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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?

Peeter Joot's Blog » Momentum of scalar field: continued.6 years ago