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I'm doing some internet research on superfluidity over my summer holiday, and was looking specifically at

^{3}He, 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