1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Deriving Klein-Gordon from Heisenberg
Loading...