suppose that the momentum operator [itex]\hat p[/itex] is acting on a momentum eigenstate [itex] | p \rangle [/itex] such that we have the eigenvalue equation [itex]\hat p | p \rangle = p| p \rangle[/itex](adsbygoogle = window.adsbygoogle || []).push({});

Now lets project [itex]\langle x | [/itex] on the equation above and use the completeness relation [itex]\int | x\rangle \langle x | dx =\hat I[/itex]

we get that [itex]\psi_{p}(x)=Ae^{ipx/\hbar}[/itex] where [itex]\psi_{p}(x)=\langle x | p \rangle[/itex]

Now let the operator [itex]\hat H[/itex] act on the eigenstate [itex]| p\rangle [/itex],such that[itex]\hat H | p \rangle=E | p \rangle[/itex]

but the two operators [itex]\hat p and \hat H [/itex] have the same eigenstate [itex]| p\rangle[/itex],since[itex]\hat H | p\rangle=\hat p^{2}/2m | p \rangle = p/2m\hat p |p\rangle =p^{2}/2m| p \rangle [/itex]

Now lets solve the eigenvalue equation [itex]\hat H | p \rangle=p^{2}/2m | p \rangle[/itex] in the [itex]X [/itex] basis, what we will certainly get is [itex]\psi_{p}(x)=\alpha e^{ipx/\hbar}+ \beta e^{-ipx/\hbar}[/itex] which is different from [itex]\psi_p(x)[/itex] above.

This is what confuses me, why when use different operators( that have same eigenstates like the energy and momentum operator)in the eigenvalue problem, mathematics gives different [itex]\psi_p(x)[/itex]?

As [itex] | p\rangle[/itex] is the same why we get different [itex]\langle x |p \rangle [/itex]s?

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# Hamiltonian and momentum operator acting on a momentum eigenstate

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