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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?

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  • $\begingroup$ Thank you for your clear and concise answer InertialObserver. Can you provide any guidance on the second question relating to commutation of the operators. Am I missing something? $\endgroup$ – user116129 Apr 26 at 6:31
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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.

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