Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Classical Hamiltonian

  1. Dec 23, 2008 #1


    User Avatar

    If the classical Hamiltonian is define as

    [tex]H = f(q, p)[/tex]

    p, q is generalized coordinates and they are time-dependent. But H does not explicitly depend on time. Can I conclude that the energy is conserved (even q, p are time-dependent implicitly)? Namely, if no matter if p, q are time-dependent or not, if H does not contains [tex]t[/tex] explicitly, I find that the Poisson bracket

    [tex]\left\{H, H\right\} \equiv 0[/tex]

    so the energy is conserved, right?

    But what about if H explicitly depend on time? According to the definition of Poisson bracket, [tex]\left\{H, H\right\} \neq 0[/tex] ????
    Last edited: Dec 23, 2008
  2. jcsd
  3. Dec 23, 2008 #2


    User Avatar
    Science Advisor
    Gold Member

    Short answer is YES!

    Long answer. Once the hamiltonian is defined the the time evolution of any quantity is given by:

    [tex] \frac{d}{dt}F = [[H,F]] + \frac{\partial}{\partial t}F[/tex]
    where [[ ]] is the Poisson bracket and the partial derivative applies to explicit time dependence. Since:
    [tex] [[H,H]] = 0[/tex]
    and there is no explicit time dependence you get:
    [tex]\frac{d H}{dt} = 0[/tex]

    Expand this in terms of p and q time dependence and you'll get a relationship which becomes the Jacobi identity when you introduce Hamilton's equations.
  4. Dec 23, 2008 #3


    User Avatar

    Thank you so much. I look up the evolution relation in some textbook, but it reads

    [tex] \frac{d}{dt}F = [[F,H]] + \frac{\partial}{\partial t}F[/tex]

    I wonder if I should exchange F, H in poisson bracket?
    Last edited: Dec 23, 2008
  5. Dec 23, 2008 #4


    User Avatar
    Science Advisor
    Gold Member

    No actually [tex] [[F,H]] = - [[H,F]][/tex]

    But there is a convention choice which is not quite uniform in the sign of the Poisson bracket. Its a matter of whether you define:

    [tex] [[A,B]] = \frac{\partial A}{\partial p}\frac{\partial B}{\partial q}-\frac{\partial B}{\partial p}\frac{\partial A}{\partial q}[/tex]
    [tex] [[A,B]] = \frac{\partial A}{\partial q}\frac{\partial B}{\partial p}-\frac{\partial B}{\partial q}\frac{\partial A}{\partial p}[/tex]
    (note p and q are reversed.)

    To conform with your reference reverse my Poisson brackets. (I recommend you stick to convention and do this.)

    I prefer to (buck convention and) reverse the sign/order so that for canonical conjugate variables U and V:

    [tex] \frac{d}{dV} F = [[U,F]][/tex]

    This fits better with the conventions for generators in Lie algebras.
  6. Dec 23, 2008 #5


    User Avatar

    Got u. Thanks a lot. X'mas
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook