我需要帮助以正确的方式写出一些带有箭头和文字的方程式，但我毫无头绪。这些是方程式：
这是代码：
\begin{DispWithArrows}[format=c]
\Psi\dl x,t\dr=A\dcos{kx-\omega t}+B\dsin{kx-\omega t} \Arrow[]{}\Arrow[jump= 2]{Obtenemos sus parciales \footnote{En este paso pretendemos buscar una forma de relacionarlos con las ecuaciones \textbf{\ref{eq:}} y \textbf{\ref{eq:}}}}\Arrow[o, tikz={text width=3.3cm}, jump= 4]{Buscamos una E.D.Lineal en $\Psi$\footnote{No puede tener términos cuadráticos y debe ser una función diferenciable} que cumpla la ecuación \textbf{\ref{eq:momcuan}}} \\
\dfrac{\partial \Psi\dl x,t\dr}{\partial t}=-\omega\left[-A\dsin{kx-\omega t}+B\dcos{kx-\omega t}\right]\propto\omega\propto E \notag \\
\dfrac{\partial \Psi\dl x,t\dr}{\partial x}=k\left[-A\dsin{kx-\omega t}+B\dcos{kx-\omega t}\right]\propto k\propto p \Arrow[o, tikz={text width=3.3cm}]{Buscamos $p^2$ \footnote{Esto se debe a que pretendemos hacer que cumpla la ecuación \textbf{\ref{eq:momcuan}}}} \notag \\
\dfrac{\partial^2 \Psi\dl x,t\dr}{\partial x^2}=-k^2\left[A\dsin{kx-\omega t}+B\dcos{kx-\omega t}\right] \propto k^2\propto p^2 \notag \\
\dmathnote{\alpha\dfrac{\partial \Psi\dl x,t\dr}{\partial t}}{E}=\dmathnote{\beta\dfrac{\partial^2 \Psi\dl x,t\dr}{\partial x^2}}{\dfrac{p^2}{2m}}+\dmathnote{V_0\Psi\dl x,t\dr}{V_0} \Arrow[jump= 2]{} \\
\textsc{Buscamos los coeficientes $A$ y $B$} \notag \\
\alpha\left[\underbrace{-\omega\left[-A\dsin{kx-\omega t}+B\dcos{kx-\omega t}\right]}_{\frac{\partial \Psi\dl x,t\dr}{\partial t}}\right]=\beta\left[\underbrace{-k^2\left[A\dsin{kx-\omega t}+B\dcos{kx-\omega t}\right]}_{\frac{\partial^2 \Psi\dl x,t\dr}{\partial x^2}}\right]+V_0\Psi\dl x,t\dr
\Arrow[down]{Reorganizando} \\
\left[-\dfrac{B}{A}\alpha\omega+\beta k^2-V_0\right]A\dcos{kx-\omega t}+\left[\dfrac{A}{B}\alpha\omega+\beta k^2-V_0\right]B\dsin{kx-\omega t}=0 \Arrow[down]{} \\
-\dfrac{B}{A}\alpha\omega+\beta k^2-V_0=\dfrac{A}{B}\alpha\omega+\beta k^2-V_0\longrightarrow B^2=-A^2\Rightarrow |B|=i|A| \footnote{La propia ecuación necesita número complejos para poder aplicar la conservación de la energía de una partícula a una onda} \\
\Psi\dl x,t\dr=A\dcos{kx-\omega t}+iA\dsin{kx-\omega t}\Rightarrow\Psi\dl x,t\dr=A \dexp{i\dl kx-\omega t\dr}
\end{DispWithArrows}
我想让箭头不要向右移动太远。另外，如果你们知道指南或更好的方法来在方程式中做笔记，我将不胜感激。谢谢。
这是我使用的代码：
\documentclass[a4paper, 8pt]{book}
% fuente de letra:
\usepackage{tgcursor}
\renewcommand*\familydefault{\ttdefault} %% Only if the base font of the document is to be typewriter style
\usepackage[T1]{fontenc}
\usepackage[mathscr]{euscript}
% matemáticas:
\usepackage{amsmath, amssymb}
% idioma:
\usepackage[utf8]{inputenc}
\usepackage[spanish, es-tabla]{babel} % 'cuadro' es el título del caption de table por defecto siguiendo indicaciones de RAE; es-tabla lo cambia a 'tabla'
% gestión de párrafos (hace innecesario indicar el salto de línea con doble barra y elimina el indent de todos los párrafos):
\usepackage{parskip}
% enlaces a URLs:
\usepackage[hidelinks]{hyperref}
% gráficos:
\usepackage{graphicx, wrapfig, caption, subcaption}
% tablas:
\usepackage{array, multirow}
\newcommand{\dl}{\left(}
\newcommand{\dr}{\right)}
\newcommand{\dsin}[1]{\sin{\left(#1\right)}}
\newcommand{\dtan}[1]{\tan{\left(#1\right)}}
\newcommand{\dcos}[1]{\cos{\left(#1\right)}}
\newcommand{\dsec}[1]{\sec{\left(#1\right)}}
\newcommand{\dcsc}[1]{\csc{\left(#1\right)}}
\newcommand{\dcot}[1]{\cot{\left(#1\right)}}
\newcommand{\darcsin}[1]{\arcsin{\left(#1\right)}}
\newcommand{\darccos}[1]{\arccos{\left(#1\right)}}
\newcommand{\darctan}[1]{\arctan{\left(#1\right)}}
\newcommand{\dsinh}[1]{\sinh{\left(#1\right)}}
\newcommand{\dcosh}[1]{\cosh{\left(#1\right)}}
\newcommand{\dtanh}[1]{\tanh{\left(#1\right)}}
\newcommand{\dln}[1]{\ln{\left(#1\right)}}
\newcommand{\dexp}[1]{e^{#1}}
\newcommand{\dsum}[2]{\displaystyle\sum_{#1}^{#2}}
\newcommand{\dlim}[2]{\lim_{#1\rightarrow#2}}
\newcommand{\dint}[2]{\displaystyle\int_{#1}^{#2}}
\newcommand{\doint}[2]{\displaystyle\oint_{#1}^{#2}}
\usepackage{thmtools}
\declaretheorem[thmbox=L]{Hipótesis}
\declaretheorem[thmbox=M]{Postulado}
\declaretheorem[thmbox=S]{Teorema}
\NewDocumentCommand{\umathnote}{ mm }
{
\overset % ❶
{% The annotation goes above
\textcolor{black!20!white}{\hbox to 0pt{\hss % ❷
$ % return to math mode
\begin{array}{c} % ❸
\displaystyle #2\\ % ❹
\Big\downarrow % ❺
\end{array}
$
\hss}
}
}
{#1} %
}
\NewDocumentCommand{\dmathnote}{ mm }
{
\underset % ❶
{% The annotation goes above
\textcolor{black!20!white}{\hbox to 0pt{\hss % ❷
$ % return to math mode
\begin{array}{c} % ❸
\Big\uparrow\\ % ❹
\displaystyle #2 % ❺
\end{array}
$
\hss}
}
}
{#1} %
}
\renewcommand{\baselinestretch}{1.35}
\usepackage{titlesec}
\makeatletter
\renewcommand\paragraph{\@startsection{paragraph}{4}{\z@}%
{-2.5ex\@plus -1ex \@minus -.25ex}%
{1.25ex \@plus .25ex}%
{\normalfont\normalsize\bfseries}}
\makeatother
\setcounter{secnumdepth}{4} % how many sectioning levels to assign numbers to
\setcounter{tocdepth}{4} % how many sectioning levels to show in ToC
\usepackage[all]{xy}
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}
\usepackage{tikz}
\usetikzlibrary{calc, arrows}
\usepackage[makeroom]{cancel}
\usepackage{pgfornament}
\NewDocumentCommand{\grand}{ m }
{\begin{center}\large\begin{tikzpicture}[every node/.style={inner sep=0pt}]
\node[text width=14cm,align=center](Text){#1} ;
\node[shift={(-1cm,1cm)},anchor=north west](CNW)
at (Text.north west) {\pgfornament[width=0cm]{88}};
\node[shift={(1cm,1cm)},anchor=north east](CNE)
at (Text.north east) {\pgfornament[width=0cm,symmetry=v]{88}};
\node[shift={(-1cm,-1cm)},anchor=south west](CSW)
at (Text.south west) {\pgfornament[width=0cm,symmetry=h]{88}};
\node[shift={(1cm,-1cm)},anchor=south east](CSE)
at (Text.south east) {\pgfornament[width=0cm,symmetry=c]{88}};
\pgfornamenthline{CNW}{CNE}{north}{88}
\pgfornamenthline{CSW}{CSE}{south}{88}
\pgfornamentvline{CNW}{CSW}{west}{88}
\pgfornamentvline{CNE}{CSE}{east}{88}
\end{tikzpicture}
\end{center}}
\usepackage[framemethod=TikZ]{mdframed}
\newcounter{dem}[chapter]\setcounter{dem}{0}
