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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?

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  • $\begingroup$ 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

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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.

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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.

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

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