Also for your other question, aparart from the physical answer given by Vanamali, you can do it by hand:



[itex] \int \frac{d^3 k}{ (2 \pi )^3 \sqrt{\omega_k}} (a_k e^{ikx} + a_{k}^{\dagger} e^{-ikx})[/itex]



this is two integrals, I take only the 2nd and write it:



[itex] \int \frac{d^3 k}{ (2 \pi )^3 \sqrt{\omega_k}} a_{k}^{\dagger} e^{-ikx}[/itex]



Do a change [itex] k \rightarrow -k [/itex]. The [itex]\omega_k = \sqrt{m^2 + |k|^2} = \omega_{-k}[/itex] (doesn't change. The [itex]d^3k \rightarrow -d^3 k[/itex] and also the integration limits ar going to be interchanged...to bring them back to what they were before, they wll take the minus of the differential .. [itex]a_k^\dagger \rightarrow a_{-k}^\dagger[/itex] and the [itex] exp(-ikx) \rightarrow exp(ikx)[/itex]. The result will then be the one you ask for...

I think the best reason to explain why you do that is bcause it makes the whole algebras easier when multiplying fields....a field multiplication like: [itex]\phi \phi[/itex] would else have 4 exponentials, while now it will have only 2....