In [1] it is stated that the creation operators of eq. 2.78

\label{eqn:scalarFieldCreationOpCommutator:20}
\alpha_k = \inv{2} \int \frac{d^3k}{(2\pi)^3} \lr{
\phi(x,0) + \frac{i}{\omega_k} \partial_0 \phi(x,0)
}
e^{-i \Bk \cdot \Bx }

associated with field operator $$\phi$$ commute. Let’s verify that.

\label{eqn:scalarFieldCreationOpCommutator:40}
\begin{aligned}
\antisymmetric{\alpha_k}{\alpha_m}
&=
\inv{4}
\frac{1}{(2\pi)^6}
\int d^3 x d^3 y
e^{-i \Bk \cdot \Bx }
e^{-i \Bm \cdot \By }
\antisymmetric
{
\phi(x,0) + \frac{i}{\omega_k} \partial_0 \phi(x,0)
}
{
\phi(y,0) + \frac{i}{\omega_m} \partial_0 \phi(y,0)
} \\
&=
\frac{i}{4}
\frac{1}{(2\pi)^6}
\int d^3 x d^3 y
e^{-i \Bk \cdot \Bx }
e^{-i \Bm \cdot \By }
\lr{
\antisymmetric{\phi(x,0)}{\inv{\omega_m} \partial_0 \phi(y,0)}
+
\antisymmetric{\inv{\omega_k} \partial_0 \phi(x,0)}{\phi(y,0)}
} \\
&=
\frac{i}{4}
\frac{1}{(2\pi)^6}
\int d^3 x d^3 y
e^{-i \Bk \cdot \Bx }
e^{-i \Bm \cdot \By }
\lr{
\frac{i}{\omega_m} \delta^3(\Bx – \By)

\frac{i}{\omega_k} \delta^3(\Bx – \By)
} \\
&=
-\frac{1}{4}
\frac{1}{(2\pi)^6}
\int d^3 x
e^{ -i (\Bk + \Bm) \cdot \Bx }
\lr{
\frac{1}{\omega_m}

\frac{1}{\omega_k}
} \\
&=
-\frac{1}{4}
\frac{1}{(2\pi)^3}
\lr{
\frac{1}{\omega_m}

\frac{1}{\omega_k}
}
\delta^3(\Bk + \Bm) \\
&=
-\frac{1}{4}
\frac{1}{(2\pi)^3}
\lr{
\frac{1}{\omega_{\Abs{-\Bk}}}

\frac{1}{\omega_{\Abs{\Bk}}}
}
\delta^3(\Bk + \Bm) \\
&=
0.
\end{aligned}

# References

[1] Michael Luke. PHY2403F Lecture Notes: Quantum Field Theory, 2015. URL https://s3.amazonaws.com/piazza-resources/i87nj8g7yie7nh/ihdwuk7wva13qq/lecturenotes.pdf?AWSAccessKeyId=AKIAIEDNRLJ4AZKBW6HA&Expires=1451803428&Signature=IF6qOjlKqOYL01FwqT%2FGV6BSDb8%3D. [Online; accessed 02-Jan-2016].