- #1
M. Kohlhaas
- 8
- 0
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?
[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?