How do i get the singlet state?

[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> = |-->

[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.

[Galileo]

These states are the right states, because they are eigenfunctions of S=S1+S2 and S^2 with the right eigenvalues. (Check it).

[Eye_in_the_Sky]

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.

... What is the thing i'm missing here?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 .

[dextercioby]

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

Daniel.

[Eye_in_the_Sky]

Have you looked at ...

... For the answer to you your question, all you need to do is understand the case in which j1 = j2 = 1/2What 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:


I can see i can derive the s=1 states by applying a lowering operator to |s=1,m=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.



Also, i can see that the singlet state is orthogonal to the other statesThat'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.



... but this doesn't help me get to it myself.
What is the thing i'm missing here?Do you still feel that something is missing?