--- jupytext: formats: ipynb,md:myst,py:light text_representation: extension: .md format_name: myst format_version: 0.13 jupytext_version: 1.16.4 kernelspec: display_name: Python 3 language: python name: python3 --- ```{code-cell} :tags: [hide-cell] import mmf_setup; mmf_setup.nbinit() import os from pathlib import Path FIG_DIR = Path(mmf_setup.ROOT) / '../Docs/_build/figures/' os.makedirs(FIG_DIR, exist_ok=True) import logging; logging.getLogger("matplotlib").setLevel(logging.CRITICAL) %matplotlib inline import numpy as np, matplotlib.pyplot as plt try: from myst_nb import glue except: glue = None ``` (sec:DeltaFunction)= Dirac Delta Function ==================== Physicists often use Dirac's [delta function][] $\delta(x)$ to express various quantities. Here are various formulations with increasing degree of rigor. ## As a Function ```{code-cell} :tags: [margin, hide-input] x = [-1, 0, 0, 0, 1] y = [0, 0, 1, 0, 0] fig, ax = plt.subplots(figsize=(3, 2)) ax.plot(x, y, 'C0-', zorder=0) ax.scatter([0], [0], edgecolors='C0', facecolors='w', zorder=1) ax.set(xlabel="$x$", xticks=[-1, 0, 1], ylabel=r"$\delta(x)$", yticks=[0,1], yticklabels=["$0$", r"$\infty$"]); ``` We often treat $\delta(x)$ as a function \begin{gather*} \delta(x) = \begin{cases} \infty & x = 0,\\ 0 & x \neq 0, \end{cases} \end{gather*} where the value $\infty$ at $x=0$ is such that the area is 1. \begin{gather*} \int_{-\epsilon}^{\epsilon} \delta(x)\d{x} = 1, \qquad \int_{-\infty}^{\infty} f(x)\delta(x-x_0)\d{x} = f(x_0). \end{gather*} Well, convenient, this representation is incomplete a $\delta(x)$ is not a function in any real sense. :::{admonition} Counterexample Consider the meaning of $[\delta(x)]^2$. ::: ```{code-cell} :tags: [margin, hide-cell] from IPython.display import HTML, Javascript, display from matplotlib import animation plt.rcParams["animation.html"] = "jshtml" epsinvs = np.arange(1, 16) fig, ax = plt.subplots(figsize=(4/2, 8/2)) def animate(epsinv): global ax eps = 1/epsinv ax.cla() ax.set(xlim=(-2, 2), xlabel="$x$", xticks=[-2, -1, 0, 1, 2], ylim=(-1, 8), ylabel=r"$\delta_{\epsilon}(x)$", title=fr"$\epsilon = 1/{epsinv}$") x = np.linspace(-2, 2, 100) y = np.exp(-(x/eps)**2)/np.sqrt(eps**2*np.pi) ax.plot(x, y, 'C0-') y = np.sin(x/eps)/np.pi/x ax.plot(x, y, 'C2-') y = np.sin((1/eps+0.5)*x)/np.sin(x/2)/2/np.pi ax.plot(x, y, 'C3-') y = 1/(1+(x/eps)**2)/eps/np.pi ax.plot(x, y, 'C4-') h = 2*eps ax.plot([-2, -h], [0, 0], 'C1-') ax.plot([-h, h], [1/h, 1/h], 'C1-') ax.plot([h, 2], [0, 0], 'C1-') ax.plot([-h, -h], [0, 1/h], 'C1:') ax.plot([h, h], [0, 1/h], 'C1:') return ax, anim = animation.FuncAnimation(fig, animate, repeat=True, frames=epsinvs, interval=500) # To save the animation using Pillow as a gif anim.save(FIG_DIR / 'delta_function1.gif', writer=animation.PillowWriter(fps=5)) ``` :::{figure} figures/delta_function1.gif --- figclass: margin --- Representing $\delta(x) = \lim_{\epsilon\rightarrow 0}\delta_{\epsilon}(x)$ as a limit of functions. The step function here has $\epsilon \rightarrow 2\epsilon$ for better visual comparison, otherwise the formula are as given in the text. ::: ## As a Limit or Distribution One way to make this idea rigorous is to think of $\delta(x)$ as a limit of appropriate functions where the limit is taken outside of the calculation. For example \begin{gather*} \delta(x) \equiv \lim_{\epsilon\rightarrow 0} \delta_{\epsilon}(x), \qquad \int\delta(x)f(x)\d{x} = \lim_{\epsilon\rightarrow 0}\int\delta_{\epsilon}(x)f(x)\d{x}. \end{gather*} Some suitable examples of $\delta_{\epsilon}(x)$ sequences are: \begin{align*} \delta_{\epsilon}(x) &= \frac{1}{\abs{\epsilon}}\begin{cases} 1 & \abs{x} < \epsilon,\\ 0 & \text{otherwise}, \end{cases},\\ \delta_{\epsilon}(x) &= \frac{1}{\abs{\epsilon}\sqrt{\pi}}e^{-(x/\epsilon)^2},\\ \delta_{\epsilon}(x) &= \frac{1}{\abs{\epsilon}\pi}\frac{1}{1+(x/\epsilon)^2},\\ \delta_{\epsilon}(x) &= \frac{\sin x/\epsilon}{\pi x} = \frac{1}{2\pi}\int_{-\epsilon^{-1}}^{\epsilon^{-1}}e^{\I k x}\d{k}. \end{align*} The last form is sometimes expressed as the Dirichlet kernel \begin{gather*} \delta_{1/n}(x) = \frac{1}{2\pi}\frac{\sin\bigl[(n+\tfrac{1}{2})x\bigr]} {\sin(\tfrac{1}{2}x)}. \end{gather*} :::{margin} What are the conditions on $\delta_{\epsilon}(x)$ and $f(x)$ so that this statement is true? ::: If the limit exists, then it will be independent of the form of $\delta_{\epsilon}(x)$. Interpreted this way, $\delta(x)$ is sometimes called a [distribution][]. ## Riemann-Stieltjes Integral Another rigorous definition is in terms of a [Riemann-Stieltjes Integral](): \begin{gather*} \int f(x)\d{g(x)} = \int f(x)g'(x)\d{x}, \end{gather*} where $g(x)$ is differentiable. The [Riemann-Stieltjes Integral]() remains valid even $g(x)$ is discontinuous, and the delta-function can be expressed in terms of the [Heaviside step function][] $H(x)$: :::{margin} The value of $H(0) = 1/2$ here is a convention that is useful in Fourier analysis. For the purposes of our discussion, it does not matter, and you will sometimes see \begin{gather*} H(x) = \begin{cases} 1 & x \geq 0,\\ 0 & x < 0. \end{cases} \end{gather*} ::: \begin{gather*} f(x_0) = \int f(x) \d{H(x-x_0)}, \qquad H(x) = \begin{cases} 1 & x >0,\\ \tfrac{1}{2} & x = 0,\\ 0 & x < 0. \end{cases} \end{gather*} Note that, informally, $\delta(x) = H'(x)$, reproducing the informal notation above. :::::{admonition} Example In class, we discussed the informal result \begin{gather*} \delta\bigl(g(x)-g_0)\bigr) = \sum_{x_i \in g^{-1}(g_0)}\frac{\delta(x-x_i)}{\abs{g'(x_i)}},\\ \int f(x)\delta\bigl(g(x)-g_0)\bigr)\d{x} = \sum_{x_i \in g^{-1}(g_0)} \frac{f(x_i)}{\abs{g'(x_i)}},\\ g^{-1}(g_0) = \{x_i \; |\; g(x_i) = g_0\}. \end{gather*} We can express this as \begin{gather*} \int f(x)\d{M(x)}, \qquad M(x) = \sum_{x_i \in g^{-1}(g_0)}\frac{H(x-x_i)}{\abs{g'(x_i)}}. \end{gather*} ::::: Kevin has some [additional notes (PDF)](VixieCommentOnDeltaFunctions.pdf). [delta function]: [Riemann-Stieltjes Integral]: [Heaviside step function]: [distribution]: