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Quantum Mechanics: Angular Momentum Operators

  1. Feb 28, 2015 #1
    1. The problem statement, all variables and given/known data

    Use the spin##-1## states ##|1,1\rangle, \ |1,0\rangle, \ |1, -1\rangle## as a basis to form the matrix representations of the angular momentum operators.


    2. Relevant equations

    ##\mathbb{\hat{S}}_+|s,m\rangle = \sqrt{s(s+1)-m(m+1)}\hbar|s,m+1\rangle##
    ##\mathbb{\hat{S}}_-|s,m\rangle = \sqrt{s(s+1)-m(m-1)}\hbar|s,m-1\rangle##


    3. The attempt at a solution

    I am wonder how exactly do I compute the equations listed above? So I have ##\langle1,1|\mathbb{\hat{S}}_+|1,1\rangle## but why would this equal ##0##.

    Also, ##\langle1,1|\mathbb{\hat{S}}_+|1,0\rangle = \sqrt{2}\hbar## and ##\langle1,0|\mathbb{\hat{S}}_+|1,0\rangle = 0##, why is that? How do I use the equations, given above, to substitute the given states?
     
  2. jcsd
  3. Feb 28, 2015 #2
    To find the matrix elements we can use these formulas (with ##s = 1##)
    ##\mathbb{\hat{S}}_+|s,m\rangle = \sqrt{s(s+1)-m(m+1)}\hbar|s,m+1\rangle##
    ##\mathbb{\hat{S}}_-|s,m\rangle = \sqrt{s(s+1)-m(m-1)}\hbar|s,m-1\rangle##
    then we "bra" them from the left by ##\langle 1, m' \mid## and use the orthogonality property
    $$\langle 1, m' \mid 1, m \rangle = \begin{cases} 1 & m = m' \\ 0 & m \neq m'\end{cases} $$

    So, for example
    ##\langle 1, 1|\mathbb{\hat{S}}_+|1,0\rangle = \sqrt{1(1+1)-0(0+1)}\hbar\langle 1, 1|1,1\rangle = \sqrt{2} \hbar##
     
  4. Feb 28, 2015 #3
    Can you elaborate please? So no matter the state, i.e. it could be a state ##\frac{3}{2}##, if ##m\ne m'## it will always be ##0##?
     
  5. Feb 28, 2015 #4
    Yes it will always be zero if they are not equal, regardless of ##s##.
     
  6. Feb 28, 2015 #5
    I see, thank you very much!
     
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