Improve knowledge baseHEADmaster

1 files changed, 12 insertions, 11 deletions

diff --git a/source/know/concept/kubo-formula/index.md b/source/know/concept/kubo-formula/index.md
index 4cb39ac..41fda3c 100644
--- a/source/know/concept/kubo-formula/index.md
+++ b/source/know/concept/kubo-formula/index.md
@@ -27,14 +27,15 @@ respectively refer to the Schrödinger
 and [interaction pictures](/know/concept/interaction-picture/):
 
 $$\begin{aligned}
-    \expval{\hat{A}}(t)
+    \expval{\hat{A}(t)}
     = \matrixel{\psi_S(t)}{\hat{A}_S}{\psi_S(t)}
     &= \matrixel{\psi_I(t)}{\hat{A}_I(t)}{\psi_I(t)}
     \\
     &= \matrixel{\psi_I(t_0)\,}{\,\hat{K}_I^\dagger(t, t_0) \hat{A}_I(t) \hat{K}_I(t, t_0)\,}{\,\psi_I(t_0)}
 \end{aligned}$$
 
-Where the time evolution operator $$\hat{K}_I(t, t_0)$$ is as follows,
+Where the [time evolution operator](/know/concept/time-evolution-operator/)
+$$\hat{K}_I(t, t_0)$$ is as follows,
 which we Taylor-expand:
 
 $$\begin{aligned}
@@ -71,18 +72,18 @@ where $$\Expval{}$$ is the expectation value for $$\Ket{\psi(t)}$$,
 and $$\Expval{}_0$$ is the expectation value for $$\Ket{\psi_I(t_0)}$$:
 
 $$\begin{aligned}
-    \expval{\hat{A}}(t)
+    \expval{\hat{A}(t)}
     = \expval{\hat{K}_I^\dagger \hat{A}_I \hat{K}_I}_0
     = \expval{\hat{A}_I(t)}_0 - \frac{i}{\hbar} \int_{t_0}^t \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
 \end{aligned}$$
 
-Now we define $$\delta\!\expval{\hat{A}}\!(t)$$
+Now we define $$\delta\!\expval{\hat{A}(t)}$$
 as the change of $$\expval{\hat{A}}$$ due to the perturbation $$\hat{H}_1$$,
-and insert $$\expval{\hat{A}}(t)$$:
+and insert $$\expval{\hat{A}(t)}$$:
 
 $$\begin{aligned}
-    \delta\!\expval{\hat{A}}\!(t)
-    \equiv \expval{\hat{A}}(t) - \expval{\hat{A}_I}_0
+    \delta\!\expval{\hat{A}(t)}
+    \equiv \expval{\hat{A}(t)} - \expval{\hat{A}_I(t)}_0
     = - \frac{i}{\hbar} \int_{t_0}^t \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
 \end{aligned}$$
 
@@ -94,7 +95,7 @@ describing the response of $$\expval{\hat{A}}$$ to first order in $$\hat{H}_1$$:
 
 $$\begin{aligned}
     \boxed{
-        \delta\!\expval{\hat{A}}\!(t)
+        \delta\!\expval{\hat{A}(t)}
         = \int_{t_0}^\infty C^R_{A H_1}(t, t') \dd{t'}
     }
 \end{aligned}$$
@@ -142,7 +143,7 @@ With this, the Kubo formula can be written as follows,
 where we have set $$t_0 = - \infty$$:
 
 $$\begin{aligned}
-    \delta\!\expval{A}\!(t)
+    \delta\!\expval{\hat{A}(t)}
     = \int_{-\infty}^\infty C^R_{A B}(t - t') f(t') \dd{t'}
     = (C^R_{A B} * f)(t)
 \end{aligned}$$
@@ -150,12 +151,12 @@ $$\begin{aligned}
 This is a convolution,
 so the [convolution theorem](/know/concept/convolution-theorem/)
 states that the [Fourier transform](/know/concept/fourier-transform/)
-of $$\delta\!\expval{\hat{A}}\!(t)$$ is simply the product
+of $$\delta\!\expval{\hat{A}(t)}$$ is simply the product
 of the transforms of $$C^R_{AB}$$ and $$f$$:
 
 $$\begin{aligned}
     \boxed{
-        \delta\!\expval{\hat{A}}\!(\omega)
+        \delta\!\expval{\hat{A}(\omega)}
         = \tilde{C}{}^R_{A B}(\omega) \: \tilde{f}(\omega)
     }
 \end{aligned}$$