---
title: "Dyson equation"
firstLetter: "D"
publishDate: 2021-11-01
categories:
- Physics
- Quantum mechanics

date: 2021-11-01T14:57:54+01:00
draft: false
markup: pandoc
---

# Dyson equation

Consider the time-dependent Schrödinger equation,
describing a wavefunction $\Psi_0(\vb{r}, t)$:

$$\begin{aligned}
    i \hbar \pdv{t} \Psi_0(\vb{r}, t)
    = \hat{H}_0(\vb{r}) \: \Psi_0(\vb{r}, t)
\end{aligned}$$

By definition, this equation's
[fundamental solution](/know/concept/fundamental-solution/)
$G_0(\vb{r}, t; \vb{r}', t')$ satisfies the following:

$$\begin{aligned}
    \Big( i \hbar \pdv{t} - \hat{H}_0(\vb{r}) \Big) G_0(\vb{r}, t; \vb{r}', t')
    = \delta(\vb{r} - \vb{r}') \: \delta(t - t')
\end{aligned}$$

From this, we define the inverse $\hat{G}{}_0^{-1}(\vb{r}, t)$
as follows, so that $\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')$:

$$\begin{aligned}
    \hat{G}{}_0^{-1}(\vb{r}, t)
    &\equiv i \hbar \pdv{t} - \hat{H}_0(\vb{r})
\end{aligned}$$

Note that $\hat{G}{}_0^{-1}$ is an operator, while $G_0$ is a function.
For the sake of consistency, we thus define
the operator $\hat{G}_0(\vb{r}, t)$
as a multiplication by $G_0$
and integration over $\vb{r}'$ and $t'$:

$$\begin{aligned}
    \hat{G}_0(\vb{r}, t) \: f
    \equiv \iint_{-\infty}^\infty G_0(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \: \dd{\vb{r}}' \dd{t'}
\end{aligned}$$

For an arbitrary function $f(\vb{r}, t)$,
so that $\hat{G}{}_0^{-1} \hat{G}_0 = \hat{G}_0 \hat{G}{}_0^{-1} = 1$.
Moving on, the Schrödinger equation can be rewritten like so,
using $\hat{G}{}_0^{-1}$:

$$\begin{aligned}
    \hat{G}{}_0^{-1}(\vb{r}, t) \: \Psi_0(\vb{r}, t)
    = 0
\end{aligned}$$

Let us assume that $\hat{H}_0$ is simple,
such that $G_0$ and $\hat{G}{}_0^{-1}$ can be found without issues
by solving the defining equation above.

Suppose we now perturb this Hamiltonian with
a possibly time-dependent operator $\hat{H}_1(\vb{r}, t)$,
in which case the corresponding fundamental solution
$G(\vb{r}, \vb{r}', t, t')$ satisfies:

$$\begin{aligned}
    \delta(\vb{r} - \vb{r}') \: \delta(t - t')
    &= \Big( i \hbar \pdv{t} - \hat{H}_0(\vb{r}) - \hat{H}_1(\vb{r}, t) \Big) G(\vb{r}, t; \vb{r}', t')
    \\
    &= \Big( \hat{G}{}_0^{-1}(\vb{r}, t) - \hat{H}_1(\vb{r}, t) \Big) G(\vb{r}, t; \vb{r}', t')
\end{aligned}$$

This equation is typically too complicated to solve,
so we would like an easier way to calculate this new $G$.
The perturbed wavefunction $\Psi(\vb{r}, t)$
satisfies the Schrödinger equation:

$$\begin{aligned}
    \Big( \hat{G}{}_0^{-1}(\vb{r}, t) - \hat{H}_1(\vb{r}, t) \Big) \Psi(\vb{r}, t)
    = 0
\end{aligned}$$

We know that $\hat{G}{}_0^{-1} \Psi_0 = 0$,
which we put on the right,
and then we apply $\hat{G}_0$ in front:

$$\begin{aligned}
    \hat{G}_0^{-1} \Psi - \hat{H}_1 \Psi
    = \hat{G}_0^{-1} \Psi_0
    \quad \implies \quad
    \Psi - \hat{G}_0 \hat{H}_1 \Psi
    &= \Psi_0
\end{aligned}$$

This equation is recursive,
so we iteratively insert it into itself.
Note that the resulting equations are the same as those from
[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/):

$$\begin{aligned}
    \Psi
    &= \Psi_0 + \hat{G}_0 \hat{H}_1 \Psi
    \\
    &= \Psi_0 + \hat{G}_0 \hat{H}_1 \Psi_0 + \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \Psi
    \\
    &= \Psi_0 + \hat{G}_0 \hat{H}_1 \Psi_0 + \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \Psi_0
    + \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \Psi_0 + \: ...
    \\
    &= \Psi_0 + \big( \hat{G}_0 + \hat{G}_0 \hat{H}_1 \hat{G}_0 +  \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \hat{G}_0 + \: ... \big) \hat{H}_1 \Psi_0
\end{aligned}$$

The parenthesized expression clearly has the same recursive pattern,
so we denote it by $\hat{G}$ and write the so-called **Dyson equation**:

$$\begin{aligned}
    \boxed{
        \hat{G}
        = \hat{G}_0 + \hat{G}_0 \hat{H}_1 \hat{G}
    }
\end{aligned}$$

Such an iterative scheme is excellent for approximating $\hat{G}(\vb{r}, t)$.
Once a satisfactory accuracy is obtained,
the perturbed wavefunction $\Psi$ can be calculated from:

$$\begin{aligned}
    \boxed{
        \Psi
        = \Psi_0 + \hat{G} \hat{H}_1 \Psi_0
    }
\end{aligned}$$

This relation is equivalent to the Schrödinger equation.
So now we have the operator $\hat{G}(\vb{r}, t)$,
but what about the fundamental solution function $G(\vb{r}, t; \vb{r}', t')$?
Let us take its definition, multiply it by an arbitrary $f(\vb{r}, t)$,
and integrate over $G$'s second argument pair:

$$\begin{aligned}
    \iint \big( \hat{G}{}_0^{-1} \!-\! \hat{H}_1 \big) G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'}
    = \iint \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'}
    = f
\end{aligned}$$

Where we have hidden the arguments $(\vb{r}, t)$ for brevity.
We now apply $\hat{G}_0(\vb{r}, t)$ to this equation
(which contains an integral over $t''$ independent of $t'$):

$$\begin{aligned}
    \hat{G}_0 f
    &= \big( \hat{G}_0 \hat{G}{}_0^{-1} - \hat{G}_0 \hat{H}_1 \big) \iint_{-\infty}^\infty G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'}
    \\
    &= \big( 1 - \hat{G}_0 \hat{H}_1 \big) \iint_{-\infty}^\infty G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'}
\end{aligned}$$

Here, the shape of Dyson's equation is clearly recognizable,
so we conclude that, as expected, the operator $\hat{G}$
is defined as multiplication by the function $G$ followed by integration:

$$\begin{aligned}
    \hat{G}(\vb{r}, t) \: f(\vb{r}, t)
    \equiv \iint_{-\infty}^\infty G(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \: \dd{\vb{r}}' \dd{t'}
\end{aligned}$$



## References
1.  H. Bruus, K. Flensberg,
    *Many-body quantum theory in condensed matter physics*,
    2016, Oxford.