Now you are in the subtree of Special Topics in Many-Body Theory, Spring 2016 project. 

Definitions including one-particle time-ordered correlation function (Green's function)

We now introduce a few formal quantities that are useful in any discussion of many-body perturbation theory. We will not be doing too much perturbation theory in this course, as the focus is on nonperturbative phenomena, but there is a philosophical underpinning of many-body perturbation theory that is actually quite important. The idea is to avoid dealing with the enormously complicated wavefunction, which is in any event cannot be measured directly. Instead we focus on correlation functions ("Green's functions") which are both easier to calculate and more directly measurable.

This idea is analogous to dealing with reduced one- and two-particle distribution functions $f_1$ and $f_2$ in classical statistical mechanics/kinetic theory, rather than the full distribution function $g$. How does the above carry over to quantum mechanics? It will still be useful to define quantum versions of the reduced distribution functions, but for the most part Boltzmann-type time evolution equations will not be used to calculate these functions: instead one usually works in frequency space because the time scales involved are so rapid (an exception is the quantum Boltzmann equation for liquid He). Instead we will come up with a diagrammatic theory for understanding small perturbations to the free Fermi or Bose gas. Another reason is that quantum systems are often studied under conditions of "linear response": when driven far from equilibrium, their behavior is not nearly as well understood as in classical systems, which already show surprising behavior (like turbulence) far from equilibrium.

Schrodinger's equation for the evolution of the wave function is
$$\begin{equation} i \hbar {\partial \psi_s(t) \over \partial t} = H(t) \psi_s(t). \end{equation}$$
Here the Hamiltonian is time-varying only if some external field applied to the system is time-varying. For an isolated system, $H$ is constant and
$$\begin{equation} \psi_s(t) = e^{-i H (t-t_0) /\hbar} \psi_s(t_0). \end{equation}$$
We can make a unitary transformation to the Heisenberg picture:
$$\begin{equation} \psi_h(t) = \psi_h = e^{i H (t-t_0)/\hbar} \psi_s(t_0). \end{equation}$$
Any operator (not just the Hamiltonian) also must transform between the two pictures:
$$\begin{equation} O_h(t) = e^{i H (t-t_0)/\hbar} O_s(t) e^{-i H (t-t_0)/\hbar} \end{equation}$$
so that expectation values, which are physical, are unchanged. Hence the time evolution of a Heisenberg operator is given by
$$\begin{equation} i \hbar {d O_h(t) \over dt} = [O_h(t),H] + i \hbar {\partial O_h(t) \over \partial t}. \end{equation}$$
Finally, in the "interaction picture" for a Hamilonian divided into $H=H_0 + H^\prime$, we treat $H_0$ in the Heisenberg rep. and $H^\prime$ in the Schrodinger rep.:
$$\begin{equation} \psi_i(t) = e^{i H_0(t-t_0)/\hbar} \psi_s(t) \end{equation}$$
$$\begin{equation} O_i(t) = e^{i H_0 (t-t_0)/\hbar} O_s(t) e^{-i H_0 (t-t_0)/\hbar}. \end{equation}$$

Schrodinger's equation in the interaction picture becomes
$$\begin{equation} i \hbar {\partial \psi_i(t) \over \partial t} = H^\prime_i(t) \psi_i(t), \end{equation}$$
where $H^\prime_i(t)$ is
$$\begin{equation} H^\prime_i(t) = e^{i H_0 (t-t_0)/\hbar} H_s^\prime e^{-i H_0 (t-t_0)/\hbar} \end{equation}$$
Henceforth in this lecture we set $\hbar=1$ and write $r_1$ for ${\bf r}_1$.

Let's define the quantum field operators for a many-particle system to be, in the Schrodinger representation (so the operators are independent of time),
$$\begin{equation} \psi_\alpha(x) = \sum_k u_{k\alpha}(x) c_{k\alpha}, \quad \psi_\alpha^\dagger(x) = \sum_k u_{k\alpha}(x)^* {c_{k\alpha}}^\dagger \end{equation}$$
Here the $u_k$ are wavefunctions in some particular basis (which we will usually just take to be plane waves of momentum $k$) and $\alpha$ is a spin index. The anticommutation relation of the quantum field operators in the Schrodinger representation is
$$\begin{equation} \{\psi_\alpha(x),\psi_\beta^\dagger(x^\prime)\} = \sum_{kk^\prime} u_{k\alpha}(x) u_{k^\prime\beta}^*(x^\prime) \{c_{k\alpha},{c_{k^\prime\beta}}^\dagger\} = \delta(x-x^\prime) \delta_{\alpha \beta}. \end{equation}$$
where we have used the orthogonality and completeness of the wavefunctions.

The local density of particles is just $\rho = \psi^\dagger(x) \psi(x)$. It is easily checked that integrating over $x$ gives just the total number of particles $N$, and that for an individual state $c^\dagger_k |0\rangle$ this definition agrees with our previous definition.
It is worth noting that the operator $H$ is sometimes defined to include a chemical potential term $\mu N$.

