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Matrix of angular momentum operator

  1. Nov 21, 2007 #1
    as known to all, we can find a matrix representation for every operator in quantum mechanics.

    for example for total angular momentum of one particle j(square) the elements are j(j+1)(square)h(bar) δmm'

    However I have stucked in two particle systems.

    for example I could not find the matrix of j1+j2- (this is a product) here j1+ is the raising operator for first particle and j2- is the lowering operator for second one.
    normally for one particle raising angular momentum operator gives the eigen value (squareroot)[j(j+1)-m(m+1)].
    but in this case as far as i know, i have to find the matrix representration of product of this two operator. but for the below conditions I could not create a matrix.
    lets say j1=2 j2=1 and the restriction is m= m1 +m2 = 2. that is m1 can take values 2,1 and coresponding m2 values are 0 and 1.
    can you help me about this?
  2. jcsd
  3. Nov 21, 2007 #2
    You must first write down your basis states, e.g. you can take the product states

    |1> = |j1=2, m1=2>|j2=1,m2=0>

    |2> = |j1=2, m1=1>|j2=1,m2=1>

    If we put A = J1^{+} J2^{-}, then the matrix elements are

    A_{i,j} = <i|A|j>


    A_{1,2} =

    <|j1=2, m1=2|<j2=1,m2=0|J1^{+} J2^{-}
    |j1=2, m1=1>|j2=1,m2=1>

    We have:

    J1^{+} J2^{-}|j1=2, m1=1>|j2=1,m2=1> =

    (J1^{+} |j1=2, m1=1>) (J2^{-}|j2=1,m2=1>) =

    2sqrt(2)h-bar^2|j1=2, m1=2>|j2=1, m2=0>

    And we see that A_{1,2} = 2sqrt(2)h-bar^2
  4. Nov 21, 2007 #3
    is that a diagonal matrix or an off diagonal matrix.
    you took the state 1 and state 2 to form the matrix of the operator A and after you applied the operators to the state of 2 you got the state of 1 then the kronecker delta gave you what? a diagonal matrix or what?

    thanks by the way for your answer.
  5. Nov 21, 2007 #4
    or could you just write the elemts of this 2x2 matrix.
    thanks alot.
  6. Nov 21, 2007 #5
    ah okey just understood
    thnaks very much for your kindness.
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