text: # How do i get the singlet state?
  
  1. Mar 18, 2005
  
  ### kingmob
  
  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:
  singlet:
  $$|s=0,m=0> = \frac {1} {\sqrt{2}}(|+-> - |-+>)$$
  
  Triplet:
  $$|s=1,m=1> = |++>$$
  $$|s=1,m=0> = \frac {1} {\sqrt{2}}(|+-> + |-+>)$$
  $$|s=1,m=-1> = |-->$$
  
  2. Mar 18, 2005
  
  ### juvenal
  
  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.
  
  3. Mar 21, 2005
  
  ### Galileo
  
  These states are the right states, because they are eigenfunctions of S=S1+S2 and S^2 with the right eigenvalues. (Check it).
  
  4. Mar 21, 2005
  
  ### Eye_in_the_Sky
  
  Have you looked at a proof of
  
  the FUNDAMENTAL THEOREM for the ADDITION OF ANGULAR MOMENTUM ?
  
  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 .
  
  5. Mar 21, 2005
  
  ### dextercioby
  
  The theorem of Clebsch-Gordan has other text.Please refer from using approximate (personally interpreted) formulations.Just pick a book.
  
  Daniel.
  
  6. Mar 22, 2005
  
  ### Eye_in_the_Sky
  
  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> = |-,-> .
  
  Next:
  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?
explanation: The document is a discussion thread from a physics forum where a user named kingmob asks about how to derive the singlet state for a two-electron system. The user provides the singlet and triplet states and expresses confusion about how to derive the singlet state. Several other users, including juvenal, Galileo, Eye_in_the_Sky, and dextercioby, provide explanations and insights to help kingmob understand the derivation.
  
  The document contains a question and multiple answers to that question.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # How do i get the singlet state?
  
  1. Mar 18, 2005
  
  ### kingmob
  
  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:
  singlet:
  $$|s=0,m=0> = \frac {1} {\sqrt{2}}(|+-> - |-+>)$$
  
  Triplet:
  $$|s=1,m=1> = |++>$$
  $$|s=1,m=0> = \frac {1} {\sqrt{2}}(|+-> + |-+>)$$
  $$|s=1,m=-1> = |-->$$
  
  2. Mar 18, 2005
  
  ### juvenal
  
  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.
  
  3. Mar 21, 2005
  
  ### Galileo
  
  These states are the right states, because they are eigenfunctions of S=S1+S2 and S^2 with the right eigenvalues. (Check it).
  
  4. Mar 21, 2005
  
  ### Eye_in_the_Sky
  
  Have you looked at a proof of
  
  the FUNDAMENTAL THEOREM for the ADDITION OF ANGULAR MOMENTUM ?
  
  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 .
  
  5. Mar 21, 2005
  
  ### dextercioby
  
  The theorem of Clebsch-Gordan has other text.Please refer from using approximate (personally interpreted) formulations.Just pick a book.
  
  Daniel.
  
  6. Mar 22, 2005
  
  ### Eye_in_the_Sky
  
  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> = |-,-> .
  
  Next:
  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?
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion thread from a physics forum where a user named kingmob asks about how to derive the singlet state for a two-electron system. The user provides the singlet and triplet states and expresses confusion about how to derive the singlet state. Several other users, including juvenal, Galileo, Eye_in_the_Sky, and dextercioby, provide explanations and insights to help kingmob understand the derivation.
  
  The document contains a question and multiple answers to that question.
  
  Label: question-and-answer