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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

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    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
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