`\tag` 和 `\Rightarrow`

Question 1

您可以使用

\renewcommand\theequation{\thesection.\arabic{equation}}

对于 Q1。对于 Q2，我将使用mathtoolsamsmath 而不是其\mathllap：

\mathllap{\Rightarrow} \qquad     \hat{H}  & = i \hbar \frac{\partial}{\partial t} \label{eq:yours}

\label此外，您可以使用andref机制并 let 来完成这项工作，而不是对方程式进行硬编码引用LaTeX。此外，最好使用\eqref而不是简单的\ref。如果您想再进一步，请阅读（学习）并使用cleveref包。

但是，正如 Sigur 所建议的，您似乎需要初步了解一下的能力LaTeX。我在这里就不多说了，我建议您尽量充分利用这位巨人的力量LaTeX。

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools}
%\usepackage{fullpage}   %% Use geometry package instead
\usepackage{secdot}
\usepackage{setspace}

\setlength\parindent{0pt}

\title{\vspace{-3em} Quantum Mechanics}  %% there are other good ways of removing space here. Search this site for examples.
\author{David G}
\renewcommand\theequation{\thesection.\arabic{equation}}
\begin{document}
\maketitle

\section{Operators In Schroedinger Equation}
\onehalfspacing
In quantum mechanics, a question that concerns us all is whether light is particles or waves. Let's first assume light is wave, so it must satisfy the wave equation:
\begin{align}
    \left (\frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}+\frac{\partial^2 \psi}{\partial z^2} \right)-
        \frac{1}{c^2}  \frac{\partial^2 \psi}{\partial t^2} & = 0 \\
    \nabla^2 \psi - \frac{1}{c^2}  \frac{\partial^2 \psi}{\partial t^2} & = 0 
\end{align}

Naturally, we construct the solution of this wave equation:
\begin{align}
\psi(x,y,z;t) = \psi_{0} \, e^{i \, [\, k (x+y+z)- \omega t + \phi_{0}\, ]} 
\end{align}

Here, we can set the initial condition $\phi_{0} = 0$, and we would like to explore the 1D case:
\begin{align}
    \psi(x,t) = \psi_{x,0} \, e^{i \, (kx- \omega t)} 
\end{align}
Now, we lay down all the foundation, and we want to express the momentum and energy operators, by using the solution of the wave function. By observing the equation:
\begin{align}\label{eq:energy}
E = \frac{p^2}{2m} + V 
\end{align}
We want to express the momentum $p$ as a differential operator.
\begin{align}
       \frac{\partial \psi}{\partial x}  &= i k \psi
                                          = i \frac{p}{h} \psi \\
                               p \, \psi &= - i \hbar \frac{\partial}{\partial x} \psi \notag\\
                                 \hat{p} &= - i \hbar \frac{\partial}{\partial x} \label{eq:momop}
\end{align}
\pagebreak

Plug equation~\eqref{eq:momop} into~\eqref{eq:energy}, we get:
\begin{align}\label{eq:mine}
        \hat{H} = - \frac{\hbar ^2}{2 m} \frac{\partial ^2}{\partial x^2} + \hat{V} 
\end{align}
, where $\hat{H}$ is Hamiltonian Operator, namely the Energy Operator.\\
To this point, we only explore the property of $\frac{\partial \psi}{\partial x}$. A natural thought right afterwards is to see $\frac{\partial \psi}{\partial t}$.
\begin{align}
         \frac{\partial \psi}{\partial t} & = - i \omega \psi \\
                                          & = - i \frac{E}{\hbar} \psi \notag \\
  \mathllap{\Rightarrow} \qquad     \hat{H}  & = i \hbar \frac{\partial}{\partial t} \label{eq:yours}
\end{align}
Now, by observing equation~\eqref{eq:mine} and~\eqref{eq:yours}, we derive that the wave function of one particle must satisfy the \textbf{Schroedinger Equation}:
\begin{align}
    \boxed{ i \hbar \frac{\partial}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial ^2}{\partial x^2} + \hat{V} } 
\end{align}
\end{document}

