1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Quantum Mechanics: Harmonic Oscillator Variance.

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data
    The problem wants me to calculate (Δx)^2 and (Δp)^2 to find the uncertainty principle. Delta x is the variance and the problem gives the formula as..

    [tex]Δx= <n|x^{2}|n>-<n|x|n>^{2}[/tex]

    2. Relevant equations
    [tex]x=\sqrt{\frac{\hbar}{2m \omega}}(A^{-}+A^{+})[/tex]
    Where A+ and A- are the raising and lowering operator respectively.

    3. The attempt at a solution

    So I plugged in this expression for x into the above expression for the variance.
    I have the answer in front of me, the math for a similar argument (when we have two different eigenvalues, k and n) is..

    http://imageshack.us/a/img266/5854/qmproblem3.png [Broken]

    Meaning, that it vanishes unless n=k+1 or k-1

    My main confusion is this..

    First we have..
    So I can distribute and get...
    But since A raises or lowers the eigenvalue...

    Why does the image I linked to change the eigenvector? I thought it only changed the eigenvalue. With the answer I got n would have to be equal to k for it not to vanish.

    If I can understand this concept I can do the problem without any problem.

    When I was trying to find [itex]<n|x^{2}|n>[/itex]
    I came to.. (Using a commutation relation)
    [tex]\frac{m \omega \hbar}{2}<n|(A^+)^2+(A^-)^2+ 2A^{+}A^{-}-1|n>[/tex]
    Which I then distributed as I did above, and it boils down to the same math.
    This is supposed to be
    [tex]\frac{m \omega \hbar}{2}(2n+1)[/tex]

    Can anyone help?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 29, 2012 #2
    Nevermind! I realized that it does indeed change the eigenvector as well as the eigenvalue. Of course this must be true because it changes the state of the system, and you cannot have an eigenvector of one state with an eigenvalue of another. With this in mind, I have gotten the problem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook