Improve knowledge baseHEADmaster

1 files changed, 19 insertions, 19 deletions

diff --git a/source/know/concept/dyson-equation/index.md b/source/know/concept/dyson-equation/index.md
index ae9eb35..03be06f 100644
--- a/source/know/concept/dyson-equation/index.md
+++ b/source/know/concept/dyson-equation/index.md
@@ -25,8 +25,8 @@ $$\begin{aligned}
     = \delta(\vb{r} - \vb{r}') \: \delta(t - t')
 \end{aligned}$$
 
-From this, we define the inverse $$\hat{G}{}_0^{-1}(\vb{r}, t)$$
-as follows, so that $$\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')$$:
+From this, we define the inverse $$\hat{G}{}_0^{-1}(\vb{r}, t)$$ as follows,
+so that $$\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')$$:
 
 $$\begin{aligned}
     \hat{G}{}_0^{-1}(\vb{r}, t)
@@ -35,16 +35,15 @@ $$\begin{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 $$t'$$:
+its operator version $$\hat{G}_0(\vb{r}, t)$$
+as a multiplication by $$G_0$$ and integration over $$\vb{r}'$$ and $$t'$$:
 
 $$\begin{aligned}
     \hat{G}_0(\vb{r}, t) \: f
-    \equiv \iint_{-\infty}^\infty G_0(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \: \dd{\vb{r}}' \dd{t'}
+    \equiv \iint_{-\infty}^\infty G_0(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}}' \dd{t'}
 \end{aligned}$$
 
-For an arbitrary function $$f(\vb{r}, t)$$,
+Where $$f(\vb{r}, t)$$ is an arbitrary function,
 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 $$\hat{G}{}_0^{-1}$$:
@@ -61,7 +60,7 @@ 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:
+$$G(\vb{r}, \vb{r}', t, t')$$ (note the lack of a $$0$$ subscript) satisfies:
 
 $$\begin{aligned}
     \delta(\vb{r} - \vb{r}') \: \delta(t - t')
@@ -72,7 +71,7 @@ $$\begin{aligned}
 
 This equation is typically too complicated to solve,
 so we would like an easier way to calculate this new $$G$$.
-The perturbed wavefunction $$\Psi(\vb{r}, t)$$
+Clearly, the perturbed wavefunction $$\Psi(\vb{r}, t)$$
 satisfies the Schrödinger equation:
 
 $$\begin{aligned}
@@ -80,9 +79,8 @@ $$\begin{aligned}
     = 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:
+We know that $$\hat{G}{}_0^{-1} \Psi_0 = 0$$ from earlier,
+which we put on the right, and then apply $$\hat{G}_0$$ to it:
 
 $$\begin{aligned}
     \hat{G}_0^{-1} \Psi - \hat{H}_1 \Psi
@@ -110,7 +108,8 @@ $$\begin{aligned}
 \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**:
+so we denote it by $$\hat{G}$$ (an operator, not the function $$G$$)
+and write the so-called **Dyson equation**:
 
 $$\begin{aligned}
     \boxed{
@@ -133,8 +132,8 @@ $$\begin{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:
+Let us take the latter's definition and multiply it by an arbitrary $$f(\vb{r}, t)$$,
+and then integrate over $$G$$'s second argument pair:
 
 $$\begin{aligned}
     \iint \big( \hat{G}{}_0^{-1} \!-\! \hat{H}_1 \big) G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'}
@@ -143,8 +142,7 @@ $$\begin{aligned}
 \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 $$t'$$):
+We apply $$\hat{G}_0(\vb{r}, t)$$ to this equation:
 
 $$\begin{aligned}
     \hat{G}_0 f
@@ -154,8 +152,10 @@ $$\begin{aligned}
 \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:
+so we conclude that the operator $$\hat{G}$$
+is defined as multiplication by the function $$G$$ followed by integration,
+exactly analogously to $$\hat{G}_0$$ and $$G_0$$,
+which should not be a big surprise:
 
 $$\begin{aligned}
     \hat{G}(\vb{r}, t) \: f(\vb{r}, t)