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

1 files changed, 44 insertions, 44 deletions

diff --git a/source/know/concept/berry-phase/index.md b/source/know/concept/berry-phase/index.md
index eedc548..d237ea5 100644
--- a/source/know/concept/berry-phase/index.md
+++ b/source/know/concept/berry-phase/index.md
@@ -8,8 +8,8 @@ categories:
 layout: "concept"
 ---
 
-Consider a Hamiltonian $\hat{H}$ that does not explicitly depend on time,
-but does depend on a given parameter $\vb{R}$.
+Consider a Hamiltonian $$\hat{H}$$ that does not explicitly depend on time,
+but does depend on a given parameter $$\vb{R}$$.
 The Schrödinger equations then read:
 
 $$\begin{aligned}
@@ -20,9 +20,9 @@ $$\begin{aligned}
     &= E_n(\vb{R}) \Ket{\psi_n(\vb{R})}
 \end{aligned}$$
 
-The general full solution $\Ket{\Psi_n}$ has the following form,
-where we allow $\vb{R}$ to evolve in time,
-and we have abbreviated the traditional phase of the "wiggle factor" as $L_n$:
+The general full solution $$\Ket{\Psi_n}$$ has the following form,
+where we allow $$\vb{R}$$ to evolve in time,
+and we have abbreviated the traditional phase of the "wiggle factor" as $$L_n$$:
 
 $$\begin{aligned}
     \Ket{\Psi_n(t)}
@@ -31,11 +31,11 @@ $$\begin{aligned}
     L_n(t) \equiv \int_0^t E_n(\vb{R}(t')) \dd{t'}
 \end{aligned}$$
 
-The **geometric phase** $\gamma_n(t)$ is more interesting.
-It is not included in $\Ket{\psi_n}$,
-because it depends on the path $\vb{R}(t)$
-rather than only the present $\vb{R}$ and $t$.
-Its dynamics can be found by inserting the above $\Ket{\Psi_n}$
+The **geometric phase** $$\gamma_n(t)$$ is more interesting.
+It is not included in $$\Ket{\psi_n}$$,
+because it depends on the path $$\vb{R}(t)$$
+rather than only the present $$\vb{R}$$ and $$t$$.
+Its dynamics can be found by inserting the above $$\Ket{\Psi_n}$$
 into the time-dependent Schrödinger equation:
 
 $$\begin{aligned}
@@ -58,8 +58,8 @@ $$\begin{aligned}
     &= \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t}
 \end{aligned}$$
 
-Front-multiplying by $i \Bra{\Psi_n}$ gives us
-the equation of motion of the geometric phase $\gamma_n$:
+Front-multiplying by $$i \Bra{\Psi_n}$$ gives us
+the equation of motion of the geometric phase $$\gamma_n$$:
 
 $$\begin{aligned}
     \boxed{
@@ -68,7 +68,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where we have defined the so-called **Berry connection** $\vb{A}_n$ as follows:
+Where we have defined the so-called **Berry connection** $$\vb{A}_n$$ as follows:
 
 $$\begin{aligned}
     \boxed{
@@ -77,9 +77,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Importantly, note that $\vb{A}_n$ is real,
-provided that $\Ket{\psi_n}$ is always normalized for all $\vb{R}$.
-To prove this, we start from the fact that $\nabla_\vb{R} 1 = 0$:
+Importantly, note that $$\vb{A}_n$$ is real,
+provided that $$\Ket{\psi_n}$$ is always normalized for all $$\vb{R}$$.
+To prove this, we start from the fact that $$\nabla_\vb{R} 1 = 0$$:
 
 $$\begin{aligned}
     0
@@ -91,14 +91,14 @@ $$\begin{aligned}
     = 2 \Imag\{ \vb{A}_n \}
 \end{aligned}$$
 
-Consequently, $\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is always real,
-because $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary.
+Consequently, $$\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is always real,
+because $$\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary.
 
-Suppose now that the parameter $\vb{R}(t)$ is changed adiabatically
+Suppose now that the parameter $$\vb{R}(t)$$ is changed adiabatically
 (i.e. so slow that the system stays in the same eigenstate)
-for $t \in [0, T]$, along a circuit $C$ with $\vb{R}(0) \!=\! \vb{R}(T)$.
-Integrating the phase $\gamma_n(t)$ over this contour $C$ then yields
-the **Berry phase** $\gamma_n(C)$:
+for $$t \in [0, T]$$, along a circuit $$C$$ with $$\vb{R}(0) \!=\! \vb{R}(T)$$.
+Integrating the phase $$\gamma_n(t)$$ over this contour $$C$$ then yields
+the **Berry phase** $$\gamma_n(C)$$:
 
 $$\begin{aligned}
     \boxed{
@@ -107,9 +107,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-But we have a problem: $\vb{A}_n$ is not unique!
+But we have a problem: $$\vb{A}_n$$ is not unique!
 Due to the Schrödinger equation's gauge invariance,
-any function $f(\vb{R}(t))$ can be added to $\gamma_n(t)$
+any function $$f(\vb{R}(t))$$ can be added to $$\gamma_n(t)$$
 without making an immediate physical difference to the state.
 Consider the following general gauge transformation:
 
