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

1 files changed, 50 insertions, 50 deletions

diff --git a/source/know/concept/hilbert-space/index.md b/source/know/concept/hilbert-space/index.md
index ef55d2b..57926ce 100644
--- a/source/know/concept/hilbert-space/index.md
+++ b/source/know/concept/hilbert-space/index.md
@@ -14,28 +14,28 @@ abstract **vector space** with a notion of length and angle.
 
 ## Vector space
 
-An abstract **vector space** $\mathbb{V}$ is a generalization of the
+An abstract **vector space** $$\mathbb{V}$$ is a generalization of the
 traditional concept of vectors as "arrows". It consists of a set of
 objects called **vectors** which support the following (familiar)
 operations:
 
-+ **Vector addition**: the sum of two vectors $V$ and $W$, denoted $V + W$.
-+ **Scalar multiplication**: product of a vector $V$ with a scalar $a$, denoted $a V$.
++ **Vector addition**: the sum of two vectors $$V$$ and $$W$$, denoted $$V + W$$.
++ **Scalar multiplication**: product of a vector $$V$$ with a scalar $$a$$, denoted $$a V$$.
 
-In addition, for a given $\mathbb{V}$ to qualify as a proper vector
+In addition, for a given $$\mathbb{V}$$ to qualify as a proper vector
 space, these operations must obey the following axioms:
 
-+ **Addition is associative**: $U + (V + W) = (U + V) + W$
-+ **Addition is commutative**: $U + V = V + U$
-+ **Addition has an identity**: there exists a $\mathbf{0}$ such that $V + 0 = V$
-+ **Addition has an inverse**: for every $V$ there exists $-V$ so that $V + (-V) = 0$
-+ **Multiplication is associative**: $a (b V) = (a b) V$
-+ **Multiplication has an identity**: There exists a $1$ such that $1 V = V$
-+ **Multiplication is distributive over scalars**: $(a + b)V = aV + bV$
-+ **Multiplication is distributive over vectors**: $a (U + V) = a U + a V$
++ **Addition is associative**: $$U + (V + W) = (U + V) + W$$
++ **Addition is commutative**: $$U + V = V + U$$
++ **Addition has an identity**: there exists a $$\mathbf{0}$$ such that $$V + 0 = V$$
++ **Addition has an inverse**: for every $$V$$ there exists $$-V$$ so that $$V + (-V) = 0$$
++ **Multiplication is associative**: $$a (b V) = (a b) V$$
++ **Multiplication has an identity**: There exists a $$1$$ such that $$1 V = V$$
++ **Multiplication is distributive over scalars**: $$(a + b)V = aV + bV$$
++ **Multiplication is distributive over vectors**: $$a (U + V) = a U + a V$$
 
-A set of $N$ vectors $V_1, V_2, ..., V_N$ is **linearly independent** if
-the only way to satisfy the following relation is to set all the scalar coefficients $a_n = 0$:
+A set of $$N$$ vectors $$V_1, V_2, ..., V_N$$ is **linearly independent** if
+the only way to satisfy the following relation is to set all the scalar coefficients $$a_n = 0$$:
 
 $$\begin{aligned}
     \mathbf{0} = \sum_{n = 1}^N a_n V_n
@@ -44,13 +44,13 @@ $$\begin{aligned}
 In other words, these vectors cannot be expressed in terms of each
 other. Otherwise, they would be **linearly dependent**.
 
-A vector space $\mathbb{V}$ has **dimension** $N$ if only up to $N$ of
+A vector space $$\mathbb{V}$$ has **dimension** $$N$$ if only up to $$N$$ of
 its vectors can be linearly indepedent. All other vectors in
-$\mathbb{V}$ can then be written as a **linear combination** of these $N$ **basis vectors**.
+$$\mathbb{V}$$ can then be written as a **linear combination** of these $$N$$ **basis vectors**.
 
-Let $\vu{e}_1, ..., \vu{e}_N$ be the basis vectors, then any
-vector $V$ in the same space can be **expanded** in the basis according to
-the unique weights $v_n$, known as the **components** of $V$
+Let $$\vu{e}_1, ..., \vu{e}_N$$ be the basis vectors, then any
+vector $$V$$ in the same space can be **expanded** in the basis according to
+the unique weights $$v_n$$, known as the **components** of $$V$$
 in that basis:
 
 $$\begin{aligned}
@@ -73,25 +73,25 @@ $$\begin{gathered}
 
 ## Inner product
 
-A given vector space $\mathbb{V}$ can be promoted to a **Hilbert space**
-or **inner product space** if it supports an operation $\Inprod{U}{V}$
+A given vector space $$\mathbb{V}$$ can be promoted to a **Hilbert space**
+or **inner product space** if it supports an operation $$\Inprod{U}{V}$$
 called the **inner product**, which takes two vectors and returns a
 scalar, and has the following properties:
 
-+ **Skew symmetry**: $\Inprod{U}{V} = (\Inprod{V}{U})^*$, where ${}^*$ is the complex conjugate.
-+ **Positive semidefiniteness**: $\Inprod{V}{V} \ge 0$, and $\Inprod{V}{V} = 0$ if $V = \mathbf{0}$.
-+ **Linearity in second operand**: $\Inprod{U}{(a V + b W)} = a \Inprod{U}{V} + b \Inprod{U}{W}$.
++ **Skew symmetry**: $$\Inprod{U}{V} = (\Inprod{V}{U})^*$$, where $${}^*$$ is the complex conjugate.
++ **Positive semidefiniteness**: $$\Inprod{V}{V} \ge 0$$, and $$\Inprod{V}{V} = 0$$ if $$V = \mathbf{0}$$.
++ **Linearity in second operand**: $$\Inprod{U}{(a V + b W)} = a \Inprod{U}{V} + b \Inprod{U}{W}$$.
 
 The inner product describes the lengths and angles of vectors, and in
 Euclidean space it is implemented by the dot product.
 
-The **magnitude** or **norm** $|V|$ of a vector $V$ is given by
-$|V| = \sqrt{\Inprod{V}{V}}$ and represents the real positive length of $V$.
+The **magnitude** or **norm** $$|V|$$ of a vector $$V$$ is given by
+$$|V| = \sqrt{\Inprod{V}{V}}$$ and represents the real positive length of $$V$$.
 A **unit vector** has a norm of 1.
 
-Two vectors $U$ and $V$ are **orthogonal** if their inner product
-$\Inprod{U}{V} = 0$. If in addition to being orthogonal, $|U| = 1$ and
-$|V| = 1$, then $U$ and $V$ are known as **orthonormal** vectors.
+Two vectors $$U$$ and $$V$$ are **orthogonal** if their inner product
+$$\Inprod{U}{V} = 0$$. If in addition to being orthogonal, $$|U| = 1$$ and
+$$|V| = 1$$, then $$U$$ and $$V$$ are known as **orthonormal** vectors.
 
 Orthonormality is desirable for basis vectors, so if they are
 not already like that, it is common to manually turn them into a new
@@ -108,15 +108,15 @@ $$\begin{gathered}
     \Inprod{V}{W} = \sum_{n = 1}^N \sum_{m = 1}^N v_n^* w_m \Inprod{\vu{e}_n}{\vu{e}_j}
 \end{gathered}$$
 
-If the basis vectors $\vu{e}_1, ..., \vu{e}_N$ are already
+If the basis vectors $$\vu{e}_1, ..., \vu{e}_N$$ are already
 orthonormal, this reduces to:
 
 $$\begin{aligned}
     \Inprod{V}{W} = \sum_{n = 1}^N v_n^* w_n
 \end{aligned}$$
 
-As it turns out, the components $v_n$ are given by the inner product
-with $\vu{e}_n$, where $\delta_{nm}$ is the Kronecker delta:
+As it turns out, the components $$v_n$$ are given by the inner product
+with $$\vu{e}_n$$, where $$\delta_{nm}$$ is the Kronecker delta:
 
 $$\begin{aligned}
     \Inprod{\vu{e}_n}{V} = \sum_{m = 1}^N \delta_{nm} v_m = v_n
@@ -125,12 +125,12 @@ $$\begin{aligned}
 
 ## Infinite dimensions
 
-As the dimensionality $N$ tends to infinity, things may or may not
-change significantly, depending on whether $N$ is **countably** or
+As the dimensionality $$N$$ tends to infinity, things may or may not
+change significantly, depending on whether $$N$$ is **countably** or
 **uncountably** infinite.
 
 In the former case, not much changes: the infinitely many **discrete**
-basis vectors $\vu{e}_n$ can all still be made orthonormal as usual,
+basis vectors $$\vu{e}_n$$ can all still be made orthonormal as usual,
 and as before:
 
 $$\begin{aligned}
@@ -141,16 +141,16 @@ A good example of such a countably-infinitely-dimensional basis are the
 solution eigenfunctions of a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/).
 
 However, if the dimensionality is uncountably infinite, the basis
-vectors are **continuous** and cannot be labeled by $n$. For example, all
-complex functions $f(x)$ defined for $x \in [a, b]$ which
-satisfy $f(a) = f(b) = 0$ form such a vector space.
-In this case $f(x)$ is expanded as follows, where $x$ is a basis vector:
+vectors are **continuous** and cannot be labeled by $$n$$. For example, all
+complex functions $$f(x)$$ defined for $$x \in [a, b]$$ which
+satisfy $$f(a) = f(b) = 0$$ form such a vector space.
+In this case $$f(x)$$ is expanded as follows, where $$x$$ is a basis vector:
 
 $$\begin{aligned}
     f(x) = \int_a^b \Inprod{x}{f} \dd{x}
 \end{aligned}$$
 
-Similarly, the inner product $\Inprod{f}{g}$ must also be redefined as
+Similarly, the inner product $$\Inprod{f}{g}$$ must also be redefined as
 follows:
 
 $$\begin{aligned}
@@ -158,17 +158,17 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The concept of orthonormality must be also weakened. A finite function
-$f(x)$ can be normalized as usual, but the basis vectors $x$ themselves
+$$f(x)$$ can be normalized as usual, but the basis vectors $$x$$ themselves
 cannot, since each represents an infinitesimal section of the real line.
 
-The rationale in this case is that action of the identity operator $\hat{I}$ must
+The rationale in this case is that action of the identity operator $$\hat{I}$$ must
 be preserved, which is given here in [Dirac notation](/know/concept/dirac-notation/):
 
 $$\begin{aligned}
     \hat{I} = \int_a^b \Ket{\xi} \Bra{\xi} \dd{\xi}
 \end{aligned}$$
 
-Applying the identity operator to $f(x)$ should just give $f(x)$ again:
+Applying the identity operator to $$f(x)$$ should just give $$f(x)$$ again:
 
 $$\begin{aligned}
     f(x) = \Inprod{x}{f} = \matrixel{x}{\hat{I}}{f}
@@ -176,9 +176,9 @@ $$\begin{aligned}
     = \int_a^b \Inprod{x}{\xi} f(\xi) \dd{\xi}
 \end{aligned}$$
 
-Since we want the latter integral to reduce to $f(x)$, it is plain to see that
-$\Inprod{x}{\xi}$ can only be a [Dirac delta function](/know/concept/dirac-delta-function/),
-i.e $\Inprod{x}{\xi} = \delta(x - \xi)$:
+Since we want the latter integral to reduce to $$f(x)$$, it is plain to see that
+$$\Inprod{x}{\xi}$$ can only be a [Dirac delta function](/know/concept/dirac-delta-function/),
+i.e $$\Inprod{x}{\xi} = \delta(x - \xi)$$:
 
 $$\begin{aligned}
     \int_a^b \Inprod{x}{\xi} f(\xi) \dd{\xi}
@@ -186,11 +186,11 @@ $$\begin{aligned}
     = f(x)
 \end{aligned}$$
 
-Consequently, $\Inprod{x}{\xi} = 0$ if $x \neq \xi$ as expected for an
-orthogonal set of vectors, but if $x = \xi$ the inner product
-$\Inprod{x}{\xi}$ is infinite, unlike earlier.
+Consequently, $$\Inprod{x}{\xi} = 0$$ if $$x \neq \xi$$ as expected for an
+orthogonal set of vectors, but if $$x = \xi$$ the inner product
+$$\Inprod{x}{\xi}$$ is infinite, unlike earlier.
 
-Technically, because the basis vectors $x$ cannot be normalized, they
+Technically, because the basis vectors $$x$$ cannot be normalized, they
 are not members of a Hilbert space, but rather of a superset called a
 **rigged Hilbert space**. Such vectors have no finite inner product with
 themselves, but do have one with all vectors from the actual Hilbert