Answer

您可以使用

\renewcommand\theequation{\thesection.\arabic{equation}}

对于 Q1。对于 Q2，我将使用mathtoolsamsmath 而不是其\mathllap：

\mathllap{\Rightarrow} \qquad     \hat{H}  & = i \hbar \frac{\partial}{\partial t} \label{eq:yours}

\label此外，您可以使用andref机制并 let 来完成这项工作，而不是对方程式进行硬编码引用LaTeX。此外，最好使用\eqref而不是简单的\ref。如果您想再进一步，请阅读（学习）并使用cleveref包。

但是，正如 Sigur 所建议的，您似乎需要初步了解一下的能力LaTeX。我在这里就不多说了，我建议您尽量充分利用这位巨人的力量LaTeX。

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools}
%\usepackage{fullpage}   %% Use geometry package instead
\usepackage{secdot}
\usepackage{setspace}

\setlength\parindent{0pt}

\title{\vspace{-3em} Quantum Mechanics}  %% there are other good ways of removing space here. Search this site for examples.
\author{David G}
\renewcommand\theequation{\thesection.\arabic{equation}}
\begin{document}
\maketitle

\section{Operators In Schroedinger Equation}
\onehalfspacing
In quantum mechanics, a question that concerns us all is whether light is particles or waves. Let's first assume light is wave, so it must satisfy the wave equation:
\begin{align}
    \left (\frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}+\frac{\partial^2 \psi}{\partial z^2} \right)-
        \frac{1}{c^2}  \frac{\partial^2 \psi}{\partial t^2} & = 0 \\
    \nabla^2 \psi - \frac{1}{c^2}  \frac{\partial^2 \psi}{\partial t^2} & = 0 
\end{align}

Naturally, we construct the solution of this wave equation:
\begin{align}
\psi(x,y,z;t) = \psi_{0} \, e^{i \, [\, k (x+y+z)- \omega t + \phi_{0}\, ]} 
\end{align}

Here, we can set the initial condition $\phi_{0} = 0$, and we would like to explore the 1D case:
\begin{align}
    \psi(x,t) = \psi_{x,0} \, e^{i \, (kx- \omega t)} 
\end{align}
Now, we lay down all the foundation, and we want to express the momentum and energy operators, by using the solution of the wave function. By observing the equation:
\begin{align}\label{eq:energy}
E = \frac{p^2}{2m} + V 
\end{align}
We want to express the momentum $p$ as a differential operator.
\begin{align}
       \frac{\partial \psi}{\partial x}  &= i k \psi
                                          = i \frac{p}{h} \psi \\
                               p \, \psi &= - i \hbar \frac{\partial}{\partial x} \psi \notag\\
                                 \hat{p} &= - i \hbar \frac{\partial}{\partial x} \label{eq:momop}
\end{align}
\pagebreak

Plug equation~\eqref{eq:momop} into~\eqref{eq:energy}, we get:
\begin{align}\label{eq:mine}
        \hat{H} = - \frac{\hbar ^2}{2 m} \frac{\partial ^2}{\partial x^2} + \hat{V} 
\end{align}
, where $\hat{H}$ is Hamiltonian Operator, namely the Energy Operator.\\
To this point, we only explore the property of $\frac{\partial \psi}{\partial x}$. A natural thought right afterwards is to see $\frac{\partial \psi}{\partial t}$.
\begin{align}
         \frac{\partial \psi}{\partial t} & = - i \omega \psi \\
                                          & = - i \frac{E}{\hbar} \psi \notag \\
  \mathllap{\Rightarrow} \qquad     \hat{H}  & = i \hbar \frac{\partial}{\partial t} \label{eq:yours}
\end{align}
Now, by observing equation~\eqref{eq:mine} and~\eqref{eq:yours}, we derive that the wave function of one particle must satisfy the \textbf{Schroedinger Equation}:
\begin{align}
    \boxed{ i \hbar \frac{\partial}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial ^2}{\partial x^2} + \hat{V} } 
\end{align}
\end{document}

