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diff --git a/content/know/concept/_index.md b/content/know/concept/_index.md new file mode 100644 index 0000000..956724a --- /dev/null +++ b/content/know/concept/_index.md @@ -0,0 +1,8 @@ +--- +title: "List of concepts" +date: 2021-02-22T20:38:58+01:00 +draft: false +layout: "know-list" +--- + +This is an alphabetical list of the concepts in this knowledge base. diff --git a/content/know/concept/blochs-theorem/index.pdc b/content/know/concept/blochs-theorem/index.pdc new file mode 100644 index 0000000..1828d8a --- /dev/null +++ b/content/know/concept/blochs-theorem/index.pdc @@ -0,0 +1,115 @@ +--- +title: "Bloch's theorem" +firstLetter: "B" +publishDate: 2021-02-22 +categories: +- Quantum mechanics + +date: 2021-02-22T20:02:14+01:00 +draft: false +markup: pandoc +--- + +# Bloch's theorem +In quantum mechanics, **Bloch's theorem** states that, +given a potential $V(\vec{r})$ which is periodic on a lattice, +i.e. $V(\vec{r}) = V(\vec{r} + \vec{a})$ +for a primitive lattice vector $\vec{a}$, +then it follows that the solutions $\psi(\vec{r})$ +to the time-independent Schrödinger equation +take the following form, +where the function $u(\vec{r})$ is periodic on the same lattice, +i.e. $u(\vec{r}) = u(\vec{r} + \vec{a})$: + +$$ +\begin{aligned} + \boxed{ + \psi(\vec{r}) = u(\vec{r}) e^{i \vec{k} \cdot \vec{r}} + } +\end{aligned} +$$ + +In other words, in a periodic potential, +the solutions are simply plane waves with a periodic modulation, +known as **Bloch functions** or **Bloch states**. + +This is suprisingly easy to prove: +if the Hamiltonian $\hat{H}$ is lattice-periodic, +then both $\psi(\vec{r})$ and $\psi(\vec{r} + \vec{a})$ +are eigenstates with the same energy: + +$$ +\begin{aligned} + \hat{H} \psi(\vec{r}) = E \psi(\vec{r}) + \qquad + \hat{H} \psi(\vec{r} + \vec{a}) = E \psi(\vec{r} + \vec{a}) +\end{aligned} +$$ + +Now define the unitary translation operator $\hat{T}(\vec{a})$ such that +$\psi(\vec{r} + \vec{a}) = \hat{T}(\vec{a}) \psi(\vec{r})$. +From the previous equation, we then know that: + +$$ +\begin{aligned} + \hat{H} \hat{T}(\vec{a}) \psi(\vec{r}) + = E \hat{T}(\vec{a}) \psi(\vec{r}) + = \hat{T}(\vec{a}) \big(E \psi(\vec{r})\big) + = \hat{T}(\vec{a}) \hat{H} \psi(\vec{r}) +\end{aligned} +$$ + +In other words, if $\hat{H}$ is lattice-periodic, +then it will commute with $\hat{T}(\vec{a})$, +i.e. $[\hat{H}, \hat{T}(\vec{a})] = 0$. +Consequently, $\hat{H}$ and $\hat{T}(\vec{a})$ must share eigenstates $\psi(\vec{r})$: + +$$ +\begin{aligned} + \hat{H} \:\psi(\vec{r}) = E \:\psi(\vec{r}) + \qquad + \hat{T}(\vec{a}) \:\psi(\vec{r}) = \tau \:\psi(\vec{r}) +\end{aligned} +$$ + +Since $\hat{T}$ is unitary, +its eigenvalues $\tau$ must have the form $e^{i \theta}$, with $\theta$ real. +Therefore a translation by $\vec{a}$ causes a phase shift, +for some vector $\vec{k}$: + +$$ +\begin{aligned} + \psi(\vec{r} + \vec{a}) + = \hat{T}(\vec{a}) \:\psi(\vec{r}) + = e^{i \theta} \:\psi(\vec{r}) + = e^{i \vec{k} \cdot \vec{a}} \:\psi(\vec{r}) +\end{aligned} +$$ + +Let us now define the following function, +keeping our arbitrary choice of $\vec{k}$: + +$$ +\begin{aligned} + u(\vec{r}) + = e^{- i \vec{k} \cdot \vec{r}} \:\psi(\vec{r}) +\end{aligned} +$$ + +As it turns out, this function is guaranteed