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How to understand momentum is the generator of translation ?

  1. Mar 16, 2014 #1
    How to understand "momentum is the generator of translation" ?

    "As it is known from classical mechanics, the momentum is the generator of translation", it is from WIKI, (http://en.wikipedia.org/wiki/Momentum_operator)
    I don't understand its meaning.
    [itex] T(ε) [/itex] is translation operator, then [itex] T(ε)f(x)=f(x-ε)=f(x)-ε\frac{\partial f}{\partial x}=(1-ε \frac{\partial }{\partial x}) f(x) [/itex] ,
    so [itex] T(ε)=1-ε \frac{\partial }{\partial x} = 1 + i ε (i \frac{\partial }{\partial x}) [/itex],
    we can say [itex] i \frac{\partial }{\partial x} [/itex] is the generator of translation, where is momentum ??
     
  2. jcsd
  3. Mar 17, 2014 #2

    maajdl

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    This is an interesting question.
    I think you will find the answer by reading about canonical transformations in classical mechanics.
    Of course you will have to think about the correspondance CM-QM, and thats' why it will be interesting.

    http://en.wikipedia.org/wiki/Canonical_transformation
     
    Last edited: Mar 17, 2014
  4. Mar 17, 2014 #3

    Fredrik

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    You're asking about a generator in classical mechanics, but your calculation involves a translation operator acting on a wavefunction. If CM has taught us to think of momentum as a generator of translations, then your result looks like a good reason to guess that momentum can be represented by -id/dx in QM. (Or +id/dx...I'm a bit puzzled by the sign, but I don't have time to think it through right now).

    If we want to argue that momentum is the generator of translations in CM, then we must do a calculation in CM, no wavefunctions involved. I don't remember the reason why momentum is said to be a generator of translations in CB (not sure I ever knew it), so I checked the mechanics books I have at home. They don't explain this, so I decided to try to figure it out. I still don't know what the answer is, but this looks like an answer:

    If
    $$S[q]=\int_0^t L(q(t),\dot q(t)) \mathrm dt,$$ where L is a Lagrangian and S is an action functional, then ##S[q+\varepsilon]=S[q]+\varepsilon p(t)## to first order in ε. This is perhaps a reason to call p a "generator of translations".

    But maybe this isn't it at all. I have a feeling that we should be looking at Noether's theorem. Maybe there's a reason to call the conserved quantities "generators". If someone has a better answer, I would like to hear it.
     
  5. Mar 17, 2014 #4
    Go to Chapter 2.8 in Shankar's QM text.

    Edit: Section attached

    He then goes on to apply this theorem to translations, but for some reason I can't clip that part without my computer freezing up.
     

    Attached Files:

    Last edited: Mar 17, 2014
  6. Mar 17, 2014 #5
    Thanks a lot, maajdl, Fredrik and HomogenousCow.

    This is my argument about the question:
    The word "generator" appears in "Group Theory" and "CM", this is the reason why we confused.
    In Group Theory, ∇ (or [itex] \pm i ∇ [/itex]) is the generator of translation, the manner we obtain it is "translating object [itex] f(x) [/itex]". Generally, we call this “active transformation”.
    In CM, canonical transformations is "coordinate transformation", that is "passive transformation". In passive infinitesimal canonical transformations,
    [itex] q'=\hat{T}(ε)(q,p)=q+ε\frac{\partial g}{\partial p} [/itex]
    [itex] p'=\hat{T}(ε)(q,p)=p-ε \frac{\partial g}{\partial q} [/itex]​
    where [itex] \hat{T}(ε) [/itex] is "transforming operator", which act on (q,p) simultaneously. we call [itex]g=g(q,p)[/itex] generator of transformation. when [itex] g(q,p)=p[/itex], corresponding transformation is spatial translation.

    Summary:
    ∇ is the generator of active transformation, g(q,p) is the generator of passive transformation.

    What do you think of my argument?

    PS:I'm from china, so I'm very grateful for someone pointing out my mistake in Physics and language. :)
    Thanks to maajdl, Fredrik and HomogenousCow again.
     
    Last edited: Mar 17, 2014
  7. Mar 17, 2014 #6

    strangerep

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    Yes, it's all about Noether's theorem (or, deeper, Lie transformation theory). For every (continuous) symmetry of the Lagrangian there's a conserved quantity which generates the symmetry.

    E.g., for a variation of a canonical coordinate ##q_k \to q_k + \delta q_k##, the conserved quantity turns out to be $$ \frac{\partial L}{\partial \dot q_k}$$which is the canonical momentum.

    Note the adjective "canonical". The correspondence with familiar everyday quantities gets trickier when dealing with non-free or relativistic situations.

    BTW, this can also be regarded as a form of the general principle that for every continuous symmetry of a (set of) dynamical equation(s) of motion (i.e., a symmetry which takes solutions to solutions), there's necessarily a generator. This is the essence of Lie transformation theory. :wink:
     
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