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

Question

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

Answer 1

The answer is yes. The operation of the operator V has eigenvalues $V(x)$ with eigenkets $|x\rangle$ with $x$ arbritary.