Schrodinger equation in 1 dimension Question details in the image?

In this image E = V + KE
KE = #1/2mv^2#
can you prove from this

#DeltaxDeltap >= h/(4pi)#

www.physlink.com
I try to solve this but instead end up proving that probablity of finding electron in all space is 1

#p^2 * ψ^2 = p^2#
This is the Schrodinger equation in 1 dimension

I'm trying to prove this from various forms of the Schrodinger equation

1 Answer
Feb 3, 2018

Starting from

#DeltaxDeltap_x >= ℏ"/"2#

isn't actually enlightening. In fact it adds hours of pain and suffering because you have to derive it from scratch to find the form

#DeltaxDeltap_x >= 1/2 |i << psi | [hatx"," hatp_x] | psi >> |#

where #[hatx, hatp_x] = hatxhatp_x - hatp_xhatx = iℏ#.

That would have shown that for normalized wave functions,

#DeltaxDeltap_x >= 1/2 |i cdot i ℏ |#

#>= ℏ/2 sqrt(i^"*"i)#

#>= ℏ/2#

Just know that the operators are:

#hatx = x#

#hatp_x = -iℏd/dx#

which satisfy #[hatx, hatp_x] = hatxhatp_x - hatp_xhatx = iℏ#.

This is consistent with any Schrodinger equation in one Cartesian dimension, and is easy to prove. In fact, I have done it several times, and here is one of those times.

https://socratic.org/questions/in-heisenberg-s-uncertainty-principle-is-the-uncertainty-in-the-position-a-2-dim?source=search

Here is an easy proof to go from here...

We know:

  • #p = mv#
  • #K = 1/2mv^2#

Thus,

#hatK = hatp^2/(2m) = (-iℏ)^2 /(2m) del^2/(delx^2) = -ℏ^2/(2m) (del^2)/(delx^2)#

So for any Schrodinger equation in one Cartesian dimension...

#hatH psi = Epsi#

#= (hatK + hatV)psi#

#= (-ℏ^2/(2m) (del^2)/(delx^2) + hatV)psi#

Thus,

#-ℏ^2/(2m) (del^2psi)/(delx^2) + hatVpsi = Epsi#

#(del^2psi)/(delx^2) - (2m)/(ℏ^2)[hatV-E]psi = 0#

#(del^2psi)/(delx^2) + (2m)/(ℏ^2)[E-hatV]psi = 0#

#color(blue)((del^2psi)/(delx^2) + (8pi^2m)/(h^2)[E-hatV]psi = 0)#

This of course does NOT tell us what the potential is. That's your responsibility to tell us.