How to apply potential operator ##V(\hat{x})##

[Kashmir]

I want some clarification on the potential operator $V(\hat{x})$. Can you please help me

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Is the action of $V(\hat{x})$ defined by its action on the position kets as $\hat{V}(x)|x\rangle=V(x)|x\rangle$?

Then we'd have for any ket $|\psi\rangle$ that $V(\hat{x})|\psi\rangle$ $=V(\hat{x}) \int d x|x\rangle\langle x \mid \psi\rangle$$=\int d x V(x)|x\rangle\langle x \mid \psi\rangle$

And $V(\hat{x}) \int d x|-x\rangle\langle x \mid \psi\rangle$ equals $\int d x V(-x)|-x\rangle\langle x \mid \psi\rangle$
Is that correct?

 

[PeroK]

Is the action of $V(\hat{x})$ defined by its action on the position kets as $\hat{V}(x)|x\rangle=V(x)|x\rangle$?

Then we'd have for any ket $|\psi\rangle$ that $V(\hat{x})|\psi\rangle$ $=V(\hat{x}) \int d x|x\rangle\langle x \mid \psi\rangle$$=\int d x V(x)|x\rangle\langle x \mid \psi\rangle$

This looks right.

And $V(\hat{x}) \int d x|-x\rangle\langle x \mid \psi\rangle$ equals $\int d x V(-x)|-x\rangle\langle x \mid \psi\rangle$

I'm not sure what this means. But, it looks right.

 

[Kashmir]

I get this expression $V(\hat{x}) \int d x|-x\rangle\langle x \mid \psi\rangle$ while doing another problem ( commutator of parity and V)