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

Is #Phi = e^(im_lphi)#?

1 Answer
Truong-Son N.
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(x) = << x | psi_n(x) >> -= << x | n >>#

and in momentum space:

#Phi(p_x) = << p_x | phi_n(p_x) >> -= << p_x | n >>#

Insert unity to obtain:

#color(blue)(psi_n(x)) = int_(0)^(oo) << x | p_x >> << p_x | n >> dp_x#

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

And similarly,

#color(blue)(Phi(p_x)) = int_(0)^(oo) << p_x | x >> << x | n >> 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.

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