Spin Triplet of Ortho Helium - Why Total Spin is 1?

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For the ortho state of Helium, the spin part of the wavefunction is symmetric.

There are 3 possible symmetric wavefunctions we can construct from the 2 electrons' spins.

(^^)

(vv)

[1/SQRT2] { (^v) + (v^) }

I am confused as to why this last state gives a total spin of one? The electrons can only be in one of these eigenstates, both of equal probability of 1/2, but the spins are anti-aligned, are they not?

Thanks guys! (I hope you'll not mind my improvisation for electron spin functions with ^ and v ! :-p )

Cheers!
 
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Master J said:
For the ortho state of Helium, the spin part of the wavefunction is symmetric.

There are 3 possible symmetric wavefunctions we can construct from the 2 electrons' spins.

(^^)

(vv)

[1/SQRT2] { (^v) + (v^) }

I am confused as to why this last state gives a total spin of one? The electrons can only be in one of these eigenstates, both of equal probability of 1/2, but the spins are anti-aligned, are they not?

Thanks guys! (I hope you'll not mind my improvisation for electron spin functions with ^ and v ! :-p )

Cheers!

1. Learn to write with LaTeX code. \left|\uparrow\downarrow\right\rangle - it looks so pretty ! :!)

2. Well, to find the spin of a certain state, you must apply the total angular momentum operator squared. When you do that, you'll get the spin 1.

Remember that it's not \mbox{m}_{\mbox{j}} who gives the spin of the state, but \mbox{j}.
 
Ah, an oversight on my part! Thanks for clearing that up!
 
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