10
$\begingroup$

There are a lot of publications dealing with d-wave and d + id superconductivity, but I found no satisfying answer what exactly makes a superconductor d + id and why they break time reversal symmetry. Is it the same with p + ip superconductors? Also, what do these tearms (p + ip, d + id) mean? Do the wavefunction of the Cooper pairs in real space or the superconducting gap in k-space have corresponding symmetries?

$\endgroup$
4
  • $\begingroup$ Most of the answer from physics.stackexchange.com/a/62364/16689 applies here as well. In addition @pawel_winzig gives the few remaining details below. $\endgroup$
    – FraSchelle
    Jun 9, 2015 at 12:00
  • $\begingroup$ Thanks @FraSchelle for the link! If I understand correctly it's the point group of the system after whose irreprducible representations the order parameter and consequently the Cooper pair wavefunction have to transform. Is there then a simple example where one can see the breaking of time reversal symmetry? Is it a spontaneous symmetry breaking? $\endgroup$ Jun 9, 2015 at 15:07
  • $\begingroup$ But triplets and singlets don't break TRS. TRS is only broken it the three triplet states aren't degenerate anymore. As long as no state of them is favoured I see no reason why triplets should break TRS. As for the singlet case, a minus sign as eigenvalue of the TRS operator doesn't break it as the state goes into it self. $\endgroup$ Jun 10, 2015 at 7:52
  • $\begingroup$ As far as I understand, the answer by pawel_winzig signifies the Hamiltonian might be of the form $h=\xi\tau_{3}+\Delta\dfrac{k_{x}^{2}-k_{y}^{2}\pm\mathbf{i}k_{x}k_{y}}{k_{F}^{2}}\sigma_{y}\tau_{y}$ from which you can not create a time-reversal operator (TRO). The TRO is defined as an anti-unitary operation which must commute with the Hamiltonian. When it's impossible to construct such an operator, the time-reversal symmetry (TRS) is broken. What I misunderstand is that particle-hole symmetry seems to be broken as well. Please give us a reference for any further clarification. $\endgroup$
    – FraSchelle
    Jun 10, 2015 at 11:54

2 Answers 2

3
$\begingroup$

A $d+id$ superconductor breaks time reversal symmetry just like a $p+ip$ one does.

In a $d+id$ superconductor, there are two coexisting $d$-wave order parameter. For example, there are two representations of $D_{4h}$ point group that have four nodes (directions where the order parameter vanishes), commonly noted as $\Delta_{d_{x^2-y^2}}\sim \cos(k_x)-\cos(k_y)$ and $\Delta_{d_{xy}}\sim \sin(k_x)\sin(k_y)$. By $d+id$, one means that both $\Delta_{d_{x^2-y^2}}$ and $\Delta_{d_{xy}}$ are nonzero, and moreover the relative phase between $\Delta_{d_{x^2-y^2}}$ and $\Delta_{d_{xy}}$ (since SC order parameters are in general complex) is $\pi/2$. Under time reversal, the relative phase becomes $-\pi/2$, and such a state can be denoted as $d-id$ state. Therefore time-reversal is broken.

Yes, in real space a $d+id$ superconductor is odd under $\pi/2$ rotations, just like a regular $d$-wave superconductor.

$\endgroup$
0
$\begingroup$

At present, the mechanism of the superconductivity in systems like Na$_{0.35}$CoO$_2$·1.3H$_2$O are of high interest. The CoO$_2$ layers have been modeled as a spin 1/2 antiferromagnetic Mott insulator on a triangular lattice. By using resonate valence bond mean-field analysis, this supports the view that the superconducting order parameter has the spin-singlet broken- time-reversal symmetry chiral $d_{x^2−y^2} \pm id_{xy}$ pairing symmetry.

$\endgroup$
1
  • 3
    $\begingroup$ Could you please elaborate your answer? There is no explanation in there of anything but buzz-words $\endgroup$ Jun 10, 2015 at 7:45

Your Answer

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

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