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Consider a system of $N$ identical harmonic oscillators in 1d. The Hamiltonian will be given by $$\mathcal{H}_N=\sum_{i=1}^N \frac{p_i^2}{2m}+\frac{m\omega^2}{2}q_i^2$$

Now supposedly the Hamiltonian incompasses all the properties of the system it describes. Therefore I would have thought we could get the phase space density from it.

I think the the phase space is $$\mathcal{H}_N=\sum_{i=1}^N \frac{p_i^2}{2m}+\frac{m\omega^2}{2}q_i^2 \leq E,$$ which is some kind of $n$-dimensional ellipse (does this have a name?)

Now do we know that the phase space density will be uniformly distributed?

I can see no explicit time dependence in the Hamiltonian given either.

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  • $\begingroup$ Are the oscillators interacting or not? $\endgroup$ – DanielSank May 5 '15 at 13:13
  • $\begingroup$ Not interacting $\endgroup$ – Permian May 5 '15 at 13:14
  • $\begingroup$ Ok, I guess that's clear from the form of the Hamiltonian. In this case, each oscillator is contrained to an ellipse in it's part of the phase space, so the volume occupied by the complete system is zero (I think). $\endgroup$ – DanielSank May 5 '15 at 13:15
  • $\begingroup$ @DanielSank How do you get 0? $\endgroup$ – Permian May 6 '15 at 10:59
  • $\begingroup$ The area of a curve in 2 dimensional space is zero. $\endgroup$ – DanielSank May 6 '15 at 19:43

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