Add "Dirac notation" + tweak "Bloch's theorem"

6 files changed, 264 insertions, 2 deletions

diff --git a/content/know/category/quantum-mechanics.md b/content/know/category/quantum-mechanics.md
index cf1d7ba..dcb6eb6 100644
--- a/content/know/category/quantum-mechanics.md
+++ b/content/know/category/quantum-mechanics.md
@@ -8,3 +8,6 @@ Alphabetical list of concepts in this category.
 
 ## B
 * [Bloch's theorem](/know/concept/blochs-theorem/)
+
+## D
+* [Dirac notation](/know/concept/dirac-notation/)
diff --git a/content/know/concept/index.md b/content/know/concept/index.md
index 5db6133..6604aba 100644
--- a/content/know/concept/index.md
+++ b/content/know/concept/index.md
@@ -8,3 +8,6 @@ Alphabetical list of concepts in this knowledge base.
 
 ## B
 * [Bloch's theorem](/know/concept/blochs-theorem/)
+
+## D
+* [Dirac notation](/know/concept/dirac-notation/)
diff --git a/latex/know/concept/blochs-theorem/source.md b/latex/know/concept/blochs-theorem/source.md
index 2307d2e..528c218 100644
--- a/latex/know/concept/blochs-theorem/source.md
+++ b/latex/know/concept/blochs-theorem/source.md
@@ -27,7 +27,7 @@ known as *Bloch functions* or *Bloch states*.
 This is suprisingly easy to prove:
 if the Hamiltonian $\hat{H}$ is lattice-periodic,
 then it will commute with the unitary translation operator $\hat{T}(\vec{a})$,
-i.e. $\comm{\hat{H}}{\hat{T}(\vec{a})} = 0$.
+i.e. $[\hat{H}, \hat{T}(\vec{a})] = 0$.
 Therefore $\hat{H}$ and $\hat{T}(\vec{a})$ must share eigenstates $\psi(\vec{r})$:
 
 $$
diff --git a/latex/know/concept/dirac-notation/source.md b/latex/know/concept/dirac-notation/source.md
new file mode 100644
index 0000000..7b384ab
--- /dev/null
+++ b/latex/know/concept/dirac-notation/source.md
@@ -0,0 +1,119 @@
+% Dirac notation
+
+
+# Dirac notation
+
+*Dirac notation* is a notation to do calculations in a Hilbert space
+without needing to worry about the space's representation. It is
+basically the *lingua franca* of quantum mechanics.
+
+In Dirac notation there are *kets* $\ket{V}$ from the Hilbert space
+$\mathbb{H}$ and *bras* $\bra{V}$ from a dual $\mathbb{H}'$ of the
+former. Crucially, the bras and kets are from different Hilbert spaces
+and therefore cannot be added, but every bra has a corresponding ket and
+vice versa.
+
+Bras and kets can only be combined in two ways: the *inner product*
+$\braket{V | W}$, which returns a scalar, and the *outer product*
+$\ket{V} \bra{W}$, which returns a mapping $\hat{L}$ from kets $\ket{V}$
+to other kets $\ket{V'}$, i.e. a linear operator. Recall that the
+Hilbert inner product must satisfy:
+
+$$\begin{aligned}
+    \braket{V | W} = \braket{W | V}^*
+\end{aligned}$$
+
+So far, nothing has been said about the actual representation of bras or
+kets. If we represent kets as $N$-dimensional columns vectors, the
+corresponding bras are given by the kets' adjoints, i.e. their transpose
+conjugates:
+
+$$\begin{aligned}
+    \ket{V} =
+    \begin{bmatrix}
+        v_1 \\ \vdots \\ v_N
+    \end{bmatrix}
+    \quad \implies \quad
+    \bra{V} =
+    \begin{bmatrix}
+        v_1^* & \cdots & v_N^*
+    \end{bmatrix}
+\end{aligned}$$
+
+The inner product $\braket{V | W}$ is then just the familiar dot product $V \cdot W$:
+
+$$\begin{gathered}
+    \braket{V | W}
+    =
+    \begin{bmatrix}
+        v_1^* & \cdots & v_N^*
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        w_1 \\ \vdots \\ w_N
+    \end{bmatrix}
+    = v_1^* w_1 + ... + v_N^* w_N
+\end{gathered}$$
+
+Meanwhile, the outer product $\ket{V} \bra{W}$ creates an $N \cross N$ matrix:
+
+$$\begin{gathered}
+    \ket{V} \bra{W}
+    =
+    \begin{bmatrix}
+        v_1 \\ \vdots \\ v_N
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        w_1^* & \cdots & w_N^*
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        v_1 w_1^* & \cdots & v_1 w_N^* \\
+        \vdots & \ddots & \vdots \\
+        v_N w_1^* & \cdots & v_N w_N^*
+    \end{bmatrix}
+\end{gathered}$$
+
+If the kets are instead represented by functions $f(x)$ of
+$x \in [a, b]$, then the bras represent *functionals* $F[u(x)]$ which
+take an unknown function $u(x)$ as an argument and turn it into a scalar
+using integration:
+
+$$\begin{aligned}
+    \ket{f} = f(x)
+    \quad \implies \quad
+    \bra{f}
+    = F[u(x)]
+    = \int_a^b f^*(x) \: u(x) \dd{x}
+\end{aligned}$$
+
+Consequently, the inner product is simply the following familiar integral:
+
+$$\begin{gathered}
+    \braket{f | g}
+    = F[g(x)]
+    = \int_a^b f^*(x) \: g(x) \dd{x}
+\end{gathered}$$
+
+However, the outer product becomes something rather abstract:
+
+$$\begin{gathered}
+    \ket{f} \bra{g}
+    = f(x) \: G[u(x)]
+    = f(x) \int_a^b g^*(\xi) \: u(\xi) \dd{\xi}
+\end{gathered}$$
+
+This result makes more sense if we surround it by a bra and a ket:
+
+$$\begin{aligned}
+    \bra{u} \!\Big(\!\ket{f} \bra{g}\!\Big)\! \ket{w}
+    &= U\big[f(x) \: G[w(x)]\big]
+    = U\Big[ f(x) \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big]
+    \\
+    &= \int_a^b u^*(x) \: f(x) \: \Big(\int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big) \dd{x}
+    \\
+    &= \Big( \int_a^b u^*(x) \: f(x) \dd{x} \Big) \Big( \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big)
+    \\
+    &= \braket{u | f} \braket{g | w}
+\end{aligned}$$
diff --git a/static/know/concept/blochs-theorem/index.html b/static/know/concept/blochs-theorem/index.html
index 6e5767c..f977739 100644
--- a/static/know/concept/blochs-theorem/index.html
+++ b/static/know/concept/blochs-theorem/index.html
@@ -59,7 +59,7 @@
 \end{aligned}
 \]</span></p>
 <p>In other words, in a periodic potential, the solutions are simply plane waves with a periodic modulation, known as <em>Bloch functions</em> or <em>Bloch states</em>.</p>
-<p>This is suprisingly easy to prove: if the Hamiltonian <span class="math inline">\(\hat{H}\)</span> is lattice-periodic, then it will commute with the unitary translation operator <span class="math inline">\(\hat{T}(\vec{a})\)</span>, i.e. <span class="math inline">\(\comm{\hat{H}}{\hat{T}(\vec{a})} = 0\)</span>. Therefore <span class="math inline">\(\hat{H}\)</span> and <span class="math inline">\(\hat{T}(\vec{a})\)</span> must share eigenstates <span class="math inline">\(\psi(\vec{r})\)</span>:</p>
+<p>This is suprisingly easy to prove: if the Hamiltonian <span class="math inline">\(\hat{H}\)</span> is lattice-periodic, then it will commute with the unitary translation operator <span class="math inline">\(\hat{T}(\vec{a})\)</span>, i.e. <span class="math inline">\([\hat{H}, \hat{T}(\vec{a})] = 0\)</span>. Therefore <span class="math inline">\(\hat{H}\)</span> and <span class="math inline">\(\hat{T}(\vec{a})\)</span> must share eigenstates <span class="math inline">\(\psi(\vec{r})\)</span>:</p>
