# 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?

• 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? Apr 26, 2019 at 6:31

## 2 Answers

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.

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 $$(a_p^\dagger a_p-b_p^\dagger b_p +a_p a_p^\dagger -b_p b_p^\dagger)$$ $$= \left[a_p a_p^\dagger-(2\pi)^3\delta^{(3)}(0) - b_pb_p^\dagger+(2\pi)^3\delta^{(3)}(0)+a_p a_p^\dagger -b_p b_p^\dagger\right]$$ $$= 2 a_pa_p^\dagger-2b_pb_p^\dagger$$