Parity of annihilation and creation operator - Real Klein-Gordon field

Question

In calculating the total-momentum operator of the real Klein-Gordon field, I end up with an equation like $$ \vec P = \frac{1}{(2\pi)^3}\int d^3p \mspace{9mu} \vec p\Big(a_pa_{-p} + a_pa^\dagger_{p}-a_{-p}^\dagger a_{-p} - a_{-p}^\dagger a_p^\dagger\Big) $$

and it seems the first and the last integrals vanish because they are antisymmetric w.r.t $ \vec p \rightarrow -\vec p $. I am unable to see this anti-symmetry. Also I am not able to see how to relate $ a_{-p} $ to $ a_p $ or $ a_p^\dagger $ other than some implicit integral relations like(which also am not very sure how to prove),

$$ \int d^3p a_p^\dagger e^{-i\vec p.\vec x} = \int d^3p a_p e^{i\vec p.\vec x}$$

Thanks for any inputs.

Answer 1

I am not a 100% sure, but I think they mean the following: \begin{equation} \begin{aligned} \mathbf{P} & \sim \int \mathrm{d}^3 p \; \mathbf{p}a_p a_{-p} \\& = \int \mathrm{d}^3 p \; \left(\frac{1}{2} \mathbf{p}a_p a_{-p} + \frac{1}{2} \mathbf{p}a_p a_{-p} \right) \\& = \int \mathrm{d}^3 p \; \left(\frac{1}{2} \mathbf{p}a_p a_{-p} - \frac{1}{2} \mathbf{p}a_{-p} a_{p} \right) \\& = \int \mathrm{d}^3 p \; \frac{1}{2} [a_p, a_{-p}] \\& = 0 \end{aligned} \end{equation} where we performed a change of integration variable $\mathbf{p} \rightarrow - \mathbf{p}$ to get to the third line. In order to understand this, note that for $\mathbf{p} \rightarrow - \mathbf{p}$ we get: \begin{equation} \int\limits_{-\infty}^{\infty} \mathrm{d}p \rightarrow \int\limits_{\infty}^{-\infty} \left(-\mathrm{d}p\right)= - \int\limits_{-\infty}^{\infty} \left(-\mathrm{d}p\right)=\int\limits_{-\infty}^{\infty} \mathrm{d}p \end{equation} I think this may also explain your second question: $a_p$ is related to $a_{-p}$ through this change of integration variable.