Improve knowledge baseHEADmaster

1 files changed, 24 insertions, 22 deletions

diff --git a/source/know/concept/interaction-picture/index.md b/source/know/concept/interaction-picture/index.md
index 8428bf3..a3bb260 100644
--- a/source/know/concept/interaction-picture/index.md
+++ b/source/know/concept/interaction-picture/index.md
@@ -39,14 +39,14 @@ Basically, any way of splitting $$\hat{H}_S$$ is valid
 as long as $$\hat{H}_{0, S}$$ is time-independent,
 but only a few ways are useful.
 
-We now define the unitary conversion operator $$\hat{U}(t)$$ as shown below.
-Note its similarity to the [time-evolution operator](/know/concept/time-evolution-operator/)
-$$\hat{K}_S(t)$$, but with the opposite sign in the exponent:
+We now define the unitary conversion operator $$\hat{U}_0(t)$$ as shown below.
+Note its similarity to the
+[time-evolution operator](/know/concept/time-evolution-operator/) $$\hat{K}_S(t)$$:
 
 $$\begin{aligned}
     \boxed{
-        \hat{U}(t)
-        \equiv \exp\!\bigg( \frac{i}{\hbar} \hat{H}_{0,S} t \bigg)
+        \hat{U}_0(t)
+        \equiv \exp\!\bigg( \!-\! \frac{i}{\hbar} \hat{H}_{0,S} t \bigg)
     }
 \end{aligned}$$
 
@@ -56,17 +56,17 @@ and operators $$\hat{L}_I(t)$$ are then defined as follows:
 $$\begin{aligned}
     \boxed{
         \Ket{\psi_I(t)}
-        \equiv \hat{U}(t) \Ket{\psi_S(t)}
+        \equiv \hat{U}_0^\dagger(t) \Ket{\psi_S(t)}
     }
     \qquad\qquad
     \boxed{
         \hat{L}_I(t)
-        \equiv \hat{U}(t) \: \hat{L}_S(t) \: \hat{U}{}^\dagger(t)
+        \equiv \hat{U}_0^\dagger(t) \: \hat{L}_S(t) \: \hat{U}{}_0(t)
     }
 \end{aligned}$$
 
 Because $$\hat{H}_{0, S}$$ is time-independent,
-it commutes with $$\hat{U}(t)$$,
+it commutes with $$\hat{U}_0$$,
 so conveniently $$\hat{H}_{0, I} = \hat{H}_{0, S}$$.
 
 
@@ -78,25 +78,25 @@ we differentiate it and multiply by $$i \hbar$$:
 
 $$\begin{aligned}
     i \hbar \dv{}{t} \Ket{\psi_I}
-    &= i \hbar \dv{\hat{U}}{t} \Ket{\psi_S} + \hat{U} \bigg( i \hbar \dv{}{t}\Ket{\psi_S} \bigg)
+    &= i \hbar \dv{\hat{U}_0^\dagger}{t} \Ket{\psi_S} + \hat{U}_0^\dagger \bigg( i \hbar \dv{}{t}\Ket{\psi_S} \bigg)
 \end{aligned}$$
 
-We insert the definition of $$\hat{U}$$ in the first term
+We insert the definition of $$\hat{U}_0$$ in the first term
 and the Schrödinger equation into the second,
-and use the fact that $$\comm{\hat{H}_{0, S}}{\hat{U}} = 0$$
+and use the fact that $$\comm{\hat{H}_{0, S}}{\hat{U}_0} = 0$$
 thanks to the time-independence of $$\hat{H}_{0, S}$$:
 
 $$\begin{aligned}
     i \hbar \dv{}{t} \Ket{\psi_I}
-    &= - \hat{H}_{0,S} \hat{U} \Ket{\psi_S} + \hat{U} \hat{H}_S \Ket{\psi_S}
+    &= - \hat{H}_{0,S} \hat{U}_0^\dagger \Ket{\psi_S} + \hat{U}_0^\dagger \hat{H}_S \Ket{\psi_S}
     \\
-    &= \hat{U} \big( \!-\! \hat{H}_{0,S} + \hat{H}_S \big) \Ket{\psi_S}
+    &= \hat{U}_0^\dagger \big( \!-\! \hat{H}_{0,S} + \hat{H}_S \big) \Ket{\psi_S}
     \\
-    &= \hat{U} \big( \hat{H}_{1,S} \big) \hat{U}{}^\dagger \hat{U} \Ket{\psi_S}
+    &= \hat{U}_0^\dagger \hat{H}_{1,S} \big( \hat{U}_0 \hat{U}_0^\dagger \big) \Ket{\psi_S}
 \end{aligned}$$
 
