# Ladder operator on momentum basis

Since in Quantum mechanics momentum operator can be written in terms of ladder operators $$\widehat{p}=-i\sqrt\frac{{\hbar m \omega}}{2}(\widehat{a}-\widehat{a}^\dagger)$$ these operators operate on energy eigen-state $|n\rangle$ of particle and we get $$\widehat{a}|n\rangle=\sqrt{n}|n-1\rangle$$ $$\widehat{a}^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$$ But what if these ladder operator or momentum operator operate on discrete momentum basis, for example $|k_x\rangle$, $|k_x-k_o\rangle$, $|k_x-2k_o\rangle$, $|k_x-3k_o\rangle$........etc.

Will it looks like $$\widehat{a}|k_x-k_o\rangle=\sqrt{k_x-k_o}|k_x-2k_o\rangle$$ and $$\widehat{a}^\dagger|k_x-k_o\rangle=\sqrt{k_x}|k_x\rangle$$