6
$\begingroup$

So far I learned that electrons of a Cooper pair should have opposite spin to account for Pauli's exclusion principle, because their other quantum numbers are the same. Is this understanding correct?

If so, I have been reading the review Linder, J., & Robinson, J. W. A. (2015). Superconducting spintronics, Nature Physics, 11(4), 307–315, also on arXiv:1510.00713. I came across how a magnetic inhomogeneity can transform single Cooper pairs into triplets by spin mixing and then by spin rotation transform the unpolarised ($S_{z} = 0$) spin triplet into spin-polarised spin triplets $\left|\uparrow\uparrow\right\rangle $ or $\left|\downarrow\downarrow\right\rangle $. 1. How can such spin rotation be achieved? Just an alignment of the spin pairs along the local direction of the inhomogenous field? 2. Does this not violate Pauli's exclusion principle?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

The spin part may have become symmetric, but no problem if the spatial part goes antisymmetric. Spatial space isn't useful for quasiparticles in condensed matter; momentum space is better. As the momentum vector makes a trek around the Fermi surface, the gap function changes sign. For the drab usual case of s-wave correlation, there is nothing much to say. For p-wave, the spatial part is taking care of Pauli for you.

Since in crystalline matter, the relation between frequency and wavevector is not simple, and is really a multiplicity of relations due to the different bands, one can find cases of odd in space, and also odd in time. Then it's back to singlets for the spin. So, I must correct how I started this answer: Either the spatial part or the time part of the pair must go antisymmetric. But not both.

As for how these things get created, each electron has its [math]B\dot \sigma[/math] term. The magnetic field is different in different places. As sure as a pothole cause my car to get bent out of shape, so does a nonuniform magnetic field bend a Cooper pair out of shape. Different torques on opposite spins. As I understand it, and I probably don't have this right, the result can be thought of as a quantum superposition of a nice straight singlet-spin Cooper pair and a triplet-spin one.

Improve this answer
$\endgroup$
1
$\begingroup$

I partially answered this question there : https://physics.stackexchange.com/a/62364/16689 : The triplet superconductivity does not violate the Pauli's exclusion principle because the wave function of the Cooper pair contains an orbital and a spin component. A triplet component is characterised by a symmetric spin part, and so the spatial component must be anti-symmetric in the exchange of the position of the electrons forming the Cooper pair. See the answer for more details.

The general way to produce spin triplet components is to associate a spin texture with a conventional (so called $s$-wave) superconducting condensate. You can do that by either

  • put several non-colinear magnetisation domains in proximity with a superconductor (for instance in Josephson junction). See Bergeret, F. S., Volkov, A. F., & Efetov, K. B. (2005). Odd triplet superconductivity and related phenomena in superconductor-ferromagnet structures. Reviews of Modern Physics, 77(4), 1321–1373.

  • put spin-dependent interfaces, see Linder, J., & Robinson, J. W. A. (2015). Superconducting spintronics. Nature Physics, 11(4), 307–315 or arXiv:1510.00713 you already cited in your question, or Eschrig, M. (2015). Spin-polarized supercurrents for spintronics: a review of current progress. Reports on Progress in Physics, 78(10), 104501 or arXiv:1509.02242 (this latter review contains more references than the Linder and Robinson's one).

  • put spin-orbit interaction in a Josephson junction, see Bergeret, F. S., & Tokatly, I. V. (2014). Spin-orbit coupling as a source of long-range triplet proximity effect in superconductor-ferromagnet hybrid structures. Physical Review B, 89(13), 134517 or arXiv:1402.1025.

Note that

  • A triplet pairing in proximity systems is not a symmetry of the condensate, it is just the expectation value of quantum correlations. In particular in Josephson junction, you may generate a non-trivial correlation between electrons on each side of the junction, without changing the symmetry of the superconducting gap in the electrodes.

  • There is no clear signature of the triplet correlations as far as I know. Only indirect measurement can be done. The reason is that singlet and triplet correlations come together in real systems. Nevertheless, the magnetic properties interesting for superconducting spintronics applications are only due to the triplet component(s).

  • In addition to the $m=-1$ triplet state $\left|\downarrow\downarrow\right\rangle $ and the $m=+1$ triplet state $\left|\uparrow\uparrow\right\rangle $, there is a third $m=0$ triplet state $\left|\uparrow\downarrow\right\rangle +\left|\downarrow\uparrow\right\rangle $ being symmetric in the exchange of spin. This last one does not participate to the magnetic properties of superconductor/ferromagnet hetero-structures, as well as the singlet $S=0$, $m=0$ state $\left|\uparrow\downarrow\right\rangle -\left|\downarrow\uparrow\right\rangle $.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ I believe one more point here is that the wavefunction for a Cooper pair has three parts to it - spin , orbital, and time reversal component. While I don't understand time-reversal very well, the presence of the additional component gives more play over the total antisymmetry requirement of the wavefunction. So one can be symmetric wrt orbit (s-wave) AND spin (triplet), but antisymmetric wrt time (odd-frequency pairing - again don't understand the physical meaning of this) and yet satisfy Pauli's exclusion principle here. Any explanations/comments/corrections, anyone? $\endgroup$
    – Gamora
    Commented Jan 10, 2017 at 19:29
  • $\begingroup$ @Gamora This is perfectly correct, you're right, thank you for the comment. In fact it's the complete correlation function $\Delta\sim\left\langle \Psi\left(\xi_{1}\right)\Psi\left(\xi_{2}\right)\right\rangle \sim-\left\langle \Psi\left(\xi_{2}\right)\Psi\left(\xi_{1}\right)\right\rangle $ which should be antisymmetric. It is often called the gap parameter, and in the previous notations, $\Psi$ is a fermionic destruction operator, and $\xi_{1,2}$ represent all necessary variable for a given problem. I was discussing only $\xi\equiv (x,\sigma)$ with $x$ the position and $\sigma$ the spin. $\endgroup$
    – FraSchelle
    Commented Jan 11, 2017 at 12:58
  • $\begingroup$ You can add the time variable if you wish (then $\xi\equiv(x,t,\sigma)$ for instance) as well as any other thing (for instance the band index or any other internal degree of freedom, like a pseudo quark-color or the (supposed-to-be the) true color in stars). $\endgroup$
    – FraSchelle
    Commented Jan 11, 2017 at 13:01

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.