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authorPrefetch2021-02-20 20:21:32 +0100
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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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