2
$\begingroup$

In the Nature article:

Spin-triplet superconductivity in $Sr_2RuO_4$ identified by ${^{17}}$O Knight shift https://www.nature.com/articles/25315

Authors state:

For a spin-triplet p-wave superconductor in the tetragonal, quasi-two-dimensional lattice like $Sr_2RuO_4$, the most stable states in the weak-coupling approximation are the equal-spin pairing states. If one can neglect the spin–orbit coupling, spin susceptibility in these states remains constant in any field direction even below $T_c$.

I understand that one consequence is that there is no Knight Shift (KS) for the triplet state.

My question is why is this the case that singlet Cooper pairs (S=0) are affected by the external magnetic field hence show the KS but triplet (S=1) does not?

$\endgroup$

1 Answer 1

2
$\begingroup$

Triplet super-fluidity is in general more complicated then this and can show a knight shift. But in the simple case of the "equal-spin pairing state", the 3 triplet pairing states are all degenerate. This means when you apply a magnetic field, you lift this degeneracy, and the whole condensate condenses in the S=1 state aligned in the direction of the magnetic field. Therefore a magnetic field can coexist with the equal-spin pairing state.

In the singlet super conductor, the cooper pairs don't have this degeneracy and can't adapt to the magnetic field. Therefore, beneath the critical field strength, they expel magnetic fields, thus the knight shift.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.