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

1. Dec 16, 2004

### Colin

Can anyone explain why?

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

thanks

2. Dec 16, 2004

### Baggio

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

3. Dec 16, 2004

### dextercioby

What is $<q|$,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 $$|p> [/itex] is an eigenstate as well,and the normalization condition reads: [tex] <q|p> =2\pi\hbar \delta(q-p)$$
,but if it'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.

So,anyway,what u've written there is wrong.

Daniel.

4. Dec 16, 2004

### 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.