to be lattice-periodic for any $\vec{k}$: + +$$ +\begin{aligned} + u(\vec{r} + \vec{a}) + &= e^{- i \vec{k} \cdot (\vec{r} + \vec{a})} \:\psi(\vec{r} + \vec{a}) + \\ + &= e^{- i \vec{k} \cdot \vec{r}} e^{- i \vec{k} \cdot \vec{a}} e^{i \vec{k} \cdot \vec{a}} \:\psi(\vec{r}) + \\ + &= e^{- i \vec{k} \cdot \vec{r}} \:\psi(\vec{r}) + \\ + &= u(\vec{r}) +\end{aligned} +$$ + +Then Bloch's theorem follows from +isolating the definition of $u(\vec{r})$ for $\psi(\vec{r})$. diff --git a/content/know/concept/convolution-theorem/index.pdc b/content/know/concept/convolution-theorem/index.pdc new file mode 100644 index 0000000..fc96f30 --- /dev/null +++ b/content/know/concept/convolution-theorem/index.pdc @@ -0,0 +1,100 @@ +--- +title: "Convolution theorem" +firstLetter: "C" +publishDate: 2021-02-22 +categories: +- Mathematics + +date: 2021-02-22T21:35:23+01:00 +draft: false +markup: pandoc +--- + +# Convolution theorem + +The **convolution theorem** states that a convolution in the direct domain +is equal to a product in the frequency domain. This is especially useful +for computation, replacing an $\mathcal{O}(n^2)$ convolution with an +$\mathcal{O}(n \log(n))$ transform and product. + +## Fourier transform + +The convolution theorem is usually expressed as follows, where +$\hat{\mathcal{F}}$ is the [Fourier transform](/know/concept/fourier-transform/), +and $A$ and $B$ are constants from its definition: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + A \cdot (f * g)(x) &= \hat{\mathcal{F}}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\ + B \cdot (\tilde{f} * \tilde{g})(k) &= \hat{\mathcal{F}}\{f(x) \: g(x)\} + \end{aligned} + } +\end{aligned}$$ + +To prove this, we expand the right-hand side of the theorem and +rearrange the integrals: + +$$\begin{aligned} + \hat{\mathcal{F}}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} + &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \exp(i s k x') \dd{x'} \Big) \exp(-i s k x) \dd{k} + \\ + &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k (x - x')) \dd{k} \Big) \dd{x'} + \\ + &= A \int_{-\infty}^\infty g(x') f(x - x') \dd{x'} + = A \cdot (f * g)(x) +\end{aligned}$$ + +Then we do the same thing again, this time starting from a product in +the $x$-domain: + +$$\begin{aligned} + \hat{\mathcal{F}}\{f(x) \: g(x)\} + &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp(- i s x k') \dd{k'} \Big) \exp(i s k x) \dd{x} + \\ + &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp(i s x (k - k')) \dd{x} \Big) \dd{k'} + \\ + &= B \int_{-\infty}^\infty \tilde{g}(k') \tilde{f}(k - k') \dd{k'} + = B \cdot (\tilde{f} * \tilde{g})(k) +\end{aligned}$$ + + +## Laplace transform + +For functions $f(t)$ and $g(t)$ which are only defined for $t \ge 0$, +the convolution theorem can also be stated using the Laplace transform: + +$$\begin{aligned} + \boxed{(f * g)(t) = \hat{\mathcal{L}}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}} +\end{aligned}$$ + +Because the inverse Laplace transform $\hat{\mathcal{L}}^{-1}$ is quite +unpleasant, the theorem is often stated using the forward transform +instead: + +$$\begin{aligned} + \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)} +\end{aligned}$$ + +We prove this by expanding the left-hand side. Note that the lower +integration limit is 0 instead