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How to define the Hamiltonian phase space for system?

  1. Oct 12, 2012 #1
    Title says it all, confused as to how I'm supposed to define the phase space of a system, in my lecture notes I have the phase space as {(q, p) ϵ ℝ2} for a 1 dimensional free particle but then for a harmonic oscillator its defined as {(q, p)}, why is the free particles phase space all squared real numbers?
     
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  3. Oct 13, 2012 #2

    vanhees71

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    Also for the harmonic oscillator the phase-space variables can take any real value. Why shouldn't it?
     
  4. Oct 14, 2012 #3
    Ok but why does the phase of the free particle take any squared real value?
     
  5. Oct 14, 2012 #4

    vanhees71

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    I do not know what you mean by this question.

    The phase space in Hamiltonian mechanics is spanned by the generalized coordinates and their canonical momenta. Let's take the harmonic oscillator in one dimension as an example.

    Usually you start with the Lagrangian:
    [tex]L=\frac{m}{2} \dot{q}^2-\frac{m \omega^2}{2} q^2.[/tex]
    The canonical momentum is
    [tex]p=\frac{\partial L}{\partial \dot{q}}=m \dot{q}.[/tex]
    The Hamiltonian is then given by
    [tex]H(q,p)=p \dot{q}-L=\frac{p^2}{2m}+\frac{m \omega^2}{2} q^2.[/tex]
    The possible values for [itex]q[/itex] and [itex]p[/itex] are all the real numbers for each of these variables since there is no singularity in the Hamiltonian for any such values. Thus the phase space for the 1D harmonic oscillator is [itex](q,p) \in \mathbb{R}^2[/itex].
     
  6. Oct 15, 2012 #5
    The [itex]\mathbb{R}^2[/itex] doesn't mean squared real numbers. It means the set of order pairs of real numbers (x,y), where x and y are real numbers. So [itex](q,p) \in \mathbb{R}^2[/itex] just means points in phase space are composed of pairs of real numbers q and p.
     
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