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Why is the spin of this state s=1 then? What am I missing?

Do hunds rules not say that the S= sum of all m_s values?

I've just come across the spin states of a two electron system. There are 4 states possible and I am a little confused as to why the state below has values of s=1 m_s=0? [1/√2]{α(1)β(2)+α(2)β(1)} where α(i) and β(i) tell us if the particle has +ve or -ve z component of spin respectively. I...

That's exactly it, thank you :). I didn't realize that you could separate the operators out. It's part of the derivation for finding the energy changes in perturbation theory.

I don't understand why the signs change at all.

Yeh that's part of the proof but that's not what I don't understand. Why do the operations flip the sign of the eigenvalues. I'd have thought <una|VA-AV|unb>=(ana-anb)<una|V|unb>

I'm going through a derivation and it shows: (dirac notation) <una|VA-AV|unb>=(anb-ana)<una|V|unb> V and A are operators that are hermition and commute with each other and ana and anb are the eigenvalues of the operator A. I imagine it is trivial and possibly doesn't even matter but why does...

Sorry I've never seen this notation before so I wasn't aware of the relationship. So why do we have L=P then?

So the angular momentum quantum number is l=1 as it's a P state? How is the value of s determined?

That's the terminology in the question yes. I don't quite understand what 4P means though? What do you mean by term symbol?

If an atom described by spin-orbit coupling is in a 4P state, why is the spin quantum number s=3/2?