Deriving the Hamiltonian of the Klein-Gordon field in terms of ladder operators (Peskin and Schroeder Eq. 2.31)

Question

In Peskin and Schroeder's QFT book they give \begin{align*} H &= \int d^3x\int \frac{d^3p d^3 p'}{(2\pi)^6}e^{i(\mathbf{p+p'})\cdot \bf x}\left\{-\frac{\sqrt{\omega_{\bf p}\omega_{\bf p'}}}{4} (a_{\bf p}-a_{\bf -p}^\dagger )(a_{\bf p'}-a_{-\bf p'}^\dagger) +\frac{-\bf p \cdot \bf p' + m^2}{4 \sqrt{\omega_{\bf p} \omega_{\bf p'}}}(a_{\bf p}+a_{-\bf p}^\dagger)(a_{\bf p'}+a_{-\bf p'}^\dagger)\right\} \\ &=\int \frac{d^3 p}{(2\pi)^3}\omega_{\bf p}\left(a_{\bf p}^\dagger a_{\bf p}+\frac 1 2 [a_{\bf p},a_{\bf p}^\dagger]\right) \end{align*} Where $$\omega_{\bf p} = \sqrt{|{\bf p}|^2+m^2}$$ I was trying to get from the first line to the second line but I'm getting stuck and was wondering if someone could nudge me in the right direction. My work: $$\delta^3(p-a)=\frac 1 {(2\pi)^3}\int e^{i(p-a)\cdot x}\,d^{3}x$$ Thus \begin{align*} H &= \int d^3x\int \frac{d^3p d^3 p'}{(2\pi)^6}e^{i(\mathbf{p+p'})\cdot \bf x}\left\{-\frac{\sqrt{\omega_{\bf p}\omega_{\bf p'}}}{4} (a_{\bf p}-a_{\bf -p}^\dagger )(a_{\bf p'}-a_{-\bf p'}^\dagger) +\frac{-\bf p \cdot \bf p' + m^2}{4 \sqrt{\omega_{\bf p} \omega_{\bf p'}}}(a_{\bf p}+a_{-\bf p}^\dagger)(a_{\bf p'}+a_{-\bf p'}^\dagger)\right\} \\ &= \int d^3x\int \frac{d^3p }{(2\pi)^3}\delta^3(p+p')\left\{-\frac{\sqrt{\omega_{\bf p}\omega_{\bf p'}}}{4} (a_{\bf p}-a_{\bf -p}^\dagger )(a_{\bf p'}-a_{-\bf p'}^\dagger) +\frac{-\bf p \cdot \bf p' + m^2}{4 \sqrt{\omega_{\bf p} \omega_{\bf p'}}}(a_{\bf p}+a_{-\bf p}^\dagger)(a_{\bf p'}+a_{-\bf p'}^\dagger)\right\} \\ &= \int \frac{d^3p }{(2\pi)^3}\left\{-\frac{\sqrt{\omega_{\bf p}\omega_{\bf -p}}}{4} (a_{\bf p}-a_{\bf -p}^\dagger )(a_{\bf -p}-a_{\bf p}^\dagger) +\frac{\bf p \cdot \bf p + m^2}{4 \sqrt{\omega_{\bf p} \omega_{-\bf p}}}(a_{\bf p}+a_{-\bf p}^\dagger)(a_{-\bf p}+a_{\bf p}^\dagger)\right\} \\ \end{align*} Since $\omega_{\bf p} = \omega_{-\bf p}$ we get \begin{align*} H &= \int \frac{d^3p }{(2\pi)^3}\left\{-\frac{\omega_{\bf p}}{4} (a_{\bf p}-a_{\bf -p}^\dagger )(a_{\bf -p}-a_{\bf p}^\dagger) +\frac{\omega_{\bf p}^2}{4 \omega_{\bf p} }(a_{\bf p}+a_{-\bf p}^\dagger)(a_{-\bf p}+a_{\bf p}^\dagger)\right\} \\ &= \int \frac{d^3p }{(2\pi)^3}\frac{\omega_{\bf p}}{4}\left\{-(a_{\bf p}-a_{\bf -p}^\dagger )(a_{\bf -p}-a_{\bf p}^\dagger) +(a_{\bf p}+a_{-\bf p}^\dagger)(a_{-\bf p}+a_{\bf p}^\dagger)\right\} \\ &= \int \frac{d^3p }{(2\pi)^3}\frac{\omega_{\bf p}}{4}\left\{-(a_{\bf p}a_{\bf -p}-a_{\bf p}a_{\bf p}^\dagger-a_{-\bf p}^\dagger a_{-\bf p}+a_{\bf -p}^\dagger a_{\bf p}^\dagger )\right.\\&\left. +(a_{\bf p}a_{-\bf p}+a_{\bf p}a_{\bf p}^\dagger +a_{-\bf p}^\dagger a_{-\bf p}+a_{\bf -p}^\dagger a_{\bf p}^\dagger )\right\} \\ &= \int \frac{d^3p }{(2\pi)^3}\frac{\omega_{\bf p}}{4}\left\{2 a_{\bf p}a_{\bf p}^\dagger + 2 a_{-\bf p}^\dagger a_{-\bf p}\right\} \\ \end{align*} So the problem reduces to showing that \begin{equation}\frac{a_{\bf p}a_{\bf p}^\dagger}{2} + \frac{a_{-\bf p}^\dagger a_{-\bf p}}{2} = a_{\bf p}^\dagger a_{\bf p}+\frac 1 2[a_{\bf p}, a_{\bf p}^\dagger] \qquad\qquad (1)\label{help}\end{equation} But because there are annihilation/creation operators for ${\bf -p}$ I can't figure out how to show this. What I think is going on is that there's some relationship between $a_{\bf p}$ and $a_{\bf -p}$ that I've missed. Did some part of the intermediary steps go wrong and if not, does anyone have any guidance as to how to show the left hand side is equal to the right hand side of (1)

Answer 1

The integral is over all $\mathbf{p}$ hence you can replace $\mathbf{p} \rightarrow -\mathbf{p}$ using that $\omega_\mathbf{p} = \omega_{-\mathbf{p}}$ the result follows.