In the Heisenberg representation, the quantum field operator is (now denoted $\Psi$)
$$\begin{equation} \Psi_\alpha(t,x) = e^{i H t} \psi_\alpha(x) e^{-i H t},\quad \Psi^\dagger_\alpha(t,x) = e^{i H t} \psi^\dagger_\alpha(x) e^{-i H t}. \end{equation}$$
Note that the Heisenberg and Schrodinger representations are identical for an operator that commutes with the Hamiltonian. So, for example, the total particle number is identical:
$$\begin{equation} N = \int\,\psi^\dagger_\alpha(x) \psi_\alpha(x)\,d^3x = \int\,\Psi^\dagger_\alpha(t,x) \Psi_\alpha(t,x)\,d^3x. \end{equation}$$

What are the commutation relations of the Heisenberg operators $\Psi$? They are actually quite nontrivial except at equal time. The equal-time commutation relations are
the same as before:
$$\begin{equation} \{\Psi_\alpha(t,x),\Psi_\beta^\dagger(t,x^\prime)\} = \delta(x-x^\prime) \delta_{\alpha \beta}. \end{equation}$$
At unequal times, we have to solve a complicated problem because of the $e^{i H t}$ factors.

The general definition of the time-ordered one-particle correlation function or Green's function is
$$\begin{equation} G_{\alpha \beta}(t_1,r_1,t_2,r_2) = -i \langle 0 | T\{ \Psi_\alpha (t_1,r_1) \Psi_\beta^\dagger(t_2,r_2) \}|0 \rangle. \end{equation}$$
Here we have introduced spin indices $\alpha,\beta$, the ground state $|0\rangle$, and the time-ordering operator $T$. The action of $T\{O_1,O_2,\ldots)\}$ is defined so that operators with earliest times act first, moving from the right. $T$ also gets a minus sign for each interchange of two operators, so the Green's function explicitly is
$$\begin{equation} G_{\alpha \beta}(t_1,r_1,t_2,r_2) = \cases{-i \langle 0 | \Psi_\alpha (t_1,r_1) \Psi_\beta^\dagger(t_2,r_2) |0 \rangle&if $t_2<t_1$\cr i \langle 0 | \Psi_\beta^\dagger(t_2,r_2) \Psi_\alpha (t_1,r_1) |0 \rangle& if $t_2 \geq t_1$}. \end{equation}$$ Note that we will be interested in the Green's function because it is more directly related to experimental quantities, as well as often easier to calculate, than the ground state $|0\rangle$. For spin-symmetric systems that are not ferromagnetic, $G_{\alpha \beta} = G \delta_{\alpha\beta}$ and the spin indices can be dropped. If there is also translational invariance in space and time, then only the coordinate differences $r = r_1 - r_2$ and $t=t_1 - t_2$ matter, and $G$ is just a function of two variables: $G(t,r)$. Now consider the following quantity, sometimes referred to as the "coordinate density matrix": $$\begin{equation} \rho_{\alpha \beta}(t,r_1,r_2) = \langle 0| \Psi_\beta^\dagger(t,r_1) \Psi_\alpha(t,r_2)|0 \rangle. \end{equation}$$ (Note that Landau and Lifshitz define this quantity with a factor $ 1 / N$.) Clearly at $r_1=r_2$ this reduces to the expectation value of the local density at point $r_1$ at time $t$. This density matrix essentially determines all one-body observables of the form $F_{\alpha\beta} = \sum_i f^i_{\alpha \beta}$ through the quantum stat. mech. relation $$\begin{equation} \langle 0| F_{\alpha \beta} |0\rangle = \int \left( f^1_{\alpha\beta} \rho_{\beta \alpha}(t,r_1,r_2)\right)_{r_1 = r_2}\,d^3x_1. \end{equation}$$ (The reason for allowing $r_1$ to be different from $r_2$ for the action of the operator, but not the integral, can be understood by considering the single-particle momentum operator, for example.) For a translationally invariant system, only the difference $r = r_1 - r_2$ matters, and with no spin-dependence, we have $\rho_{\alpha \beta} = \rho \delta_{\alpha \beta}$ and $$\begin{equation} \rho(r) = -i G(0^-,r). \end{equation}$$ So the equal-time correlation function has a natural interpretation in terms of the particle density. At $r=0$, for a translationally invariant system the density is just the mean density $N/V$: $$\begin{equation} N/V = 2 \rho(0) = -2 i G(0^-,0). \end{equation}$$ We'll often use the Green's function in the momentum representation, which is $$\begin{equation} G(\omega,p) = \int G(t,r) e^{-i (p \cdot r - \omega t)}\,dt\,d^3x. \end{equation}$$ The inverse relation is $$\begin{equation} G(t,r)=\int G(\omega,p) e^{i (p \cdot r - \omega t)}\,{d\omega\,d^3p \over (2 \pi)^4}. \end{equation}$$