1. Not finding help here? Sign up for a free 30min 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!

Is this right? Re: Finding expectation value of L_z

  1. Aug 7, 2009 #1
    Okay, so I'm now reviewing ladder operators (no, not homework).

    While reviewing a quantum problem involving the L_z operator at this website (http://quantummechanics.ucsd.edu/ph130a/130_notes/node219.html#example:expectLz"), I found myself confused.

    Okay, here's my question: don't we need to take the complex conjugate of the wavefunction for the bra? They have -i for both the bra and the ket...

    Am I wrong? It's been a while since I took quantum, so it's possible I'm not recalling some stuff properly...

    Thanks,
    Geez

    PS...Sorry, I don't know how to TeX (but feel free to teach me or show me some good resources to learn), so I don't know how else to show you what I'm talking about.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Aug 7, 2009 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Yes, they do and you are right. However, the final answer is correct but inconsistent with the work shown. If you multiply things out (i.e. do the angular integrals) according to the second line, you should get 2/3+1/3 not 2/3-1/3 because (-i)*(+i) = +1 for the second term. It's an honest mistake that does not affect the bottom line.
     
  4. Aug 7, 2009 #3
    But shouldn't the final answer, then, be 2/3 h-bar + 1/3 h-bar, which is just one unit of h-bar?
     
  5. Aug 7, 2009 #4

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    No, the final answer is (2/3 - 1/3) h-bar = (1/3)h-bar, as stated. They should have +i in the ket. Then this should become -i*hbar when operated on by Lz.

    When terms are multiplied out, the second term becomes

    (-i)(-i)(1/3)hbar = -(1/3)hbar​

    And the +(1/3)hbar final answer makes sense, since in the original wavefunction there is probability (2/3) to measure Lz=+hbar, and probability (1/3) to measure -hbar. (So the expectation value is +(1/3)hbar.)
     
  6. Aug 7, 2009 #5
    Okay. I see my error now. I had multiplied the "i's" correctly, but had neglected to apply the m_l, which is -1.

    Thanks, everyone!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Is this right? Re: Finding expectation value of L_z
Loading...