# Identical particles: Why is triplet oxygen more stable than singlet oxygen?

Quoting from Eisberg Resnick Quantum Physics:

If we consider space variables of two electrons (identical particles) to have almost the same values, then their wavefunctions are 'almost' identical if they are in the same quantum state, ie, $\psi_{a}(1)~ \simeq~\psi_{a}(2)$ and $\psi_{b}(1)~\simeq~\psi_{b}(2)$ [the label 1 and 2 denote the spatial co-ordinates of the electron '1' and '2' i.e. ($x_1,y_1,z_1$) and ($x_2,y_2,z_2$), and the labels a and b for the wavefunction denote the three quantum numbers $n,l,m$ of two different quantum states].

In this case, the antisymmetric space eigenfunction describing the system of two electrons is

$$\frac{1}{\sqrt2}[\psi_{a}(1)\psi_{b}(2) - \psi_{a}(2) \psi_{b}(1)]\simeq\frac{1}{\sqrt2}[\psi_{b}(1)\psi_{a}(2) - \psi_{b}(1) \psi_{a}(2)]\simeq 0$$

and the symmetric space eigenfunction is

$$\frac{1}{\sqrt2}[\psi_{a}(1)\psi_{b}(2) + \psi_{a}(2) \psi_{b}(1)]\simeq\frac{1}{\sqrt2}[\psi_{b}(1)\psi_{a}(2) + \psi_{b}(1) \psi_{a}(2)]\simeq\sqrt2\psi_{b}(1) \psi_{a}(2).$$

Clearly symmetric space eigenfunction has more probability density, implying when two electrons have 'almost' same spatial co-ordinates, symmetric space eigenfunction and antisymmetric spin state(the singlet state) is more favourable as opposed to antisymmetric space eigenfunction and symmetric spin state(the triplet state), since the total eigenfunction has to be antisymmetric. This implies it is energetically more favourable for electrons to have symmetric space eigenfunctions which is why in solids, when atoms are brought together from infinite distance apart, energy splitting takes place such that in the energy band that forms, the lowest energy is that of symmetric space eigenfunction.

My question then, is, why is ortho oxygen (triplet oxygen) energetically more favourable (source: Wikipedia)? The reason I am asking is that ortho and para states of oxygen form from the unpaired electrons and not from the nucleons. So then why is it not the case that singlet state is more favourable in case of oxygen molecule when their electrons have 'almost' the same spatial co-ordinates?

PS: I did read about the molecular orbitals and I know the Hund's rules but the contradiction still bugs me.

• Possible duplicate of What is the physical basis for Hund's first rule? – orthocresol Oct 2 '16 at 9:00
• And to directly answer your issue: electrons repel each other more when they have almost the same spatial coordinate. This repulsion is a destabilising effect - so we want to avoid it as much as possible. So, the singlet is higher energy. – orthocresol Oct 2 '16 at 9:01
• So simple coulumb repulsion is the answer? What about all i quoted about the symmetric and antisymmetric space wavefunctions playing a role? – Prasad Mani Oct 2 '16 at 9:04
• Yes, simply larger Coulomb repulsion in singlet state, because electrons have higher probability of being close to each other (as can be seen from the eqns you wrote). Your sentence "...implying when two electrons have 'almost' same spatial co-ordinates, symmetric space eigenfunction... is more favourable as opposed to antisymmetric space eigenfunction..." is wrong. It is precisely the other way round, because same spatial coordinates means greater repulsion and that is not something we like having. Greater value of PDF does not imply greater stability, it means it is more likely to happen. – orthocresol Oct 2 '16 at 9:09
• Much clearer. one question though. I got that greater probability density does not mean greater stability. But when it does mean it is more likely to happen, why isnt it happening? I mean, in other words, that configuration which has less energy is 'more likely to happen' – Prasad Mani Oct 2 '16 at 9:35