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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 $$\widehat{a}^\dagger|k_x-k_o\rangle=\sqrt{k_x}|k_x\rangle$$

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No, because the momentum eigenstates are not energy eigenstates. The ladder operators have that action only on energy eigenstates. If you want to find the action of the ladder operators on momentum states then the easiest way to do that is to write the ladder operators in terms of momentum (and position) and evaluate their action on the momentum states.

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