帮助从长方程发展的角度看

Question

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}

Answer 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}