I'm interested in the link between the single-particles Green's function in imaginary time $\tau$ and the Lagrangian.
I guessed that the Lagrangian $L$ can be expressed by the inverse of the Green's function $G_{ij}(\tau)$. This argument follows from the following guess/solution of the quadratic/bilinear Hamiltonian $H_0$:
Some preliminaries are
$H_0 = \sum_{ij} \phi^\dagger_i \,\mathrm{H}_{ij} \phi_j$ using second quantization creation and annihilation operators $\phi^{(\dagger)}.$
$G_{ij}(\tau) = -\langle \mathcal{T} \phi_i(\tau) \phi^\dagger_j(0)\rangle$ as the definition of the Green's function.
$(-\partial_\tau \delta_{ij} - \mathrm{H}_{ij})G_{jk}(\tau) = \delta(\tau)\delta_{ik}$ as the Green's function solves the (quadratic and unperturbated) Schrödinger equation $i\partial_\tau \psi = H_0 \psi$ as $t \rightarrow -i\tau \implies \partial_t \rightarrow i\partial_\tau.$
Now I use the connection between the creation and annihilaton operators and the coordinates and momenta for all coordinates $q_i$: \begin{align} q &= 1/\sqrt{2}\ (\phi^\dagger + \phi) \\ \Pi_q &=-i/\sqrt{2}\ (\phi^\dagger - \phi). \end{align}
Starting with expressing the Lagrangian through the Hamiltonian and the creation and annihilation operator (I use the thermodynamics convention not the analytic mechanical one) \begin{align} L = H_0 - \Pi_q \partial_t q = H_0 - i \frac{1}{2}(\phi^\dagger \partial_t \phi - \phi\partial_t \phi^\dagger +\phi^\dagger \partial_t \phi^\dagger -\phi\partial_t \phi). \end{align} Now i partially integrate the second part and drop the total derivatives $2\phi^{(\dagger)}\partial_t\phi^{(\dagger)} = \partial_t {\phi^{(\dagger)}}^2$ and switch to imaginary time $i\partial_t \phi = - \partial_\tau \phi$: $$ L = H_0 + \phi^\dagger \partial_\tau \phi = \sum_{ij}\phi^\dagger_i (\mathrm{H}_{ij} + \delta_{ij}\partial_\tau)\phi_j,$$ which is simply $$ L(\tau) = \sum_{ij}\phi^\dagger_i [-G^{-1}_{ij}(\tau)]\phi_j.$$ And the action $S$ would be $S = \int^\beta_0 d\tau L(\tau)$. So two questions:
Is this correct or when does this hold? This includes the formulas for $q$ and $\Pi_q$
And
How do I show this in general? Or verify this in a more rigorous manner?
Of course this is just a restatment of the problem, as it is the goal of the Lagrangian and the Green'S function to solve the Schrödinger equation, which is in any case still the task at hand.
I can verify from the Lagrangian the time evolution of $\phi^{(\dagger)}$ as $\partial_\tau \phi_i^{(\dagger)} = [H_0, \phi_i^{(\dagger)}]$ using $[H_0, \phi_i] = -\sum_{j} \mathrm{H}_{ij} \phi_j$ and $[H_0, \phi^\dagger_i] = \sum_{j} \mathrm{H}_{ji} \phi_j$, so this is something that gives me some confidence in the steps before.
In the best case, if I found the right solution,I would like to see an argument why the inverse of "$\langle \phi_i \phi_j^\dagger\rangle$" appears in $L$.
