TITLE: Issue with a sign in the commutator calculation of field operators for a real scalar field
QUESTION [0 upvotes]: In the derivation for the commutator (real scalar field, Klein-Gordon equation)
$$[\phi(x),\phi(y)]=0$$
I have solved up to
$$[\phi(x),\phi(y)]=\frac{1}{2(2\pi)^3}\int \frac{d^3 p}{\omega_p}[e^{ip(x-y)}-e^{ip(y-x)}]$$
The resources I have consulted say that I have to make a change from $p$ to  $-p$ in the second term. So as $\omega_p=\omega_p(p^2)$
$$-\int \frac{d^3 (-p)}{\omega_{(-p)}}e^{i(-p)(y-x)}=-\int \frac{d^3 (-p)}{\omega_{p}}e^{ip(x-y)}=\int \frac{d^3 p}{\omega_{p}}e^{ip(x-y)}$$
I think $d^3(-p)$would become $-d^3 (p)$ because $\vec{p}=(p_x,p_y,p_z)$ and $\vec{-p}=(-p_x,-p_y,-p_z)$ which would make $d^3 p=dp_x dp_y dp_z$ as $d^3 (-p)=d(-p_x) d(-p_y) d(-p_z)=-d^3p$. If this is true, we don't get the right answer. Where did I go wrong?

REPLY [1 votes]: You're integrating over all possible values of the momenta, so in this case the minus sign is not needed.
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