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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 @@ +<!DOCTYPE html> +<html xmlns="http://www.w3.org/1999/xhtml" lang="" xml:lang=""> +<head> + <meta charset="utf-8" /> + <meta name="generator" content="pandoc" /> + <meta name="viewport" content="width=device-width, initial-scale=1.0, user-scalable=yes" /> + <title>Prefetch | Dirac notation</title> + <link rel="icon" href="data:,"> + <style> + body { + background:#ddd; + color:#222; + max-width:80ch; + text-align:justify; + margin:auto; + padding:1em 0; + font-family:sans-serif; + line-height:1.3; + } + a {text-decoration:none;color:#00f;} + h1,h2,h3 {text-align:center} + h1 {font-size:200%;} + h2 {font-size:160%;} + h3 {font-size:120%;} + .nav {height:3rem;font-size:250%;} + .nav a:link,a:visited {color:#222;} + .nav a:hover,a:focus,a:active {color:#00f;} + .navl {width:30%;float:left;text-align:left;} + .navr {width:70%;float:left;text-align:right;} + pre {filter:invert(100%);} + @media (prefers-color-scheme: dark) { + body {background:#222;filter:invert(100%);} + } </style> + <script> + MathJax = { + loader: {load: ["[tex]/physics"]}, + tex: {packages: {"[+]": ["physics"]}} + }; + </script> + <script src="/mathjax/tex-svg.js" type="text/javascript"></script> + </head> +<body> +<div class="nav"> +<div class="navl"><a href="/">PREFETCH</a></div> +<div class="navr"> +<a href="/blog/">blog</a>  +<a href="/code/">code</a>  +<a href="/know/">know</a> +</div> +</div> +<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">\(\mathbb{H}'\)</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> +<hr> +© "Prefetch". Licensed under <a href="https://creativecommons.org/licenses/by-sa/4.0/">CC BY-SA 4.0</a>. +</body> +</html> |