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Klein-gordan Hamiltonian time-independent?

  1. Jul 31, 2012 #1
    How can you tell if the Klein-Gordan Hamiltonian, [tex]H=\int d^3 x \frac{1}{2}(\partial_t \phi \partial_t \phi+\nabla^2\phi+m^2\phi^2) [/tex] is time-independent? Don't you have to plug in the expression for the field to show this? But isn't the only way you know how the field evolves with time is through [tex]\partial_t \phi=i[H,\phi] [/tex], and in order to evaluate this you have to assume the Hamiltonian is independent of time to use the equal-time commutation relations?
  2. jcsd
  3. Aug 1, 2012 #2


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    You should look at a simple problem in classical mechanics.

    A Hamiltonian is (rather sloppy) called "time-independent" if it is not "explicitly time-dependent", i.e. if ∂H/∂t = 0; that means that the time-dependence is hidden in the canonical variables. For the simple harmonic oscillator H ~ p² + x² the canonical variables p,x carry (implicitly) time-dependence via the e.o.m., but ∂H/∂t = 0 and b/c of conservation of energy dH/dt = 0.

    The Klein-Gordon-Hamiltonian is "time-independent" in the same sense, i.e. ∂H/∂t = 0.

    Remark: what you have written down is not really the Hamiltonian, which generates the e.o.m., but only the energy. In a Hamiltonian you have to eliminate the velocities and express everything in terms of generalized positions and momenta, i.e. in terms of φ(x,t) and π(x,t) ~ ∂0φ(x,t). Therefore H = H[φ,π] and the e.o.m. for φ and π are derived via the commutators [H,φ] and [H,π], respectively.

    For a time-independent system the canonical e.o.m. of an operator A are generated solely via commutators [H,A]. Therefore the time-dependence of H i.e. dH/dt is generated via [H,H] which is zero, of course.
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