of $-\infty$, because we set both $f(t)$ +and $g(t)$ to zero for $t < 0$: + +$$\begin{aligned} + \hat{\mathcal{L}}\{(f * g)(t)\} + &= \int_0^\infty \Big( \int_0^\infty g(t') f(t - t') \dd{t'} \Big) \exp(- s t) \dd{t} + \\ + &= \int_0^\infty \Big( \int_0^\infty f(t - t') \exp(- s t) \dd{t} \Big) g(t') \dd{t'} +\end{aligned}$$ + +Then we define a new integration variable $\tau = t - t'$, yielding: + +$$\begin{aligned} + \hat{\mathcal{L}}\{(f * g)(t)\} + &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'} + \\ + &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s \tau) \dd{\tau} \Big) g(t') \exp(- s t') \dd{t'} + \\ + &= \int_0^\infty \tilde{f}(s) g(t') \exp(- s t') \dd{t'} + = \tilde{f}(s) \: \tilde{g}(s) +\end{aligned}$$ diff --git a/content/know/concept/dirac-delta-function/index.pdc b/content/know/concept/dirac-delta-function/index.pdc new file mode 100644 index 0000000..3982afc --- /dev/null +++ b/content/know/concept/dirac-delta-function/index.pdc @@ -0,0 +1,109 @@ +--- +title: "Dirac delta function" +firstLetter: "D" +publishDate: 2021-02-22 +categories: +- Mathematics +- Physics + +date: 2021-02-22T21:35:38+01:00 +draft: false +markup: pandoc +--- + +# Dirac delta function + +The **Dirac delta function** $\delta(x)$, often just called the **delta function**, +is an infinitely narrow discontinuous "spike" at $x = 0$ whose area is +defined to be 1: + +$$\begin{aligned} + \boxed{ + \delta(x) = + \begin{cases} + +\infty & \mathrm{if}\: x = 0 \\ + 0 & \mathrm{if}\: x \neq 0 + \end{cases} + \quad \mathrm{and} \quad + \int_{-\varepsilon}^\varepsilon \delta(x) \dd{x} = 1 + } +\end{aligned}$$ + +It is sometimes also called the **sampling function**, due to its most +important property: the so-called **sampling property**: + +$$\begin{aligned} + \boxed{ + \int f(x) \: \delta(x - x_0) \: dx = \int f(x) \: \delta(x_0 - x) \: dx = f(x_0) + } +\end{aligned}$$ + +$\delta(x)$ is thus an effective weapon against integrals. This may not seem very +useful due to its "unnatural" definition, but in fact it appears as the +limit of several reasonable functions: + +$$\begin{aligned} + \delta(x) + = \lim_{n \to +\infty} \!\Big\{ \frac{n}{\sqrt{\pi}} \exp(- n^2 x^2) \Big\} + = \lim_{n \to +\infty} \!\Big\{ \frac{n}{\pi} \frac{1}{1 + n^2 x^2} \Big\} + = \lim_{n \to +\infty} \!\Big\{ \frac{\sin(n x)}{\pi x} \Big\} +\end{aligned}$$ + +The last one is especially important, since it is equivalent to the +following integral, which appears very often in the context of +[Fourier transforms](/know/concept/fourier-transform/): + +$$\begin{aligned} + \boxed{ + \delta(x) + %= \lim_{n \to +\infty} \!\Big\{\frac{\sin(n x)}{\pi x}\Big\} + = \frac{1}{2\pi} \int_{-\infty}^\infty \exp(i k x) \dd{k} + \:\:\propto\:\: \hat{\mathcal{F}}\{1\} + } +\end{aligned}$$ + +When the argument of $\delta(x)$ is scaled, the delta function is itself scaled: + +$$\begin{aligned} + \boxed{ + \delta(s x) = \frac{1}{|s|} \delta(x) + } +\end{aligned}$$ + +*__Proof.__ Because it is symmetric, $\delta(s x) = \delta(|s| x)$. Then by +substituting $\sigma = |s| x$:* + +$$\begin{aligned} + \int \delta(|s| x) \dd{x} + &= \frac{1}{|s|} \int \delta(\sigma) \dd{\sigma} = \frac{1}{|s|} +\end{aligned}$$ + +*__Q.E.D.