\langle x|p \rangle =e^{i\mathbf{p.x}}

Consider

U = e^{i\mathbf{X.p}} (Capitals for Operators)

\langle x|e^{i\mathbf{X.p'}}|p\rangle= \int dx' \langle x|e^{i\mathbf{X.p'}}|x'\rangle \langle x'|p\rangle

= e^{i\mathbf{(p+p').x}} (skipped the delta function integ. step)

does this imply that position operator generates momentum?

Posted By : Venkateshan

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