# psi(x) = 1/(sqrt(2pibarh))intPhi(p)e^(1/hpx)dp Is Phi = e^(im_lphi)?

## Is $\Phi = {e}^{i {m}_{l} \phi}$?

Feb 8, 2018

I assume you are looking at the Fourier transform from momentum space to position space. $\Phi$ is NOT what you say it is.

A plane wave is written as the overlap of the momentum ket vector with a complete set of position states:

<< x | p_x >> = 1/sqrt(2piℏ) e^(ip_x x//ℏ)

The wave function in position space in the Schrodinger formulation is written in the Heisenberg formulation as:

${\psi}_{n} \left(x\right) = \left\langlex | {\psi}_{n} \left(x\right)\right\rangle \equiv \left\langlex | n\right\rangle$

and in momentum space:

$\Phi \left({p}_{x}\right) = \left\langle{p}_{x} | {\phi}_{n} \left({p}_{x}\right)\right\rangle \equiv \left\langle{p}_{x} | n\right\rangle$

Insert unity to obtain:

$\textcolor{b l u e}{{\psi}_{n} \left(x\right)} = {\int}_{0}^{\infty} \left\langlex | {p}_{x}\right\rangle \left\langle{p}_{x} | n\right\rangle {\mathrm{dp}}_{x}$

= color(blue)(1/sqrt(2piℏ)int_(0)^(oo) e^(ip_x x//ℏ) Phi(p_x) dp_x)

And similarly,

$\textcolor{b l u e}{\Phi \left({p}_{x}\right)} = {\int}_{0}^{\infty} \left\langle{p}_{x} | x\right\rangle \left\langlex | n\right\rangle \mathrm{dx}$

= color(blue)(int_(0)^(oo) e^(-ip_x x//ℏ) psi_n(x) dx)

This has absolutely NOTHING whatsoever to do with the hydrogen atom. Nothing. This is a Fourier transform between momentum and position space eigenfunctions.