Solving Hydrogen Molecule Basis Functions

[rasko]

A molecule of hydrogen. We assume that the only possible states of
its two electrons (indistinguishable) are $$|A\uparrow\rangle,<br /> |A\downarrow\rangle, |B\uparrow\rangle$$ and $$|B\downarrow\rangle$$
(A=1s-orbital at atom A, B=1s-orbital at atom B).
Formulate the 6 basis functions using the four possible single
particle states above. (Don't forget the Pauli-principle!)


Here is my solution:

Total spin S=1 or 0.
$$|1,1\rangle=|A\uparrow\rangle|B\uparrow\rangle$$,
$$|1,0\rangle=\frac{1}{\sqrt{2}}(|A\uparrow\rangle|B\downarrow\rangle+|A\downarrow\rangle|B\uparrow\rangle)$$,
$$|1,-1\rangle=|A\downarrow\rangle|B\downarrow\rangle$$,
$$|0,0\rangle=\frac{1}{\sqrt{2}}(|A\uparrow\rangle|B\downarrow\rangle-|A\downarrow\rangle|B\uparrow\rangle)$$.

Are they basis functions? There should be 6 but I got only 4.

 

[Mute]

You've written down spin triplet and spin singlet states, and those would be all you had if your single particle states were $\left|\uparrow\right>$ and $\left|\downarrow\right>$, but your states aren't labelled by just the spin - you have labels A and B too. However, all of the states you have written are pairs A and B. What about AA or BB pairs?

 

[rasko]

Hi, Mute. I think there should be no AA or BB pairs. Because at the same time A or B has only one state.

 

[2Tesla]

Why do you think that? It's not given anywhere in the problem statement that each atom can only have one electron.

Remember that the Pauli exclusion principle only rules out two indist. particles in the *exact* same state...

 

[rasko]

2Tesla and Mute, thanks. I understand now.

|0,0>=|A up>|A down>
|0,0>=|B up>|B down>

Thanks u for ur tips.