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The conventional way of writing down the gap in a superconducting material $\Delta(k) = \langle \hat{c}_\alpha({\bf k}) \hat{c}_\beta({\bf -k}) \rangle$ is

$$\Delta(k) = (\Delta_0(k) + {\bf d(k) \sigma}) (i \sigma_2)_{\alpha\beta}$$

where $\Delta({\bf k}) = \Delta({\bf -k})$ is the spin-singlet contribution and ${\bf d(k)} = {\bf - d(-k)}$ is the spin-triplet contribution. For more details I refer to this well written post by FraSchelle.

My question refers to the spin part $i \sigma_2$. Why is there an $i$ in front of this term? To me it seems that the intention is to make it symmetric with respect to time-reversal. (Which I think in this case would be complex conjugation)

Even if it is only convention, was there a certain history that led to it?

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  • Due to the spin-statistics property, one usually imposes $\left\langle c_{\uparrow}c_{\downarrow}\right\rangle \propto-\left\langle c_{\downarrow}c_{\uparrow}\right\rangle $, and since the most-used convention reads $\sigma_{2}=\left(\begin{array}{cc} 0 & -\mathbf{i}\ \mathbf{i} & 0 \end{array}\right)$, the spin-statistics reads $\mathbf{i}\sigma_{2}$ in the basis $\left(\begin{array}{c} c_{\uparrow}\ c_{\downarrow} \end{array}\right)$. But you're right, one writes $\mathbf{i}\sigma_{2}$ as in the time-reversal operation. But everything is just a convention. – FraSchelle Aug 24 '15 at 19:02
  • Time reversal operation reads $T=K\mathbf{i}\sigma_{2}=-\mathbf{i}K\sigma_{2}=\mathbf{i}\sigma_{2}K$ with $K$ the complex conjugation. One more time the $\mathbf{i}$ is just there to make $\mathbf{i}\sigma_{2}$ real. – FraSchelle Aug 24 '15 at 19:03

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