3
$\begingroup$

Lets say we have a non-interacting translation invariant Hamiltonian $H$ in a matrix form:

$$H=\Psi^{\dagger}\bar{H}\Psi$$

where $\Psi^{\dagger}=(\Psi_{\uparrow}^{\dagger}\quad \Psi_{\downarrow}^{\dagger} \quad\Psi_{\downarrow}\quad-\Psi_{\uparrow})$ is the extended nambu spinor.

In such a basis what is the meaning of a particular element of the matrix representation of the retarded green function:

$$\bar{G}^{R}(\omega)=\frac{1}{\omega-\bar{H}+i\delta}$$

what is $\bar{G}^{R}_{ij}$?

$\endgroup$
3
  • $\begingroup$ Since that $i G^{Ret}(xt,x't') = \langle GS \vert \text{T} \psi_m(xt) \psi^\dagger_{m'}(x't') \vert GS \rangle$ so $G^{Ret}_{ij}(\omega)$ is the propagator in the $\omega$ space. $\endgroup$
    – Algebrato
    Commented Jul 6, 2017 at 22:45
  • $\begingroup$ What do you mean what is $\tilde{G}^R_{ij}$? You ask about the physical meaning? Or do you want a mathematical equation in terms of specific operators? $\endgroup$
    – gingras.ol
    Commented Jul 9, 2017 at 3:41
  • $\begingroup$ @gingras.ol Well, a mathematical equation would do. $\endgroup$ Commented Jul 9, 2017 at 11:37

1 Answer 1

2
+50
$\begingroup$

As a general recipe, the retarded Green's function is defined as $$ G^{R}_{ij}(t,t') = -i \theta(t,t') \langle \{\psi_{i}, \psi_{j}^{\dagger} \} \rangle, $$ where $\theta$ is the step-function, and $\{ ,\}$ denotes anticommutator. $\psi_{k}$ based on your suggested language is k-th elemnt of $\Psi$. For instance, $\psi_{1}= \Psi_{\uparrow}$. The full retarded Green's function would be $$ G^{R}(t,t')= -i \theta(t,t') \begin{pmatrix} \langle \{ \Psi_{\uparrow}(t) , \Psi^{\dagger}_{\uparrow}(t') \} \rangle& \langle \{ \Psi_{\uparrow}(t), \Psi^{\dagger}_{\downarrow}(t') \} \rangle& \langle \{ \Psi_{\uparrow}(t), \Psi_{\downarrow}(t') \} \rangle & -\langle \{ \Psi_{\uparrow}(t), \Psi_{\uparrow}(t') \} \rangle\\ \langle \{ \Psi_{\downarrow}(t) , \Psi_{\uparrow}^{\dagger}(t') \} \rangle & \langle \{ \Psi_{\downarrow}(t) , \Psi_{\downarrow}^{\dagger}(t') \}\rangle & \langle \{ \Psi_{\downarrow}(t) , \Psi_{\downarrow}(t') \} \rangle & -\langle \{ \Psi_{\downarrow}(t) ,\Psi_{\uparrow}(t') \} \rangle \\ \langle \{ \Psi^{\dagger}_{\downarrow}(t) , \Psi_{\uparrow}^{\dagger}(t') \} \rangle & \langle \{ \Psi^{\dagger}_{\downarrow}(t) , \Psi_{\downarrow}^{\dagger}(t') \} \rangle& \langle \{ \Psi^{\dagger}_{\downarrow}(t) , \Psi_{\downarrow}(t') \}\rangle & -\langle \{ \Psi^{\dagger}_{\downarrow}(t) ,\Psi_{\uparrow}(t') \} \rangle\\ -\langle \{ \Psi^{\dagger}_{\uparrow}(t) , \Psi_{\uparrow}^{\dagger}(t') \} \rangle& -\langle \{ \Psi^{\dagger}_{\uparrow}(t) , \Psi_{\downarrow}^{\dagger}(t') \} \rangle& -\langle \{ \Psi^{\dagger}_{\uparrow}(t) , \Psi_{\downarrow}(t') \} \rangle& \langle \{ \Psi^{\dagger}_{\uparrow}(t) ,\Psi_{\uparrow}(t') \} \rangle \end{pmatrix} $$

The suggested generic definition of the retarded Green's function in your question is just the Fourier transform of the above matrix in the basis where $\overline{H}$is defined.

$\endgroup$
4
  • $\begingroup$ Thanks. I don't fully understand the relation between Green the "function" and Green the "operator". Could you give me a source so that I can get this cleared up? $\endgroup$ Commented Jul 11, 2017 at 17:08
  • 1
    $\begingroup$ Please remember that the Green's function is a particular two-point "function" of "operators". Most likely a review on real-time Green's functions can be illustrative. $\endgroup$
    – Shasa
    Commented Jul 12, 2017 at 8:23
  • $\begingroup$ Could you please suggest something specific? I have read the usual treatments in Doniac and Walecka, but neither of them seems to treat operator-wise aspects of the Green's function. $\endgroup$ Commented Jul 12, 2017 at 18:14
  • 1
    $\begingroup$ Reading "Nonequilibrium Many-Body Theory of quantum systems" by Stefanucci and van Leeuwen is most likely helpful. You can also try to learn the "Keldysh Green's function" formalism. $\endgroup$
    – Shasa
    Commented Jul 12, 2017 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.