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Normal ordering for total momentum

  1. May 13, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm reading David Tong's QFT lecture notes. And he explains how to drive the free theory total momentum by quantum field operators from the classical field theory.
    But I'm confusing on the normal ordering process a little bit.

    2. Relevant equations

    In his notation, the free field [tex]\phi(\vec{x})[/tex] and its momentum conjugate [tex]\pi(\vec{x})[/tex] are

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

    And he calculates the normal ordered total momentum from the classical one (the conserved momentum from the energy momentum tensor)
    and says the normal ordered total momentum is :
    [tex]\vec{P}=\int \frac{d^3p}{(2\pi)^3}\vec{p}{a^\dag}_{\vec{p}}a_{\vec{p}}.[/tex]

    3. The attempt at a solution

    But the actual calculation goes like this:
    [tex]\vec{P}=-\int d^3x \pi \nabla\phi[/tex]
    [tex]=\frac{1}{2}\int \frac{d^3p}{(2\pi)^3}\vec{p}(a_{\vec{p}}a_{-\vec{p}}+{a^\dag}_{\vec{p}}a_{\vec{p}}+a_{\vec{p}}{a^\dag}_{\vec{p}}+{a^\dag}_{\vec{p}}{a^\dag}_{-\vec{p}})[/tex]
    [tex]=\frac{1}{2}\int \frac{d^3p}{(2\pi)^3}\vec{p}(a_{\vec{p}}a_{-\vec{p}}+2{a^\dag}_{\vec{p}}a_{\vec{p}}+(2\pi)^3\delta(\vec{0})+{a^\dag}_{\vec{p}}{a^\dag}_{-\vec{p}})[/tex]

    So (my question is) what about the other terms? :
    [tex]\frac{1}{2}\int \frac{d^3p}{(2\pi)^3}\vec{p}(a_{\vec{p}}a_{-\vec{p}}+{a^\dag}_{\vec{p}}{a^\dag}_{-\vec{p}})[/tex]

    Thanks in advance.
    Last edited: May 13, 2010
  2. jcsd
  3. May 14, 2010 #2


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    Homework Helper
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    HInt: What happens when you integrate an odd function over the reals?:wink:
  4. May 16, 2010 #3
    I got it. Thank you very much.
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