Question 2

这里有一种方法，使用chngcntr包，将节号放在方程式编号中，mathtools更改标记的格式（粗体数字）cleveref并将对象类型（此处为方程式）合并到标签中，而无需键入单词“方程式”。我没有使用不必要的secdot包（我的系统上没有安装），而是用fullpage（同样的原因）替换。我还用（我认为尺寸更好）geometry替换了几乎所有\hat命令；同样，我用在序言中定义的命令替换了最后一个命令。在文档的末尾，我使用了包中的命令（中等大小的分数），因为我认为文本中的分数确实太小了。\widehat\boxed\widebox\mfracnccmath

对于第二个问题，我认为最好写一些短语，例如whence:。将其放在等式的左边一点，使用 \ flalign environment and the\rlap` 命令使等式保持居中。

对于 seconf \documentclass[a4paper,12pt]{article} \usepackage{mathtools} \newtagform{bold}[\textbf]{(}{)} \usetagform{bold} \usepackage{nccmath} \usepackage[margin = 1.5cm]{geometry} %\usepackage{fullpage} \usepackage{chngcntr} \counterwithin{equation}{section} %\usepackage{secdot} \usepackage{setspace}

\usepackage[noabbrev]{cleveref}
\creflabelformat{equation}{#1#2#3}
\newcommand \widebox[1]{\setlength\fboxsep{6pt}\boxed {\enspace#1\enspace}}
\setlength\parindent{0pt}

\title{\vspace{-3em} Quantum Mechanics}
\author{David G}

\begin{document}
\maketitle

\section{Operators In Schroedinger Equation}
\onehalfspacing
In quantum mechanics, a question that concerns us all is whether light is particles or waves. Let's first assume light is wave, so it must satisfy the wave equation:
\begin{align}
    \left (\frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}+\frac{\partial^2 \psi}{\partial z^2} \right)-
        \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} & = 0 \\
    \nabla^2 \psi - \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} & = 0
\end{align}

Naturally, we construct the solution of this wave equation:
\begin{equation}
  \psi(x,y,z;t) = \psi_{0} \, e^{i [ k (x+y+z)- \omega t + \phi_{0} ]}
\end{equation}

Here, we can set the initial condition $\phi_{0} = 0$, and we would like to explore the 1D case:
\begin{equation}
    \psi(x,t) = \psi_{x,0} \, e^{i (kx- \omega t)}
\end{equation}
Now, we lay down all the foundation, and we want to express the momentum and energy operators, by using the solution of the wave function. By observing the equation:
\begin{equation}\label{energy}
E = \frac{p^2}{2m} + V
\end{equation}
We want to express the momentum $p$ as a differential operator.
\begin{align}
       \frac{\partial \psi}{\partial x} &= i k \psi = i \frac{p}{h} \psi \\
                               p \,\psi &= - i \hbar \frac{\partial}{\partial x} \psi \notag \\%
                                 \hat{p} &= - i \hbar \frac{\partial}{\partial x} \label{momentum}
\end{align}
\pagebreak

Plug \cref{momentum} into \ref{energy}, we get:
\begin{equation}\label{ham}
        \widehat{H} = - \frac{\hbar ^2}{2 m} \frac{\partial ^2}{\partial x^2} + \widehat{V},
\end{equation}
where $\widehat{H}$ is Hamiltonian Operator, namely the Energy Operator.



    To this point, we only explore the property of $\mfrac{\partial \psi}{\partial x}$. A natural thought right afterwards is to see $\mfrac{\partial \psi}{\partial t}$.
    \begin{align}
   & & \frac{\partial \psi}{\partial t} & = - i \omega \psi = - i \frac{E}{\hbar} \psi & & \\
 \text{\rlap{whence: }} &{} & \widehat{H} & = i \hbar \frac{\partial}{\partial t} \label{newham}
    \end{align}
    Now, by observing \cref{ham,newham}, we derive that the wave function of one particle must satisfy the \textbf{Schroedinger Equation}:
    \begin{equation}
        \widebox{ i \hbar \frac{\partial}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial ^2}{\partial x^2} + \widehat{V}. }
\end{equation}


