# Issue with a sign in the commutator calculation of field operators for a real scalar field

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?

• the change of variables also flips the limits of integration, $\int_{-\infty}^{+\infty}\to\int_{+\infty}^{-\infty}$. Feb 10 at 16:14
• @AccidentalFourierTransform yes, I did get that from the answer that I accepted. Thank you never the less. Also, does anyone know what should I do with the question now? I feel that the issue I highlighted is really merger and might not help anyone in the future. So should I delete the question or let it be? Feb 10 at 16:28