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Linear combination of states

  1. Apr 23, 2003 #1
    The following linear combination of states is considered in almost all quantum mechanics textbooks when they try to explain the addition of spin 1/2 and orbital angular momentum. The thing I don't understand is how the left hand side is equal to the right. Please, if you can, explain how.

    |j,m + 1/2> = a|lm,+> + b|lm,-> where |+> is spin up and |-> is spin down.


    Thanks,

    Sam
     
  2. jcsd
  3. Apr 23, 2003 #2

    Tom Mattson

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    Write the spin and orbital states separately, and the total as a direct product of the two, as follows:

    |J,MJ>=|L,ML>*|1/2,1/2>

    Then use the ladder operators to work your way down.

    Example:

    L=1, S=1/2

    |3/2,3/2>=|1,1>*|1/2,1/2>

    Apply the ladder operator J-=L-+S-, noting that on the RHS the operator only acts on its respective ket.

    J-|3/2,3/2>=(L-|1,1>)*|1/2,1/2>+|1,1>*(S-|1/2,1/2>)

    and keep working down.

    Try to work out the combination for |J=3/2,MJ=1/2>. If you get stuck, post what you came up with and I'll help you through it.
     
  4. Apr 23, 2003 #3

    Tom Mattson

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    What the hell, I'm feeling ambitious!

    J-|3/2,3/2>=(L-|1,1>)*|1/2,1/2>+|1,1>*(S-|1/2,1/2>)

    RHS:
    J-|3/2,3/2>=31/2(hbar)|3/2,1,2>

    LHS:
    (L-|1,1>)*|1/2,1/2>+|1,1>*(S-|1/2,1/2>)=
    21/2(hbar)|1,0>*|1/2,1/2>+(hbar)|1,1>*|1/2,-1/2>

    Putting them together and solving for |3/2,1/2> yields:

    |3/2,1/2>=(2/3)1/2|1,0>*|1/2,1/2>+(1/3)1/2|1,1>*|1/2,-1/2>

    Try the rest, and let me know if you need help.

    Remember: Once you get to J=1/2, you will have a parity change. See me if you need help on that, too.
     
  5. Apr 23, 2003 #4
    Thanks Tom for your time and help.

    Sam
     
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