# Relationship between Dyson equations from different problems

Recently, I noticed that the Dyson equation $$G=G_0+G_0\Sigma G$$ is used not only in quantum field theory but in some other branches of physics. For instance:

### 1. Wave equation

From the wave equation in a random media $$\mathcal{L}G(x,x_0)\equiv \left[\Delta_x+k^2 (1+\epsilon(x))\right]G(x,x_0)=\delta(x-x_0)$$ and $$\mathcal{L}_0G_0(x,x_0)\equiv\left[\Delta_x+k^2 \right]G_0(x,x_0)=\delta(x-x_0)$$ , we get $$\mathcal{L}_0G(x,x_0)=-k^2\epsilon(x)G(x,x_0)+\delta(x-x_0)$$ $$G(x,x_0)=\int dx_1\,\,G_0(x,x_1)\mathcal{L}_0G(x_1,x_0)=G_0(x,x_0)-\int dx_1\,G_0(x,x_1)k^2\epsilon(x_1)G(x_1,x_0)$$ , which obviously has the form of the Dyson equation.

### 2. Single-particle propagator - 1

Let $$U(t,t_0)$$ be the time evolution operator. The single particle propagator $$K(xt;x't')=\langle x|U(t,t_0)|x'\rangle$$ satisfies follwoing equations: $$\left[i\hbar\partial_t -\mathcal{H}\right]K(xt;x't')=\delta(x-x')\delta(t-t')$$ $$\left[i\hbar\partial_t -\mathcal{H}_0\right]K_0(xt;x't')=\delta(x-x')\delta(t-t')$$ where $$\mathcal{H}=\mathcal{H}_0+V$$. Those equations are mathematically identical to the wave equations above, i.e. $$G\leftrightarrow K,\,-k^2\epsilon\leftrightarrow V$$.

### 3. Single-particle propagator - 2

Using the Dyson series for $$U$$ in the interaction representation, $$U=1+\sum_{n=1}^{\infty}\frac{1}{n!}\left(-\frac{i}{\hbar}\right)^n\int\int\cdots \mathcal{T}\left[V(\tau_1)\cdots V(\tau_n)\right]$$ , we can also get the Dyson equation; strictly speaking, we can get the Dyson expansion of $$K$$ if we expand the Dyson equation in an iterative manner: $$K=K_0+K_0\sum(K_0+K_0\sum(\cdots))$$.

### 4. Green function in condensed matter physics

At zero temperature, the casual Green function is $$iG_{\alpha \alpha'}=\frac{\langle \psi_0|\mathcal{T}U(\infty,-\infty) \Psi_{\alpha}(x,t)\Psi_{\alpha'}^{\dagger}(x',t')|\psi_0 \rangle}{\langle \psi_0 |U(\infty,-\infty)|\psi_0 \rangle }$$ and it is not a Green function in a mathematical manner, i.e. $$\left[i\hbar\partial_t -\mathcal{H}_0\right] iG_{\alpha \alpha'} \neq \delta$$. Indeed, the unperturbed Green function $$iG_{0,\alpha \alpha'}$$ is a mathematical Green function. Using the Dyson expansion for $$U$$ in the interaction picture and the Wick's theorem, we can express $$iG_{\alpha \alpha'}$$ in terms of $$V$$ and $$iG_{0,\alpha \alpha'}$$. Such expression is known to satisfy the Dyson equation: $$G=G_0+G_0\sum(G_0+G_0\sum(\cdots))$$.

## Question

Obviously, the case 1 and 2 are closely related to each other. The case 3 and 4 are connected, either. However, I am not sure if there is a mathematical explanation about the reason that the Dyson expansion (case 3 & 4) leads to the result identical to the simple algebra (case 1 & 2).

I think the fact that the Dyson expansion produces not the Dyson equation itself $$G=G_0+G_0\Sigma G$$ but its iterative version $$G=G_0+G_0\sum(G_0+G_0\sum(\cdots))$$ would be a clue; but I have not made any progress so far.

## Green's functions and example 1

First off, you are using the symbol $$G$$ and $$K$$ in 1 & 2 for Green's functions.

So let's define a Green's function first as $$G(t,u)$$ such as: $$\mathcal{L}\,G(t,u) = \delta(t-u),$$ where $$\mathcal{L}$$ is a linear differential operator governing the evolution of the system.

The Green's function is used in solving the dynamics $$x(t)$$ caused by a source $$f(u)$$ by integrating over it: $$\mathcal{L}\,x(t) = \int \mathrm{d}u\,\mathcal{L}\,G(t,u)\, f(u) = \int \mathrm{d}u \ \delta(t-u)f(u) = f(t).$$

These are widely used in classical physics, such as in example 1.

## Example 2 & Quantum mechanics

Quantum mechanics (first quantisation) is described by the Schrödinger equation, i.e. just a linear differential operator $$\mathcal{L}$$ like above. Similarly, then, one has:

$$\phi(x,t_x) = \int \mathrm{d}y\, G^+(x,t_x,y,t_y) \phi(y,t_y),$$

where the Green's function $$G^+$$ propagates the particle, described by a wavefunction $$\phi$$, from position $$y$$ and time $$t_y$$ to a position $$x$$ at time $$t_x$$. The $$+$$ sign is to prevent particles from travelling back in time, so technically $$G^+ = \theta(t_x-t_y)G$$ is the retarted Green's function.

