With reference to the Nambu (or famously, Nambu-Gorkov) transformation in this paper, could someone explain the reason behind using the 3rd Pauli matrix in the Lagrangian after equation (2.3) (would essentially like to understand how to do a Nambu-Gorkov transformation for any arbitrary (Lorentz-invariant)Lagrangian)?

P.S - another reference that's usually given along with the above is here, though I'm primarily following 1.


It's because $$ \psi^\dagger \psi= \frac 12 (\psi^\dagger \psi-\psi \psi^\dagger)+const.\\ = \frac 12 (\psi^\dagger,\psi)\left[\matrix{1&0\cr0&-1}\right] \left[\matrix{\psi\cr \psi^\dagger}\right]+const\\ =\frac 12 \Psi^\dagger \tau_3 \Psi+const. $$

  • $\begingroup$ I understood the above calculation & see why 3rd Pauli matrix is used (in the 2nd term) of the transformed Lagrangian. Now, expanding (only) the first term in eq. (2.1) for i=1 to 2, we see that the time-derivative term (for i=2 spinor) in the transformed Lagrangian (the one with Nambu-Gorkov bispinors) comes out differently from the corresponding term in eq. (2.1). In other words, the transformed Lagrangian doesn't seem equivalent to the original Lagrangian. Why is that? Thanks. $\endgroup$ – user263315 Oct 8 '20 at 13:46
  • $\begingroup$ There's no minus needed for the time derivative so no $\tau_3$, because $\psi^\dagger$ comes with a time derivative with opposite sign that arises from the integration by parts that takes the time derivative off $\psi$ and onto $\psi^\dagger$.. $\endgroup$ – mike stone Oct 8 '20 at 14:02
  • $\begingroup$ $\Psi ^+ (p)\Big( i\partial /\partial t \Big)\Psi (p) = i [\psi ^+ \psi] \Big[ \dot {\psi _1} \dot{\psi _2 ^+} \Big]^T = i\psi_1 ^+\dot{\psi _1} + i\psi _2 \dot{\psi _2 ^+}$ from the (first term of the) transformed Lagrangian. While the first term in the original Lagrangian (eq. (2.1)) is $i\psi _1 ^+ \dot{\psi _1} + i\psi _2 ^+\dot{\psi _2 }$. Both are different. Why? $\endgroup$ – user263315 Oct 11 '20 at 11:52
  • $\begingroup$ Because $\psi_2^\dagger \dot \psi_2 = - \dot \psi_2^\dagger\psi_2 $ up to a total deriative, and $- \dot \psi_2^\dagger\psi_2 = \psi_2 \dot \psi_2^\dagger$ because the fields anticommute. The constants and total derivative do not affect Lagrange's equations. $\endgroup$ – mike stone Oct 11 '20 at 12:06
  • $\begingroup$ thanks, now I get it. $\endgroup$ – user263315 Oct 12 '20 at 6:11

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