1 files changed, 28 insertions, 27 deletions
diff --git a/source/know/concept/equation-of-motion-theory/index.md b/source/know/concept/equation-of-motion-theory/index.md
index f62fb56..02ed856 100644
--- a/source/know/concept/equation-of-motion-theory/index.md
+++ b/source/know/concept/equation-of-motion-theory/index.md
@@ -13,9 +13,9 @@ is a method to calculate the time evolution of a system's properties
 using [Green's functions](/know/concept/greens-functions/).
 
 Starting from the definition of
-the retarded single-particle Green's function $G_{\nu \nu'}^R(t, t')$,
-we simply take the $t$-derivative
-(we could do the same with the advanced function $G_{\nu \nu'}^A$):
+the retarded single-particle Green's function $$G_{\nu \nu'}^R(t, t')$$,
+we simply take the $$t$$-derivative
+(we could do the same with the advanced function $$G_{\nu \nu'}^A$$):
 
 $$\begin{aligned}
     i \hbar \pdv{G^R_{\nu \nu'}(t, t')}{t}
@@ -27,22 +27,22 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where we have used that the derivative
-of a [Heaviside step function](/know/concept/heaviside-step-function/) $\Theta$
-is a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$.
+of a [Heaviside step function](/know/concept/heaviside-step-function/) $$\Theta$$
+is a [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta$$.
 Also, from the [second quantization](/know/concept/second-quantization/),
-$\expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}$
-for $t = t'$ is zero when $\nu \neq \nu'$.
+$$\expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}$$
+for $$t = t'$$ is zero when $$\nu \neq \nu'$$.
 
 Since we are in the [Heisenberg picture](/know/concept/heisenberg-picture/),
-we know the equation of motion of $\hat{c}_\nu(t)$:
+we know the equation of motion of $$\hat{c}_\nu(t)$$:
 
 $$\begin{aligned}
     \dv{\hat{c}_\nu(t)}{t}
     = \frac{i}{\hbar} \comm{\hat{H}_0(t)}{\hat{c}_\nu(t)} + \frac{i}{\hbar} \comm{\hat{H}_\mathrm{int}(t)}{\hat{c}_\nu(t)}
 \end{aligned}$$
 
-Where the single-particle part of the Hamiltonian $\hat{H}_0$
-and the interaction part $\hat{H}_\mathrm{int}$
+Where the single-particle part of the Hamiltonian $$\hat{H}_0$$
+and the interaction part $$\hat{H}_\mathrm{int}$$
 are assumed to be time-independent in the Schrödinger picture.
 We thus get:
 
@@ -52,8 +52,8 @@ $$\begin{aligned}
     \Expval{\Comm{\comm{\hat{H}_0}{\hat{c}_\nu} + \comm{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}}
 \end{aligned}$$
 
-The most general form of $\hat{H}_0$, for any basis,
-is as follows, where $u_{\nu' \nu''}$ are constants:
+The most general form of $$\hat{H}_0$$, for any basis,
+is as follows, where $$u_{\nu' \nu''}$$ are constants:
 
 $$\begin{aligned}
     \hat{H}_0
@@ -68,7 +68,7 @@ $$\begin{aligned}
 <label for="proof-commH0">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-commH0">Proof.</label>
-Using the commutator identity for $\comm{A B}{C}$,
+Using the commutator identity for $$\comm{A B}{C}$$,
 we decompose it like so:
 
 $$\begin{aligned}
@@ -105,11 +105,12 @@ $$\begin{aligned}
     - 2 \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
     = - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''}
 \end{aligned}$$
+
 </div>
 </div>
 
-Substituting this into $G_{\nu \nu'}^R$'s equation of motion,
-we recognize another Green's function $G_{\nu'' \nu'}^R$:
+Substituting this into $$G_{\nu \nu'}^R$$'s equation of motion,
+we recognize another Green's function $$G_{\nu'' \nu'}^R$$:
 
 $$\begin{aligned}
     i \hbar \pdv{G^R_{\nu \nu'}}{t}
@@ -132,9 +133,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where $D_{\nu \nu'}^R$ represents a correction due to interactions $\hat{H}_\mathrm{int}$,
+Where $$D_{\nu \nu'}^R$$ represents a correction due to interactions $$\hat{H}_\mathrm{int}$$,
 and also has the form of a retarded Green's function,
-but with $\hat{c}_{\nu}$ replaced by $\comm{-\hat{H}_\mathrm{int}}{\hat{c}_\nu}$:
+but with $$\hat{c}_{\nu}$$ replaced by $$\comm{-\hat{H}_\mathrm{int}}{\hat{c}_\nu}$$:
 
 $$\begin{aligned}
     \boxed{
@@ -143,19 +144,19 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Unfortunately, calculating $D_{\nu \nu'}^R$
-might still not be doable due to $\hat{H}_\mathrm{int}$.
-The key idea of equation-of-motion theory is to either approximate $D_{\nu \nu'}^R$ now,
-or to differentiate it again $i \hbar \idv{D_{\nu \nu'}^R}{t}$,
+Unfortunately, calculating $$D_{\nu \nu'}^R$$
+might still not be doable due to $$\hat{H}_\mathrm{int}$$.
+The key idea of equation-of-motion theory is to either approximate $$D_{\nu \nu'}^R$$ now,
+or to differentiate it again $$i \hbar \idv{D_{\nu \nu'}^R}{t}$$,
 and try again for the resulting corrections,
 until a solvable equation is found.
 There is no guarantee that that will ever happen;
 if not, one of the corrections needs to be approximated.
 
-For non-interacting particles $\hat{H}_\mathrm{int} = 0$,
-so clearly $D_{\nu \nu'}^R$ trivially vanishes then.
-Let us assume that $\hat{H}_0$ is also time-independent,
-such that $G_{\nu'' \nu'}^R$ only depends on the difference $t - t'$:
+For non-interacting particles $$\hat{H}_\mathrm{int} = 0$$,
+so clearly $$D_{\nu \nu'}^R$$ trivially vanishes then.
+Let us assume that $$\hat{H}_0$$ is also time-independent,
+such that $$G_{\nu'' \nu'}^R$$ only depends on the difference $$t - t'$$:
 
 $$\begin{aligned}
         \sum_{\nu''} \Big( i \hbar \delta_{\nu \nu''} \pdv{}{t} - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(t - t')
@@ -163,14 +164,14 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We take the [Fourier transform](/know/concept/fourier-transform/)
-$(t \!-\! t') \to (\omega + i \eta)$, where $\eta \to 0^+$ ensures convergence:
+$$(t \!-\! t') \to (\omega + i \eta)$$, where $$\eta \to 0^+$$ ensures convergence:
 
 $$\begin{aligned}
         \sum_{\nu''} \Big( \hbar \delta_{\nu \nu''} (\omega + i \eta) - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(\omega)
         = \delta_{\nu \nu'}
 \end{aligned}$$
 
-If we assume a diagonal basis $u_{\nu \nu''} = \varepsilon_\nu \delta_{\nu \nu''}$,
+If we assume a diagonal basis $$u_{\nu \nu''} = \varepsilon_\nu \delta_{\nu \nu''}$$,
 this reduces to the following:
 
 $$\begin{aligned}