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Hamiltonian For The Simple Harmonic Oscillator

  1. Feb 10, 2014 #1
    I am reading an article on the "energy surface" of a Hamiltonian. For a simple harmonic oscillator, I am assuming this "energy surface" has one (1) degree of freedom. For this case, the article states that the "dimensionality of phase space" = 2N = 2 and "dimensionality of the energy surface" = 2N-1 = 1.

    Adjusting x gives H = ω (p + x). That's H = ω(p*p + x*x).

    When I plot the energy surface for H, I get a 3-dimensional paraboloid plotted against the x-p plane. Or, lines of constant H are 2-dimensional circles in the p-x plane. What energy surface would be 2N-1 = one (1)-dimensional here? Am not trying to quibble about terminology here. Just want to know if I am missing something.

    Thanks for reading.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 10, 2014 #2
    Hi.
    I assume N refers to the number of spatial dimensions, so in the case of a one-dimensional oscillator your phase space is indeed two-dimensional while the energy "surface" is a line (i.e. one-dimensional).
    In general, you understand how phase space volume would be 2N-dimensional; now the harmonic oscillator equation always determines a hypersphere (E = x^2 + y^2 +...), so the dimensionality of the hyper-surface is naturally one dimension less than the volume: 2N–1...
     
  4. Feb 10, 2014 #3
    I get it now. Thanks so much for your fast reply.
    Ted
     
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