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

1 files changed, 79 insertions, 79 deletions

diff --git a/source/know/concept/jellium/index.md b/source/know/concept/jellium/index.md
index c951fd7..5c50f80 100644
--- a/source/know/concept/jellium/index.md
+++ b/source/know/concept/jellium/index.md
@@ -21,7 +21,7 @@ Let us start by neglecting electron-electron interactions.
 This is clearly a dubious assumption, but we will stick with it for now.
 For an infinitely large sample of jellium,
 the single-electron states are simply plane waves.
-We consider an arbitrary cube of volume $V$,
+We consider an arbitrary cube of volume $$V$$,
 and impose periodic boundary conditions on it,
 such that the single-particle orbitals are (suppressing spin):
 
@@ -33,22 +33,22 @@ $$\begin{aligned}
     \vb{k} = \frac{2 \pi}{V^{1/3}} (n_x, n_y, n_z)
 \end{aligned}$$
 
-Where $n_x, n_y, n_z \in \mathbb{Z}$.
+Where $$n_x, n_y, n_z \in \mathbb{Z}$$.
 This is a discrete (but infinite) set of independent orbitals,
 so it is natural to use the
 [second quantization](/know/concept/second-quantization/)
-to write the non-interacting Hamiltonian $\hat{H}_0$,
-where $\hbar^2 |\vb{k}|^2 / (2 m)$ is the kinetic energy
-of the orbital with wavevector $\vb{k}$, and $s$ is the spin:
+to write the non-interacting Hamiltonian $$\hat{H}_0$$,
+where $$\hbar^2 |\vb{k}|^2 / (2 m)$$ is the kinetic energy
+of the orbital with wavevector $$\vb{k}$$, and $$s$$ is the spin:
 
 $$\begin{aligned}
     \hat{H}_0
     = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}
 \end{aligned}$$
 
-Assuming that the temperature $T = 0$,
-the $N$-electron ground state of this Hamiltonian
-is known as the **Fermi sea** or **Fermi sphere** $\Ket{\mathrm{FS}}$,
+Assuming that the temperature $$T = 0$$,
+the $$N$$-electron ground state of this Hamiltonian
+is known as the **Fermi sea** or **Fermi sphere** $$\Ket{\mathrm{FS}}$$,
 and is constructed by filling up the single-electron states
 starting from the lowest energy:
 
@@ -57,9 +57,9 @@ $$\begin{aligned}
     = \prod_{s} \prod_{j = 1}^{N/2} \hat{c}_{s,\vb{k}_j}^\dagger \Ket{0}
 \end{aligned}$$
 
-Because $T = 0$, all the electrons stay in their assigned state.
-The energy and wavenumber $|\vb{k}|$ of the highest filled orbital
-are called the **Fermi energy** $\epsilon_F$ and **Fermi wavenumber** $k_F$,
+Because $$T = 0$$, all the electrons stay in their assigned state.
+The energy and wavenumber $$|\vb{k}|$$ of the highest filled orbital
+are called the **Fermi energy** $$\epsilon_F$$ and **Fermi wavenumber** $$k_F$$,
 and obey the expected kinetic energy relation:
 
 $$\begin{aligned}
@@ -69,13 +69,13 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The Fermi sea can be visualized in $\vb{k}$-space as a sphere with radius $k_F$.
-Because $\vb{k}$ is discrete, the sphere's surface is not smooth,
-but in the limit $V \to \infty$ it becomes perfect.
+The Fermi sea can be visualized in $$\vb{k}$$-space as a sphere with radius $$k_F$$.
+Because $$\vb{k}$$ is discrete, the sphere's surface is not smooth,
+but in the limit $$V \to \infty$$ it becomes perfect.
 
 Now, we would like a relation between the system's parameters,
-e.g. $N$ and $V$, and the resulting values of $\epsilon_F$ or $k_F$.
-The total population $N$ must be given by:
+e.g. $$N$$ and $$V$$, and the resulting values of $$\epsilon_F$$ or $$k_F$$.
+The total population $$N$$ must be given by:
 
 $$\begin{aligned}
     N
@@ -83,11 +83,11 @@ $$\begin{aligned}
     = \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} \dd{\vb{k}}
 \end{aligned}$$
 
-Where we have turned the sum over $\vb{k}$ into an integral with a constant factor,
-by using that each orbital exclusively occupies a volume $(2 \pi)^3 / V$ in $\vb{k}$-space.
+Where we have turned the sum over $$\vb{k}$$ into an integral with a constant factor,
+by using that each orbital exclusively occupies a volume $$(2 \pi)^3 / V$$ in $$\vb{k}$$-space.
 