__* + +An even more impressive property is the behaviour of the derivative of +$\delta(x)$: + +$$\begin{aligned} + \boxed{ + \int f(\xi) \: \delta'(x - \xi) \dd{\xi} = f'(x) + } +\end{aligned}$$ + +*__Proof.__ Note which variable is used for the +differentiation, and that $\delta'(x - \xi) = - \delta'(\xi - x)$:* + +$$\begin{aligned} + \int f(\xi) \: \dv{\delta(x - \xi)}{x} \dd{\xi} + &= \dv{x} \int f(\xi) \: \delta(x - \xi) \dd{x} + = f'(x) +\end{aligned}$$ + +*__Q.E.D.__* + +This property also generalizes nicely for the higher-order derivatives: + +$$\begin{aligned} + \boxed{ + \int f(\xi) \: \dv[n]{\delta(x - \xi)}{x} \dd{\xi} = \dv[n]{f(x)}{x} + } +\end{aligned}$$ diff --git a/content/know/concept/dirac-notation/index.pdc b/content/know/concept/dirac-notation/index.pdc new file mode 100644 index 0000000..f624574 --- /dev/null +++ b/content/know/concept/dirac-notation/index.pdc @@ -0,0 +1,129 @@ +--- +title: "Dirac notation" +firstLetter: "D" +publishDate: 2021-02-22 +categories: +- Quantum mechanics +- Physics + +date: 2021-02-22T21:35:46+01:00 +draft: false +markup: pandoc +--- + +# 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 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/content/know/concept/fourier-transform/index.pdc b/content/know/concept/fourier-transform/index.pdc new file mode 100644 index 0000000..6d8901a --- /dev/null +++ b/content/know/concept/fourier-transform/index.pdc @@ -0,0 +1,117 @@ +--- +title: "Fourier transform" +firstLetter: "F" +publishDate: 2021-02-22 +categories: +- Mathematics +- Physics + +date: 2021-02-22T21:35:54+01:00 +draft: false +markup: pandoc +--- + +# Fourier transform + +The **Fourier transform** (FT) is an integral transform which converts a +function $f(x)$ into its frequency representation $\tilde{f}(k)$. +Great volumes have already been written about this subject, +so let us focus on the aspects that are useful to physicists. + +The **forward** FT is defined as follows, where $A$, $B$, and $s$ are unspecified constants +(for now): + +$$\begin{aligned} + \boxed{ + \tilde{f}(k) + = \hat{\mathcal{F}}\{f(x)\} + = A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x} + } +\end{aligned}$$ + +The **inverse Fourier transform** (iFT) undoes the forward FT operation: + +$$\begin{aligned} + \boxed{ + f(x) + = \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\} + = B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k x) \dd{k} + } +\end{aligned}$$ + +Clearly, the inverse FT of the forward FT of $f(x)$ must equal $f(x)$ +again. Let us verify this, by rearranging the integrals to get the +[Dirac delta function](/know/concept/dirac-delta-function/) $\delta(x)$: + +$$\begin{aligned} + \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{f(x)\}\} + &= A B \int_{-\infty}^\infty \exp(-i s k x) \int_{-\infty}^\infty f(x') \exp(i s k x') \dd{x'} \dd{k} + \\ + &= 2 \pi A B \int_{-\infty}^\infty f(x') \Big(\frac{1}{2\pi} \int_{-\infty}^\infty \exp(i s k (x' - x)) \dd{k} \Big) \dd{x'} + \\ + &= 2 \pi A B \int_{-\infty}^\infty f(x') \: \delta(s(x' - x)) \dd{x'} + = \frac{2 \pi A B}{|s|} f(x) +\end{aligned}$$ + +Therefore, the constants $A$, $B$, and $s$ are subject to the following +constraint: + +$$\begin{aligned} + \boxed{\frac{2\pi A B}{|s|} = 1} +\end{aligned}$$ + +But that still gives a lot of freedom. The exact choices of $A$ and $B$ +are generally motivated by the [convolution