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<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">\(\mathbb{H}'\)</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">\(\ket{V'}\)</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">\[\begin{aligned}
\ket{V} =
\begin{bmatrix}
v_1 \\ \vdots \\ v_N
\end{bmatrix}
\quad \implies \quad
\bra{V} =
\begin{bmatrix}
v_1^* & \cdots & v_N^*
\end{bmatrix}
\end{aligned}\]</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">\[\begin{gathered}
\braket{V | W}
=
\begin{bmatrix}
v_1^* & \cdots & v_N^*
\end{bmatrix}
\cdot
\begin{bmatrix}
w_1 \\ \vdots \\ w_N
\end{bmatrix}
= v_1^* w_1 + ... + v_N^* w_N
\end{gathered}\]</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">\[\begin{gathered}
\ket{V} \bra{W}
=
\begin{bmatrix}
v_1 \\ \vdots \\ v_N
\end{bmatrix}
\cdot
\begin{bmatrix}
w_1^* & \cdots & w_N^*
\end{bmatrix}
=
\begin{bmatrix}
v_1 w_1^* & \cdots & v_1 w_N^* \\
\vdots & \ddots & \vdots \\
v_N w_1^* & \cdots & v_N w_N^*
\end{bmatrix}
\end{gathered}\]</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">\[\begin{aligned}
\bra{u} \!\Big(\!\ket{f} \bra{g}\!\Big)\! \ket{w}
&= U\big[f(x) \: G[w(x)]\big]
= U\Big[ f(x) \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big]
\\
&= \int_a^b u^*(x) \: f(x) \: \Big(\int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big) \dd{x}
\\
&= \Big( \int_a^b u^*(x) \: f(x) \dd{x} \Big) \Big( \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big)
\\
&= \braket{u | f} \braket{g | w}
\end{aligned}\]</span></p>
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