# Understanding Klein-Gordon

1. Sep 22, 2007

### M. Kohlhaas

I'm just reading the schroeder/peskin introduction to quantum field theory. On Page 21 there is the equation

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

and in the next step:

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

with $$\omega_{\vec{p}}=\sqrt{|\vec{p}|^2+m^2}$$

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

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

What's wrong with me?

2. Sep 22, 2007

### OOO

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").

3. Sep 22, 2007

### meopemuk

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

Eugene.

4. Sep 22, 2007

Thanks.