theorem](/know/concept/convolution-theorem/) +and [Parseval's theorem](/know/concept/parsevals-theorem/). + +The choice of $|s|$ depends on whether the frequency variable $k$ +represents the angular ($|s| = 1$) or the physical ($|s| = 2\pi$) +frequency. The sign of $s$ is not so important, but is generally based +on whether the analysis is for forward ($s > 0$) or backward-propagating +($s < 0$) waves. + + +## Derivatives + +The FT of a derivative has a very interesting property. +Below, after integrating by parts, we remove the boundary term by +assuming that $f(x)$ is localized, i.e. $f(x) \to 0$ for $x \to \pm \infty$: + +$$\begin{aligned} + \hat{\mathcal{F}}\{f'(x)\} + &= A \int_{-\infty}^\infty f'(x) \exp(i s k x) \dd{x} + \\ + &= A \big[ f(x) \exp(i s k x) \big]_{-\infty}^\infty - i s k A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x} + \\ + &= (- i s k) \tilde{f}(k) +\end{aligned}$$ + +Therefore, as long as $f(x)$ is localized, the FT eliminates derivatives +of the transformed variable, which makes it useful against PDEs: + +$$\begin{aligned} + \boxed{ + \hat{\mathcal{F}}\{f'(x)\} = (- i s k) \tilde{f}(k) + } +\end{aligned}$$ + +This generalizes to higher-order derivatives, as long as these +derivatives are also localized in the $x$-domain, which is practically +guaranteed if $f(x)$ itself is localized: + +$$\begin{aligned} + \boxed{ + \hat{\mathcal{F}} \Big\{ \dv[n]{f}{x} \Big\} + = (- i s k)^n \tilde{f}(k) + } +\end{aligned}$$ + +Derivatives in the frequency domain have an analogous property: + +$$\begin{aligned} + \boxed{ + \dv[n]{\tilde{f}}{k} + = A \int_{-\infty}^\infty (i s x)^n f(x) \exp(i s k x) \dd{x} + = \hat{\mathcal{F}}\{ (i s x)^n f(x) \} + } +\end{aligned}$$ diff --git a/content/know/concept/gram-schmidt-method/index.pdc b/content/know/concept/gram-schmidt-method/index.pdc new file mode 100644 index 0000000..88488dd --- /dev/null +++ b/content/know/concept/gram-schmidt-method/index.pdc @@ -0,0 +1,47 @@ +--- +title: "Gram-Schmidt method" +firstLetter: "G" +publishDate: 2021-02-22 +categories: +- Mathematics + +date: 2021-02-22T21:36:08+01:00 +draft: false +markup: pandoc +--- + +# Gram-Schmidt method + +Given a set of linearly independent non-orthonormal vectors +$\ket*{V_1}, \ket*{V_2}, ...$ from a [Hilbert space](/know/concept/hilbert-space/), +the **Gram-Schmidt method** +turns them into an orthonormal set $\ket*{n_1}, \ket*{n_2}, ...$ as follows: + +1. Take the first vector $\ket*{V_1}$ and normalize it to get $\ket*{n_1}$: + + $$\begin{aligned} + \ket*{n_1} = \frac{\ket*{V_1}}{\sqrt{\braket*{V_1}{V_1}}} + \end{aligned}$$ + +2. Begin loop. Take the next non-orthonormal vector $\ket*{V_j}$, and + subtract from it its projection onto every already-processed vector: + + $$\begin{aligned} + \ket*{n_j'} = \ket*{V_j} - \ket*{n_1} \braket*{n_1}{V_j} - \ket*{n_2} \braket*{n_2}{V_j} - ... - \ket*{n_{j-1}} \braket*{n_{j-1}}{V_{j-1}} + \end{aligned}$$ + + This leaves only the part of $\ket*{V_j}$ which is orthogonal to + $\ket*{n_1}$, $\ket*{n_2}$, etc. This why the input vectors must be + linearly independent; otherwise $\ket{n_j'}$ may become zero at some + point. + +3. Normalize the resulting ortho*gonal* vector $\ket*{n_j'}$ to make it + ortho*normal*: + + $$\begin{aligned} + \ket*{n_j} = \frac{\ket*{n_j'}}{\sqrt{\braket*{n_j'}{n_j'}}} + \end{aligned}$$ + +4. Loop back to step 2, taking the next vector $\ket*{V_{j+1}}$. + +If you are unfamiliar with this notation, take a look at [Dirac notation](/know/concept/dirac-notation/). diff --git a/content/know/concept/hilbert-space/index.pdc b/content/know/concept/hilbert-space/index.pdc new file mode 100644 index 0000000..1faf08a --- /dev/null +++ b/content/know/concept/hilbert-space/index.pdc @@ -0,0 +1,202 @@ +--- +title: "Hilbert space" +firstLetter: "H" +publishDate: 2021-02-22 +categories: +- Mathematics +- Quantum mechanics + +date: 2021-02-22T21:36:24+01:00 +draft: false +markup: pandoc +--- + +# Hilbert space + +A **Hilbert space**, also known as an **inner product space**, is an +abstract **vector space** with a notion of length and angle. + + +## Vector space + +An abstract **vector space** $\mathbb{V}$ is a generalization of the +traditional concept of vectors as "arrows". It consists of a set of +objects called **vectors** which support the following (familiar) +operations: + ++ **Vector addition**: the sum of two vectors $V$ and $W$, denoted $V + W$. ++ **Scalar multiplication**: product of a vector $V$ with a scalar $a$, denoted $a V$. + +In addition, for a given $\mathbb{V}$ to qualify as a proper vector +space, these operations must obey the following axioms: + ++ **Addition is associative**: $U + (V + W) = (U + V) + W$ ++ **Addition is commutative**: $U + V = V + U$ ++ **Addition has an identity**: there exists a $\mathbf{0}$ such that $V + 0 = V$ ++ **Addition has an inverse**: for every $V$ there exists $-V$ so that $V + (-V) = 0$ ++ **Multiplication is associative**: $a (b V) = (a b) V$ ++ **Multiplication has an identity**: There exists a $1$ such that $1 V = V$ ++ **Multiplication is distributive over scalars**: $(a + b)V = aV + bV$ ++ **Multiplication is distributive over vectors**: $a (U + V) = a U + a V$ + +A set of $N$ vectors $V_1, V_2, ..., V_N$ is **linearly independent** if +the only way to satisfy the following relation is to set all the scalar coefficients $a_n = 0$: + +$$\begin{aligned} + \mathbf{0} = \sum_{n = 1}^N a_n V_n +\end{aligned}$$ + +In other words, these vectors cannot be expressed in terms of each +other. Otherwise, they would be **linearly dependent**. + +A vector space $\mathbb{V}$ has **dimension** $N$ if only up to $N$ of +its vectors can be linearly indepedent. All other vectors in +$\mathbb{V}$ can then be written as a **linear combination** of these $N$ **basis vectors**. + +Let $\vu{e}_1, ..., \vu{e}_N$ be the basis vectors, then any +vector $V$ in the same space can be **expanded** in the basis according to +the unique weights $v_n$, known as the **components** of $V$ +in that basis: + +$$\begin{aligned} + V = \sum_{n = 1}^N v_n \vu{e}_n +\end{aligned}$$ + +Using these, the vector space operations can then be implemented as follows: + +$$\begin{gathered} + V = \sum_{n = 1} v_n \vu{e}_n + \quad + W = \sum_{n = 1} w_n \vu{e}_n + \\ + \quad \implies \quad + V + W = \sum_{n = 1}^N (v_n + w_n) \vu{e}_n + \qquad + a V = \sum_{n = 1}^N a v_n \vu{e}_n +\end{gathered}$$ + + +## Inner product + +A given vector space $\mathbb{V}$ can be promoted to a **Hilbert space** +or **inner product space** if it supports an operation $\braket{U}{V}$ +called the **inner product**, which takes two