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

Kashmir
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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?
 
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Kashmir said:
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.
Kashmir said:
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.
 
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)
 
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