Hence the connection between a Green's function and a propagator in quantum mechanics. Generally, the propagator $$G^+$$ can be written: $$G^+(x,t_x,y,t_y) = \theta(t_x-t_y)\, \langle x(t_x) | y(t_y)\rangle,$$ i.e. the probability amplitude that a particle in state $$|y\rangle$$ at time $$t_y$$ ends ip in state $$|x\rangle$$ at time $$t_x$$.

The fact that the propagator is non-zero for causally disconnected events, i.e. events that are space-like separated, is the failure of first quantisation and the need of second quantisation (quantum field theory)

## Dyson's equation

Green's function in the energy domain can be written as:

$$G^+(x,y,E) = \sum_n \frac{i\phi_n(x) \, \phi_n^\ast (y)}{E-E_n} \propto \frac{1}{E-H_0}.$$

easily justified by taking the Laplace equation for the electric potential $$\phi$$ in vacuo $$\nabla^2\phi = 0$$. The Green's function solution is $$\epsilon_0 \nabla^2V(\mathbf{r}) = - \delta^{(3)}(\mathbf{r}-\mathbf{r}_0)$$, to which you know the solution to be a point charge with $$V(\mathbf{r}) = 1/|\mathbf{r}-\mathbf{r}_0|$$

Green's functions allow us to interpret a perturbation problem in terms of a propagating particle. A perturbation interrupts the propagation via a scattering process, following which free-particle propagation resumes.

A generic Hamiltonian $$H = H_0 + V$$ is decomposed in to its solvable, free-particle propagator-giving solution $$H_0$$ and in a non-analytically solvable perturbation $$V$$.

The Green's function is given by $$G \propto \frac{1}{E-H} = \frac{1}{E-H_0 -V},$$ which would be called the full propagator to distinguish it from the free propagator $$G_0 \propto 1/(E-H_0)$$.

In a perturbative approach, where $$V \ll H_0$$, the expression above can be written as:

$$G = \frac{1}{E-H_0-V} = \frac{1}{E-H_0} + \frac{1}{E-H_0}V\frac{1}{E-V_0}+\frac{1}{E-H_0}V\frac{1}{E-V_0}V\frac{1}{E-V_0} + \dots$$ or $$G = G_0 + G_0 V G_0 + G_0 V G_0 V G_0 + \dots$$

which is a geometric series, rewritten as:

$$G = G_0(1 + VG_0 + VG_0VG_0 + \dots) = \frac{G_0}{1-V G_0} = \frac{1}{G_0^{-1}-V},$$

which is known as Dyson's equation.

## Example 3

The Schrödinger equation for a generic Hamiltonian $$H$$ is:

$$i\hbar \frac{\mathrm{d}}{\mathrm{d}}|\psi_t\rangle = H |\psi_{t_0}\rangle.$$ $$\implies |\psi_t\rangle = U(t,t_0) |\psi_{t_0}\rangle,$$

where $$U$$ is the time-evolution operator.

In the interaction picture, the Hamiltonian is split between its free part $$H_0$$ and interacting part $$H'$$ such that the time-evolution of the interaction picture Haimltonian $$H_I$$ is given by $$H_I(t) = e^{iH_0 t/\hbar}H'e^{-iH_0t/\hbar}$$.

The time evolution operator in this picture then becomes \$U_I = e^{iH_0t/\hbar}Ue^{-iH_0t/\hbar}, and satisfies the following equation of motion:

$$i\hbar \frac{\mathrm{d}}{\mathrm{d}t}U_I = H_I U_I.$$

The solution to the above is : $$U_I(t,t_0) = 1 - \frac{i}{\hbar} \int_{t_0}^t \mathrm{d}t' H_I(t', t_0)\,U_I(t', t_0) ,$$

and it can be iterated by keeping plugging in the expression for the $$U_I$$ in the integral:

$$U_I(t,t_0) = 1 - \frac{i}{\hbar} \int_{t_0}^t \mathrm{d}t' H_I(t', t_0)\ + \left (-\frac{i}{\hbar} \right )^2 \int_{t_0}^t \mathrm{d}t'' \int_{t_0}^t \mathrm{d}t''' H_I(t'', t_0)\,H_I(t''', t_0) +\dots$$

Going back to $$U = e^{-iH_0t}U_Ie^{iH_0t}$$, idenfitying $$G_0 = e^{iH_0t}$$ and $$H_I = V$$, we get the Dyson's equation as per the point above.

Hence $$U$$ is also called the Dyson's operator and the perturbative expansion known as Dyson series/expansion.

The time ordering operator $$T$$ should also be added.

## Quantum field theory & example 4

While the first quantisation propagator is the same as a Green's function, in second quantisation the propagator is only termed as such by analogy.

Indeed: $$G^+(x,y) = \theta(x^0 - y^0) \langle \Omega| \hat{\phi}(x)\hat{\phi}^\dagger (y) |\Omega \rangle,$$ where $$|\Omega\rangle$$ is the interacting vacuum, $$x$$ and $$y$$ are now four-vectors in spacetime, and $$\hat{\phi^\dagger}(y)$$ is a field operator creating a particle at $$(y^0, \mathbf{y})$$.

Your expression in just obtained by requiring the overlap $$A$$ between an incoming and outcoming state to be expressed in terms of the free, non-interacting states: $$A = _{\mathrm{interacting}}\langle q|p\rangle_{\mathrm{interacting}} = _{\mathrm{free}}\langle q|S|p\rangle_{\mathrm{free}},$$ with $$S$$ being the scattering matrix.

The scattering matrix $$S$$ is related to the evolution operator $$U$$ above by $$S = \lim_{t\rightarrow \infty, t_0 \rightarrow -\infty} U(t,t_0)$$.