- #1

- 2,637

- 785

This might be trivial for some people but this has been bothering lately.

If P is momentum operator and p its eigenvalue then the eigenfunction is u

While it can also be proved that <x|p> = exp(ipx/h) (omitting constant pre factor), where |p> is also momentum operator eigenfunction in different notation, in this sense one would say that u

I will appreciate anyone who is trying to enlighten me in this problem. Thanks in advance.

If P is momentum operator and p its eigenvalue then the eigenfunction is u

_{p}(x) = exp(ipx/h). where h is the reduced Planck constant (sorry can't find a way to make the proper notation).While it can also be proved that <x|p> = exp(ipx/h) (omitting constant pre factor), where |p> is also momentum operator eigenfunction in different notation, in this sense one would say that u

_{p}(x) = |p>. But why is <x|p> = u_{p}(x) instead?I will appreciate anyone who is trying to enlighten me in this problem. Thanks in advance.

Last edited: