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

1 files changed, 36 insertions, 36 deletions

diff --git a/source/know/concept/landau-quantization/index.md b/source/know/concept/landau-quantization/index.md
index 82ea86e..592d266 100644
--- a/source/know/concept/landau-quantization/index.md
+++ b/source/know/concept/landau-quantization/index.md
@@ -8,23 +8,23 @@ categories:
 layout: "concept"
 ---
 
-When a particle with charge $q$ is moving in a homogeneous
+When a particle with charge $$q$$ is moving in a homogeneous
 [magnetic field](/know/concept/magnetic-field/),
 quantum mechanics decrees that its allowed energies split
 into degenerate discrete **Landau levels**,
 a phenomenon known as **Landau quantization**.
 
-Starting from the Hamiltonian $\hat{H}$ for a particle with mass $m$
-in a vector potential $\vec{A}(\hat{Q})$:
+Starting from the Hamiltonian $$\hat{H}$$ for a particle with mass $$m$$
+in a vector potential $$\vec{A}(\hat{Q})$$:
 
 $$\begin{aligned}
     \hat{H}
     &= \frac{1}{2 m} \big( \hat{p} - q \vec{A} \big)^2
 \end{aligned}$$
 
-We choose $\vec{A} = (- \hat{y} B, 0, 0)$,
-yielding a magnetic field $\vec{B} = \nabla \times \vec{A}$
-pointing in the $z$-direction with strength $B$.
+We choose $$\vec{A} = (- \hat{y} B, 0, 0)$$,
+yielding a magnetic field $$\vec{B} = \nabla \times \vec{A}$$
+pointing in the $$z$$-direction with strength $$B$$.
 The Hamiltonian becomes:
 
 $$\begin{aligned}
@@ -32,18 +32,18 @@ $$\begin{aligned}
     &= \frac{\big( \hat{p}_x - q B \hat{y} \big)^2}{2 m}  + \frac{\hat{p}_y^2}{2 m} + \frac{\hat{p}_z^2}{2 m}
 \end{aligned}$$
 
-The only position operator occurring in $\hat{H}$ is $\hat{y}$,
-so $[\hat{H}, \hat{p}_x] = [\hat{H}, \hat{p}_z] = 0$.
-Because $\hat{p}_z$ appears in an unmodified kinetic energy term,
-and the corresponding $\hat{z}$ does not occur at all,
-the particle has completely free motion in the $z$-direction.
-Likewise, because $\hat{x}$ does not occur in $\hat{H}$,
-we can replace $\hat{p}_x$ by its eigenvalue $\hbar k_x$,
-although the motion is not free, due to $q B \hat{y}$.
+The only position operator occurring in $$\hat{H}$$ is $$\hat{y}$$,
+so $$[\hat{H}, \hat{p}_x] = [\hat{H}, \hat{p}_z] = 0$$.
+Because $$\hat{p}_z$$ appears in an unmodified kinetic energy term,
+and the corresponding $$\hat{z}$$ does not occur at all,
+the particle has completely free motion in the $$z$$-direction.
+Likewise, because $$\hat{x}$$ does not occur in $$\hat{H}$$,
+we can replace $$\hat{p}_x$$ by its eigenvalue $$\hbar k_x$$,
+although the motion is not free, due to $$q B \hat{y}$$.
 
-Based on the absence of $\hat{x}$ and $\hat{z}$,
-we make the following ansatz for the wavefunction $\Psi$:
-a plane wave in the $x$ and $z$ directions, multiplied by an unknown $\phi(y)$:
+Based on the absence of $$\hat{x}$$ and $$\hat{z}$$,
+we make the following ansatz for the wavefunction $$\Psi$$:
+a plane wave in the $$x$$ and $$z$$ directions, multiplied by an unknown $$\phi(y)$$:
 
 $$\begin{aligned}
     \Psi(x, y, z)
@@ -51,15 +51,15 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Inserting this into the time-independent Schrödinger equation gives,
-after dividing out the plane wave exponential $\exp(i k_x x + i k_z z)$:
+after dividing out the plane wave exponential $$\exp(i k_x x + i k_z z)$$:
 
 $$\begin{aligned}
     E \phi
     &= \frac{1}{2 m} \Big( (\hbar k_x - q B y)^2 + \hat{p}_y^2 + \hbar^2 k_z^2 \Big) \phi
 \end{aligned}$$
 
