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.

  • $\begingroup$ This is a silly question I know, but have you fully expanded the commutator, and if you did, what result did you get? A similar question applies , with trying to apply the creation operator to the state vector indicated. I ask because I don't know the subject very well, but it is not clear to me why you have not done that. It is probably more complicated than I am aware of, my apologies if it is. $\endgroup$ – user108787 Sep 18 '16 at 2:08
  • $\begingroup$ Why wouldn't it be zero? $\endgroup$ – aquirdturtle Sep 18 '16 at 2:40
  • $\begingroup$ I just wasn't sure as to why the negative momentum would come into it. Hypothetically, if it was just $\delta_{\textbf{|p|,|q|}}$ then this would not be an issue. It would be obvious that $\delta_{\textbf{|p|,|-p|}} \neq 0$. $\endgroup$ – Tweej Sep 18 '16 at 17:58

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.

  • $\begingroup$ Sometimes you just need someone to give you a solid yes/no answer. Very helpful and clarifying. Thank you. $\endgroup$ – Tweej Sep 20 '16 at 2:14

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