How do we calculate spin functions and determine singlet and triplet states?

[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!

 

[vela]

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, S_z must be 0. When S=1, you can have S_z=-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.

 

[lemma]

with a little help from http://quantummechanics.ucsd.edu/ph130a/130_notes/node326.html your answer was helpful.

thank you very much!

 

[McKendrigo]

Hi,

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

Does this mean that the triplet state can have three degenerate m_s 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)?

 

[vela]

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