    However there is a reason those commutation type relations tend to occur:
    http://bolvan.ph.utexas.edu/~vadim/classes/2013s/brackets.pdf

    Thanks
    Bill
     
  4. Aug 7, 2014 #3

    atyy

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    In addition to the relationship between the quantum commutator and the classical Poisson bracket that bhobba metions, another (related) way to see that these classical sounding names are appropriate for the quantum observables is by Ehrenfest's theorem http://en.wikipedia.org/wiki/Ehrenfest_theorem, where the average values of the variables obey the classical equations of motion.

    One way to understand the meaning of the commutator is to derive the uncertainty principle from it http://www.eng.fsu.edu/~dommelen/quantum/style_a/commute.html. The uncertainty principle says that if one measures position accurately from an ensemble prepared in a particular quantum state, and if one measures momentum accurately from an ensemble prepared in the same quantum state, the product of the standard deviations of the position results and momentum results is greater than zero.
     
    Last edited: Aug 7, 2014
  5. Aug 7, 2014 #4
    I have come to understand it's the order in which you measure them in. If they commute it makes no difference, but if they don't you won't get the same result if you measure p then x or x then p. Generally the standard deviation of this difference is hbar.
     
  6. Aug 7, 2014 #5
    There's not a single experiment. It is based on the general observation that in quantum experiments there exists certain pairs of quantities wherein the order in which you measure them affects the final outcome.
     
  7. Aug 8, 2014 #6

    Jano L.

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    There is no experiment. It was the other way around. Heisenberg played with matrices and came to the conclusion that his infinite matrices ##P## (connected to momentum) and ##X## (connected to position) should obey algebraic relation

    $$
    XP - PX = i\hbar.
    $$

    In those days this was thought to imply that this description does not allow knowledge of both position and momentum at the same time. So Heisenberg attempted to save the new formalism by interpreting this in his thought experiments by saying something as "look, if we measure position, then we disturb the particle so that momentum afterwards is not determined but will vary case to case so we cannot know both position and momentum after the measurement" and similarly for measuring position first and then momentum.

    Nowadays, the above commutation relation is viewed differently among physicists than Heisenberg tried to explain it, but most probably most of them agree that the relation has little to do with disturbance caused by measurements. One prosaic view is that all that is needed to derive the commutation relation is the calculus relation
    $$
    x (\partial_x \psi) -\partial_x(x\psi) = -\psi
    $$
    and the operator for momentum
    $$
    \hat{p}_x =- i\hbar \partial_x
    $$
    introduced by Schroedinger. Schroedinger also showed how the Heisenberg matrices and their commutation relation follow from the Schroedinger equation and these formulae in his papers from 1925-1927.
     
  8. Aug 15, 2014 #7
    Thank you guys, You gave Me food for thought.
     
  9. Aug 16, 2014 #8
    If you consult a book on group theory (for physicists), you might find some very interesting discussions which lead to the canonical commutation relations. It seems the more you learn the more things become "obvious".
     
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