# \tag and \Rightarrow

\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}


I have two questions about my code:

1. Is there an easy way to \tag my equation? Like, some command automatically tag (1.1), (1.2)...and so on. I don't want to type each time those equation number.

2. \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*}

In this code, I want to add \Rightarrow at the leftmost side in the line of equation 1.10, how to do it? Besides, I am not sure if the position of the leftmost will be the nicest, you are welcome to adjust the right position for me, as long as it looks nice.

1. Any other suggestions on my code, like redundancies or improvement, are very welcome. I want to improve every time!
-
For Q1, use \begin{equation}\end{equation}. – Sigur Jun 24 '14 at 23:15
@Sigur But it will not show (1.1) but(1). – Lawerance Jun 24 '14 at 23:16
You can set up this. \numberwithin{equation}{section} in preamble. – Sigur Jun 24 '14 at 23:17
@Sigur Can you refer me to an actual example that contains the way of setting this up? – Lawerance Jun 24 '14 at 23:20
I also recommend you to learn how to use cross references. You don't need to cite the equation number by hand. Read some basic tutorial. There are dozens over the Internet. – Sigur Jun 24 '14 at 23:25

You can use

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


For Q1. For Q2, I would use mathtools instead of amsmath and use its \mathllap:

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


Further, instead of hard coding references to equations, you can use the \label and ref mechanism and let LaTeX do the job. Also, it is good to use \eqref instead of plain \ref. If you want go go an extra mile, read (study) and use cleveref package.

However, it seems that some initial reading about capabilities of LaTeX would be essential for you as suggested by Sigur. I will stop here with the advice that you should try to make full use of the powers of a giant called 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}

-
Great help, thanks! – Lawerance Jun 25 '14 at 2:43
For your comment, why you think \vspace{-3em} is not good? – Lawerance Jun 25 '14 at 2:44
@Lawerance I meant there are other better ways semantically! – Harish Kumar Jun 26 '14 at 1:10

Here is a way, with the chngcntr package, to have the section number in the equation number, mathtools to change the formatting of tags (boldface numbers) and cleveref to incorporate te type of object (here equations) into the label, without having to type the word ‘equation’. I didn't use the unessential secdot package (not installed on my system), and replaced fullpage with geometry (same reason). I also replaced almost all \hat commands with \widehat (better sized in my opinion); likewise I replaced the final \boxed command with a \widebox defined in preamble. Towards the end of your document, I used the \mfrac command (medium-sized fractions), from the nccmath package, as I think in-text fractions are really too small.

For the second question, I think it's better to write some short phrase, such as whence:. Placed it slightly on the left of the equation, using the \ flalignenvironment and the\rlap command to keep the equations centred.

For the 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:
$$\psi(x,y,z;t) = \psi_{0} \, e^{i [ k (x+y+z)- \omega t + \phi_{0} ]}$$

Here, we can set the initial condition $\phi_{0} = 0$, and we would like to explore the 1D case:
$$\psi(x,t) = \psi_{x,0} \, e^{i (kx- \omega t)}$$
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:
$$\label{energy} E = \frac{p^2}{2m} + V$$
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:
$$\label{ham} \widehat{H} = - \frac{\hbar ^2}{2 m} \frac{\partial ^2}{\partial x^2} + \widehat{V},$$
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}:
$$\widebox{ i \hbar \frac{\partial}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial ^2}{\partial x^2} + \widehat{V}. }$$

\end{document}
`

-
Great help! But can you reedit your post, because you left some code text format. – Lawerance Jun 25 '14 at 2:38