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What is the hamiltonian in Klein-Gordon equation?

  1. Sep 8, 2010 #1
    since the time derivative is second order, the KG equation can not be put in the form

    i \dot{\psi}= H \psi

    so there is no H in KG equation?

    and no heisenberg picture for KG equation?
     
  2. jcsd
  3. Sep 8, 2010 #2
    But ofcourse there is a Hamiltonian.

    You can always start off with a Lagrangian which produces the equations of motion. From this Lagrangian you can define a "coordinate" q + its "conjugate momentum" p and use these to switch to a Hamiltonian approach.
    Specifically, tThe Hamiltonian is the usual definition of

    [tex]H = p(\partial_t q) - L[/tex]

    Now, in field theory the "coordinates" and "momentum" are replaced by fields. In this case we have the scalar KG field [tex]\phi(x,t)[/tex] and its conjugate field [tex]\pi(x,t)[/tex], which happens to be the time derivative of the field, [tex]\pi(x,t) = \partial_t \phi(x,t)[/tex]. The Hamiltonian density at a given timeslice [tex]t[/tex] is then:

    [tex]\mathcal{H} = \pi(x,t)\partial_t\phi(x,t) - \mathcal{L}[/tex]

    You can look up the Lagrangian for the KG theory and come up with the definite form of H. The full Hamiltonian is given by integrating over all of space for a given timeslice t,

    [tex]H = \int d^3x \mathcal{H}[/tex]
     
  4. Sep 8, 2010 #3
    As for the Heisenberg picture... it's still there! You can switch to a Heisenberg picture through a unitary transformation. In this picture the operators evolve through time as:

    [tex]A_H(t) = e^{iHt} A_H(0) e^{-iHt}[/tex]

    The operator [tex]e^{-iHt)[/tex] is the time-evolution operator. In the Schroedinger picture the states transform in the usual way:

    [tex]|\psi_S(t)\rangle = e^{-iHt}|\psi_S(t)\rangle[/tex]

    And here is something you do not hear every day (but is certainly still true!). The equation of motion of the time evolution operator is nothing but Schroedingers equation!

    [tex]\frac{d}{dt} e^{-iHt} = -i H e^{-iHt}[/tex]

    Yes, the Schroedinger equation still plays a role in relativistic quantum field theory!

    In fact, for fields operators we can apply the time evolution for operators. Take for instance the conjugate field [tex]\pi(x,t)[/tex]

    [tex]\pi(x,t) = e^{iHt} \pi(x,0) e^{-iHt}[/tex]

    which -- when taking the time derivative -- reduces to Heisenbergs equations of motion:
    [tex]\partial_t \phi(x,t) = i[H,\pi(x,t)][/tex]

    But for a free scalar field theory, i.e. KG theory, this commutator can be computed explicitly. Namely, the Hamiltonian is given by:

    [tex]H = \frac{1}{2} \int d^3 x' \left(\pi(x,0)^2 + (\nabla\phi(x,t))^2 +m^2\phi(x,t)^2\right)[/tex]

    Now, the equal-time commutation relations should be familiar
    [tex] [\phi(x,t),\pi(x',t)] = i \delta(x-x')[/tex]

    with on the right hand side the three dimensional spatial delta function. All other commutators are zero. The commutator we are after is then...

    [tex] [H,\pi(x,t)] = i\int d^3x' \left(\nabla^2 \phi(x',t) + m^2\phi(x',t)\right)\delta(x-x') = \nabla^2 \phi(x',t) + m^2\phi(x',t)[/tex]

    You need to apply some partial integration to obtain this, but it's quite straightforward. Anyways, now, this commutator determines the time-evolution of the conjugate field. So we have:

    [tex]\partial_t \pi(x,t) = \nabla^2 \phi(x,t) - m^2\phi(x,t)[/tex]

    This is simply the Heisenberg equation of motion for the conjugate field. But remember, the conjugate field [tex]\pi(x,t)[/tex] is nothing but the time derivative of the original KG field [tex]\phi(x,t)[/tex]. And so we end up with...

    [tex]\partial_t^2 \phi(x,t) - \nabla^2 \phi(x,t) + m^2\phi(x,t) = 0[/tex]

    Ha, that should look familiar!
     
  5. Sep 8, 2010 #4
    Another way of looking at it:

    K-G equation is of 2-order. Its Hilbert space of (positive energy) solutions can be parametrized by the Cauchy data - the function itself and its time derivative at, say, [tex]t=0.[/tex] On this Hilbert space you have a natural unitary representation of the Poincare group. The generator of time translations is the Hamiltonian.

    Formally:

    [tex]H=(-\nabla^2+m^2)^\frac12[/tex]
     
    Last edited: Sep 8, 2010
  6. Sep 9, 2010 #5

    Demystifier

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    The Klein-Gordon equation can be viewed as a Hamiltonian constraint
    [tex]H \psi(x) =0[/tex]
    where
    [tex]H = -p^{\mu}p_{\mu}+m^2[/tex]
    This Hamiltonian is a scalar (not a time component of a vector), it is not equal to energy, and it generates evolution with respect to the proper time. For more details see, e.g.,
    http://xxx.lanl.gov/abs/1006.1986
     
    Last edited: Sep 9, 2010
  7. Sep 9, 2010 #6
    Some people call it super-Hamiltonian in order to distinguish it from the Hamiltonian?
     
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