Coulomb scattering of spin-zero particle (QFT)

[eudo]

I'm looking at Aitchison and Hey's QFT book, trying to verify Eq. 8.27 (which is in fact problem 8.2). It asks us to verify that the matrix element for the scattering of a charged spin zero particle ($s^+$) is
$$<s^+,p'|j^\mu_{em,s}|s^+,p> = e(p+p')^\mu e^{-i(p-p')\cdot x}$$
where
$$|s^+,p>=\sqrt{2E}\hat{a}^\dagger(p)|0>$$
and
$$j^\mu_{em,s}=ie(\phi^\dagger\partial^\mu \phi - (\partial^\mu \phi^\dagger)\phi$$
Now, it turns out the solutions for this problem are online here

But I have a question about one of the steps. They expand the $\phi$'s in terms of $\hat{a}$ and $\hat{b}$ operators and note that the $\hat{a}$'s and $\hat{b}$'s commute, so that we can move all the $\hat{b}$'s to the right where they will give zero when acting on |0>. So we're only left with a term that has the $\hat{a}$ operators.

But it seems to me we still have a term

$$ie\sqrt{4EE'}<0|\hat{a}(p')\int\frac{d^3\boldsymbol{k'}}{(2\pi)^3\sqrt{2\omega'}}\hat{b}(k')e^{-ik'\cdot x}\int \frac{d^3\boldsymbol{k}}{(2\pi)^3\sqrt{2\omega}}ik^\mu \hat{b}^\dagger(k)e^{ik\cdot x}\hat{a}^\dagger(p)|0>$$

Where did this term go? The $\hat{b}$ and $\hat{b}^\dagger$ operators do not commute.

 

[NateD]

The answer is that this term vanishes because the integral over k' is antisymmetric in k' and k, so that the integrand is odd. Then, since the integrand is odd, its integral over all k' vanishes.