Understanding <q|p>=exp(ip.x): The Explanation You've Been Searching For

[Colin]

Can anyone explain why?

<q|p>=exp(ip.x)

thanks

 

[Baggio]

Depends on the nature of your function to begin with.. more detail needed.. probably due to orthogonality condition

 

[dextercioby]

Can anyone explain why?

<q|p>=exp(ip.x)

thanks

What is &lt;q|,an eigenstate of the momentum operator,or of the position operator (actually of their adjoints,but,because they are selfadjoint,you can think that way as well)?
If it's for the momentum operator,then |p&gt; [/itex] is an eigenstate as well,and the normalization condition reads:<br /> &amp;lt;q|p&amp;gt; =2\pi\hbar \delta(q-p)<br /> ,but if it&#039;s for the coordinate operator,then the scalar product is zero,since one of them is an eigenvector from a Hilbert space and the other is a linear functional over another Hilbert space.<br /> <br /> So,anyway,what u&#039;ve written there is wrong.<br /> <br /> Daniel.

 

[Colin]

correction <x|p>=exp(ip.x)

sorry I meant <x|p>=exp(ip.x)

and I think I have it

P|p>=p|p>

<x|P|p> =p<x|p>=-id/dx<x|p>
solution
<x|p>=exp(ipx)

It's a long time since I did QM, and I started reading Prof Zee's Quantum field theory in a nut shell. He quotes this very early on without proof and I couldn't see where it came from. Thanks for comments.