2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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, 2016 at 2:08
  • $\begingroup$ Why wouldn't it be zero? $\endgroup$ Sep 18, 2016 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, 2016 at 17:58

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Sometimes you just need someone to give you a solid yes/no answer. Very helpful and clarifying. Thank you. $\endgroup$
    – Tweej
    Sep 20, 2016 at 2:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.