# Symmetry of the Green-Keldysh-Nambu function

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• Paul159

#### Paul159

Hello,

I would like to understand a relation of this article by Volkov (eq. 4).
Let's define the Green function $$G^{ij}_{ab} (1,2) = -i \langle T_c \Psi_a (1_i) \Psi_b (2_j) \rangle$$ where ##a,b = (1,2)## are the spin indices and ##i,j = (1,2) ## are the indices for the Keldysh contour ; ##\Psi_1 = \Psi_\uparrow##, ##\Psi_2 = \Psi_\downarrow^\dagger##. ##T_c## is the contour ordering.

Now I cite Volkov : "if the system contains no fields acting directly on the spins and if the spin-orbit interaction can be neglected we have :
$$G^{ij*}_{ab} (1,2) = G^{\bar{ij}}_{\bar{ab}} (1,2)$$ where ##\bar{1} = 2, \bar{2} = 1##. "

I don't understand why this last symmetry is true. For example I get with the last equation that
$$\langle \Psi_\uparrow (1_1)^\dagger \Psi_\uparrow(2_1) \rangle = \langle \Psi_\downarrow (1_2)^\dagger \Psi_\downarrow(2_2) \rangle ,$$ where 1 is after 2 on the first branch.

Thanks in advance for any help.