\end{document}

在此处输入图片描述

Answer

这里有一种方法，使用chngcntr包，将节号放在方程式编号中，mathtools更改标记的格式（粗体数字）cleveref并将对象类型（此处为方程式）合并到标签中，而无需键入单词“方程式”。我没有使用不必要的secdot包（我的系统上没有安装），而是用fullpage（同样的原因）替换。我还用（我认为尺寸更好）geometry替换了几乎所有\hat命令；同样，我用在序言中定义的命令替换了最后一个命令。在文档的末尾，我使用了包中的命令（中等大小的分数），因为我认为文本中的分数确实太小了。\widehat\boxed\widebox\mfracnccmath

对于第二个问题，我认为最好写一些短语，例如whence:。将其放在等式的左边一点，使用 \ flalign environment and the\rlap` 命令使等式保持居中。

对于 seconf \documentclass[a4paper,12pt]{article} \usepackage{mathtools} \newtagform{bold}[\textbf]{(}{)} \usetagform{bold} \usepackage{nccmath} \usepackage[margin = 1.5cm]{geometry} %\usepackage{fullpage} \usepackage{chngcntr} \counterwithin{equation}{section} %\usepackage{secdot} \usepackage{setspace}

\usepackage[noabbrev]{cleveref}
\creflabelformat{equation}{#1#2#3}
\newcommand \widebox[1]{\setlength\fboxsep{6pt}\boxed {\enspace#1\enspace}}
\setlength\parindent{0pt}

\title{\vspace{-3em} Quantum Mechanics}
\author{David G}

\begin{document}
\maketitle

\section{Operators In Schroedinger Equation}
\onehalfspacing
In quantum mechanics, a question that concerns us all is whether light is particles or waves. Let's first assume light is wave, so it must satisfy the wave equation:
\begin{align}
    \left (\frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}+\frac{\partial^2 \psi}{\partial z^2} \right)-
        \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} & = 0 \\
    \nabla^2 \psi - \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} & = 0
\end{align}

Naturally, we construct the solution of this wave equation:
\begin{equation}
  \psi(x,y,z;t) = \psi_{0} \, e^{i [ k (x+y+z)- \omega t + \phi_{0} ]}
\end{equation}

Here, we can set the initial condition $\phi_{0} = 0$, and we would like to explore the 1D case:
\begin{equation}
    \psi(x,t) = \psi_{x,0} \, e^{i (kx- \omega t)}
\end{equation}
Now, we lay down all the foundation, and we want to express the momentum and energy operators, by using the solution of the wave function. By observing the equation:
\begin{equation}\label{energy}
E = \frac{p^2}{2m} + V
\end{equation}
We want to express the momentum $p$ as a differential operator.
\begin{align}
       \frac{\partial \psi}{\partial x} &= i k \psi = i \frac{p}{h} \psi \\
                               p \,\psi &= - i \hbar \frac{\partial}{\partial x} \psi \notag \\%
                                 \hat{p} &= - i \hbar \frac{\partial}{\partial x} \label{momentum}
\end{align}
\pagebreak

Plug \cref{momentum} into \ref{energy}, we get:
\begin{equation}\label{ham}
        \widehat{H} = - \frac{\hbar ^2}{2 m} \frac{\partial ^2}{\partial x^2} + \widehat{V},
\end{equation}
where $\widehat{H}$ is Hamiltonian Operator, namely the Energy Operator.



    To this point, we only explore the property of $\mfrac{\partial \psi}{\partial x}$. A natural thought right afterwards is to see $\mfrac{\partial \psi}{\partial t}$.
    \begin{align}
   & & \frac{\partial \psi}{\partial t} & = - i \omega \psi = - i \frac{E}{\hbar} \psi & & \\
 \text{\rlap{whence: }} &{} & \widehat{H} & = i \hbar \frac{\partial}{\partial t} \label{newham}
    \end{align}
    Now, by observing \cref{ham,newham}, we derive that the wave function of one particle must satisfy the \textbf{Schroedinger Equation}:
    \begin{equation}
        \widebox{ i \hbar \frac{\partial}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial ^2}{\partial x^2} + \widehat{V}. }
\end{equation}


\end{document}

在此处输入图片描述

`\tag` 和 `\Rightarrow`

答案1

答案2

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