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Understanding Klein-Gordon

  1. Sep 22, 2007 #1
    I'm just reading the schroeder/peskin introduction to quantum field theory. On Page 21 there is the equation

    [tex]\phi(x)=\int\frac{d^3 p}{(2\pi)^3}\frac{1}{ \sqrt{2\omega_{\vec{p}}} }

    (a_{\vec{p}} e^{i \vec{p} \cdot \vec{x}}

    +a^{+}_{\vec{p}} e^{-i \vec{p} \cdot \vec{x}}
    )[/tex]

    and in the next step:

    [tex]\phi(x)=\int\frac{d^3 p}{(2\pi)^3}\frac{1}{ \sqrt{2\omega_{\vec{p}}} }

    (a_{\vec{p}}



    +a^{+}_{\vec{-p}}
    )e^{i \vec{p} \cdot \vec{x}}[/tex]

    with [tex]\omega_{\vec{p}}=\sqrt{|\vec{p}|^2+m^2}[/tex]

    I don't understand that. When I substitute [tex]\vec{p}[/tex] for [tex]-\vec{p}[/tex] shouldn't the Jacobi-determinant then put a minus sign such that:

    [tex]\phi(x)=\int\frac{d^3 p}{(2\pi)^3}\frac{1}{ \sqrt{2\omega_{\vec{p}}} }

    (a_{\vec{p}}



    -a^{+}_{\vec{-p}}
    )e^{i \vec{p} \cdot \vec{x}}[/tex]

    What's wrong with me?
     
  2. jcsd
  3. Sep 22, 2007 #2

    OOO

    User Avatar

    The transformation theorem for integrals involves the modulus of the jacobian, not the jacobian itself, so the sign drops out. E.g. if you think of a single integral as the area under a curve, then it doesn't make a difference when you mirror the function at the vertical axis (as long as you keep the same orientation for the integration "volume").
     
  4. Sep 22, 2007 #3
    Remember that you also need to change integration limits in 3 integrals. This leads to another change of the total sign.

    Eugene.
     
  5. Sep 22, 2007 #4
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