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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.