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Way back in lecture 8, it was claimed that
\begin{equation}\label{eqn:momentum:20}
P^k = \int d^3 x \hat{\pi} \partial^k \hat{\phi} = \int \frac{d^3 p}{(2\pi)^3} p^k a_\Bp^\dagger a_\Bp.
\end{equation}
If I compute this, I get a normal ordered variation of this operator, but also get some time dependent terms. Here’s the computation (dropping hats)
\begin{equation}\label{eqn:momentum:40}
\begin{aligned}
P^k
&= \int d^3 x \hat{\pi} \partial^k \phi \\
&= \int d^3 x \partial_0 \phi \partial^k \phi \\
&= \int d^3 x \frac{d^3 p d^3 q}{(2 \pi)^6} \inv{\sqrt{2 \omega_p 2 \omega_q} }
\partial_0
\lr{
a_\Bp e^{-i p \cdot x}
+
a_\Bp^\dagger e^{i p \cdot x}
}
\partial^k
\lr{
a_\Bq e^{-i q \cdot x}
+
a_\Bq^\dagger e^{i q \cdot x}
}.
\end{aligned}
\end{equation}
The exponential derivatives are
\begin{equation}\label{eqn:momentum:60}
\begin{aligned}
\partial_0 e^{\pm i p \cdot x}
&=
\partial_0 e^{\pm i p_\mu x^\mu} \\
&=
\pm i p_0
\partial_0 e^{\pm i p \cdot x},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:momentum:80}
\begin{aligned}
\partial^k e^{\pm i p \cdot x}
&=
\partial^k e^{\pm i p^\mu x_\mu} \\
&=
\pm i p^k e^{\pm i p \cdot x},
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:momentum:100}
\begin{aligned}
P^k
&=
-\int d^3 x \frac{d^3 p d^3 q}{(2 \pi)^6} \inv{\sqrt{2 \omega_p 2 \omega_q} }
p_0 q^k
\lr{
-a_\Bp e^{-i p \cdot x}
+
a_\Bp^\dagger e^{i p \cdot x}
}
\lr{
-a_\Bq e^{-i q \cdot x}
+
a_\Bq^\dagger e^{i q \cdot x}
} \\
&=
-\inv{2} \int d^3 x \frac{d^3 p d^3 q}{(2 \pi)^6} \sqrt{\frac{\omega_p}{\omega_q}} q^k
\lr{
a_\Bp a_\Bq e^{-i (p + q) \cdot x}
+ a_\Bp^\dagger a_\Bq^\dagger e^{i (p + q) \cdot x}
– a_\Bp a_\Bq^\dagger e^{i (q – p) \cdot x}
– a_\Bp^\dagger a_\Bq e^{i (p – q) \cdot x}
} \\
&=
\inv{2} \int \frac{d^3 p d^3 q}{(2 \pi)^3} \sqrt{\frac{\omega_p}{\omega_q}} q^k
\lr{
– a_\Bp a_\Bq e^{- i(\omega_\Bp + \omega_\Bq) t} \delta^3(\Bp + \Bq)
– a_\Bp^\dagger a_\Bq^\dagger e^{i(\omega_\Bp + \omega_\Bq) t} \delta^3(-\Bp – \Bq)
+ a_\Bp a_\Bq^\dagger e^{i(\omega_\Bq – \omega_\Bp) t} \delta^3(\Bp – \Bq)
+ a_\Bp^\dagger a_\Bq e^{i(\omega_\Bp – \omega_\Bq) t} \delta^3(\Bq – \Bp)
} \\
&=
\inv{2} \int \frac{d^3 p }{(2 \pi)^3} p^k
\lr{
a_\Bp^\dagger a_\Bp
+ a_\Bp a_\Bp^\dagger
– a_\Bp a_{-\Bp} e^{- 2 i \omega_\Bp t}
– a_\Bp^\dagger a_{-\Bp}^\dagger e^{2 i \omega_\Bp t}
}.
\end{aligned}
\end{equation}

What is the rationale for ignoring those time dependent terms? Does normal ordering also implicitly drop any non-paired creation/annihilation operators? If so, why?