= Solution
For the <Quantum harmonic oscillator> define
$$
\hat a=\frac{m\omega\hat x+i\hat p}{\sqrt{2m\hbar\omega}},\qquad\hat a^\dagger=\frac{m\omega\hat x-i\hat p}{\sqrt{2m\hbar\omega}}.
$$
The <Canonical commutation relation> gives $[\hat a,\hat a^\dagger]=1$, and
$$
\boxed{\hat H=\hbar\omega(\hat a^\dagger\hat a+\tfrac12),\qquad[\hat H,\hat a^\dagger]=\hbar\omega\hat a^\dagger,\qquad[\hat H,\hat a]=-\hbar\omega\hat a.}
$$
Thus the <creation operator> and <annihilation operator> raise or lower the energy by one quantum. Positivity of $\hat a^\dagger\hat a$ and termination of downward stepping give $E_n=\hbar\omega(n+1/2)$, $n=0,1,\ldots$. Normalized actions are $\hat a|n\rangle=\sqrt n|n-1\rangle$ and $\hat a^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$. The annihilated <ground state> obeys
$$
\psi_0(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^2/(2\hbar)},\qquad|n\rangle=\frac{(\hat a^\dagger)^n}{\sqrt{n!}}|0\rangle.
$$
The nonzero <Ground state> energy is the <zero-point energy>, consistent with position-momentum <quantum uncertainty>. Since $\hat x=\sqrt{\hbar/(2m\omega)}(\hat a+\hat a^\dagger)$, a perturbation linear in <position> connects only neighbouring number states. The algebra is especially useful for <matrix elements> and <selection rules>.<Quantum Mechanics by David Tong>[Section 2: harmonic oscillator ladder construction]{ref}
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2016/paper-1-oscillator-ladder.png]
{title=Creation and annihilation connect equally spaced oscillator levels while the ground-state lowering action vanishes}
{source=paper-1-oscillator-ladder.py}
Back to article page