vectors and returns a +scalar, and has the following properties: + ++ **Skew symmetry**: $\braket{U}{V} = (\braket{V}{U})^*$, where ${}^*$ is the complex conjugate. ++ **Positive semidefiniteness**: $\braket{V}{V} \ge 0$, and $\braket{V}{V} = 0$ if $V = \mathbf{0}$. ++ **Linearity in second operand**: $\braket{U}{(a V + b W)} = a \braket{U}{V} + b \braket{U}{W}$. + +The inner product describes the lengths and angles of vectors, and in +Euclidean space it is implemented by the dot product. + +The **magnitude** or **norm** $|V|$ of a vector $V$ is given by +$|V| = \sqrt{\braket{V}{V}}$ and represents the real positive length of $V$. +A **unit vector** has a norm of 1. + +Two vectors $U$ and $V$ are **orthogonal** if their inner product +$\braket{U}{V} = 0$. If in addition to being orthogonal, $|U| = 1$ and +$|V| = 1$, then $U$ and $V$ are known as **orthonormal** vectors. + +Orthonormality is desirable for basis vectors, so if they are +not already like that, it is common to manually turn them into a new +orthonormal basis using e.g. the [Gram-Schmidt method](/know/concept/gram-schmidt-method). + +As for the implementation of the inner product, it is given by: + +$$\begin{gathered} + V = \sum_{n = 1}^N v_n \vu{e}_n + \quad + W = \sum_{n = 1}^N w_n \vu{e}_n + \\ + \quad \implies \quad + \braket{V}{W} = \sum_{n = 1}^N \sum_{m = 1}^N v_n^* w_m \braket{\vu{e}_n}{\vu{e}_j} +\end{gathered}$$ + +If the basis vectors $\vu{e}_1, ..., \vu{e}_N$ are already +orthonormal, this reduces to: + +$$\begin{aligned} + \braket{V}{W} = \sum_{n = 1}^N v_n^* w_n +\end{aligned}$$ + +As it turns out, the components $v_n$ are given by the inner product +with $\vu{e}_n$, where $\delta_{nm}$ is the Kronecker delta: + +$$\begin{aligned} + \braket{\vu{e}_n}{V} = \sum_{m = 1}^N \delta_{nm} v_m = v_n +\end{aligned}$$ + + +## Infinite dimensions + +As the dimensionality $N$ tends to infinity, things may or may not +change significantly, depending on whether $N$ is **countably** or +**uncountably** infinite. + +In the former case, not much changes: the infinitely many **discrete** +basis vectors $\vu{e}_n$ can all still be made orthonormal as usual, +and as before: + +$$\begin{aligned} + V = \sum_{n = 1}^\infty v_n \vu{e}_n +\end{aligned}$$ + +A good example of such a countably-infinitely-dimensional basis are the +solution eigenfunctions of a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/). + +However, if the dimensionality is uncountably infinite, the basis +vectors are **continuous** and cannot be labeled by $n$. For example, all +complex functions $f(x)$ defined for $x \in [a, b]$ which +satisfy $f(a) = f(b) = 0$ form such a vector space. +In this case $f(x)$ is expanded as follows, where $x$ is a basis vector: + +$$\begin{aligned} + f(x) = \int_a^b \braket{x}{f} \dd{x} +\end{aligned}$$ + +Similarly, the inner product $\braket{f}{g}$ must also be redefined as +follows: + +$$\begin{aligned} + \braket{f}{g} = \int_a^b f^*(x) \: g(x) \dd{x} +\end{aligned}$$ + +The concept of orthonormality must be also weakened. A finite function +$f(x)$ can be normalized as usual, but the basis vectors $x$ themselves +cannot, since each represents an infinitesimal section of the real line. + +The rationale in this case is that action of the identity operator $\hat{I}$ must +be preserved, which