An elementary question regarding scalar field creation/annihilation operators under reflection I was studying a qft text when this equality was taken for granted. When I try to do $p\rightarrow -p$ by hand, I get an overall factor of $-1$ from the differential. How is this justified?
$$\phi(x)=\iiint_{-\infty}^\infty \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2p^0}} (a_{\vec{p}}  +a_{-\vec{p}}^\dagger)e^{ip\cdot x} = \iiint_{-\infty}^\infty \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2p^0}} (a_{\vec{p}}\,e^{ip\cdot x}  +a_{\vec{p}}^\dagger\,e^{-ip\cdot x})$$
 A: First, a quick sanity check. Consider the integral
$$I = \int_{-\infty}^\infty dx\, f(x)$$
where $f(x)$ is a positive function, so that clearly $I>0$. Under a change to $x' = -x$, the integral is the same. The integrand will still be $f$ of something ($f(-x')$ in this case), which is positive, so we cannot have a negative sign outside. It gets canceled by the flipping of the bounds:
$$\int_{-\infty}^\infty dx = \int_{\infty}^{-\infty} (-dx') = \int_{-\infty}^\infty dx'.$$
The same reasoning applies in your integral, only in three dimensions. Written out it's just
$$\int_{-\infty}^\infty dp_x \int_{-\infty}^\infty dp_y \int_{-\infty}^\infty dp_z,$$
and each of the integral bounds + differential remains the same, as we just saw.
In general, there are two ways of doing coordinate changes. You can write the bounds of the integral explicitly, giving the values between which your variables will run. In this case, the order matters of course. When you do a change of variables, you have to include the Jacobian of the transformation, which in your case is -1.
Or, you can write the integral as being over some region ($\mathbb{R}$ or $\mathbb{R}^3$ in our examples), without specifying an orientation. And if you use this system, you use the absolute value of the Jacobian. Both are equivalent, of course, but in many cases the second is more convenient, because the flipping of the bounds is taken care of automatically.