\renewcommand{\thedem}{\Roman{dem}}
\newenvironment{dem}[2][]{%
\refstepcounter{dem}%
\ifstrempty{#1}%
{\mdfsetup{%
frametitle={%
\tikz[baseline=(current bounding box.east),outer sep=0pt]
\node[anchor=east,rectangle,fill=white]
{\strut Demostración~\thedem};}}
}%
{\mdfsetup{%
frametitle={%
\tikz[baseline=(current bounding box.east),outer sep=0pt]
\node[anchor=east,rectangle,fill=white]
{\begin{minipage}{0.99\linewidth}Demostración~\thedem~#1\end{minipage}};}}%
}%
\mdfsetup{innertopmargin=10pt,linecolor=black,%
linewidth=0.5pt,topline=true,%
frametitleaboveskip=\dimexpr-\ht\strutbox\relax
}
\begin{mdframed}[]\relax%
\label{#2}}{\end{mdframed}}
\usepackage{nccmath}
\usepackage{tcolorbox}
\tcbuselibrary{most}
\newtcolorbox[auto counter,
number within=chapter,
list inside=ej
]{ej}[1][]{%
enhanced, breakable,
title={{\begin{minipage}{\linewidth}\textbf{Ejercicio}~\thetcbcounter.~\textit{#1}\end{minipage}}},
halign title=left,
sharp corners,
colback=white,
coltitle=black,
colbacktitle=white,
boxrule=0pt,frame hidden,
underlay unbroken and first={%
\ifnumequal{\tcbsegmentstate}{0}{
\draw[black,double] (interior.north west)--(interior.south west);
}{\ifnumequal{\tcbsegmentstate}{1}{
\draw[black,double] (interior.north west)--(segmentation.west);
\begin{tcbclipinterior}
\draw[help lines, step=2.1mm, black!10!white](segmentation.south west) grid (frame.south east);
\end{tcbclipinterior}
}{\ifnumequal{\tcbsegmentstate}{2}{
\begin{tcbclipinterior}
\draw[help lines, step=2.1mm, black!10!white](interior.north west) grid (interior.south east);
\end{tcbclipinterior}
}}}
},
underlay middle and last={%
\ifnumequal{\tcbsegmentstate}{0}{
\draw[black,double] (interior.north west)--(interior.south west);
}{\ifnumequal{\tcbsegmentstate}{1}{
\draw[black,double] (interior.north west)--(segmentation.west);
\begin{tcbclipinterior}
\draw[help lines, step=2.1mm, black!10!white](segmentation.south west) grid (frame.south east);
\end{tcbclipinterior}
}{\ifnumequal{\tcbsegmentstate}{2}{
\begin{tcbclipinterior}
\draw[help lines, step=2.1mm, black!10!white](interior.north west) grid (interior.south east);
\end{tcbclipinterior}
}}}
},
boxed title style={%
colframe=white,
boxrule=0pt,
colback=white,
left=0pt,
right=0pt},
attach boxed title to top left={xshift={-5pt}},
lower separated=false,
before lower = {\tcbsubtitle[colback=white, opacityback=0, colframe=black, opacityframe=0, boxrule=1pt, height=1cm, width=2.55cm, valign=center]{\textbf{Solución:}}}
}
\newtcolorbox[auto counter,
number within=chapter,
list inside=defi
]{defi}[1][]{%
enhanced,
title={{\begin{minipage}{0.99\linewidth}\textbf{\textit{#1}}\end{minipage}}},
,
halign title=left,
sharp corners,
colback=white,
coltitle=black,
colbacktitle=white,
boxrule=0pt,frame hidden,
overlay unbroken={%
\draw[black,double] (interior.north west)--(interior.south west);%
},
boxed title style={%
colframe=white,
boxrule=0pt,
colback=white,
left=0pt,
right=0pt},
attach boxed title to top left={xshift={-5pt}},
}
\usepackage{float}
\usepackage{fancyhdr}
\usepackage[Conny]{fncychap}
\usepackage[bottom=2cm,top=2cm,right=1.5cm,left=1cm,binding=1cm]{geometry}
\usepackage[footnote]{witharrows}
答案1
也许是这样的：
\documentclass[a4paper, 8pt]{book}
\usepackage[T1]{fontenc}
\usepackage[spanish, es-tabla]{babel}
\usepackage{fancyhdr}
\usepackage[bottom=2cm,top=2cm,right=1.5cm,left=1cm]{geometry}
\newcommand{\dl}{\left(}
\newcommand{\dr}{\right)}
\newcommand{\dcos}[1]{\cos{\left(#1\right)}}
\newcommand{\dsin}[1]{\sin{\left(#1\right)}}
\usepackage{mathtools}
\usepackage[footnote]{witharrows}
\usetikzlibrary{calc}
\NewDocumentCommand{\dmathnote}{ mm }
{
\underset % ❶
{% The annotation goes above
\textcolor{black!20!white}{\hbox to 0pt{\hss % ❷
$ % return to math mode
\begin{array}{c} % ❸
\Big\uparrow\\ % ❹
\displaystyle #2 % ❺
\end{array}
$
\hss}
}
}
{#1} %
}
\newcommand{\dexp}[1]{e^{#1}}
\begin{document}
\begin{DispWithArrows}[format=l,groups]
\Psi\dl x,t\dr=A\dcos{kx-\omega t}+B\dsin{kx-\omega t} \Arrow[]{}\Arrow[jump= 2]{Obtenemos\\ sus parciales \footnote{En este paso pretendemos buscar una forma de relacionarlos con las ecuaciones \textbf{\ref{eq:}} y \textbf{\ref{eq:}}}}\Arrow[o,tikz={text width=3.3cm}, jump= 4]{Buscamos una E.D.Lineal en $\Psi$\footnote{No puede tener términos cuadráticos y debe ser una función diferenciable} que cumpla la ecuación \textbf{\ref{eq:momcuan}}} \\
\dfrac{\partial \Psi\dl x,t\dr}{\partial t}=-\omega\left[-A\dsin{kx-\omega t}+B\dcos{kx-\omega t}\right]\propto\omega\propto E \notag \\
\dfrac{\partial \Psi\dl x,t\dr}{\partial x}=k\left[-A\dsin{kx-\omega t}+B\dcos{kx-\omega t}\right]\propto k\propto p
\Arrow{Buscamos $p^2$ \footnote{Esto se debe a que pretendemos hacer que cumpla la ecuación \textbf{\ref{eq:momcuan}}}} \notag \\
\dfrac{\partial^2 \Psi\dl x,t\dr}{\partial x^2}=-k^2\left[A\dsin{kx-\omega t}+B\dcos{kx-\omega t}\right] \propto k^2\propto p^2 \notag \\
\dmathnote{\alpha\dfrac{\partial \Psi\dl x,t\dr}{\partial t}}{E}=\dmathnote{\beta\dfrac{\partial^2 \Psi\dl x,t\dr}{\partial x^2}}{\dfrac{p^2}{2m}}+\dmathnote{V_0\Psi\dl x,t\dr}{V_0} \\
\textsc{Buscamos los coeficientes $A$ y $B$} \notag \\
\alpha\left[\underbrace{-\omega\left[-A\dsin{kx-\omega t}+B\dcos{kx-\omega t}\right]}_{\frac{\partial \Psi\dl x,t\dr}{\partial t}}\right]=\beta\left[\underbrace{-k^2\left[A\dsin{kx-\omega t}+B\dcos{kx-\omega t}\right]}_{\frac{\partial^2 \Psi\dl x,t\dr}{\partial x^2}}\right]+V_0\Psi\dl x,t\dr
\Arrow[down]{Reorganizando} \\
\left[-\dfrac{B}{A}\alpha\omega+\beta k^2-V_0\right]A\dcos{kx-\omega t}+\left[\dfrac{A}{B}\alpha\omega+\beta k^2-V_0\right]B\dsin{kx-\omega t}=0 \Arrow[down]{} \\
-\dfrac{B}{A}\alpha\omega+\beta k^2-V_0=\dfrac{A}{B}\alpha\omega+\beta k^2-V_0\longrightarrow B^2=-A^2\Rightarrow |B|=i|A| \footnote{La propia ecuación necesita número complejos para poder aplicar la conservación de la energía de una partícula a una onda} \\
\Psi\dl x,t\dr=A\dcos{kx-\omega t}+iA\dsin{kx-\omega t}\Rightarrow\Psi\dl x,t\dr=A \dexp{i\dl kx-\omega t\dr}
\end{DispWithArrows}
\end{document}