is given here in [Dirac notation](/know/concept/dirac-notation/): + +$$\begin{aligned} + \hat{I} = \int_a^b \ket{\xi} \bra{\xi} \dd{\xi} +\end{aligned}$$ + +Applying the identity operator to $f(x)$ should just give $f(x)$ again: + +$$\begin{aligned} + f(x) = \braket{x}{f} = \matrixel{x}{\hat{I}}{f} + = \int_a^b \braket{x}{\xi} \braket{\xi}{f} \dd{\xi} + = \int_a^b \braket{x}{\xi} f(\xi) \dd{\xi} +\end{aligned}$$ + +Since we want the latter integral to reduce to $f(x)$, it is plain to see that +$\braket{x}{\xi}$ can only be a [Dirac delta function](/know/concept/dirac-delta-function/), +i.e $\braket{x}{\xi} = \delta(x - \xi)$: + +$$\begin{aligned} + \int_a^b \braket{x}{\xi} f(\xi) \dd{\xi} + = \int_a^b \delta(x - \xi) f(\xi) \dd{\xi} + = f(x) +\end{aligned}$$ + +Consequently, $\braket{x}{\xi} = 0$ if $x \neq \xi$ as expected for an +orthogonal set of basis vectors, but if $x = \xi$ the inner product +$\braket{x}{\xi}$ is infinite, unlike earlier. + +Technically, because the basis vectors $x$ cannot be normalized, they +are not members of a Hilbert space, but rather of a superset called a +**rigged Hilbert space**. Such vectors have no finite inner product with +themselves, but do have one with all vectors from the actual Hilbert +space. diff --git a/content/know/concept/legendre-transform/index.pdc b/content/know/concept/legendre-transform/index.pdc new file mode 100644 index 0000000..8a0d3e3 --- /dev/null +++ b/content/know/concept/legendre-transform/index.pdc @@ -0,0 +1,89 @@ +--- +title: "Legendre transform" +firstLetter: "L" +publishDate: 2021-02-22 +categories: +- Mathematics +- Physics + +date: 2021-02-22T21:36:35+01:00 +draft: false +markup: pandoc +--- + +# Legendre transform + +The **Legendre transform** of a function $f(x)$ is a new function $L(f')$, +which depends only on the derivative $f'(x)$ of $f(x)$, and from which +the original function $f(x)$ can be reconstructed. The point is, +analogously to other transforms (e.g. [Fourier](/know/concept/fourier-transform/)), +that $L(f')$ contains the same information as $f(x)$, just in a different form. + +Let us choose an arbitrary point $x_0 \in [a, b]$ in the domain of +$f(x)$. Consider a line $y(x)$ tangent to $f(x)$ at $x = x_0$, which has +a slope $f'(x_0)$ and intersects the $y$-axis at $-C$: + +$$\begin{aligned} + y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C +\end{aligned}$$ + +The Legendre transform $L(f')$ is defined such that $L(f'(x_0)) = C$ (or +sometimes $-C$ instead) for all $x_0 \in [a, b]$, where $C$ is the +constant corresponding to the tangent line at $x = x_0$. This yields: + +$$\begin{aligned} + L(f'(x)) = f'(x) \: x - f(x) +\end{aligned}$$ + +We want this function to depend only on the derivative $f'$, but +currently $x$ still appears here as a variable. We fix that problem in +the easiest possible way: by assuming that $f'(x)$ is invertible for all +$x \in [a, b]$. If $x(f')$ is the inverse of $f'(x)$, then $L(f')$ is +given by: + +$$\begin{aligned} + \boxed{ + L(f') = f' \: x(f') - f(x(f')) + } +\end{aligned}$$ + +The only requirement for the existence of the Legendre transform is thus +the invertibility of $f'(x)$ in the target interval $[a,b]$, which can +only be true if $f(x)$ is either convex or concave, i.e. its derivative +$f'(x)$ is monotonic. + |