Editing Quantum harmonic oscillator (section)

===Natural length and energy scales===
{{see also|Path integral formulation#Simple harmonic oscillator}}

The quantum harmonic oscillator possesses natural scales for length and energy, which can be used to simplify the problem. These can be found by [[nondimensionalization#Quantum harmonic oscillator|nondimensionalization]].

The result is that, if   ''energy'' is measured in units of  {{math|''ħω''}} and ''distance'' in units of {{math|{{sqrt|''ħ''/(''mω'')}}}}, then the Hamiltonian simplifies to
<math display="block"> H = -\frac{1}{2} {d^2 \over dx^2} +\frac{1}{2}  x^2 ,</math>
while the energy eigenfunctions and eigenvalues simplify to Hermite functions and integers offset by a half,
<math display="block">\psi_n(x)= \left\langle x \mid n \right\rangle = {1 \over \sqrt{2^n n!}}~ \pi^{-1/4} \exp(-x^2 / 2)~ H_n(x),</math>
<math display="block">E_n = n + \tfrac{1}{2} ~,</math>
where {{math|''H''<sub>''n''</sub>(''x'')}} are the [[Hermite polynomials]].

To avoid confusion,  these "natural units"   will mostly not be adopted in this article. However, they frequently come in handy when performing calculations, by bypassing clutter.

For example, the [[fundamental solution]] ([[Propagator#Basic_examples:_propagator_of_free_particle_and_harmonic_oscillator|propagator]]) of {{math|''H'' − ''i∂<sub>t</sub>''}}, the time-dependent Schrödinger operator for this oscillator, simply boils down to the [[Mehler kernel]],<ref>[[Wolfgang Pauli|Pauli, W.]]  (2000), ''Wave Mechanics: Volume 5 of Pauli Lectures on Physics'' (Dover Books on Physics). {{ISBN|978-0486414621}} ; Section 44.</ref><ref>[[Edward Condon|Condon, E. U.]]  (1937). "Immersion of the Fourier transform in a continuous group of functional transformations", ''Proc. Natl. Acad. Sci. USA''  '''23''', 158–164. [https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1076889/pdf/pnas01779-0028.pdf online]</ref>
<math display="block">\langle x \mid \exp (-itH) \mid y \rangle \equiv K(x,y;t)= \frac{1}{\sqrt{2\pi i \sin t}} \exp \left(\frac{i}{2\sin t}\left ((x^2+y^2)\cos t - 2xy\right )\right )~,</math>
where {{math|1= ''K''(''x'',''y'';0) = ''δ''(''x'' − ''y'')}}. The most general solution for a given initial configuration {{math|''ψ''(''x'',0)}} then is simply
<math display="block">\psi(x,t)=\int dy~ K(x,y;t) \psi(y,0) \,.</math>