# A question about wavefunction

1. Oct 5, 2005

### Eole

Here's a derivation of wavefunction of State Ψ in representations of coordinates and momentum
Ψ (x)=<x|Ψ >=<x|∫dp|p><p|Ψ >=∫dp<x|p><p|Ψ>=∫dp exp{ipx/h}Ψ(p)
Ψ (p)=<p|Ψ >=∫dx exp{-ipx/h}Ψ(x)

Ψ (x)=<x|Ψ >=<x|∫dp|p><p|Ψ >=∫dp<x|p><p|Ψ>=∫dp exp{ipx/h}Ψ(p)
i don't understand how ∫dp<x|p><p|Ψ> become ∫dp exp{ipx/h}Ψ(p)
Could you please tell me the drivation of this formula?

and another question is why Ψ (x) could be denoted as <x|Ψ >?

2. Oct 5, 2005

### big man

I probably won't be able to help, but I'm sure that if I'm struggling to make sense of that, then other people are too. Some things just can't be written without proper equations. Maybe try writing the equation on Microsoft word's equation tool and attach the document.

3. Oct 12, 2005

### Theoretician

It seems to me that you are asking why
<x|p>=exp{ipx/h}
but surely you must have this derivation in your notes? BTW, the above equation isn't actually quite right (needs a factor in order to normalize it) and, in my notes anyway, the 'h' is actually an 'h-bar'.

As regards your other question, I can only assume that it is very sloppy notation because I have never seen a state use the same symbol as the wave function (although I am still an undergraduate so my experience is rather limited).

4. Oct 13, 2005

### Galileo

$\psi(x) = \langle x | \psi \rangle$ by definition. It is the component of |x> in the expansion of $| \psi \rangle$ in the position basis.

∫dp<x|p><p|Ψ> = ∫dp exp{ipx/h}Ψ(p), because $\langle x|p \rangle =\exp{ipx/\hbar}$ (apart from a constant factor) and <p|Ψ>=Ψ(p), again by definition.