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Hamiltonian and Commuting operators

  1. Dec 27, 2011 #1
    A general question..
    In analytical mechanics, we take a given hamiltonian and re-write it in term of generalzed coordinates. In a way- we recode the hamiltonian to concern only the "essence" of the problem.
    However, it seems to me, that in QM we do the opposite- we look for operators that commute with the hamiltonian and recode the hamiltonian in terms of these operators.
    But.. isn't using conserved quantities the exact opposite of using with generalzed coordinates (idea-wise)?
    And more specifically - Given any hamiltonian , I can simply choose an operator that has nothing to do with the problem (say.. choose the spin operator in a spinless problem). ofcourse this operator will commute with the hamiltonian, but it won't help in solving the problem (ie finding the energy values). This I dont understand- why do some commuting operators help in solving the problem, and some don't? I mean, formally, what is the differece btwn those who do help us, and those who don't?
  2. jcsd
  3. Dec 27, 2011 #2


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    Actually, it's not really different in QM. In the Hamilton-Jacobi method of classical mechanics, one attempts to find "enough" conserved quantities so that the full dynamical solution becomes easy. But the passage from canonical coordinates [itex](p,q)[/itex] to [itex](J,q_0)[/itex], where J is a conserved quantity, and [itex]q_0[/itex] is an initial condition, is simply a canonical transformation. That's just what generalized coordinates are, i.e., new coordinates obtained from the old ones via a canonical transformation. A specific case of this is the so-called Action-Angle variables in periodic motion.

    To solve the full dynamics, it often helps to substitute the expressions for the conserved quantities back into the equations of motion -- because the conserved quantities have zero time-derivative.

    More precisely, we search for a set of conserved quantities which are "in involution" with each other. (meaning that all the Poisson brackets between those quantities vanish). Typically, energy is one of the conserved quantities, so the Hamiltonian is one of the quantities in the set.

    Passing to QM, the phrase "mutually in involution" in CM becomes "mutually commuting" in QM.

    In the spinless case, the corresponding conserved quantity (angular momentum) is trivial (i.e., 0). In the language of classical mechanics, this means your system is "degenerate", in that some of the conserved quantities are trivial and the corresponding degrees of freedom can be factored out of the dynamics. We do something similar when solving the Kepler problem -- motion occurs in a plane, so rotation outside that plane is trivial and we can factor out those degenerate degrees of freedom to get a simpler set of equations to solve.

    Hopefully this is now answered by the distinction between degenerate and non-degenerate degrees of freedom in the problem?

    [For further reading, try Goldstein, or Jose & Saletan.]
    Last edited: Dec 27, 2011
  4. Dec 29, 2011 #3
    Thank you! Very helpful
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