 <p><span class="math display">\[
 \begin{aligned}
     \hat{H} \:\psi(\vec{r}) = E \:\psi(\vec{r})
diff --git a/static/know/concept/dirac-notation/index.html b/static/know/concept/dirac-notation/index.html
new file mode 100644
index 0000000..74aa0b4
--- /dev/null
+++ b/static/know/concept/dirac-notation/index.html
@@ -0,0 +1,137 @@
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+<hr>
+<h1 id="dirac-notation">Dirac notation</h1>
+<p><em>Dirac notation</em> is a notation to do calculations in a Hilbert space without needing to worry about the space’s representation. It is basically the <em>lingua franca</em> of quantum mechanics.</p>
+<p>In Dirac notation there are <em>kets</em> <span class="math inline">\(\ket{V}\)</span> from the Hilbert space <span class="math inline">\(\mathbb{H}\)</span> and <em>bras</em> <span class="math inline">\(\bra{V}\)</span> from a dual <span class="math inline">\(\mathbb{H}&#39;\)</span> of the former. Crucially, the bras and kets are from different Hilbert spaces and therefore cannot be added, but every bra has a corresponding ket and vice versa.</p>
+<p>Bras and kets can only be combined in two ways: the <em>inner product</em> <span class="math inline">\(\braket{V | W}\)</span>, which returns a scalar, and the <em>outer product</em> <span class="math inline">\(\ket{V} \bra{W}\)</span>, which returns a mapping <span class="math inline">\(\hat{L}\)</span> from kets <span class="math inline">\(\ket{V}\)</span> to other kets <span class="math inline">\(\ket{V&#39;}\)</span>, i.e. a linear operator. Recall that the Hilbert inner product must satisfy:</p>
+<p><span class="math display">\[\begin{aligned}
+    \braket{V | W} = \braket{W | V}^*
+\end{aligned}\]</span></p>
+<p>So far, nothing has been said about the actual representation of bras or kets. If we represent kets as <span class="math inline">\(N\)</span>-dimensional columns vectors, the corresponding bras are given by the kets’ adjoints, i.e. their transpose conjugates:</p>
+<p><span class="math display">\[\begin{aligned}
+    \ket{V} =
+    \begin{bmatrix}
+        v_1 \\ \vdots \\ v_N
+    \end{bmatrix}
+    \quad \implies \quad
+    \bra{V} =
+    \begin{bmatrix}
+        v_1^* &amp; \cdots &amp; v_N^*
+    \end{bmatrix}
+\end{aligned}\]</span></p>
+<p>The inner product <span class="math inline">\(\braket{V | W}\)</span> is then just the familiar dot product <span class="math inline">\(V \cdot W\)</span>:</p>
+<p><span class="math display">\[\begin{gathered}
+    \braket{V | W}
+    =
+    \begin{bmatrix}
+        v_1^* &amp; \cdots &amp; v_N^*
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        w_1 \\ \vdots \\ w_N
+    \end{bmatrix}
+    = v_1^* w_1 + ... + v_N^* w_N
+\end{gathered}\]</span></p>
+<p>Meanwhile, the outer product <span class="math inline">\(\ket{V} \bra{W}\)</span> creates an <span class="math inline">\(N \cross N\)</span> matrix:</p>
+<p><span class="math display">\[\begin{gathered}
+    \ket{V} \bra{W}
+    =
+    \begin{bmatrix}
+        v_1 \\ \vdots \\ v_N
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        w_1^* &amp; \cdots &amp; w_N^*
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        v_1 w_1^* &amp; \cdots &amp; v_1 w_N^* \\
+        \vdots &amp; \ddots &amp; \vdots \\
+        v_N w_1^* &amp; \cdots &amp; v_N w_N^*
+    \end{bmatrix}
+\end{gathered}\]</span></p>
+<p>If the kets are instead represented by functions <span class="math inline">\(f(x)\)</span> of <span class="math inline">\(x \in [a, b]\)</span>, then the bras represent <em>functionals</em> <span class="math inline">\(F[u(x)]\)</span> which take an unknown function <span class="math inline">\(u(x)\)</span> as an argument and turn it into a scalar using integration:</p>
+<p><span class="math display">\[\begin{aligned}
+    \ket{f} = f(x)
+    \quad \implies \quad
+    \bra{f}
+    = F[u(x)]
+    = \int_a^b f^*(x) \: u(x) \dd{x}
+\end{aligned}\]</span></p>
+<p>Consequently, the inner product is simply the following familiar integral:</p>
+<p><span class="math display">\[\begin{gathered}
+    \braket{f | g}
+    = F[g(x)]
+    = \int_a^b f^*(x) \: g(x) \dd{x}
+\end{gathered}\]</span></p>
+<p>However, the outer product becomes something rather abstract:</p>
+<p><span class="math display">\[\begin{gathered}
+    \ket{f} \bra{g}
+    = f(x) \: G[u(x)]
+    = f(x) \int_a^b g^*(\xi) \: u(\xi) \dd{\xi}
+\end{gathered}\]</span></p>
+<p>This result makes more sense if we surround it by a bra and a ket:</p>
+<p><span class="math display">\[\begin{aligned}
+    \bra{u} \!\Big(\!\ket{f} \bra{g}\!\Big)\! \ket{w}
+    &amp;= U\big[f(x) \: G[w(x)]\big]
+    = U\Big[ f(x) \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big]
+    \\
+    &amp;= \int_a^b u^*(x) \: f(x) \: \Big(\int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big) \dd{x}
+    \\
+    &amp;= \Big( \int_a^b u^*(x) \: f(x) \dd{x} \Big) \Big( \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big)
+    \\
+    &amp;= \braket{u | f} \braket{g | w}
+\end{aligned}\]</span></p>
+<hr>
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