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: Angular momentum

  1. Apr 9, 2007 #1
    1. The problem statement, all variables and given/known data
    For the non stationary state

    [tex] \Psi = \frac{1}{\sqrt{2}} \left(Psi_{100}+\Psi_{110}\right) = \frac{1}{\sqrt{2}} \left(R_{10}Y_{00}e^{-iE_{10}t/\hbar}+R_{11} Y_{10}e^{-iE_{11}t/\hbar}[/tex]

    find [itex] <L^2>,<(L^2)^2>,<L_{z}>,<L_{z}^2>,\Delta L^2, \Delta L_{z}[/itex]

    2. Relevant equations
    [tex] <L^2>=\hbar^2 l(l+1)[/tex]
    [tex] <(L^2)^2>=(\hbar^2 l(l+1))^2[/tex]
    [tex] <L_{z}>=\hbar m_{l}[/tex]
    [tex] <L_{z}^2>=(\hbar m_{l})^2[/tex]
    [tex] \Delta x = \sqrt{<x^2>-<x>^2} [/tex]

    3. The attempt at a solution
    [tex]<L^2>=\frac{\hbar^2}{2} \left(0(0+1) + 1(1+1)\right)= \frac{3}{2} \hbar^2 [/tex]
    the answer is supposed to be hbar^2 ... what am i doing wrong...

    [tex] <(L^2)^2> = \frac{\hbar^4}{4} \left((0(0+1))^2+(1(1+1))^2\right) =\frac{5}{4} \hbar^2[/tex]

    [tex] <L_{z}> = \frac{\hbar}{2} (0+0) = 0 [/tex]

    [tex] <L_{z}^2} = \frac{hbar^2}{4} (0) = 0 [/tex]

    are the values right... i fear that i am sorely mistaken about how to calculate the expectation values

    any help would be greatly appreciated!!
  2. jcsd
  3. Apr 10, 2007 #2

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    For your first answer, try 1*2=2, and 0*1=0.
    Also 0*1=0 for the second answer.
    Back to arithmetic 101.
    Last edited: Apr 10, 2007
  4. Apr 10, 2007 #3

    stupid me
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook