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Angular momentum commutation relations

  1. Jun 1, 2017 #1
    1. The problem statement, all variables and given/known data
    Show that ##|l, m\rangle## for ##l=1## vanishes for the commutator ##[l_i^2, l_j^2]##.

    2. Relevant equations
    ##L^2 = l_1^2 + l_2^2 + l_3^2## and ##[l_i^2,L^2]=0##

    3. The attempt at a solution
    I managed to so far prove that ##[l_1^2, l_2^2] = [l_2^2, l_3^2] = [l_3^2, l_1^2]##. I know that for ##l=1##, I have ##m=-1,0,1## but I'm not really sure how to proceed from here though. Any tips?
     
  2. jcsd
  3. Jun 1, 2017 #2

    BvU

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    What is the complete problem statement ? Or: what do you mean when you write ##|l, m\rangle## vanishes ??
     
  4. Jun 1, 2017 #3
    That is the full question. I believe it means that the commutator acting on the state ##|l,m\rangle## i.e. ##\langle m, l| [l_i^2, l_j^2] |l, m \rangle = 0##. Of course ##l## and ##m## are the usual quantum numbers corresponding to ##L^2## and ##l_3##
     
  5. Jun 2, 2017 #4

    BvU

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    Ah, clear. Sorry to have pulled you off the unanswred threads list :oops:

    So the thing to do is work out ##\left ( l_i^2 l_j^2 - l_j^2 l_i^2 \right ) \left | 1,m \right \rangle ## and conclude that it's zero :rolleyes:|
    But you knew that already.

    I tried working them out but had a hard time keeping it from getting longer and longer. Still looking for a shorter path through.
     
  6. Jun 2, 2017 #5
    That's okay, thank you for helping. I guess the idea is to somehow write ##l_1## and ##l_2## in terms of the ladder operators but some basic manipulations haven't really led me anywhere.
     
  7. Jun 2, 2017 #6

    BvU

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    Same idea here. I'm looking at $$ \begin{align*} L_- L_+ & = L_x^2 + L_y^2 +i \left ( L_x L_y - L_y L_x \right ) \\ & = L^2 -L_z^2 -\hbar L_z \end{align*}$$which is from Ballentine (7.9) but haven't worked it towards our commutator yet.
     
  8. Jun 2, 2017 #7

    DrClaude

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    You can prove that ##\langle l, m| [l_i^2, l_j^2] |l, m \rangle = 0## when i or j is equal to 3 in about two lines by calculating explicitly the commutator and using the fact that ##|l, m \rangle## is an eigenfunction of ##l_3##.

    Combining that with
    should complete the proof.
     
  9. Jun 2, 2017 #8
    I feel silly for not trying the obvious. Thank you Dr. Claude.
     
  10. Jun 3, 2017 #9

    BvU

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    For the admiring spectators (including me :rolleyes:): can you show it ?
     
  11. Jun 3, 2017 #10
    Of course. Let's start with the ##[l_2, l_3]## commutator

    ##
    \begin{align*}
    \langle m, l | [l_2^2, l_3^2]| l,m \rangle &= \langle m, l | l_2^2l_3^2 - l_3^2l_2^2| l,m \rangle \\
    &= \langle m, l | l_2^2l_3^2| l,m \rangle - \langle m, l | l_3^2l_2^2| l,m \rangle \\
    \end{align*}
    ##

    Utilize the fact that ##l_i## are Hermitian and the eigenvalue equation ##l_3| l,m \rangle = m| l,m \rangle##, we get
    ##
    \begin{align*}
    \langle m, l | l_2^2l_3^2| l,m \rangle - \langle m, l | l_3^2l_2^2| l,m \rangle &= m^2(\langle m, l | l_2^2| l,m \rangle - \langle m, l | l_2^2| l,m \rangle) \\
    & = 0
    \end{align*}
    ##

    As for the commutator result, we have
    ##
    \begin{align*}
    [l_1^2, l_2^2] &= [l_1^2, L^2 - l_1^2 - l_3^2] \\
    &= -[l_1^2, l_3^2] \\
    & = [l_3^2, l_1^2],
    \end{align*}
    ##
    where we use that any component of angular momentum commutes with the total angular momentum. A similar proof can be done for the ##[l_2^2, l_3^2]## commutator.

    Also, correct me if I am wrong but it appears that this is true for all ##|l,m\rangle##, not just the specific case of ##l=1##.
     
    Last edited: Jun 3, 2017
  12. Jun 7, 2017 #11

    DrClaude

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    You indeed never need to make use of that. I wonder if this is a red herring or if the the person who wrote the question had another proof in mind, involving ladder operators (where square ladder operators only need to involve two states, ##|l, l\rangle## and ##|l, -l\rangle##).
     
  13. Jun 7, 2017 #12

    TSny

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    Maybe this means to show that ##[l_i^2, l_j^2] \, |l, m\rangle = 0## for ## l = 1##.

    You might consider using the matrix representation of the angular momentum operators for ##l = 1##.
    http://quantummechanics.ucsd.edu/ph130a/130_notes/node247.html
     
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