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Spin states of the addition of 2 spin 1/2 particles

  1. Feb 9, 2010 #1
    Hi everyone,

    A QM question that im stuck on:

    Consider the addition of the spin of 2 electrons, s1=s2=½. Show
    that there are one ms=1; two ms=0 and one ms=-1 states. Show the
    4 spin states are given by:

    a1a2
    b1b2
    (1/(sqrt(2)))(a1b2+b1a2)
    (1/(sqrt(2)))(a1b2-b1a2)

    I've used a and b in place of alpha and beta, which represent the spin up and spin down states.

    I can see why there are one ms=1 and one ms=-1 states, and 2 ms=0 states, from adding the values of ms for individual particles, but am not sure on how to proceed to the second part.

    Any nudge in the right direction would be great.

    Cheers,

    Simon.
     
  2. jcsd
  3. Feb 9, 2010 #2

    SpectraCat

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    Science Advisor

    Check out this thread: https://www.physicsforums.com/showthread.php?t=376142"

    and the one referred to in it.
     
    Last edited by a moderator: Apr 24, 2017
  4. Feb 9, 2010 #3

    Avodyne

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    That is because you are not given enough information. Any linear combination of the last two states is a state with ms=0.

    Now, it happens that the first state has s=1 and the second state has s=0, and that is why these two particular linear combinations of ms=0 states are important. But the question does say anything about the value of s.
     
  5. Feb 9, 2010 #4
    Cheers for the help.


    The next part says:

    Indicate which values of S and ms correspond to each of the 4 states.

    So far what i've done is substituted the spin up (1 0 ) (<< vertically! dont know how to format that on here!) and spin down (0 1) into the states it says im looking for .

    e.g, (1/sqrt2)*(1 -1)

    hmmm...
     
  6. Feb 10, 2010 #5
    This area is very small to present these big issues. As you search on this topic you will go in deep. Online is best help for you. There you will find several e-books to solve your problems.
     
  7. Feb 10, 2010 #6

    DrDu

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    For the states aa and bb it should be clear that they can only belong to s=1. As they form a irreducible representation of the rotation group, the third state with s=1 should be obtainable from any of the two by applying a suitable rotation. How does a single spin transform under a rotation of 90 degrees? How does the product aa transform under the rotation?
     
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