Question on angular momentum of triplet and singlet wavefunctions

[The_Foetus]

Hi, I'm relatively new to QM so just a basic explanation of my problem would be amazing!

I'm doing some internet research on superfluidity over my summer holiday, and was looking specifically at ³He, and the way it forms Cooper pairs. Having read a classical analogy to why the relative angular momentum of the two He atoms must not be 0, I then read that this excludes the possibility of having the wavefunction (|+,-> - |-,+>)/√2, where + and - represent up and down spin respectively.

This leaves then a possible triplet wavefunction of
ψ = a*|+,+> + b*(|+,-> + |-,+>)/√2 + c*|-,->
where a, b, c are constants.

My question is why the |+,-> + |-,+> state has a non-zero angular momentum, but the |+,-> - |-,+> state has l = 0.

Simple question I know, but would really appreciate an explanation.
Regards,
Foetus

 

[atyy]

I'm not sure there is a short explanation. It has to do with angular momentum or spin addition, and something called the Clebsch-Gordon coefficients. There is an explanation in http://homepage.univie.ac.at/reinhold.bertlmann/pdfs/T2_Skript_Ch_7.pdf , where the answer to your question is reached in Eq 7.58 and 7.59.

 

[Meir Achuz]

My question is why the |+,-> + |-,+> state has a non-zero angular momentum, but the |+,-> - |-,+> state has l = 0.

The raising operator J_+ gives zero when acting on that state.

 

[Avodyne]

But that's not enough information to conclude that the total angular momentum is zero. (In QM jargon, not intended for the OP, J_+ annihilates any state with m=+j.)

However, it's also true that J_x, J_y, and J_z all give zero when acting on this state. So it is an eigenstate of each component of the angular momentum operator with eigenvalue zero.