The Harmonic Oscillator
The Hamiltonian for the Harmonic Oscillator is
where p is the momentum operator and x is the postition operator. We know that the Schrödinger prescription is
while , as usual. This means, as we well know, that the commutator of and is non-zero.
(Note that we've dropped the operator subscript.)
The commutator:
where ,
which leads to
which means, of course,
The Ladder Operators Construction
For the Harmonic Oscillator, we form the two operators
and
which differ solely by that intervening sign
(Remember that ).
The entire derivation now hinges on the properties of these two operators.
We start with the elementary question, what is the commutator of and
?
<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>
No comments:
Post a Comment