Simple Programming Examples: September 2019

Testing MathJax Example 2 (autonumbering) This is the test of equation \ref{num1} (but ( eqref(num1) )\eqref{num1} also works) \begin{equation} \sqrt{b^2} = \pm b \label{num1} \end{equation} In equation \eqref{eq:sample}, we find the value of an interesting integral: \begin{equation} \int_0^\infty \frac{x^3}{e^x-1}\,dx = \frac{\pi^4}{15} \label{eq:sample} \end{equation}


<head> 
  <title>Testing MathJax Example 2 (autonumbering)</title>
  <script src="https://polyfill.io/v3/polyfill.min.js?features=es6"></script>
  <script>
  MathJax = {
    tex: {
      tags: 'ams'  // should be 'ams', 'none', or 'all'
    }
  };
  </script>
  <script id="MathJax-script" async src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-chtml.js"></script>
</head>

<body>
This is the test of equation \ref{num1} (but  ( eqref(num1) )\eqref{num1} also works)
\begin{equation}
  \sqrt{b^2} = \pm b 
  \label{num1}
\end{equation}

In equation \eqref{eq:sample}, we find the value of an
interesting integral:

\begin{equation}
  \int_0^\infty \frac{x^3}{e^x-1}\,dx = \frac{\pi^4}{15} 
  \label{eq:sample}
\end{equation}


<hr>

The Harmonic Oscillator The Hamiltonian for the Harmonic Oscillator is \begin{equation}\frac{p^2}{2\mu} + \frac{k}{2} x^2 \end{equation} where p is the momentum operator and x is the postition operator. We know that the Schrödinger prescription is \begin{equation}p \rightarrow - \boldsymbol{i} \hbar \frac{\partial}{\partial x} \end{equation} while $x \rightarrow x$, as usual. This means, as we well know, that the commutator of $x$ and $p$ is non-zero. (Note that we've dropped the operator subscript.)

The commutator: $[\alpha,\beta]$

\begin{equation*} [p,x]f =( px - xp)f = -\boldsymbol{i} \hbar \frac{\partial}{\partial x}xf - x(-\boldsymbol{i} \hbar \frac{\partial}{\partial x}f) \end{equation*} where $f(x) \mapsto f$, which leads to \begin{equation*} [p,x]f =( px - xp)f = -f\boldsymbol{i} \hbar \frac{\partial x}{\partial x} + x \left (-\boldsymbol{i} \hbar \frac{\partial}{\partial x}\right )f - x\left (-\boldsymbol{i} \hbar \frac{\partial}{\partial x}f\right ) \end{equation*} which means, of course, \begin{equation} [p,x] = px - xp = -\boldsymbol{i} \hbar \end{equation}

The Ladder Operators Construction

For the Harmonic Oscillator, we form the two operators \begin{equation} a^+ = p + \boldsymbol{i} \mu\omega x \end{equation} and \begin{equation} a^- = p - \boldsymbol{i} \mu\omega x \end{equation} which differ solely by that intervening sign (Remember that $\omega = \sqrt{\frac{k}{\mu}}$). The entire derivation now hinges on the properties of these two operators. We start with the elementary question, what is the commutator of $a^+$ and $a^-$?

<html>
<head>
<title>The Harmonic Oscillator</title>
  <script src="https://polyfill.io/v3/polyfill.min.js?features=es6"></script>
  <script>
  MathJax = {
    tex: {
      tags: 'ams'  // should be 'ams', 'none', or 'all'
    }
  };
  </script>
  <script async="" id="MathJax-script" src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-chtml.js"></script>
<script>
MathJax = {
  tex: {
    inlineMath: [['$', '$'], ['\\(', '\\)']],
tags:'ams'
  },
  svg: {
    fontCache: 'global'
  }
};
</script>




</head>

<body>
The Hamiltonian for the Harmonic Oscillator is
\begin{equation}\frac{p^2}{2\mu} + \frac{k}{2} x^2
\end{equation}

where p is the momentum operator and x is the postition operator. We know that the Schrödinger prescription is
\begin{equation}p \rightarrow - \boldsymbol{i} \hbar \frac{\partial}{\partial x}
\end{equation}

while $x \rightarrow x$, as usual. This means, as we well know, that the commutator of $x$ and $p$ is non-zero. 
(Note that we've dropped the operator subscript.)


<h2>
The commutator: $[\alpha,\beta]$</h2>
\begin{equation*}
[p,x]f =( px - xp)f = -\boldsymbol{i} \hbar \frac{\partial}{\partial x}xf -
x(-\boldsymbol{i} \hbar \frac{\partial}{\partial x}f)
\end{equation*}
where $f(x) \mapsto f$,
which leads to
\begin{equation*}
[p,x]f =( px - xp)f = -f\boldsymbol{i} \hbar \frac{\partial x}{\partial x} +
x \left (-\boldsymbol{i} \hbar \frac{\partial}{\partial x}\right )f - 
x\left (-\boldsymbol{i} \hbar \frac{\partial}{\partial x}f\right )
\end{equation*}
which means, of course,
\begin{equation}
[p,x] = px - xp = -\boldsymbol{i} \hbar
\end{equation}
<br />
<h2>
The Ladder Operators Construction</h2>
For the Harmonic Oscillator, we form the two operators
\begin{equation}
a^+ = p + \boldsymbol{i}  \mu\omega x
\end{equation}
and
\begin{equation}
a^- = p - \boldsymbol{i}  \mu\omega x
\end{equation}
which differ solely by that intervening sign
(Remember that $\omega = \sqrt{\frac{k}{\mu}}$).
The entire derivation now hinges on the properties of these two operators.
We start with the elementary question, what is the commutator of $a^+$ and
$a^-$?


</body>
</html>

Simple Programming Examples

Tuesday, September 17, 2019

MathJax example 2

Monday, September 16, 2019

MathJax Example using Latex and Automatic Equation Numbering

The commutator: $[\alpha,\beta]$

The Ladder Operators Construction

Wednesday, September 11, 2019

MathJax example with Latex

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