-At zero temperature, this inner product can only be $0$ or $1$,
-depending on whether $\vb{k}$ is outside or inside the Fermi sphere.
+At zero temperature, this inner product can only be $$0$$ or $$1$$,
+depending on whether $$\vb{k}$$ is outside or inside the Fermi sphere.
 We can therefore rewrite using a
 [Heaviside step function](/know/concept/heaviside-step-function/):
 
@@ -98,10 +98,10 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where we realized that spin does not matter,
-and replaced the sum over $s$ by a factor $2$.
+and replaced the sum over $$s$$ by a factor $$2$$.
 In order to evaluate this 3D integral,
 we go to [spherical coordinates](/know/concept/spherical-coordinates/)
-$(|\vb{k}|, \theta, \varphi)$:
+$$(|\vb{k}|, \theta, \varphi)$$:
 
 $$\begin{aligned}
     N
@@ -112,7 +112,7 @@ $$\begin{aligned}
     = \frac{V}{3 \pi^2} k_F^3
 \end{aligned}$$
 
-Using that the electron density $n = N/V$,
+Using that the electron density $$n = N/V$$,
 we thus arrive at the following relation:
 
 $$\begin{aligned}
@@ -122,15 +122,15 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-This result also justifies our assumption that $T = 0$:
-we can accurately calculate the density $n$ for many conducting materials,
-and this relation then gives $k_F$ and $\epsilon_F$.
-It turns out that $\epsilon_F$ is usually very large
-compared to the thermal energy $k_B T$ at reasonable temperatures,
+This result also justifies our assumption that $$T = 0$$:
+we can accurately calculate the density $$n$$ for many conducting materials,
+and this relation then gives $$k_F$$ and $$\epsilon_F$$.
+It turns out that $$\epsilon_F$$ is usually very large
+compared to the thermal energy $$k_B T$$ at reasonable temperatures,
 so we can conclude that thermal fluctuations are negligible.
 
-Now, $\epsilon_F$ is the highest single-electron energy,
-but about the total $N$-particle energy $E^{(0)}$?
+Now, $$\epsilon_F$$ is the highest single-electron energy,
+but about the total $$N$$-particle energy $$E^{(0)}$$?
 
 $$\begin{aligned}
     E^{(0)}
@@ -138,7 +138,7 @@ $$\begin{aligned}
     = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}
 \end{aligned}$$
 
-Once again, we turn the sum over $\vb{k}$ into an integral,
+Once again, we turn the sum over $$\vb{k}$$ into an integral,
 and recognize the spin's irrelevance:
 
 $$\begin{aligned}
@@ -150,7 +150,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 In spherical coordinates,
-we evaluate the integral and find that $E^{(0)}$ is proportional to $k_F^5$:
+we evaluate the integral and find that $$E^{(0)}$$ is proportional to $$k_F^5$$:
 
 $$\begin{aligned}
     E^{(0)}
@@ -163,8 +163,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 In general, it is more useful to consider
-the average kinetic energy per electron $E^{(0)} / N$,
-which we find to be as follows, using that $k_F^3 = 3 \pi^2 n$:
+the average kinetic energy per electron $$E^{(0)} / N$$,
+which we find to be as follows, using that $$k_F^3 = 3 \pi^2 n$$:
 
 $$\begin{aligned}
     \boxed{
@@ -175,9 +175,9 @@ $$\begin{aligned}
     \:\sim\: n^{2/3}
 \end{aligned}$$
 
-Traditionally, this is expressed using a dimensionless parameter $r_s$,
+Traditionally, this is expressed using a dimensionless parameter $$r_s$$,
 defined as the radius of a sphere containing a single electron,
-measured in Bohr radii $a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m)$:
+measured in Bohr radii $$a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m)$$:
 
 $$\begin{aligned}
         \frac{4 \pi}{3} (a_0 r_s)^3
@@ -206,14 +206,14 @@ To include Coulomb interactions, let us try
 Clearly, this will give better results when the interaction is relatively weak, if ever.
 
 The Coulomb potential is proportional to the inverse distance,
-and the average electron spacing is roughly $n^{-1/3}$,
-so the interaction energy $E_\mathrm{int}$ should scale as $n^{1/3}$.
-We already know that the kinetic energy $E_\mathrm{kin} = E^{(0)}$ scales as $n^{2/3}$,
+and the average electron spacing is roughly $$n^{-1/3}$$,
+so the interaction energy $$E_\mathrm{int}$$ should scale as $$n^{1/3}$$.
+We already know that the kinetic energy $$E_\mathrm{kin} = E^{(0)}$$ scales as $$n^{2/3}$$,
 meaning perturbation theory should be reasonable
-if $1 \gg E_\mathrm{int} / E_\mathrm{kin} \sim n^{-1/3}$,
-so in the limit of high density $n \to \infty$.
+if $$1 \gg E_\mathrm{int} / E_\mathrm{kin} \sim n^{-1/3}$$,
+so in the limit of high density $$n \to \infty$$.
 
