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What is the reason behind using the weird minus sign in the fourth component of the extended Nambu basis? $$\Psi^{\dagger}=(\Psi_{i\uparrow}^{\dagger}\quad \Psi_{i\downarrow}^{\dagger} \quad \Psi_{i\downarrow}\quad-\Psi_{i\uparrow})$$ While it does allow neat expressions for certain tight-binding Hamiltonians in terms of pauli matrices(in either spin or p-h space) I am not sure if this really is the motivation. Also, the spinor is not even the tensor product of particle-hole and spin components, so another question would be why the Nambu basis at all?

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  • $\begingroup$ it looks eerily like a Minkowskian metric, but there is no obvious connection $\endgroup$ – lurscher Feb 22 '18 at 15:25
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The minus sign in your spinor, sometimes known as the spinor with SO(4) symmetry, is inserted to enforce both the SU(2) spin and pseudospin symmetries. Basically to impose the $SU(2)_{s} \times SU(2)_{p}$ one should employ two doublets which based on your language are $D_{1}= (\Psi_{i \uparrow}^{\dagger} \quad \Psi_{i \downarrow})$ and $D_{2}= (\Psi_{i \downarrow}^{\dagger} \quad -\Psi_{i \uparrow})$ which can be also represented in a $2\times2$ matrix form, see also here.

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The Nambu basis is just a useful representation for superconducting Hamiltonians. Is is, in fact, a tensor product between spin and particle-hole spaces, but not a trivial one. Take a look at here to see some of the usefulness and construction of such basis.

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  • $\begingroup$ This doesn't seem to really answer the question. $\endgroup$ – Kyle Kanos Feb 22 '18 at 14:04
  • $\begingroup$ it looks eerily like a Minkowskian metric, but there is no obvious connection $\endgroup$ – lurscher Feb 22 '18 at 15:25
  • $\begingroup$ Maybe it doesn't seem to answer the question because it is not immediate to observe how this Nambu basis arrives. But if you fix a representation for time-reversal transformations for spin-half particles, you conclude that, in order to describe the Hamiltonian as in the section "Important and useful basis change" of the above link you must write the spinors as: $$ \Psi^{\dagger} = (\xi^{\dagger}\ \ i \sigma_y \xi) $$ with $\xi = (\psi_{\uparrow}\ \ \psi_{\downarrow})$. And there is your minus sign. $\endgroup$ – Antônio Lucas Rigotti Manesco Feb 23 '18 at 19:12
  • $\begingroup$ After the similarity transformation, the superconducting order parameter can be rewritten in a useful fashion with a singlet term proportional to $\sigma_0$ and the three triplet states proportional to the three Pauli matrices $\{\sigma_i\}_{i=1}^3$. $\endgroup$ – Antônio Lucas Rigotti Manesco Feb 23 '18 at 19:23

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