 Which leads to an analogue of the Schrödinger equation,
-with $$\hat{H}_{1,I} = \hat{U} \hat{H}_{1,S} \hat{U}{}^\dagger$$:
+with $$\hat{H}_{1,I} = \hat{U}_0^\dagger \hat{H}_{1,S} \hat{U}_0$$:
 
 $$\begin{aligned}
     \boxed{
@@ -110,11 +110,11 @@ in order to describe its evolution in time:
 
 $$\begin{aligned}
     \dv{\hat{L}_I}{t}
-    &= \dv{\hat{U}}{t} \hat{L}_S \hat{U}{}^\dagger + \hat{U} \hat{L}_S \dv{\hat{U}{}^\dagger}{t}
-    + \hat{U} \dv{\hat{L}_S}{t} \hat{U}{}^\dagger
+    &= \dv{\hat{U}_0^\dagger}{t} \hat{L}_S \hat{U}_0 + \hat{U}_0^\dagger \hat{L}_S \dv{\hat{U}_0}{t}
+    + \hat{U}_0^\dagger \dv{\hat{L}_S}{t} \hat{U}_0
     \\
-    &= \frac{i}{\hbar} \hat{U} \hat{H}_{0,S} \big( \hat{U}{}^\dagger \hat{U} \big) \hat{L}_S \hat{U}{}^\dagger
-    - \frac{i}{\hbar} \hat{U} \hat{L}_S \big( \hat{U}{}^\dagger \hat{U} \big) \hat{H}_{0,S} \hat{U}{}^\dagger
+    &= \frac{i}{\hbar} \hat{U}_0^\dagger \hat{H}_{0,S} \big( \hat{U}_0 \hat{U}_0^\dagger \big) \hat{L}_S \hat{U}_0
+    - \frac{i}{\hbar} \hat{U}_0^\dagger \hat{L}_S \big( \hat{U}_0 \hat{U}_0^\dagger \big) \hat{H}_{0,S} \hat{U}_0
     + \bigg( \dv{\hat{L}_S}{t} \bigg)_I
     \\
     &= \frac{i}{\hbar} \hat{H}_{0,I} \hat{L}_I
@@ -144,13 +144,15 @@ can be solved in isolation in a kind of Schrödinger picture.
 What about the time evolution operator $$\hat{K}_S(t)$$?
 Its interaction version $$\hat{K}_I(t)$$
 is unsurprisingly obtained by the standard transform
-$$\hat{K}_I = \hat{U} \hat{K}_S \hat{U}^\dagger$$:
+$$\hat{K}_I = \hat{U}_0^\dagger \hat{K}_S \hat{U}_0$$:
 
 $$\begin{aligned}
     \Ket{\psi_I(t)}
-    &= \hat{U}(t) \Ket{\psi_S(t)}
+    &= \hat{U}_0^\dagger(t) \Ket{\psi_S(t)}
     \\
-    &= \hat{U}(t) \: \hat{K}_S(t) \: \hat{U}^\dagger(t) \Ket{\psi_I(0)}
+    &= \hat{U}_0^\dagger(t) \: \hat{K}_S(t) \Ket{\psi_S(0)}
+    \\
+    &= \hat{U}_0^\dagger(t) \: \hat{K}_S(t) \: \hat{U}_0(t) \: \hat{U}_0^\dagger(t) \Ket{\psi_S(0)}
     \\
     &\equiv \hat{K}_I(t) \Ket{\psi_I(0)}
 \end{aligned}$$