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Question on angular momentum of triplet and singlet wavefunctions

  1. Aug 22, 2014 #1
    Hi, I'm relatively new to QM so just a basic explanation of my problem would be amazing!

    I'm doing some internet research on superfluidity over my summer holiday, and was looking specifically at 3He, and the way it forms Cooper pairs. Having read a classical analogy to why the relative angular momentum of the two He atoms must not be 0, I then read that this excludes the possibility of having the wavefunction (|+,-> - |-,+>)/√2, where + and - represent up and down spin respectively.

    This leaves then a possible triplet wavefunction of
    ψ = a*|+,+> + b*(|+,-> + |-,+>)/√2 + c*|-,->
    where a, b, c are constants.

    My question is why the |+,-> + |-,+> state has a non-zero angular momentum, but the |+,-> - |-,+> state has l = 0.

    Simple question I know, but would really appreciate an explanation.
  2. jcsd
  3. Aug 22, 2014 #2


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    Last edited: Aug 22, 2014
  4. Aug 22, 2014 #3

    Meir Achuz

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    The raising operator J_+ gives zero when acting on that state.
  5. Aug 22, 2014 #4


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    But that's not enough information to conclude that the total angular momentum is zero. (In QM jargon, not intended for the OP, J_+ annihilates any state with m=+j.)

    However, it's also true that J_x, J_y, and J_z all give zero when acting on this state. So it is an eigenstate of each component of the angular momentum operator with eigenvalue zero.
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