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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Mimic on August 18, 2021, 04:20:35 PM

Suppose we consider the first excited state of the helium atom. We know that the first excited state of helium can exist as a triplet or singlet. The possible functions related to the spin of the two electrons in the triplet state are
[tex]\alpha(1)\alpha(2)[/tex][tex]\beta(1)\beta(2)[/tex][tex]\dfrac{1}{\sqrt{2}} [\alpha(1)\beta(1) + \beta(1)\alpha(2) ][/tex]
while the one for the singlet state is [tex]\dfrac{1}{\sqrt{2}} [\alpha(1)\beta(1)  \beta(1)\alpha(2) ][/tex]
The triplet state predicts that the spins of the two electrons are parallel, but according to this equation[tex]\dfrac{1}{\sqrt{2}} [\alpha(1)\beta(1) + \beta(1)\alpha(2) ] [/tex]
there is a 50% probability that electron 1 is in the alpha state and a 50% probability that it is in the beta state: the same goes for electron 2.
So, if this function predicts that the two spins are antiparallel, why is it part of one of the triplet states?

It's called a triplet state because it includes three degenerate states that split into three nondegenerate states when placed in an external field  one of them is aligned parallel with the field, one of them aligned antiparallel with the field, and one that doesn't interact with the field (or, the interactions of the two electrons cancel out). Three microstates is also required by the wavefunction solutions to the wave equation.
Another way to look at it is that the three microstates comprising the triplet state all have the same symmetry (symmetric) whereas the one that makes up the singlet state is antisymmetric. This also has implications on the symmetry of the spatial wavefunctions and their energy eigenvalues.

I try to explain better: can you see this image?
(https://upload.wikimedia.org/wikipedia/commons/4/49/ISC_excited_states.png)
If one of the triplet excited states has one alpha and one beta electron, shouldn't this be the configuration? 1s :spinup: 2s :spindown:

The diagram is only a handy guide and should not be taken too literally. The arrows help to understand what the total spin is but it doesn't provide any information on the component states of the system. The triplet state has a total spin of S = 1 and includes three microstates that have, respectively, m_{s} = 1, 0, and 1.

The triplet state predicts that the spins of the two electrons are parallel, but according to this equation
[tex]\dfrac{1}{\sqrt{2}} [\alpha(1)\beta(1) + \beta(1)\alpha(2) ] [/tex]
there is a 50% probability that electron 1 is in the alpha state and a 50% probability that it is in the beta state: the same goes for electron 2.
Why ? Second electron is not assigned to the beta state at all.
It should be not rather a:
[tex]\dfrac{1}{\sqrt{2}} [\alpha(1)\beta(2) + \alpha(2)\beta(1)][/tex]
?
If one of the triplet excited states has one alpha and one beta electron, shouldn't this be the configuration? 1s :spinup: 2s :spindown:
I found that singlet state is symmetric combination of 1s :spinup: 2s :spindown: and 1s :spindown: 2s :spinup: while triplet  antisymmetric combination of the same (whatever this means).
The triplet state has a total spin of S = 1 and includes three microstates that have, respectively, m_{s} = 1, 0, and 1.
Why ? m_{s} should not be restricted to +1/2 and 1/2 ?

I found that singlet state is symmetric combination of 1s :spinup: 2s :spindown: and 1s :spindown: 2s :spinup: while triplet  antisymmetric combination of the same (whatever this means).
I need to correct  singlet is antisymmetric combination, while triplet  symmetric.
But you are right  for triplet state there is the same probability that electron 1 and 2 have alfa or beta sign.

@Orcio
There are two electrons. Each one has possible spin of ±1/2, so the total spin for the state is S = 1 and M_{s} values of 1, 0, 1.
For this configuration there is also a singlet state. The best way to see this is using term symbols after considering all the possible micro states, but that may be beyond the OPs interest

But you are right  for triplet state there is the same probability that electron 1 and 2 have alfa or beta sign.
Perhaps some (small) excess of one over the other (look into Boltzmann distributions etc) but very little in it?