Hamiltonian For The Simple Harmonic Oscillator

[morangta]

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.

 

[Goddar]

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...

 

[morangta]

I get it now. Thanks so much for your fast reply.
Ted