Let $\phi(\vec{x},t)$ denote the canonical fields and $\pi(\vec{x},t)$ denote the canonical impulses where they're given by:
\begin{equation}
\phi(x)=\int\frac{d^3\vec{p}}{(2\pi)^3\sqrt{2\omega_{\vec{p}}}}(a_{\vec{p}}e^{-ipx}+a_{\vec{p}}^{\dagger}e^{ipx}) \tag{2.78/81}
\end{equation}
\begin{equation}
\pi(x)=-i\int\frac{d^3\vec{p}}{(2\pi)^3}\sqrt{\frac{\omega_{\vec{p}}}{2}}(a_{\vec{p}}e^{-ipx}-a_{\vec{p}}^{\dagger}e^{ipx}) \tag{2.91}
\end{equation}
where (I'll omit the vectors in the subscripts now because I can't align them properly)
\begin{equation}
[a_{p},a_{p'}^{\dagger}]=(2\pi)^{3}\delta^{3}(\vec{p}-\vec{p}'),\qquad [a_{p},a_{p'}]=0=[a_{p}^{\dagger},a_{p'}^{\dagger}].\tag{2.69}
\end{equation}
I am trying to prove the commutation relations:
\begin{equation}
[\phi(\vec{x},t),\phi(\vec{y},t)]=0=[\pi(\vec{x},t),\pi(\vec{y},t)].\tag{2.90}
\end{equation}
With a straightforward calculation I arrive at:
\begin{equation}
[\phi(\vec{x},t),\phi(\vec{y},t)]=\int\frac{d^{3}\vec{p}}{(2\pi)^{3}}\frac{1}{\omega_{p}}(e^{ip(y-x)}-e^{ip(x-y)}).\tag{2.89}
\end{equation}
And I must admit I see no way how this is zero. I then tried to consult books on QFT and I came across a "proof" in Schwartz' Quantum Field Theory and the Standard Model, on the bottom of p. 24:
Since the integral measure and $\omega_{p}=\sqrt{\vec{p}^{2}+m^{2}}$ are symmetric under $\vec{p}\to-\vec{p}$ we can flip the sign on the exponent of one of the terms to see that the commutator vanishes.
But to me this just doesn't seem right. How can we flip the sign on only one part of the function? This comes out just as wishful thinking. Is there an alternative way to prove these relations? Or did we "observe" that these commutation relations are correct so we just go with the flow?
Best Answer
There's no "wishful thinking" here, just calculus. As an example let's consider the integral $$I = \int_{-1}^1 \cos(x) - \cos(-x) \, dx.$$ We evaluate it as $$I = \int_{-1}^1 \cos(x) \, dx - \int_{-1}^1 \cos(-x) \, dx = \int_{-1}^1 \cos(x) \, dx + \int_{1}^{-1} \cos(u) \, du$$ where we substituted $u = -x$. By exchanging the limits of integration, we get $$I = \int_{-1}^1 \cos(x) \, dx - \int_{-1}^{1} \cos(u) \, du.$$ Finally, by renaming $u$ to $x$, we have $$I = \int_{-1}^1 \cos(x) \, dx - \int_{-1}^{1} \cos(x) \, dx = \int_{-1}^1 \cos(x) - \cos(x) \, dx = \int_{-1}^1 0 \, dx = 0.$$ Schwartz's reasoning is the same, just replace $dx$ with $d^3\vec{p}$ and the cosine with the exponential. Since this is usually taught in calculus class, Schwartz didn't both filling in the intermediate steps.