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

1 files changed, 72 insertions, 72 deletions

diff --git a/source/know/concept/time-independent-perturbation-theory/index.md b/source/know/concept/time-independent-perturbation-theory/index.md
index ba1e89d..4d1ac2f 100644
--- a/source/know/concept/time-independent-perturbation-theory/index.md
+++ b/source/know/concept/time-independent-perturbation-theory/index.md
@@ -19,11 +19,11 @@ $$\begin{aligned}
     \hat{H} = \hat{H}_0 + \lambda \hat{H}_1
 \end{aligned}$$
 
-Where $\hat{H}_0$ is a Hamiltonian for which the time-independent
-Schrödinger equation has a known solution, and $\hat{H}_1$ is a small
-perturbing Hamiltonian. The eigenenergies $E_n$ and eigenstates
-$\Ket{\psi_n}$ of the composite problem are expanded in the
-perturbation "bookkeeping" parameter $\lambda$:
+Where $$\hat{H}_0$$ is a Hamiltonian for which the time-independent
+Schrödinger equation has a known solution, and $$\hat{H}_1$$ is a small
+perturbing Hamiltonian. The eigenenergies $$E_n$$ and eigenstates
+$$\Ket{\psi_n}$$ of the composite problem are expanded in the
+perturbation "bookkeeping" parameter $$\lambda$$:
 
 $$\begin{aligned}
     \Ket{\psi_n}
@@ -33,7 +33,7 @@ $$\begin{aligned}
     &= E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + ...
 \end{aligned}$$
 
-Where $E_n^{(1)}$ and $\ket{\psi_n^{(1)}}$ are called the **first-order
+Where $$E_n^{(1)}$$ and $$\ket{\psi_n^{(1)}}$$ are called the **first-order
 corrections**, and so on for higher orders. We insert this into the
 Schrödinger equation:
 
@@ -49,7 +49,7 @@ $$\begin{aligned}
     &\qquad + \lambda^2 \big( E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}} \big) + ...
 \end{aligned}$$
 
-If we collect the terms according to the order of $\lambda$, we arrive
+If we collect the terms according to the order of $$\lambda$$, we arrive
 at the following endless series of equations, of which in practice only
 the first three are typically used:
 
@@ -68,58 +68,58 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The first equation is the unperturbed problem, which we assume has
-already been solved, with eigenvalues $E_n^{(0)} = \varepsilon_n$ and
-eigenvectors $\ket{\psi_n^{(0)}} = \Ket{n}$:
+already been solved, with eigenvalues $$E_n^{(0)} = \varepsilon_n$$ and
+eigenvectors $$\ket{\psi_n^{(0)}} = \Ket{n}$$:
 
 $$\begin{aligned}
     \hat{H}_0 \Ket{n} = \varepsilon_n \Ket{n}
 \end{aligned}$$
 
 The approach to solving the other two equations varies depending on
-whether this $\hat{H}_0$ has a degenerate spectrum or not.
+whether this $$\hat{H}_0$$ has a degenerate spectrum or not.
 
 
 ## Without degeneracy
 
 We start by assuming that there is no degeneracy, in other words, each
-$\varepsilon_n$ corresponds to one $\Ket{n}$. At order $\lambda^1$, we
+$$\varepsilon_n$$ corresponds to one $$\Ket{n}$$. At order $$\lambda^1$$, we
 rewrite the equation as follows:
 
 $$\begin{aligned}
     (\hat{H}_1 - E_n^{(1)}) \Ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} = 0
 \end{aligned}$$
 
-Since $\Ket{n}$ form a complete basis, we can express
-$\ket{\psi_n^{(1)}}$ in terms of them:
+Since $$\Ket{n}$$ form a complete basis, we can express
+$$\ket{\psi_n^{(1)}}$$ in terms of them:
 
 $$\begin{aligned}
     \ket{\psi_n^{(1)}} = \sum_{m \neq n} c_m \Ket{m}
 \end{aligned}$$
 
-Importantly, $n$ has been removed from the summation to prevent dividing
+Importantly, $$n$$ has been removed from the summation to prevent dividing
 by zero later. We are allowed to do this, because
-$\ket{\psi_n^{(1)}} - c_n \Ket{n}$ also satisfies the order-$\lambda^1$
-equation for any value of $c_n$, as demonstrated here:
+$$\ket{\psi_n^{(1)}} - c_n \Ket{n}$$ also satisfies the order-$$\lambda^1$$
+equation for any value of $$c_n$$, as demonstrated here:
 
 $$\begin{aligned}
     (\hat{H}_1 - E_n^{(1)}) \Ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} - (\varepsilon_n - \varepsilon_n) c_n \Ket{n} = 0
 \end{aligned}$$
 
-Where we used $\hat{H}_0 \Ket{n} = \varepsilon_n \Ket{n}$.
-We insert the series form of $\ket{\psi_n^{(1)}}$ into the $\lambda^1$-equation:
+Where we used $$\hat{H}_0 \Ket{n} = \varepsilon_n \Ket{n}$$.
+We insert the series form of $$\ket{\psi_n^{(1)}}$$ into the $$\lambda^1$$-equation:
 
 $$\begin{aligned}
     (\hat{H}_1 - E_n^{(1)}) \Ket{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \Ket{m} = 0
 \end{aligned}$$
 
-We then put an arbitrary basis vector $\Bra{k}$ in front of this
+We then put an arbitrary basis vector $$\Bra{k}$$ in front of this
 equation to get:
 
 $$\begin{aligned}
     \matrixel{k}{\hat{H}_1}{n} - E_n^{(1)} \Inprod{k}{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \Inprod{k}{m} = 0
 \end{aligned}$$
 
-Suppose that $k = n$. Since $\Ket{n}$ form an orthonormal basis, we end
+Suppose that $$k = n$$. Since $$\Ket{n}$$ form an orthonormal basis, we end
 up with:
 
 $$\begin{aligned}
@@ -128,20 +128,20 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-In other words, the first-order energy correction $E_n^{(1)}$ is the
-expectation value of the perturbation $\hat{H}_1$ for the unperturbed
-state $\Ket{n}$.
+In other words, the first-order energy correction $$E_n^{(1)}$$ is the
+expectation value of the perturbation $$\hat{H}_1$$ for the unperturbed
+state $$\Ket{n}$$.
 
-Suppose now that $k \neq n$, then only one term of the summation
+Suppose now that $$k \neq n$$, then only one term of the summation
 survives, and we are left with the following equation, which tells us
-$c_l$:
+$$c_l$$:
 
 $$\begin{aligned}
     \matrixel{k}{\hat{H}_1}{n} + c_k (\varepsilon_k - \varepsilon_n) = 0
 \end{aligned}$$
 
-We isolate this result for $c_k$ and insert it into the series form of
-$\ket{\psi_n^{(1)}}$ to get the full first-order correction to the wave
+We isolate this result for $$c_k$$ and insert it into the series form of
+$$\ket{\psi_n^{(1)}}$$ to get the full first-order correction to the wave
 function:
 
 $$\begin{aligned}
@@ -154,27 +154,27 @@ $$\begin{aligned}
 Here it is clear why this is only valid in the non-degenerate case:
 otherwise we would divide by zero in the denominator.
 
-Next, to find the second-order energy correction $E_n^{(2)}$,
-we take the corresponding equation and put $\Bra{n}$ in front of it:
+Next, to find the second-order energy correction $$E_n^{(2)}$$,
+we take the corresponding equation and put $$\Bra{n}$$ in front of it:
 
 $$\begin{aligned}
     \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} + \matrixel{n}{\hat{H}_0}{\psi_n^{(2)}}
     &= E_n^{(2)} \Inprod{n}{n} + E_n^{(1)} \inprod{n}{\psi_n^{(1)}} + \varepsilon_n \inprod{n}{\psi_n^{(2)}}
 \end{aligned}$$
 
-Because $\hat{H}_0$ is Hermitian, we know that
-$\matrixel{n}{\hat{H}_0}{\psi_n^{(2)}} = \varepsilon_n \inprod{n}{\psi_n^{(2)}}$,
+Because $$\hat{H}_0$$ is Hermitian, we know that
+$$\matrixel{n}{\hat{H}_0}{\psi_n^{(2)}} = \varepsilon_n \inprod{n}{\psi_n^{(2)}}$$,
 i.e. we apply it to the bra, which lets us eliminate two terms. Also,
-since $\Ket{n}$ is normalized, we find:
+since $$\Ket{n}$$ is normalized, we find:
 
 $$\begin{aligned}
     E_n^{(2)}
     = \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \inprod{n}{\psi_n^{(1)}}
 \end{aligned}$$
 
-We explicitly removed the $\Ket{n}$-dependence of $\ket{\psi_n^{(1)}}$,
+We explicitly removed the $$\Ket{n}$$-dependence of $$\ket{\psi_n^{(1)}}$$,
 so the last term is zero. By simply inserting our result for
-$\ket{\psi_n^{(1)}}$, we thus arrive at:
+$$\ket{\psi_n^{(1)}}$$, we thus arrive at:
 
 $$\begin{aligned}
     \boxed{
@@ -188,8 +188,8 @@ In practice, it is not particulary useful to calculate more corrections.
 
 ## With degeneracy
 
-If $\varepsilon_n$ is $D$-fold degenerate, then its eigenstate could be
-any vector $\Ket{n, d}$ from the corresponding $D$-dimensional
+If $$\varepsilon_n$$ is $$D$$-fold degenerate, then its eigenstate could be
+any vector $$\Ket{n, d}$$ from the corresponding $$D$$-dimensional
 eigenspace:
 
 $$\begin{aligned}
@@ -199,28 +199,28 @@ $$\begin{aligned}
     = \sum_{d = 1}^{D} c_{d} \Ket{n, d}
 \end{aligned}$$
 
-In general, adding the perturbation $\hat{H}_1$ will *lift* the
+In general, adding the perturbation $$\hat{H}_1$$ will *lift* the
 degeneracy, meaning the perturbed states will be non-degenerate. In the
-limit $\lambda \to 0$, these $D$ perturbed states change into $D$
-orthogonal states which are all valid $\Ket{n}$.
+limit $$\lambda \to 0$$, these $$D$$ perturbed states change into $$D$$
+orthogonal states which are all valid $$\Ket{n}$$.
 
-However, the $\Ket{n}$ that they converge to are not arbitrary: only
-certain unperturbed eigenstates are "good" states. Without $\hat{H}_1$,
+However, the $$\Ket{n}$$ that they converge to are not arbitrary: only
+certain unperturbed eigenstates are "good" states. Without $$\hat{H}_1$$,
 this distinction is irrelevant, but in the perturbed case it will turn
 out to be important.
 
-For now, we write $\Ket{n, d}$ to refer to any orthonormal set of
-vectors in the eigenspace of $\varepsilon_n$ (not necessarily the "good"
-ones), and $\Ket{n}$ to denote any linear combination of these. We then
-take the equation at order $\lambda^1$ and prepend an arbitrary
-eigenspace basis vector $\Bra{n, \delta}$:
+For now, we write $$\Ket{n, d}$$ to refer to any orthonormal set of
+vectors in the eigenspace of $$\varepsilon_n$$ (not necessarily the "good"
+ones), and $$\Ket{n}$$ to denote any linear combination of these. We then
+take the equation at order $$\lambda^1$$ and prepend an arbitrary
+eigenspace basis vector $$\Bra{n, \delta}$$:
 
 $$\begin{aligned}
     \matrixel{n, \delta}{\hat{H}_1}{n} + \matrixel{n, \delta}{\hat{H}_0}{\psi_n^{(1)}}
     &= E_n^{(1)} \Inprod{n, \delta}{n} + \varepsilon_n \inprod{n, \delta}{\psi_n^{(1)}}
 \end{aligned}$$
 
-Since $\hat{H}_0$ is Hermitian, we use the same trick as before to
+Since $$\hat{H}_0$$ is Hermitian, we use the same trick as before to
 reduce the problem to:
 
 $$\begin{aligned}
@@ -228,8 +228,8 @@ $$\begin{aligned}
     &= E_n^{(1)} \Inprod{n, \delta}{n}
 \end{aligned}$$
 
-We express $\Ket{n}$ as a linear combination of the eigenbasis vectors
-$\Ket{n, d}$ to get:
+We express $$\Ket{n}$$ as a linear combination of the eigenbasis vectors
+$$\Ket{n, d}$$ to get:
 
 $$\begin{aligned}
     \sum_{d = 1}^{D} c_d \matrixel{n, \delta}{\hat{H}_1}{n, d}
@@ -238,13 +238,13 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Let us now interpret the summation terms as matrix elements
-$M_{\delta, d}$:
+$$M_{\delta, d}$$:
 
 $$\begin{aligned}
     M_{\delta, d} = \matrixel{n, \delta}{\hat{H}_1}{n, d}
 \end{aligned}$$
 
-By varying the value of $\delta$ from $1$ to $D$, we end up with
+By varying the value of $$\delta$$ from $$1$$ to $$D$$, we end up with
 equations of the form:
 
 $$\begin{aligned}
@@ -262,10 +262,10 @@ $$\begin{aligned}
     \end{bmatrix}
 \end{aligned}$$
 
