Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

1 files changed, 28 insertions, 28 deletions

diff --git a/source/know/concept/imaginary-time/index.md b/source/know/concept/imaginary-time/index.md
index 1dbdf11..245c3c1 100644
--- a/source/know/concept/imaginary-time/index.md
+++ b/source/know/concept/imaginary-time/index.md
@@ -8,7 +8,7 @@ categories:
 layout: "concept"
 ---
 
-Let $\hat{A}_S$ and $\hat{B}_S$ be time-independent in the Schrödinger picture.
+Let $$\hat{A}_S$$ and $$\hat{B}_S$$ be time-independent in the Schrödinger picture.
 Then, in the [Heisenberg picture](/know/concept/heisenberg-picture/),
 consider the following expectation value
 with respect to thermodynamic equilibium
@@ -19,9 +19,9 @@ $$\begin{aligned}
     &= \frac{1}{Z} \Tr\!\Big( \exp(-\beta \hat{H}_{0,S}(t)) \: \hat{A}_H(t) \: \hat{B}_H(t') \Big)
 \end{aligned}$$
 
-Where the "simple" Hamiltonian $\hat{H}_{0,S}$ is time-independent.
-Suppose a (maybe time-dependent) "difficult" $\hat{H}_{1,S}$ is added,
-so that the total Hamiltonian is $\hat{H}_S = \hat{H}_{0,S} + \hat{H}_{1,S}$.
+Where the "simple" Hamiltonian $$\hat{H}_{0,S}$$ is time-independent.
+Suppose a (maybe time-dependent) "difficult" $$\hat{H}_{1,S}$$ is added,
+so that the total Hamiltonian is $$\hat{H}_S = \hat{H}_{0,S} + \hat{H}_{1,S}$$.
 Then it is easier to consider the expectation value
 in the [interaction picture](/know/concept/interaction-picture/):
 
@@ -30,16 +30,16 @@ $$\begin{aligned}
     &= \frac{1}{Z} \Tr\!\Big( \exp(-\beta \hat{H}_S(t)) \: \hat{K}_I(0, t) \hat{A}_I(t) \hat{K}_I(t, t') \hat{B}_I(t') \hat{K}_I(t', 0) \Big)
 \end{aligned}$$
 
-Where $\hat{K}_I(t, t_0)$ is the time evolution operator of $\hat{H}_{1,S}$.
-In front, we have $\exp(-\beta \hat{H}_S(t))$,
-while $\hat{K}_I$ is an exponential of an integral of $\hat{H}_{1,I}$, so we are stuck.
+Where $$\hat{K}_I(t, t_0)$$ is the time evolution operator of $$\hat{H}_{1,S}$$.
+In front, we have $$\exp(-\beta \hat{H}_S(t))$$,
+while $$\hat{K}_I$$ is an exponential of an integral of $$\hat{H}_{1,I}$$, so we are stuck.
 Keep in mind that exponentials of operators
 cannot just be factorized, i.e. in general
-$\exp(\hat{A} \!+\! \hat{B}) \neq \exp(\hat{A}) \exp(\hat{B})$
+$$\exp(\hat{A} \!+\! \hat{B}) \neq \exp(\hat{A}) \exp(\hat{B})$$
 
 To get around this, a useful mathematical trick is
-to use an **imaginary time** variable $\tau$ instead of the real time $t$.
-Fixing a $t$, we "redefine" the interaction picture along the imaginary axis:
+to use an **imaginary time** variable $$\tau$$ instead of the real time $$t$$.
+Fixing a $$t$$, we "redefine" the interaction picture along the imaginary axis:
 
 $$\begin{aligned}
     \boxed{
@@ -48,13 +48,13 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Ironically, $\tau$ is real; the point is that this formula
-comes from the real-time definition by replacing $t \to -i \tau$.
+Ironically, $$\tau$$ is real; the point is that this formula
+comes from the real-time definition by replacing $$t \to -i \tau$$.
 The Heisenberg and Schrödinger pictures can be redefined in the same way.
 
-In fact, by substituting $t \to -i \tau$,
+In fact, by substituting $$t \to -i \tau$$,
 all the key results of the interaction picture can be updated,
-for example the Schrödinger equation for $\Ket{\psi_S(\tau)}$ becomes:
+for example the Schrödinger equation for $$\Ket{\psi_S(\tau)}$$ becomes:
 
 $$\begin{aligned}
     \hbar \dv{}{t}\Ket{\psi_S(\tau)}
@@ -64,7 +64,7 @@ $$\begin{aligned}
     = \exp\!\bigg( \!-\! \frac{\tau \hat{H}_S}{\hbar} \bigg) \Ket{\psi_H}
 \end{aligned}$$
 
-And the interaction picture's time evolution operator $\hat{K}_I$
+And the interaction picture's time evolution operator $$\hat{K}_I$$
 turns out to be given by:
 
 $$\begin{aligned}
@@ -74,9 +74,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where $\mathcal{T}$ is the
+Where $$\mathcal{T}$$ is the
 [time-ordered product](/know/concept/time-ordered-product/)
-with respect to $\tau$.
+with respect to $$\tau$$.
 This operator works as expected:
 
 $$\begin{aligned}
@@ -84,7 +84,7 @@ $$\begin{aligned}
     = \hat{K}_I(\tau, \tau_0) \Ket{\psi_I(\tau_0)}
 \end{aligned}$$
 
-Where $\Ket{\psi_I(\tau)}$ is related to
+Where $$\Ket{\psi_I(\tau)}$$ is related to
 the Schrödinger and Heisenberg pictures as follows:
 
 $$\begin{aligned}
@@ -94,7 +94,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 It is interesting to combine this definition
-with the action of time evolution $\hat{K}_I(\tau, \tau_0)$:
+with the action of time evolution $$\hat{K}_I(\tau, \tau_0)$$:
 
 $$\begin{aligned}
     \Ket{\psi_I(\tau)}
@@ -105,7 +105,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Rearranging this leads to the following useful
-alternative expression for $\hat{K}_I(\tau, \tau_0)$:
+alternative expression for $$\hat{K}_I(\tau, \tau_0)$$:
 
 $$\begin{aligned}
     \boxed{
@@ -117,8 +117,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Returning to our initial example,
-we can set $\tau = \hbar \beta$ and $\tau_0 = 0$,
-so $\hat{K}_I(\tau, \tau_0)$ becomes:
+we can set $$\tau = \hbar \beta$$ and $$\tau_0 = 0$$,
+so $$\hat{K}_I(\tau, \tau_0)$$ becomes:
 
 $$\begin{aligned}
     \hat{K}_I(\hbar \beta, 0)
@@ -130,7 +130,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Using the easily-shown fact that
-$\hat{K}_I(\hbar \beta, 0) \hat{K}_I(0, \tau) = \hat{K}_I(\hbar \beta, \tau)$,
+$$\hat{K}_I(\hbar \beta, 0) \hat{K}_I(0, \tau) = \hat{K}_I(\hbar \beta, \tau)$$,
 we can therefore rewrite the thermodynamic expectation value like so:
 
 $$\begin{aligned}
@@ -139,8 +139,8 @@ $$\begin{aligned}
     \hat{A}_I(\tau) \hat{K}_I(\tau, \tau') \hat{B}_I(\tau') \hat{K}_I(\tau', 0) \!\Big)
 \end{aligned}$$
 
-We now introduce a time-ordering $\mathcal{T}$,
-letting us reorder the (bosonic) $\hat{K}_I$-operators inside,
+We now introduce a time-ordering $$\mathcal{T}$$,
+letting us reorder the (bosonic) $$\hat{K}_I$$-operators inside,
 and thereby reduce the expression considerably:
 
 $$\begin{aligned}
@@ -151,9 +151,9 @@ $$\begin{aligned}
     &= \frac{1}{Z} \Tr\!\Big( \mathcal{T}\Big\{ \hat{K}_I(\hbar \beta, 0) \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\} \exp(-\beta \hat{H}_{0,S}) \Big)
 \end{aligned}$$
 
-Where $Z = \Tr\!\big(\exp(-\beta \hat{H}_S)\big) = \Tr\!\big(\hat{K}_I(\hbar \beta, 0) \exp(-\beta \hat{H}_{0,S})\big)$.
-If we now define $\Expval{}_0$ as the expectation value with respect
-to the unperturbed equilibrium involving only $\hat{H}_{0,S}$,
+Where $$Z = \Tr\!\big(\exp(-\beta \hat{H}_S)\big) = \Tr\!\big(\hat{K}_I(\hbar \beta, 0) \exp(-\beta \hat{H}_{0,S})\big)$$.
+If we now define $$\Expval{}_0$$ as the expectation value with respect
+to the unperturbed equilibrium involving only $$\hat{H}_{0,S}$$,
 we arrive at the following way of writing this time-ordered expectation:
 
 $$\begin{aligned}