1 files changed, 11 insertions, 59 deletions
diff --git a/content/know/concept/kubo-formula/index.pdc b/content/know/concept/kubo-formula/index.pdc
index f0208da..f5430da 100644
--- a/content/know/concept/kubo-formula/index.pdc
+++ b/content/know/concept/kubo-formula/index.pdc
@@ -35,7 +35,7 @@ $$\begin{aligned}
     = \matrixel{\psi_S(t)}{\hat{A}_S}{\psi_S(t)}
     &= \matrixel{\psi_I(t)}{\hat{A}_I(t)}{\psi_I(t)}
     \\
-    &= \matrixel{\psi_I(t_0)}{\hat{K}_I^\dagger(t, t_0) \hat{A}_I(t) \hat{K}_I(t, t_0)}{\psi_I(t_0)}
+    &= \matrixel{\psi_I(t_0)\,}{\,\hat{K}_I^\dagger(t, t_0) \hat{A}_I(t) \hat{K}_I(t, t_0)\,}{\,\psi_I(t_0)}
 \end{aligned}$$
 
 Where the time evolution operator $\hat{K}_I(t, t_0)$ is as follows,
@@ -116,9 +116,12 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-This result applies to bosonic operators,
-whereas for fermionic operators
-the commutator would be replaced by an anticommutator.
+Note that observables are bosonic,
+because in the [second quantization](/know/concept/second-quantization/)
+they consist of products of even numbers
+of particle creation/annihiliation operators.
+Therefore, this correlation function
+is a two-particle [Green's function](/know/concept/greens-functions/).
 
 A common situation is that $\hat{H}_1$ consists of
 a time-independent operator $\hat{B}$
@@ -133,67 +136,16 @@ $$\begin{aligned}
     = C^R_{A B}(t, t') f(t')
 \end{aligned}$$
 
-Conveniently, it can be shown that in this case
-$C^R_{AB}$ only depends on the difference $t - t'$,
-if we assume that the system was initially in thermodynamic equilibrium:
+Since $C_{AB}^R$ is a Green's function,
+we know that it only depends on the difference $t - t'$,
+as long as the system was initially in thermodynamic equilibrium,
+and $\hat{H}_{0,S}$ is time-independent:
 
 $$\begin{aligned}
     C^R_{A B}(t, t')
     = C^R_{A B}(t - t')
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-time-difference"/>
-<label for="proof-time-difference">Proof</label>
-<div class="hidden">
-<label for="proof-time-difference">Proof.</label>
-This is trivial for $\Theta(t\!-\!t')$,
-so the challenge is to prove that
-$\expval*{\comm*{\hat{A}_I(t)}{B_I(t')}}$
-depends only on the time difference $t - t'$.
-
-Suppose that the system started in thermodynamic equilibrium
-(see [canonical ensemble](/know/concept/canonical-ensemble/)),
-so that its (unnormalized) [density operator](/know/concept/density-operator/)
-$\hat{\rho}$ was as follows before $\hat{H}_{1,I}$ was applied:
-
-$$\begin{aligned}
-    \hat{\rho} = \exp\!(- \beta \hat{H}_{0,S})
-\end{aligned}$$
-
-Let us assume that the perturbation $\hat{H}_{1,I}$
-does not affect the distribution of states,
-but only their individual evolutions in time.
-Note that, in general, this is not equilibrium.
-
-In that case, the expectation value of the product
-of two time-independent observables $\hat{A}$ and $\hat{B}$
-can be calculated as follows,
-where $Z_0 \equiv \Tr\!(\hat{\rho})$ is the partition function:
-
-$$\begin{aligned}
-    \expval*{\hat{A} \hat{B}}
-    = \frac{1}{Z_0} \Tr\!\big( \hat{\rho} \hat{A}_I \hat{H}_{1,I} \big)
-    = \frac{1}{Z_0} \Tr\!\Big( e^{-\beta \hat{H}_0} e^{i t \hat{H}_0 / \hbar} \hat{A}_S e^{-i t \hat{H}_0 / \hbar}
-    e^{i t' \hat{H}_0 / \hbar} \hat{B}_S e^{-i t' \hat{H}_0 / \hbar} \Big)
-\end{aligned}$$
-
-Where $\hat{H}_0 \equiv \hat{H}_{0,S}$ for brevity.
-Using that the trace $\Tr$ is invariant
-under cyclic permutations of its argument,
-and that functions of $\hat{H}_{0,S}$ always commute, we find:
-
-$$\begin{aligned}
-    \expval*{\hat{A} \hat{B}}
-    = \frac{1}{Z_0} \Tr\!\Big( e^{-\beta \hat{H}_0} e^{i (t - t') \hat{H}_0 / \hbar} \hat{A}_S
-    e^{-i (t - t') \hat{H}_0 / \hbar} \hat{B}_S \Big)
-\end{aligned}$$
-
-As expected, this clearly only depends on the time difference $t - t'$,
-because $\hat{H}_{0,S}$ is time-independent by assumption.
-</div>
-</div>
-
 With this, the Kubo formula can be written as follows,
 where we have set $t_0 = - \infty$: