Greens Function formalism for the independent boson model

Question

This may be a very specific question, but since almost everyone seems to be quoting this book, I want to understand the derivation(s) of the solution for the independent boson model (IBM) from Mahan's book 'Many-particle physics', specifically the one using the Greens function (GF) formalism. My problem in understanding is mainly in the beginning, where he starts by writing down the GF as \begin{equation} G(t) = -i \langle T c(t) c^\dagger(0) \rangle = -i \theta(t) \langle e^{iH_0 t} c^\dagger e^{-iHt}c \rangle.\tag{4.376} \end{equation} $T$ is the time-ordering operator, $H_0$ the free Hamiltonian, $H$ the full Hamiltonian and $c$ the annihilation operator. I get that the first part is just the definition of the GF, but I don't get the second equal sign at all. Why do exponentials with both $H_0$ and $H$ appear? Why is $c$ to the right? He writes:

'Since there is only one particle in the band, the creation operator must be to the right, so $t>0$'.

But $c$ is to the right, and not $c^\dagger$? At least $t>0$ is enforced by the step function. Did he skip some step so this makes sense? This is all from chapter 4, equation (4.376)

Answer 1

It looks like we are in the interaction picture, and the time dependence of an operator $O$ is defined as $$ O(t) = e^{iH_0t}\, O \, e^{-iH_0t}. $$ The perturbation $V$ also has a corresponding $V(t)$ defined as above.

Then the quantity that gets averaged in the second equality is really: $$ e^{iH_0t} c^{\dagger} e^{iH t} c = c^{\dagger}(t) \, T\!\exp\left(i\int_0^{t} dt' \, V(t') \right) \, c(0) $$ The time-ordered exponential in between evolves ket-state from zero to $t$.

But then I do suspect $c$ and $c^{\dagger}$ are swapped by mistake.

--edit--

The specific model is described in (4.385-388) in Mahan. It is non-relativistic and without any negative energy electron band. You can verify that the zero-particle vacuum state $\vert 0 \rangle$ satisfies $$ H_0 \vert 0 \rangle = V \vert 0 \rangle = 0. $$ It is the exact ground state even in the presence of interaction.

All that S-matrix of $t\rightarrow \pm \infty$ stuff came from the fact that we don't know the ground state with interaction in general. We have to adiabatically evolve the non-interacting ground state from and to $t \rightarrow \pm \infty$. But here you can just do $$ \langle 0 \vert e^{iHt} c^{\dagger} e^{-iHt} c \vert 0 \rangle $$ directly because we know the exact ground state.

Or you can also go back to your old formulas, and just note that $S(t, t')\vert0\rangle = \vert0\rangle$ for any $t, t'$.

The above is true for any non-relativistic gas.

--edit--

Of course this feature doesn't persist at finite temperature.

$\vert 0 \rangle$ being the exact $T=0$ ground state is equivalent to saying that there is no virtual pair production and no vacuum bubble diagram correction toward the ground state. Draw a few vacuum diagrams and convince yourself that:

1. At zero temperature they are all identically zero due to the pole structure of fermion propagator

2. The same is obviously not true for Matsubara sum at $T\neq 0$

But ultimately it hardly matters. The vacuum contribution is added to and then subtracted from the connected correlation function anyway. You will calculate the exact same Feynman diagrams using the exact same Feynman rules in any case.