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Deriving Klein-Gordon from Heisenberg

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Sort of stuck deriving the Klein-Gordon equation from Heisenberg equation of motion

    [tex] \dot{\varphi} = i [H, \varphi ] [/tex]

    2. Relevant equations

    [tex] \dot{\varphi} = \frac{\partial\varphi}{\partial t} [/tex]

    [tex] H = \int d^3x \mathcal{H} [/tex]

    [tex] \Pi (x) = \dot{\varphi}(x) [/tex]

    [tex] \mathcal{H} = \Pi \dot{\varphi} - \mathcal{L} [/tex]

    3. The attempt at a solution

    [tex] \dot{\varphi} = \int d^3x [ \dot{\varphi}^2 - \mathcal{L}, \varphi ] [/tex]

    if I expand that, then it becomes a real mess, not sure if I'm on the right track?
    Last edited: Apr 7, 2009
  2. jcsd
  3. May 24, 2011 #2
    first, from the condition i*d(phi)/dt=[H,phi], you can arrive an equation, which makes [intgral(delta_phi)^2,phi]=0, then you differentiate the Heisenberg, equation, get d^2(phi)/dt^2=[H,[H,phi]], and substitute back. Then you get K-G equation immediately
  4. May 24, 2011 #3
    \dot{\varphi} = i [H, \varphi ], we obtain

    \int d^3x [ \delta{\varphi}^2, \varphi ]=0 Eq(1)

    differentiate Heisenberg Eq. again, we have

    \dot\dot{\varphi} = i [H, \dot{\varphi} ]=- [H, [H, \dot{\varphi} ] ]

    substitute \Pi (x) = \dot{\varphi}(x) back,

    using the specific form of H and use Eq.1, you'll get the answer of KG equation.
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