QFT Complex scalar field and commutators The conserved charge is $$Q=i\int\ d^3x(\phi\pi-\phi^\dagger\pi^\dagger)$$
Expressing this in terms of creation annihilation operators gives
$$Q=i\int d^3 x \frac{d^3p d^3k}{(2\pi)^3(2\pi)^3}\frac{i\sqrt{E_p}}{2\sqrt{E_k}}(a_k e^{ikx}+b_k^\dagger e^{-ikx})(a_p^\dagger e^{-ipx}+b_p e^{ipx}).$$ 
Expanding this out gives $$Q=i\int d^3x \frac{d^3p d^3k}{(2\pi)^3(2\pi)^3}\frac{i\sqrt{E_p}}{2\sqrt{E_k}} (a_k^\dagger a_p-b_k^\dagger b_p)e^{i(k-p)x} +(a_ka_p^\dagger -b_k b_p^\dagger)e^{-i(k-p)x}+(a_k^\dagger b_p^\dagger -b_k^\dagger a_p^\dagger)e^{i(k+p)x} +(a_k b_p -b_k a_p)e^{-i(k+p)x}.$$
Now do the integral over all space gives two delta functions from the exponentials on each of the terms in brackets.   One gives $p=k$ and the other $p=-k$.  The first is straight forward but the second I need to understand .  My first question is how does the change from $k$ to $-k$ affect the creation and annihilation operators and the integral.  If I ignore the minus, the commutators make the terms in the last two brackets go to zero (which is what I want) but I read somewhere that the integral over $k$ to$-k$ is odd and goes to zero anyway.  Can anyone give me a definitive answer or point me to a relevant text book which explains which way it is?
My second question relates to the terms in the first two brackets after the application of the delta function.  It gives $$Q=-\int \frac{d^3k}{(2\pi)^3} (a_p^\dagger a_p-b_p^\dagger b_p +a_pa_p^\dagger -b_p b_p^\dagger).$$  Do the $a_p^\dagger a_p$ and $b_p^\dagger b_p$ commute? They do (because that gives the answer) but I seem to recall from QM that they don't.  Why do these commute? Doesn't $N=a^\dagger a$ and $[a,a^\dagger ]=1$?  Can someone give me some guidance here please?
 A: Note that 
$$ Q \sim \int d^3k \int d^3p\int d^3x \  (a_k^\dagger b_p^\dagger -b_k^\dagger a_p^\dagger)e^{i(k+p)x} +(a_k b_p -b_k a_p)e^{-i(k+p)x}\\
=\int d^3k \int d^3p (a_k^\dagger b_p^\dagger -b_k^\dagger a_p^\dagger)\delta^3(p+k) +(a_k b_p -b_k a_p)\delta^3(p+k)\\
=\int d^3p \ a_{-p}^\dagger b_p^\dagger -b_{-p}^\dagger a_p^\dagger +a_{-p} b_p -b_{-p} a_p
 $$
Now imagine that we distribute out the integral over each term so that we have 4 integrals. and suppose that in the 2nd and 4th terms we make the change of variables $p \to -p$ Then we have that 
$$
b^\dagger_{-p}a^\dagger_p \to b^\dagger_p a^\dagger_{-p}\\
b_{-p}a_p\to b_p a_{-p}\\
\int_{-\infty}^\infty \to  - \int_{\infty}^{-\infty} = \int_{-\infty}^\infty
$$
Since $[a_p,b_k]= 0$ and writing it again as a single integral we see that this term vanishes. 
A: With the commutation relations 
$$
[a_p,a_q^\dagger]=(2\pi)^3\delta^{(3)}(\vec{p}-\vec{q})$$and
$$
[b_p,b_q^\dagger]=(2\pi)^3\delta^{(3)}(\vec{p}-\vec{q})
$$
We have $$(a_p^\dagger a_p-b_p^\dagger b_p +a_p a_p^\dagger -b_p b_p^\dagger)
$$
$$=
\left[a_p a_p^\dagger-(2\pi)^3\delta^{(3)}(0) - b_pb_p^\dagger+(2\pi)^3\delta^{(3)}(0)+a_p a_p^\dagger -b_p b_p^\dagger\right]
$$
$$=
2 a_pa_p^\dagger-2b_pb_p^\dagger
$$
