\documentclass[a4paper,12pt]{article}
\usepackage{amsmath}
\usepackage{fullpage}
\usepackage{secdot}
\usepackage{setspace}
\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 \tag{\textbf{1.1}}\\
\nabla^2 \psi - \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} & = 0 \tag{\textbf{1.2}}
\end{align*}
Naturally, we construct the solution of this wave equation:
\[
\psi(x,y,z;t) = \psi_{0} \, e^{i \, [\, k (x+y+z)- \omega t + \phi_{0}\, ]} \tag{\textbf{1.3}}
\]
Here, we can set the initial condition $\phi_{0} = 0$, and we would like to explore the 1D case:
\[
\psi(x,t) = \psi_{x,0} \, e^{i \, (kx- \omega t)} \tag{\textbf{1.4}}
\]
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:
\[
E = \frac{p^2}{2m} + V \tag{\textbf{1.5}}
\]
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 \tag{\textbf{1.6}} \\
p \, \psi &= - i \hbar \frac{\partial}{\partial x} \psi \\
\hat{p} &= - i \hbar \frac{\partial}{\partial x} \tag{\textbf{1.7}}
\end{align*}
\pagebreak
Plug equation $1.7$ into $1.5$, we get:
\[
\hat{H} = - \frac{\hbar ^2}{2 m} \frac{\partial ^2}{\partial x^2} + \hat{V} \tag{\textbf{1.8}}
\]
, 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 \tag{\textbf{1.9}}
= - i \frac{E}{\hbar} \psi \\
\hat{H} & = i \hbar \frac{\partial}{\partial t} \tag{\textbf{1.10}}
\end{align*}
Now, by observing equation $1.8$ and $1.10$, we derive that the wave function of one particle must satisfy the \textbf{Schroedinger Equation}:
\[
\boxed{ i \hbar \frac{\partial}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial ^2}{\partial x^2} + \hat{V} } \tag{\textbf{1.11}}
\]
\end{document}
我对我的代码有两个疑问：
有没有一种简单的方法可以计算
\tag我的方程式？比如，一些命令会自动计算tag(1.1)、(1.2)...等等。我不想每次都输入那些方程式编号。\begin{align*} \frac{\partial \psi}{\partial t} & = - i \omega \psi \tag{\textbf{1.9}} = - i \frac{E}{\hbar} \psi \\ \hat{H} & = i \hbar \frac{\partial}{\partial t} \tag{\textbf{1.10}} \end{align*}
在这段代码中，我想\Rightarrow在等式 1.10 的最左边添加一个，该怎么做？另外，我也不确定最左边的位置是不是最好看，大家可以帮我调整一下位置，只要好看就行。
- 任何关于我的代码的建议，比如冗余或改进，都欢迎。我希望每次都能有所进步！
答案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}
答案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}
