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---
title: "Dyson equation"
date: 2021-11-01
categories:
- Physics
- Quantum mechanics
layout: "concept"
---
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 add a more complicated and
possibly time-dependent term $\hat{H}_1(\vb{r}, t)$,
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.
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