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?

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    $\begingroup$ the change of variables also flips the limits of integration, $\int_{-\infty}^{+\infty}\to\int_{+\infty}^{-\infty}$. $\endgroup$ Feb 10, 2021 at 16:14
  • $\begingroup$ @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? $\endgroup$ Feb 10, 2021 at 16:28

1 Answer 1


You're integrating over all possible values of the momenta, so in this case the minus sign is not needed.


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