@@ -118,9 +118,9 @@ $$\begin{aligned}
     \equiv \exp(i f(\vb{R})) \: \Ket{\psi_n(\vb{R})}
 \end{aligned}$$
 
-To find $\vb{A}_n$ for a particular choice of $f$,
+To find $$\vb{A}_n$$ for a particular choice of $$f$$,
 we need to evaluate the inner product
-$\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$:
+$$\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$$:
 
 $$\begin{aligned}
     \inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}
@@ -131,16 +131,16 @@ $$\begin{aligned}
     &= i \nabla_\vb{R} f + \inprod{\psi_n}{\nabla_\vb{R} \psi_n}
 \end{aligned}$$
 
-Unfortunately, $f$ does not vanish as we would have liked,
-so $\vb{A}_n$ depends on our choice of $f$.
+Unfortunately, $$f$$ does not vanish as we would have liked,
+so $$\vb{A}_n$$ depends on our choice of $$f$$.
 
 However, the curl of a gradient is always zero,
-so although $\vb{A}_n$ is not unique,
-its curl $\nabla_\vb{R} \cross \vb{A}_n$ is guaranteed to be.
-Conveniently, we can introduce a curl in the definition of $\gamma_n(C)$
+so although $$\vb{A}_n$$ is not unique,
+its curl $$\nabla_\vb{R} \cross \vb{A}_n$$ is guaranteed to be.
+Conveniently, we can introduce a curl in the definition of $$\gamma_n(C)$$
 by applying Stokes' theorem, under the assumption
-that $\vb{A}_n$ has no singularities in the area enclosed by $C$
-(fortunately, $\vb{A}_n$ can always be chosen to satisfy this):
+that $$\vb{A}_n$$ has no singularities in the area enclosed by $$C$$
+(fortunately, $$\vb{A}_n$$ can always be chosen to satisfy this):
 
 $$\begin{aligned}
     \boxed{
@@ -149,10 +149,10 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where we defined $\vb{B}_n$ as the curl of $\vb{A}_n$.
-Now $\gamma_n(C)$ is guaranteed to be unique.
-Note that $\vb{B}_n$ is analogous to a magnetic field,
-and $\vb{A}_n$ to a magnetic vector potential:
+Where we defined $$\vb{B}_n$$ as the curl of $$\vb{A}_n$$.
+Now $$\gamma_n(C)$$ is guaranteed to be unique.
+Note that $$\vb{B}_n$$ is analogous to a magnetic field,
+and $$\vb{A}_n$$ to a magnetic vector potential:
 
 $$\begin{aligned}
     \vb{B}_n(\vb{R})
@@ -160,9 +160,9 @@ $$\begin{aligned}
     = \Imag\!\Big\{ \nabla_\vb{R} \cross \Inprod{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})} \Big\}
 \end{aligned}$$
 
-Unfortunately, $\nabla_\vb{R} \psi_n$ is difficult to evaluate explicitly,
-so we would like to rewrite $\vb{B}_n$ such that it does not enter.
-We do this as follows, inserting $1 = \sum_{m} \Ket{\psi_m} \Bra{\psi_m}$ along the way:
+Unfortunately, $$\nabla_\vb{R} \psi_n$$ is difficult to evaluate explicitly,
+so we would like to rewrite $$\vb{B}_n$$ such that it does not enter.
+We do this as follows, inserting $$1 = \sum_{m} \Ket{\psi_m} \Bra{\psi_m}$$ along the way:
 
 $$\begin{aligned}
     i \vb{B}_n
@@ -172,9 +172,9 @@ $$\begin{aligned}
     &= \sum_{m} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\psi_m}{\nabla_\vb{R} \psi_n}
 \end{aligned}$$
 
-The fact that $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary
+The fact that $$\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary
 means it is parallel to its complex conjugate,
-and thus the cross product vanishes, so we exclude $n$ from the sum:
+and thus the cross product vanishes, so we exclude $$n$$ from the sum:
 
 $$\begin{aligned}
     \vb{B}_n
@@ -200,8 +200,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Which only involves $\nabla_\vb{R} \hat{H}$,
-and is therefore easier to evaluate than any $\Ket{\nabla_\vb{R} \psi_n}$.
+Which only involves $$\nabla_\vb{R} \hat{H}$$,
+and is therefore easier to evaluate than any $$\Ket{\nabla_\vb{R} \psi_n}$$.