# Homework Help: Singlet and triplet states

1. Aug 4, 2010

### lemma

Hi there,

I would be most grateful if someone would explain how do we calculate the spin function
χ(s_1,s_2 )=1/√2 [α(s_1 )β(s_2 )±α(s_2 )β(s_1 )]

both the symmetric and antisymmetric
α(s_1 )β(s_2 )+α(s_2 )β(s_1 ) = ?
α(s_1 )β(s_2 )-α(s_2 )β(s_1 ) = ?

knowing that α( + 1 / 2) = β( − 1 / 2) = 1 and α( − 1 / 2) = β( + 1 / 2) = 0.

And how do we get singlet (S=0, S_z=0) and triplet (S=1, S_z=+1,0,-1) states from this?

I have not found this explained explicitly anywhere, everybody seems to take it for granted, and it is really bothering me.

Thank you very much for possible clarification!

2. Aug 4, 2010

### vela

Staff Emeritus
The rules for the addition of angular momenta tell you the allowed values for S. Since you have two spin-1/2 particles, the allowed values for S are 0 and 1. When S=0, Sz must be 0. When S=1, you can have Sz=-1, 0, or 1.

To see how the $|s_1,m_1; s_2,m_2\rangle$ states, which I'll label using arrows below, combine to form the $|S,S_z\rangle$ states, what you do is start with the $|1,1\rangle$ state. For the z-component to be conserved, both particles must be spin-up.

$$|1,1\rangle = |\uparrow\,\uparrow\,\rangle$$

Now to find $|1,0\rangle$, you apply the lowering operator to get

$$|1,0\rangle = (|\uparrow\,\downarrow\,\rangle + |\downarrow\,\uparrow\,\rangle)/\sqrt{2}$$

and if you apply the lowering operator again, you get

$$|1,-1\rangle = |\downarrow\,\downarrow\,\rangle$$

The remaining state $|0,0\rangle= (|\uparrow\,\downarrow\,\rangle - |\downarrow\,\uparrow\,\rangle)/\sqrt{2}$ is determined by requiring it to be orthogonal to the other three.

P.S. I don't know if applying the lowering operator actually results in the correct normalization, but if it doesn't, just assume I normalized the result after lowering the state.

Last edited: Aug 4, 2010
3. Aug 5, 2010

4. Jan 5, 2011

### McKendrigo

Hi,

I'm getting myself confused about the physical meaning of Sz = -1, 0, 1

Does this mean that the triplet state can have three degenerate ms energy levels, with spin quantum number = -1 (i.e. two "spin down" electrons), 1 (two "spin up electrons") or 0 (spin up/spin down on the HOMO and spin down/spin up on the LUMO)?

And the singlet state can have only one energy level where the spin quantum number = 0 (i.e. two electrons with opposite spin)?

5. Jan 5, 2011

### vela

Staff Emeritus
If you're asking if the quantum number Sz is another way to write ms, the answer is yes. Whether the states are degenerate depends on the Hamiltonian.