-This is an eigenvalue problem for $E_n^{(1)}$, where $c_d$ are the
+This is an eigenvalue problem for $$E_n^{(1)}$$, where $$c_d$$ are the
 components of the eigenvectors which represent the "good" states.
-After solving this, let $\Ket{n, g}$ be the resulting "good" states.
-Then, as long as $E_n^{(1)}$ is a non-degenerate eigenvalue of $M$:
+After solving this, let $$\Ket{n, g}$$ be the resulting "good" states.
+Then, as long as $$E_n^{(1)}$$ is a non-degenerate eigenvalue of $$M$$:
 
 $$\begin{aligned}
     \boxed{
@@ -283,36 +283,36 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-This works because the matrix $M$ is diagonal in the $\Ket{n, g}$-basis,
-such that when $\Ket{m}$ is any vector $\Ket{n, \gamma}$ in the
-$\Ket{n}$-eigenspace (except for $\Ket{n,g}$, which is
+This works because the matrix $$M$$ is diagonal in the $$\Ket{n, g}$$-basis,
+such that when $$\Ket{m}$$ is any vector $$\Ket{n, \gamma}$$ in the
+$$\Ket{n}$$-eigenspace (except for $$\Ket{n,g}$$, which is
 explicitly excluded), then the corresponding numerator
-$\matrixel{n, \gamma}{\hat{H}_1}{n, g} = M_{\gamma, g} = 0$, so the term
+$$\matrixel{n, \gamma}{\hat{H}_1}{n, g} = M_{\gamma, g} = 0$$, so the term
 does not contribute.
 
-If any of the eigenvalues $E_n^{(1)}$ of $M$ are degenerate, then there
-is still information missing about the components $c_d$ of the
+If any of the eigenvalues $$E_n^{(1)}$$ of $$M$$ are degenerate, then there
+is still information missing about the components $$c_d$$ of the
 "good" states, in which case we must find them some other way.
 
 Such an alternative way of determining these "good" states is also of
-interest even if there is no degeneracy in $M$, since such a shortcut would
+interest even if there is no degeneracy in $$M$$, since such a shortcut would
 allow us to use the formulae from non-degenerate perturbation theory
 straight away.
 
-The trick is to find a Hermitian operator $\hat{L}$ (usually using
-symmetries of the system) which commutes with both $\hat{H}_0$ and $\hat{H}_1$:
+The trick is to find a Hermitian operator $$\hat{L}$$ (usually using
+symmetries of the system) which commutes with both $$\hat{H}_0$$ and $$\hat{H}_1$$:
 
 $$\begin{aligned}
     \comm{\hat{L}}{\hat{H}_0} = \comm{\hat{L}}{\hat{H}_1} = 0
 \end{aligned}$$
 
-So that it shares its eigenstates with $\hat{H}_0$ (and $\hat{H}_1$),
-meaning all the vectors of the $D$-dimensional
-$\Ket{n}$-eigenspace are also eigenvectors of $\hat{L}$.
+So that it shares its eigenstates with $$\hat{H}_0$$ (and $$\hat{H}_1$$),
+meaning all the vectors of the $$D$$-dimensional
+$$\Ket{n}$$-eigenspace are also eigenvectors of $$\hat{L}$$.
 
-The crucial part, however, is that $\hat{L}$ must be chosen such that
-$\Ket{n, d_1}$ and $\Ket{n, d_2}$ have distinct eigenvalues
-$\ell_1 \neq \ell_2$ for $d_1 \neq d_2$:
+The crucial part, however, is that $$\hat{L}$$ must be chosen such that
+$$\Ket{n, d_1}$$ and $$\Ket{n, d_2}$$ have distinct eigenvalues
+$$\ell_1 \neq \ell_2$$ for $$d_1 \neq d_2$$:
 
 $$\begin{aligned}
     \hat{L} \Ket{n, d_1} = \ell_1 \Ket{n, d_1}
@@ -320,9 +320,9 @@ $$\begin{aligned}
     \hat{L} \Ket{n, d_2} = \ell_2 \Ket{n, d_2}
 \end{aligned}$$
 
-When this condition holds for any orthogonal choice of $\Ket{n, d_1}$ and
-$\Ket{n, d_2}$, then these specific eigenvectors of $\hat{L}$ are the
-"good states", for any valid choice of $\hat{L}$.
+When this condition holds for any orthogonal choice of $$\Ket{n, d_1}$$ and
+$$\Ket{n, d_2}$$, then these specific eigenvectors of $$\hat{L}$$ are the
+"good states", for any valid choice of $$\hat{L}$$.