In this unit the derivation of energy levels of a harmonic oscillator is explained using
commutation relations.
$\newcommand{\ket}[1]{|#1\rangle}$
Introduction Milestones References [[nid:592]] [[nid:591]] [[nid:590]] Main
A mass attached to a spring, when stretched and released, executes a simple harmonic motion. The energy of a harmonic oscillator is given by
\begin{equation}
H = \frac{p^2}{2m} + \frac{1}{2}m \omega^2 x^2
\end{equation}
where $p$ is the momentum of the particle at position $x$. Classically the energy depends on the amplitude and all positive values are allowed. Quantum mechanically, energy is quantized and all energies are not allowed. The allowed values are the eigenvalues of the Hamiltonian operator.
\begin{equation}
\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m \omega^2 \hat{x}^2 \label{EQ02}
\end{equation}
Here $\hat{x},\hat{p}$ denote the position and momentum operators and $\hat{H}$ has been written down by replacing $x,p$ in \EqRef{EQ01} by the corresponding operators $\hat{x}$ and $\hat{p}$. The energy levels can be derived by solving the Schr\"{o}dinger equation or using commutation relations. The use of algebraic methods, employing the commutation relations only, followed here underlines the power of the canonical quantization rule listed as one of the postulates.
Goals
$$ E_n = (n+\tfrac{1}{2}) \hbar \omega $$
\begin{eqnarray}
a \ket{n} &=& \sqrt{n} \ket{n-1}\\
a^\dagger \ket{n} &=&\sqrt{n+1}\ket{n+1}
\end{eqnarray}
$$ [ \hat{x},\hat{p}] = i \hbar; $$
\begin{eqnarray}
a &=& \frac{1} {\sqrt{2m\omega \hbar}}( p-im\omega x) ,\\
a^\dagger &=& \frac{1}{\sqrt{2m\omega \hbar}}( p+im\omega x).\\
N &=& a^\dagger a.
\end{eqnarray}
\begin{equation}
[ a, a^\dagger ] =1,\qquad [N,a]=-a,\qquad
[N,a^\dagger]=a^\dagger.
\end{equation}
\begin{equation}
H = \frac{p^2}{2m} + \frac{1}{2}\, m \omega^2 q^2 .
\end{equation}
expressed in terms of $a$ and $a^\dagger$ takes the form
\begin{equation} H = ( a^\dagger a +\frac{1}{2} ) \hbar \omega .\end{equation}
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