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authorPrefetch2021-02-20 14:55:33 +0100
committerPrefetch2021-02-20 14:55:33 +0100
commit5999e8682785cc397e266122fba91fafa8b48269 (patch)
treedd76e2a0249253b33f021d4ed1163f80ad8780aa /static/know/concept/dirac-notation
parente71c14aa725d71a2ea7310c69b3d11a8bc12c0b0 (diff)
Add "Dirac notation" + tweak "Bloch's theorem"
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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">\(\mathbb{H}&#39;\)</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>
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