-The two-body Coulomb interaction operator $\hat{W}$
+The two-body Coulomb interaction operator $$\hat{W}$$
 is as follows in second-quantized form:
 
 $$\begin{aligned}
@@ -222,7 +222,7 @@ $$\begin{aligned}
     \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1}
 \end{aligned}$$
 
-The first-order correction $E^{(1)}$ to the ground state (i.e. Fermi sea) energy
+The first-order correction $$E^{(1)}$$ to the ground state (i.e. Fermi sea) energy
 is then given by:
 
 $$\begin{aligned}
@@ -235,14 +235,14 @@ $$\begin{aligned}
 \end{aligned}$$
 
 This inner product can only be nonzero
-if the two creation operators $\hat{c}^\dagger$
-are for the same orbitals as the two annihilation operators $\hat{c}$.
-Since $\vb{q} \neq 0$, this means that $s_1 = s_2$,
-and that momentum is conserved: $\vb{k}_2 = \vb{k}_1 \!+\! \vb{q}$.
-And of course both $\vb{k}_1$ and $\vb{k}_1 \!+\! \vb{q}$
+if the two creation operators $$\hat{c}^\dagger$$
+are for the same orbitals as the two annihilation operators $$\hat{c}$$.
+Since $$\vb{q} \neq 0$$, this means that $$s_1 = s_2$$,
+and that momentum is conserved: $$\vb{k}_2 = \vb{k}_1 \!+\! \vb{q}$$.
+And of course both $$\vb{k}_1$$ and $$\vb{k}_1 \!+\! \vb{q}$$
 must be inside the Fermi sphere,
 to avoid annihilating an empty orbital.
-Let $s = s_1$ and $\vb{k} = \vb{k}_1$:
+Let $$s = s_1$$ and $$\vb{k} = \vb{k}_1$$:
 
 $$\begin{aligned}
     E^{(1)}
@@ -260,11 +260,11 @@ $$\begin{aligned}
     \Theta(k_F - |\vb{k}|) \:\Theta(k_F - |\vb{k} \!+\! \vb{q}|)
 \end{aligned}$$
 
-Next, we convert the sum over $\vb{q}$ into an integral in spherical coordinates.
-Clearly, $\vb{q}$ is the "jump" made by an electron from one orbital to another,
+Next, we convert the sum over $$\vb{q}$$ into an integral in spherical coordinates.
+Clearly, $$\vb{q}$$ is the "jump" made by an electron from one orbital to another,
 so the largest possible jump
 goes from a point on the Fermi surface to the opposite point,
-and thus has length $2 k_F$.
+and thus has length $$2 k_F$$.
 This yields the integration limit, and therefore leads to:
 
 $$\begin{aligned}
@@ -277,11 +277,11 @@ $$\begin{aligned}
     \int_0^{2 k_F} \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \dd{|\vb{q}|}
 \end{aligned}$$
 
-Where we have used that the direction of $\vb{q}$,
-i.e. $(\theta_q,\varphi_q)$, is irrelevant,
-as long as we define $\theta_k$ as
-the angle between $\vb{q}$ and $\vb{k} \!+\! \vb{q}$
-when we go to spherical coordinates $(|\vb{k}|, \theta_k, \varphi_k)$ for $\vb{k}$:
+Where we have used that the direction of $$\vb{q}$$,
+i.e. $$(\theta_q,\varphi_q)$$, is irrelevant,
+as long as we define $$\theta_k$$ as
+the angle between $$\vb{q}$$ and $$\vb{k} \!+\! \vb{q}$$
+when we go to spherical coordinates $$(|\vb{k}|, \theta_k, \varphi_k)$$ for $$\vb{k}$$:
 
 $$\begin{aligned}
     E^{(1)}
@@ -296,10 +296,10 @@ $$\begin{aligned}
 
 Unfortunately, this last step function is less easy to translate into integration limits.
 In effect, we are trying to calculate the intersection volume of two spheres,
-both with radius $k_F$, one centered on the origin (for $\vb{k}$),
-and the other centered on $\vb{q}$ (for $\vb{k} \!+\! \vb{q}$).
-Imagine a triangle with side lengths $|\vb{k}|$, $|\vb{q}|$ and $|\vb{k} \!+\! \vb{q}|^2$,
-where $\theta_k$ is the angle between $|\vb{k}|$ and $|\vb{k} \!+\! \vb{q}|$.
+both with radius $$k_F$$, one centered on the origin (for $$\vb{k}$$),
+and the other centered on $$\vb{q}$$ (for $$\vb{k} \!+\! \vb{q}$$).
+Imagine a triangle with side lengths $$|\vb{k}|$$, $$|\vb{q}|$$ and $$|\vb{k} \!+\! \vb{q}|^2$$,
+where $$\theta_k$$ is the angle between $$|\vb{k}|$$ and $$|\vb{k} \!+\! \vb{q}|$$.
 The *law of cosines* then gives the following relation:
 