-By defining the cyclotron frequency $\omega_c \equiv q B / m$ and rearranging,
-we can turn this into a 1D quantum harmonic oscillator in $y$,
+By defining the cyclotron frequency $$\omega_c \equiv q B / m$$ and rearranging,
+we can turn this into a 1D quantum harmonic oscillator in $$y$$,
 with a couple of extra terms:
 
 $$\begin{aligned}
@@ -67,8 +67,8 @@ $$\begin{aligned}
     &= \bigg( \frac{1}{2} m \omega_c^2 \Big( y - \frac{\hbar k_x}{m \omega_c} \Big)^2 + \frac{\hat{p}_y^2}{2 m} \bigg) \phi
 \end{aligned}$$
 
-The potential minimum is shifted by $y_0 = \hbar k_x / (m \omega_c)$,
-and a plane wave in $z$ contributes to the energy $E$.
+The potential minimum is shifted by $$y_0 = \hbar k_x / (m \omega_c)$$,
+and a plane wave in $$z$$ contributes to the energy $$E$$.
 In any case, the energy levels of this type of system are well-known:
 
 $$\begin{aligned}
@@ -77,29 +77,29 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-And $\Psi_n$ is then as follows,
-where $\phi$ is the known quantum harmonic oscillator solution:
+And $$\Psi_n$$ is then as follows,
+where $$\phi$$ is the known quantum harmonic oscillator solution:
 
 $$\begin{aligned}
     \Psi_n(x, y, z)
     = \phi_n(y - y_0) \exp(i k_x x + i k_z z)
 \end{aligned}$$
 
-Note that this wave function contains $k_x$ (also inside $y_0$),
-but $k_x$ is absent from the energy $E_n$.
+Note that this wave function contains $$k_x$$ (also inside $$y_0$$),
+but $$k_x$$ is absent from the energy $$E_n$$.
 This implies degeneracy:
-assuming periodic boundary conditions $\Psi(x\!+\!L_x) = \Psi(x)$,
-then $k_x$ can take values of the form $2 \pi n / L_x$, for $n \in \mathbb{Z}$.
+assuming periodic boundary conditions $$\Psi(x\!+\!L_x) = \Psi(x)$$,
+then $$k_x$$ can take values of the form $$2 \pi n / L_x$$, for $$n \in \mathbb{Z}$$.
 
-However, $k_x$ also occurs in the definition of $y_0$, so the degeneracy
-is finite, since $y_0$ must still lie inside the system,
-or, more formally, $y_0 \in [0, L_y]$:
+However, $$k_x$$ also occurs in the definition of $$y_0$$, so the degeneracy
+is finite, since $$y_0$$ must still lie inside the system,
+or, more formally, $$y_0 \in [0, L_y]$$:
 
 $$\begin{aligned}
     0 \le y_0 = \frac{\hbar k_x}{m \omega_c} = \frac{\hbar 2 \pi n}{q B L_x} \le L_y
 \end{aligned}$$
 
-Isolating this for $n$, we find the following upper bound of the degeneracy:
+Isolating this for $$n$$, we find the following upper bound of the degeneracy:
 
 $$\begin{aligned}
     \boxed{
@@ -108,12 +108,12 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where $A \equiv L_x L_y$ is the area of the confinement in the $(x,y)$-plane.
-Evidently, the degeneracy of each level increases with larger $B$,
-but since $\omega_c = q B / m$, the energy gap between each level increases too.
+Where $$A \equiv L_x L_y$$ is the area of the confinement in the $$(x,y)$$-plane.
+Evidently, the degeneracy of each level increases with larger $$B$$,
+but since $$\omega_c = q B / m$$, the energy gap between each level increases too.
 In other words: the [density of states](/know/concept/density-of-states/)
 is a constant with respect to the energy,
-but the states get distributed across the $E_n$ differently depending on $B$.
+but the states get distributed across the $$E_n$$ differently depending on $$B$$.