Commutation of creation and annihilation operators with opposite momentum

Question

So we have the rules for creation and annihilation operators.

\begin{equation} \left[ a_\textbf{p}, a_\textbf{q}^\dagger \right] = \delta_{\textbf{p},\textbf{q}} \end{equation} and \begin{equation} \left[ a_\textbf{p}, a_\textbf{q} \right] = \left[ a^\dagger_\textbf{p}, a_\textbf{q}^\dagger \right] = 0 \end{equation}

I just have the question, does this mean that \begin{equation} \left[ a_\textbf{p}, a_\textbf{-p}^\dagger \right] = 0 \end{equation}

Despite having the same momentum just, in the opposite direction?

What if I was to apply the creation operator $a^\dagger_\textbf{-p}$ on the state $\left|n, \textbf{p} \right>$?

If the commutator does equal zero, is this an example of right handed and left handed particles (opposite helicity) not interacting?

Thank you for reading.

Answer 1

Yes it's zero since $\delta(\textbf{p},-\textbf{p})=0$. There is nothing special about $\textbf{p}$ and $-\textbf{p}$. Described in relatively moving coordinates these states would not have opposite momenta.

"What if I was to apply the creation operator $a^\dagger_\textbf{-p}$ on the state $\left|n, \textbf{p} \right>$?"

You would get the $n+1$ particle state $\left|1, -\textbf{p} \right>\left|n, \textbf{p} \right>$. Each momentum state can be thought of as an independent harmonic oscillator.