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
$a^-$?
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<title>The Harmonic Oscillator</title>
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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
$a^-$?
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