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Potential Invariant under translation

  1. Dec 5, 2011 #1
    When I was learning translational symmetry I saw that for translation invariance, i.e
    the momentum P needs to be conserved
    This momentum is actually the generator of small translations defined as
    Now, I was solving some problems and I met one which is interesting. I am given a potential
    V(x)=Asin(2πx/ε) (where A is a constant)​
    That potential is actually invariant under the translation defined above. In that problem they say that consequently the "momentum is not conserved". Can anybody tell me why?
    I tried to understand it by saying that, if I can show that the Hamiltonian is not conserved hence the momentum is not. But it looks that I don't have enough information about H.
  2. jcsd
  3. Dec 5, 2011 #2


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    In classical and quantum mechanics, we typically write the Hamiltonian as the sum of kinetic and potential energy terms, denoted like this:
    H ~=~ K(p) + V(x)
    E.g., in nonrelativistic mechanics, [itex]K(p) = p^2/2m[/itex].

    If a naive quantization is possible by simply promoting the classical momentum p and position x to operators (P and X) satisfying the usual canonical commutation rules, then it's easy to see that:
    [H,P] ~=~ [V(X), P]
    which is nonzero in general because X and P don't commute.

    (A similar thing happens in the classical case of course, but we must use a Poisson bracket instead of a commutator.)
  4. Dec 6, 2011 #3


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    Your translation

    is one specific translation for one specific parameter ε.

    There are two definitions of translation symmetry:

    1) continuous, i.e. the system is invariant w.r.t.


    for all ε (the fact that ε is small is only required to identify the momentum as the generator of translations; in general nothing prevents you from a translation with ε' ≠ ε which is small, too; or from a translation where ε is large)

    2) discrete (in a lattice or a crystal), i.e. the system is invariant w.r.t.


    where xL is a lattice vector and n is an integer.

    In both cases a) and b) it's not the case that you first fix ε (or n) and then fix the potentials for that specific parameter, but that you have a potential and study translations for arbitrary parameter ε (or n).
  5. Dec 6, 2011 #4
    Thank you very much. I calculated
    in x-basis and I found a cosine function. Hence the momentum is not always conserved and hence is conserved at certain values of x.
  6. Dec 6, 2011 #5


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    This conclusion is wrong!

    Think about a ball moving in a potential V(x) ~ sin(x); of course its momentum is not conserved (you could say that at certain points it is conserved for an infinitesimal short time, but of course in QM the particle will never be exactly located at these points)

    For momentum conservation

    [P, H] = 0

    is required (as an operator identity). Anything else but zero on the r.h.s. means that momentum is not conserved. Please have a look at classical Poisson brackets and Heisenberg's equation of motion

    Last edited: Dec 6, 2011
  7. Dec 6, 2011 #6
    Ok, thanks a lot, now this makes sense
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