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Total Momentum Operator for Free Scalar Field

  1. Oct 18, 2009 #1

    nicksauce

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    1. The problem statement, all variables and given/known data
    I want to show that
    [tex]\mathbf{P} = -\int d^{3}x}\pi(x)\nabla\phi(x) = \int{\frac{d^{3}p}{(2\pi)^3}\mathbf{p}a_{p}^{\dagger}a_p[/tex]

    for the KG field.


    2. Relevant equations
    [tex]\phi(x) = \int{\frac{d^{3}p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p + a_{-p}^{\dagger})e^{ipx}[/tex]

    [tex]\pi(x) = -i\int{\frac{d^{3}p}{(2\pi)^3}\sqrt{\frac{\omega_k}{2}}(a_p - a_{-p}^{\dagger})e^{ipx}[/tex]


    3. The attempt at a solution
    I'm having trouble seeing why this is true. What happens to the [itex]a_pa_k[/tex] and [itex]a_p^{\dagger}a_k^{\dagger}[/itex]-like cross terms?
     
  2. jcsd
  3. Oct 20, 2009 #2

    gabbagabbahey

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    I'd write [itex]\pi(x)[/itex] and [itex]\phi(x)[/itex] as

    [tex]\pi(\textbf{x}) = -i\int{\frac{d^{3}\textbf{p}'}{(2\pi)^3}\sqrt{\frac{\omega_{p'}} {2}}(a_{\textbf{p}'} - a_{-\textbf{p}'}^{\dagger})e^{i\textbf{p}'\cdot\textbf{x}}[/tex]

    [tex]\phi(\textbf{x}) = \int{\frac{d^{3}\textbf{p}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\textbf{p}}}}(a_{\textbf{p}} + a_{-\textbf{p}}^{\dagger})e^{i\textbf{p}\cdot\textbf{x}}[/tex]

    Then just calculate [itex]\mathbf{\nabla}\phi[/itex], substitute everything in and do the integration over [itex]\textbf{x}[/itex] first. You should get something with a delta function like [itex]\delta^3(\textbf{p}+\textbf{p}')[/itex], which allows you to get rid of all the [tex]a_{\textbf{p}}a^{\dagger}_{-\textbf{p}'}[/itex]-like cross terms just by integrating over [itex]\textbf{p}'[/itex].
     
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