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How do i get the singlet state?

  1. Mar 18, 2005 #1
    In all my books the singlet and triplet state of a two-electron system seem to be postulated as obvious. The problem is that the sollutions somehow aren't obvious to me at all. I can see i can derive the s=1 states by applying a lowering operator to |s=1,m=1> =|++> But it doesn't help me in my understanding of the matter.
    Also, i can see that the singlet state is orthogonal to the other states, but this doesn't help me get to it myself.
    What is the thing i'm missing here?

    For completeness:
    [tex]|s=0,m=0> = \frac {1} {\sqrt{2}}(|+-> - |-+>)[/tex]

    [tex]|s=1,m=1> = |++>[/tex]
    [tex]|s=1,m=0> = \frac {1} {\sqrt{2}}(|+-> + |-+>)[/tex]
    [tex]|s=1,m=-1> = |-->[/tex]
  2. jcsd
  3. Mar 18, 2005 #2
    The whole point is that the singlet state is invariant under "rotations", unlike the triplet state. So invoking orthogonality is the only way to find it, I believe.
  4. Mar 21, 2005 #3


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    These states are the right states, because they are eigenfunctions of S=S1+S2 and S^2 with the right eigenvalues. (Check it).
  5. Mar 21, 2005 #4
    Have you looked at a proof of


    This is how the theorem goes:

    In the (2j1+1)(2j2+1)-dimensional space spanned by the vectors |j1,m1>|j2,m2> (with j1,j2 fixed, and m1,m2 variable), the possible values of j are

    j1+ j2 , j1+ j2-1 , ... , |j1- j2| ,

    and to each of these values there corresponds one, and only one, sequence of 2j+1 eigenvectors |j,mj> , mj = -j,...,j .

    For the answer to you your question, all you need to do is understand the case in which j1 = j2 = 1/2 .
  6. Mar 21, 2005 #5


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    The theorem of Clebsch-Gordan has other text.Please refer from using approximate (personally interpreted) formulations.Just pick a book.

  7. Mar 22, 2005 #6
    What does the theorem look like for the case j1 = j2 = ½ ? It looks like:

    In the 4-dimensional space spanned by the vectors |j1=½,m1>|j2=½,m2> (with m1,m2 = ± ½), the possible values of j are

    1,0 ,

    and to the j=1 value there corresponds one, and only one, sequence of 3 eigenvectors |1,1>, |1,0>, |1,-1>, and to the j=0 value there corresponds one, and only one, sequence consisting of a single eigenvector |0,0>.

    The members of a given sequence are related to one another by means of the raising or lowering operators. That is, for example, using the lowering operator J, we have for the j=1 sequence

    J|1,1> = √2 |1,0> ,

    J|1,0> = √2 |1,-1> ,

    J|1,-1> = 0 ;

    whereas, for the j=0 sequence, we have

    J|0,0> = 0 .

    These four |j,m> vectors form an orthonormal basis of the 4-dimensional joint spin-space. However, an alternative orthonormal basis is given by the original set of vectors { |j1=½,m1>|j2=½,m2> ; m1,m2 = ± ½ }, which in simplified notation is nothing but { |+,+>, |+,->, |-,+>, |-,-> }.

    It is trivial to check, and true in general, that the vector |j1,m1>|j2,m2> is an eigenvector of (total) Jz with eigenvalue m=m1+m2. This means that, when each of m1 and m2 takes on its largest allowed value (i.e. m1=j1, m2=j2), then the associated vector must correspond to the |j,m> vector given by |jmax,mmax>. Specifically, for the case at hand, this fact is expressed by

    |+,+> ↔ |1,1> .

    There is no difficulty in writing the above relationship as an actual equality. Therefore, we can write

    |1,1> = |+,+> .

    By similar reasoning, we are also able to write

    |1,-1> = |-,-> .

    So, then you can see

    |1,0> = (1/√2) J|1,1>

    = (1/√2) ( J1– + J2– ) |+,+>

    = (1/√2) ( J1–|+,+> + J2–|+,+> )

    = (1/√2) ( |-,+> + |+,-> ) .

    We have now, therefore, solved for the triplet.

    That's good. We've already used up 3 out of 4 dimensions, and since the normalized vector

    (1/√2) ( |-,+> – |+,-> )

    is orthogonal to all of the triplets, it must correspond to |0,0>, the singlet.

    And we are done.

    Do you still feel that something is missing?
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