 $$\begin{aligned}
@@ -307,10 +307,10 @@ $$\begin{aligned}
     = |\vb{q}|^2 + |\vb{k} \!+\! \vb{q}|^2 - 2 |\vb{q}| |\vb{k} \!+\! \vb{q}| \cos(\theta_k)
 \end{aligned}$$
 
-We already know that $|\vb{k}| < k_F$ and $0 < |\vb{q}| < 2 k_F$,
-so by isolating for $\cos(\theta_k)$,
-we can obtain bounds on $\theta_k$ and $|\vb{k}|$.
-Let $|\vb{k}| \to k_F$ in both cases, then:
+We already know that $$|\vb{k}| < k_F$$ and $$0 < |\vb{q}| < 2 k_F$$,
+so by isolating for $$\cos(\theta_k)$$,
+we can obtain bounds on $$\theta_k$$ and $$|\vb{k}|$$.
+Let $$|\vb{k}| \to k_F$$ in both cases, then:
 
 $$\begin{aligned}
     \cos(\theta_k)
@@ -322,16 +322,16 @@ $$\begin{aligned}
     = 1
 \end{aligned}$$
 
-Meaning that $0 < \theta_k < \arccos{|\vb{q}| / (2 k_F)}$.
-To get a lower limit for $|\vb{k}|$, we "cheat" by artificially demanding
-that $\vb{k}$ does not cross the halfway point between the spheres,
-with the result that $|\vb{k}| \cos(\theta_k) > |\vb{q}|/2$.
+Meaning that $$0 < \theta_k < \arccos{|\vb{q}| / (2 k_F)}$$.
+To get a lower limit for $$|\vb{k}|$$, we "cheat" by artificially demanding
+that $$\vb{k}$$ does not cross the halfway point between the spheres,
+with the result that $$|\vb{k}| \cos(\theta_k) > |\vb{q}|/2$$.
 Then, thanks to symmetry (both spheres have the same radius),
-we just multiply the integral by $2$,
-for $\vb{k}$ on the other side of the halfway point.
+we just multiply the integral by $$2$$,
+for $$\vb{k}$$ on the other side of the halfway point.
 
-Armed with these integration limits, we return to calculating $E^{(1)}$,
-substituting $\xi \equiv \cos(\theta_k)$:
+Armed with these integration limits, we return to calculating $$E^{(1)}$$,
+substituting $$\xi \equiv \cos(\theta_k)$$:
 
 $$\begin{aligned}
     E^{(1)}
@@ -345,7 +345,7 @@ $$\begin{aligned}
     \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|}
 \end{aligned}$$
 
-Where we have used that $\varphi_k$ does not appear in the integrand.
+Where we have used that $$\varphi_k$$ does not appear in the integrand.
 Evaluating these integrals:
 
 $$\begin{aligned}
@@ -369,7 +369,7 @@ $$\begin{aligned}
     = -\frac{3 e^2 N}{16 \pi^2 \varepsilon_0} k_F
 \end{aligned}$$
 
-Per particle, the first-order energy correction $E^{(1)}$
+Per particle, the first-order energy correction $$E^{(1)}$$
 is therefore found to be as follows:
 
 $$\begin{aligned}
@@ -379,7 +379,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-This can also be written using the parameter $r_s$ introduced above, leading to:
+This can also be written using the parameter $$r_s$$ introduced above, leading to:
 
 $$\begin{aligned}
     \frac{E^{(1)}}{N}
@@ -387,8 +387,8 @@ $$\begin{aligned}
     = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 r_s}
 \end{aligned}$$
 
-Consequently, for sufficiently high densities $n$,
-the total energy $E$ per particle is given by:
+Consequently, for sufficiently high densities $$n$$,
+the total energy $$E$$ per particle is given by:
 
 $$\begin{aligned}
     \boxed{
@@ -398,7 +398,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Unfortunately, this is as far as we can go.
-In theory, the second-order energy correction $E^{(2)}$ is as shown below,
+In theory, the second-order energy correction $$E^{(2)}$$ is as shown below,
 but it turns out that it (and all higher orders) diverge:
 
 $$\begin{aligned}