Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Minimum-uncertainty wave function -- contradiction?

  1. Dec 22, 2015 #1
    The following equation (5.122) is the minimum-uncertainty wave function, which is a Gaussian wave packet. Since it is Gaussian in ##x##, we may get ##\Delta x## directly from the ##\sigma## of the Gaussian distribution: ##(\Delta x)^2=\frac{\hbar^2}{2(\Delta p_x)^2}##. We have ##\Delta x\Delta p_x=\frac{\hbar}{\sqrt2}##, which contradicts the fact that the wave function has minimum uncertainty, ie., ##\Delta x\Delta p_x=\frac{\hbar}{2}##.

    Derivation of (5.122):
    Screen Shot 2015-12-23 at 5.08.19 am.png
    Screen Shot 2015-12-23 at 5.08.42 am.png

    Gaussian distribution:
    Screen Shot 2015-12-23 at 5.09.04 am.png
     
  2. jcsd
  3. Dec 22, 2015 #2

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    You should first calculate ##|\psi(x)|^2## and then extract ##\Delta x## from there, not from ##\psi(x)##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Minimum-uncertainty wave function -- contradiction?
Loading...