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-ih/2pi d/dx =P operator

  1. Jan 25, 2010 #1
    friends, this may appear a silly question to experts, but still......
    why/ on what physical reasons am i associationizing -ih/2 pi d/dx with the momentum operator while working in X basis?
    i mean, ok. in x basis the momentum operator takes the structurtum of the first derivative of the dirac delta but why scientists associated this mathematical thing with something as as momentum operator?
     
  2. jcsd
  3. Jan 25, 2010 #2

    SpectraCat

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    I am not sure anyone knows the answer to this ... it is commonly taught as one of the postulates of Q.M. I have run into other treatments in which it is derived from a different set of postulates, but as I understand it that turns it into a chicken-egg problem where one ends up questioning the other postulates.
     
  4. Jan 25, 2010 #3
    I haven't studied it myself but many books like Ballentine apparently derive the commutation relation of x and p from the symmetry under space translation (and other dynamical variables from other symmetries) and a bit of dimensional analysis to get the hbar. In short, it is the momentum operator because it associates naturally with space translational symmetry. Perhaps someone here can give a quick proof/explanation.

    Edit: SpectraCat, is this kind of explanation the chicken and egg problem you mentioned?
     
  5. Jan 25, 2010 #4
    this should be found in any page on google or textbook, that the momentum operator has the form it has in x-space due to the fact that momentum is the generator of translations (Truecrimson)

    apart from the "i" this is the same as in classical mechanics....

    you might want to learn classical mechanics before doing QM ;)
     
  6. Jan 25, 2010 #5

    SpectraCat

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    I am not sure ... like you I haven't studied this myself. Anyway, what I was trying to say is that, however you formulate it, Q.M. requires that you take some things as postulates ... the particular choices depend on the interpretation. So, one always ends up questioning a postulate, and thus reaches a dead end. That is what I meant by the chicken-egg problem. In the case you are talking about, I think it is the translational symmetry of space that is the postulate.

    @ Truecrimson: actually, the wikipedia entry for "momentum operator" gives a short derivation that might interest you.
     
  7. Jan 25, 2010 #6

    Mentz114

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    For a wave function of this form

    [tex]
    \psi(x,t) = A\exp\left[ \frac{i}{h}(Et+px)\right]
    [/tex]
    [tex]
    -ih\frac{d}{dx}\psi(x,t)=p\psi(x,t)
    [/tex]
    Note that -ihd/dt is the energy operator by the same logic.
     
  8. Jan 25, 2010 #7
    Introduction to Quantum Mechanics, David Jeffrey Griffiths, page 14. While it might not be the best description, it suffices to answer if you are new to QM.
    Take the mean value of X, and differentiate it in order to time. Multiply by the mass of the particle. Now look at what you get inside the integral, that's the momentum.
     
  9. Jan 25, 2010 #8
    yes but HOW do you arrive at that WF without using that p is the generator of translations?....
     
  10. Jan 25, 2010 #9
    translation symmetry of space is done in classical mechanics as well...
     
  11. Jan 25, 2010 #10
    Yeah, the use of symmetry argument looks fine to me i.e. no chicken-egg problem. It's the same as when we define the relativistic momentum so that it is conveniently a conserved quantity.
     
  12. Jan 25, 2010 #11

    SpectraCat

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    Of course you are correct, so that cannot be a special postulate for Q.M. Instead, I'll bet it is the factor of i[tex]\hbar[/tex] that arises from a fundamental postulate. (I'll dig out my Griffiths and read it when I get home.)
     
  13. Jan 25, 2010 #12
    You won't find any derivation for the quantum momentum as a consequence of translation symmetry of space in the Griffiths Introduction to Quantum Mechanics I believe.
     
  14. Jan 25, 2010 #13
    I find the answers very unsatisfying so far. It is true that if you accept that the exponential waves are momentum eigenstates, then in the position basis you can justify differentiation as being the momentum operator. That comes from properties of the fourier transform: multiplication by p in the momentum domain becomes differentiation by x in the x domain.

    This falls short of being a satisfying physical explanation.
     
  15. Jan 25, 2010 #14
    You can flesh it out a bit further, since the properties of the fourier transform in turn come from the symmetry of space, although it still falls a bit short I think.

    If x is a label for the elements of an abelian group (the real line under addition, or in general a position in homogenoeous euclidean space), then functions of the form [tex]\exp(i p x)[/tex] are irreducible unitary representations of the translation group. The representations are labelled by 'p', which we call momentum. Fourier transformations are really a decomposition of a function into contributions from different representations of the translational symmetry group.

    If a hamiltonian has a symmetry, then its eigenstates must transform under a representation of that symmetry, so eigenstates of translationally invariant hamiltonians have eigenstates which are eigenstates of momentum - and must be plane waves of the form [tex]\exp(i p x)[/tex].
     
  16. Jan 25, 2010 #15

    Mentz114

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    No one has referred to Dirac's original derivation on pages 100-103 of 'Principles...'.
    he starts by wanting the superposition principle to be invariant under 'displacement' and ends up showing that d/dx has the same commutation relations as the momentum.

    If I get time I'll transcribe the highlights.
     
  17. Jan 25, 2010 #16
    Hi. Previous authors has already answered well. I am afraid there is nothing new in my reply.

    Physically a memontum eigenstate is a plane wave in coordinate space and a coordinate eigenstate is a plane wave in momentum space.
    In short, <x'|p">=exp(i 2pi x'p"/h)/sqrt of h.

    p= -i hbar d/dx holds the commutation relation required for coorcdinate and momentum, xp-px = i hbar.
    Momentum can be p= -i hbar d/dx + f(x). f(x) vanishes by properly choosing the phase factor. The plane waves above are derived from this mathematical relation.
    Regards.
     
  18. Jan 25, 2010 #17
    In addition to the already mentioned method used in textbooks that derives the momentum operator as a translation generator, I'd like to say that if the form of the momentum operator were different it would probably make predictions that conflict with the correspondence principle.
     
  19. Jan 25, 2010 #18
    thank you, friends. i have more/less understood it. but i think i still have a long way to go-about 50% of the posts basically went over my head. i will try my best to understand all of them- if i get stuck ,it is my earnest request to you all to show me the right way.

    thanks friends, you really are masters of quantum mechanics.:redface:
     
  20. Jan 25, 2010 #19
    but that was really surprising!!! i mean, X will always commute with f(x). that means momentum operator can be anything, not a definitum -ih/2 pi d/dx
     
  21. Jan 25, 2010 #20
    thanks, nunamjar. i will ask my professor what these generators are in my next class to understand the other posts. in classical mech, we have just completed lagrangian formulation.
     
  22. Jan 26, 2010 #21
    Hi, Mr confusion.

    Dirac IV 22 helps you to understand that f(x) affects only on phase factor of states thus causes no effects on physics.
    Regards.
     
  23. Jan 26, 2010 #22

    dextercioby

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    I'm quite surprised that nobody mentioned the Stone-von Neumann theorem in connection to a certain representation of the momentum operator in a space of functions. This is the answer to everything.
     
  24. Jan 26, 2010 #23

    SpectraCat

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    No no .. I know that one ... it's 42.
     
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