Questions about the Dyson equation

Question

I'm studying finite temperature many-body perturbation theory, and am trying to understand The Dyson equation. In particular, I'm using Mattuck - A guide to Feynman diagrams in the many body problem.

There appear to be two types of Dyson equation: one which is valid in the special case of no external potential and with diagrams calculated in $(k,w)$-space,

\begin{equation} G(k,w) = \frac{1}{G_0 (k,w)^{-1} - \Sigma (k,w)}. \end{equation}

And one valid at other times which is an integral equation for $G$ which can be different depending on what space you're in and whether you have an external potential or not, but is something like

\begin{equation} G(k,\tau-\tau') = G_0(k,\tau-\tau') + \int_0^\beta d\tau_1 d\tau_2 G_0(k,\tau- \tau_2) \Sigma(k,\tau_2-\tau_1) G(k,\tau_1-\tau'). \end{equation}

So, my questions:

1. What is an external potential? I can't find an actual definition in the book. At first I assumed it meant the part of the system you want to treat perturbatively, I.e. the Hamiltonian is $H=H_0+H_1$ where $H_1$ is small and can there for be considered a perturbation. But obviously this can't be the case otherwise there's no expansion to use in a Dyson equation.

2. Why is the second Dyson equation useful? - How does it allow partial summations?

3. Every book I've looked at seems to give slightly different rules for calculating the Feynman diagrams. Any thoughts on which is correct? (for example, Negele - Quantum many-particle systems is different to Mattuck).

Answer 1

The two equations are equivalent (at least in the physicist's sense, I won't try to make the derivation rigorous).

Second equation, written in the condensed form reads $$G = G_0 + G_0 \Sigma G$$ or, when iterated (plug in the RHS into the second $G$, ad infinitum), $$G = G_0 + G_0 \Sigma G_0 + G_0 \Sigma G_0 \Sigma G_0 + \cdots$$

This last form is actually what one starts with when solving the evolution equation $$(i\partial_t - H_0 - \Sigma) G = \delta$$ If the $\Sigma$ wouldn't depend on time, you'd just write $G = \exp(-it(H_0 + \Sigma))$ but if it does, you have to integrate -- either by iterative process above, or by time ordered exponential, which is equivalent but I won't go into that.

In any case, one can formally sum the above infinite series to obtain $$G = G_0 {1 \over 1 - \Sigma G_0} = {1 \over G_0^{-1} - \Sigma}$$ This relation actually doesn't say anything else than what we have already seen in the evolution equation for $G$, namely that it is inverse to the evolution operator.

Now, to your questions:

1. External potential means part of the potential that is not self-energy. Consider the Fermi liquid. There electrons interact by Coulomb potential and you obtain lots of terms for self-energy when you try to compute the Feynman diagrams (Hartree, Fock exchange, potential screening, etc., etc.). By external potential we usually understand the rest of the world. The primary representative would be a laser pulse, which in the simplest case you can model as classical monochromatic EM plane wave perturbation. Or you could include photons + full QED. Depends on what you need.

2. This should be obvious now: iterate. To any given order (in powers of $\Sigma$) you need. But to higher order you get, the more mess you'll receive. The complexity can be tamed a little by Feynman diagrams but even number of those grows fast and in practice you'll see terms like LO, NLO and at most NNLO, meaning leading order, next-to-leading order, and so on.

3. Not sure what you mean, you'd need to be more explicit. Obviously you are free to move lots of complex $i$s and minuses between various places (i.e. whether to put them into propagators, vertices, integrals, etc.) and also various combinatorial factors (coming from graph counting). You won't really understand why this works unless you'll derive the whole machinery yourself. I mean, try to calculate some process by hand (meaning evaluating integrals and vacuum expectations explicitly) and then compare that with the diagramatic approach. In any case be sure that diagrams are nothing else than shorthand notation for the real calculations.

Answer 2

First, the first equation (differential form) is fully equivalent to the second equation (integral form). One can see this connection just after taking the Fourier transform tor the integral equation: \begin{equation} \int_{0}^{\beta} d(\tau-\tau')e^{i\omega_l(\tau-\tau')} G(k,\tau-\tau') = \int_{0}^{\beta} d(\tau-\tau')e^{i\omega_l(\tau-\tau')}G_0(k,\tau-\tau') + \int_{0}^{\beta} d(\tau-\tau')e^{i\omega_l(\tau-\tau')} \int_0^\beta d\tau_1 d\tau_2 G_0(k,\tau- \tau_2) \Sigma(k,\tau_2-\tau_1) G(k,\tau_1-\tau'). \tag{1} \end{equation}

\begin{align} \Rightarrow G(k,\omega_l) & = G_0(k,\omega_l) + \int_{0}^{\beta} d(\tau-\tau')\int_0^\beta d\tau_1 \int_0^\beta d\tau_2 e^{i\omega_l(\tau-\tau')} G_0(k,\tau- \tau_2) \Sigma(k,\tau_2-\tau_1) G(k,\tau_1-\tau') \\ & = G_0(k,\omega_l) + (\dfrac{1}{\beta})^3\int_{0}^{\beta} d(\tau-\tau')\int_0^\beta d\tau_1 \int_0^\beta d\tau_2 e^{i\omega_l(\tau-\tau')} \left[ \sum_a e^{-i\omega_a(\tau-\tau_1)} G_0(k,\omega_a) \right] \\ & \qquad \qquad \qquad \qquad \qquad \quad \times \left[ \sum_b e^{-i\omega_b(\tau_1-\tau_2)} \Sigma (k,\omega_b) \right] \left[ \sum_c e^{-i\omega_c(\tau_2-\tau')} G(k,\omega_c) \right] \\ & = G_0(k,\omega_l) + (\dfrac{1}{\beta})^3 \sum_{a}\sum_b\sum_c G_0(k,\omega_a)\Sigma(k,\omega_b)G(k,\omega_c) \int_0^{\beta} d\tau(\tau-\tau')\int_0^\beta d\tau_1 \int_0^\beta d\tau_2 \\ & \qquad \qquad \qquad \left[ e^{i(\omega_l-\omega_a)\tau} e^{i(\omega_a-\omega_b)\tau_1} e^{i(\omega_b-\omega_c)\tau_2} e^{i(\omega_c-\omega_l)\tau'} \right] \\ & = G_0(k,\omega_l) + G(k,\omega_l)\Sigma(k,\omega_l)G_0(k,\omega_l) \tag{2} \end{align} Then you can see the equivalence clearly. On the last line we use the fact that \begin{equation} \dfrac{1}{\beta}\int_0^\beta d\tau e^{i(\omega_m-\omega_n)\tau} = \delta_{mn} \end{equation} the discrete delta function will kill the summation and also note that the integral order for $(2)$ will be choosed as $$\tau_1\rightarrow\tau_2\rightarrow(\tau-\tau').$$