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Momentum as the generator of translation

  1. Feb 14, 2008 #1
    Hi all,

    I'm trying to convince myself that momentum is the correct infinitesimal generator for translations.

    In his book, Sakurai justifies this assertion by making an analogy between the translation operator acting on the ket-space [itex]\mathcal{T}(d\mathbf{x}) : \mathcal{H} \to \mathcal{H} ; | \mathbf{x} \rangle \mapsto | \mathbf{x} + d\mathbf{x} \rangle[/itex] and the type-2 generating function which gives the canonical translation of coordinates.

    Is there any stronger mathematical justification other than ``they look similar'', or is this one of those fundamental postulates that can only be proven by experiment?

    Thanks.
     
  2. jcsd
  3. Feb 14, 2008 #2
    uhhmmm... let me think???

    i think another good way too see what u are askig is:

    Look what galileo/Poincarè Group is Locally:
    the "generator" of the group is P!

    This for just pure space-time translation....

    I think that its just an algebraic propertie that reflect the common life experience;

    regards.
    matco.
     
  4. Feb 15, 2008 #3

    jambaugh

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    There is an empirical manifestation. Eigen-modes of the momentum operator, i.e. physical systems with observed momentum, [itex]P|k\rangle = k|k\rangle[/itex], will be translation invariant up to a phase factor [tex]e^{ikx}[/tex] which may be confirmed in interference experiments.

    [ I'm using units where [tex]\hbar=1[/tex] ]

    In short the action of the exponentiated momentum operator [tex]e^{iP\xi}[/tex] on eigen-modes will of necessity be to translate them a distance [tex]\xi[/tex].
    (The act of translation being equivalent to multiplying by the phase factor.)
    Since the momentum eigen-modes form a basis this must also be the case for any mode.

    You might view this as a "mathematical" argument and you wouldn't be wrong. But the mathematical postulate I invoke is simply the eigen-value principle and superposition.
    The phase factor and momentum value are empirical quantities.

    The eigen-value principle dictates that indeed momentum generates translation. But it does not do this uniquely. You can add on another (gauge) operator which commutes with position and it too will generate translations.
    Example: [tex] P' = P + \hat{\imath} A(x)[/tex].

    This is the covariant momentum for a gauge theory of e.g. electrodynamics.

    When the generator [tex]\hat{\imath}[/tex] acts differently on quanta with different charges:
    [tex]\hat{\imath}\mid q,x\rangle = iq\mid q,x \rangle[/tex]
    they will translate differently i.e. experience a net gauge force.
     
  5. Feb 15, 2008 #4
    You can find the operator that transforms psi(x) into psi(x+dx) with Taylor's expansion:

    psi(x+dx)=psi(x)+d/dx(psi(x))*dx

    psi(x+dx)= (1+dx*d/dx) (psi(x))= (1+i*dx*2*Pi*px/h)(psi(x))


    (we translated the function by dx using momentum operator px)
     
    Last edited: Feb 15, 2008
  6. Feb 16, 2008 #5

    reilly

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    It's really simple, given the notion of the exponential of an operator, which goes back way before QM.

    EXP(ad/dx) = 1 + ad/dx + ....+(1/n!) (ad/dx)^^n +....

    (There are many ways to prove this expansion.)

    S0, EXP(ad/dx) F(x) = F(x+a), per the standard Taylor series.

    And, d/dx = ip, and EXP(iap) is a unitary xform for translation by a -- in configuration space.

    Regards,
    Reilly Atkinson
     
  7. Feb 16, 2008 #6
    You're both assuming here that we know in advance that the x-space representation of momentum is p = - i d/dx.

    Sakurai actually derives this relation using the fact that momentum generates translation.

    I totally agree that p generates translation [itex]\iff p = - id/dx[/itex].

    But how do you know that either of these are true to begin with?
     
  8. Feb 17, 2008 #7

    reilly

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    Best to look in Dirac's book. That p = -i d/dx is about as basic QM as it gets -- it's an assumption, and can't be proved without other assumptions, more or less equal in content. Also look in Lanczos or Goldstein to see how translations work with contact transformations in classical mechanics.
    Regards,
    Reilly Atkinson

     
  9. Feb 17, 2008 #8

    jambaugh

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    This ultimately comes down to the "proof"/assumption of the Heisenberg relation:
    [tex] [\mathbf{p},\mathbf{x}] = -i\mathbf{1}[/tex]

    I'll repeat that this is again empirically verified up to some degree of precision by interference experiments on quanta in momentum eigen-states. The fact that momentum eigenstates behave as waves with frequency in proportion to their momentum values (as shown in double slit and crystal diffraction experiments) then leads via the linear algebra to the momentum operator being proportional to the coordinate derivatives i.e. translation generators.
     
  10. Feb 28, 2008 #9
    Another way to look at it is to consider under what symmetry momentum is associated with.
    Translational symmetry is essentially an abelian translation group, the irreducible representations of which are then 1-dimensional, unitary, i.e e^(ikx) each representation labelled by k.
    To extract the quantum number (i.e, the label of the representation, the momentum k) we can use -id/dx
     
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