# 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?