What is the Significance of Substituting p for -p in the Klein-Gordon Equation?

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M. Kohlhaas
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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}}} } <br /> <br /> (a_{\vec{p}} e^{i \vec{p} \cdot \vec{x}}<br /> <br /> +a^{+}_{\vec{p}} e^{-i \vec{p} \cdot \vec{x}}<br /> )[/tex]

and in the next step:

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

What's wrong with me?
 
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M. Kohlhaas said:
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

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").
 
Remember that you also need to change integration limits in 3 integrals. This leads to another change of the total sign.

Eugene.