When we deal with an infinite-dimensional basis,the normalization condition of this basis becomes <x|x'>=δ(x-x')(here for example the position basis).Same thing for momentum eigenstates <p|p'>=δ(p-p').(adsbygoogle = window.adsbygoogle || []).push({});

Lets look now on the eigenvalue problem of the momentum operator:

[itex]\hat p | p \rangle =p | p\rangle [/itex]

projecting [itex]\langle x |[/itex] on both sides this will yield to a differential equation of the form :

[itex]-iħ\frac{d\psi_{p}(x)}{dx}=p\psi_{p}(x)[/itex]

where [itex]\psi_{p}(x)=\langle x | p\rangle[/itex]

this will finally yield that [itex]\psi_{p}(x)=\frac{1}{√(2πħ)}e^{ipx/ħ}[/itex]

so [itex] | p \rangle \leftrightarrow \frac{1}{√2πħ} (e^{ ipx_{1} } | x_{1} \rangle +e^{ ipx_{2} } |x_{2} \rangle+ . . . . . +e^{ ipx_{n} } | x_{n} \rangle)[/itex] where n goes to infinity and [itex]x_{n}=x_{n-1}+dx[/itex]

but doesn't this violate<x|x'>=δ(x-x')? since if we project,for example,<x1| on |p> and we take by consideration δ(x-x')=<x|x'>, the result will be <x|p>=∞ and not [itex]\frac{1}{√2πħ}e^{ ipx_{1} }[/itex] ???

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# Confusion related to momentum eigenstates

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