Editing Uncertainty principle (section)

=== Quantum harmonic oscillators with Gaussian initial condition ===
{{multiple image
| align = right
| direction = vertical
| footer =
Position (blue) and momentum (red) probability densities for an initial Gaussian distribution. From top to bottom, the animations show the cases {{nowrap|1=Ω = ''ω''}}, {{nowrap|1=Ω = 2''ω''}}, and {{nowrap|1=Ω = ''ω''/2}}. Note the tradeoff between the widths of the distributions.
| width1 = 360
| image1 = Position_and_momentum_of_a_Gaussian_initial_state_for_a_QHO,_balanced.gif
| width2 = 360
| image2 = Position_and_momentum_of_a_Gaussian_initial_state_for_a_QHO,_narrow.gif
| width3 = 360
| image3 = Position_and_momentum_of_a_Gaussian_initial_state_for_a_QHO,_wide.gif
}}

In a quantum harmonic oscillator of characteristic angular frequency ''ω'', place a state that is offset from the bottom of the potential by some displacement ''x''<sub>0</sub> as
<math display="block">\psi(x)=\left(\frac{m \Omega}{\pi \hbar}\right)^{1/4} \exp{\left( -\frac{m \Omega (x-x_0)^2}{2\hbar}\right)},</math>
where Ω describes the width of the initial state but need not be the same as ''ω''. Through integration over the [[Propagator#Basic examples: propagator of free particle and harmonic oscillator|propagator]], we can solve for the {{Not a typo|full time}}-dependent solution. After many cancelations, the probability densities reduce to
<math display="block">|\Psi(x,t)|^2 \sim \mathcal{N}\left( x_0 \cos{(\omega t)} , \frac{\hbar}{2 m \Omega} \left( \cos^2(\omega t) + \frac{\Omega^2}{\omega^2} \sin^2{(\omega t)} \right)\right)</math>
<math display="block">|\Phi(p,t)|^2 \sim \mathcal{N}\left( -m x_0 \omega \sin(\omega t), \frac{\hbar m \Omega}{2} \left( \cos^2{(\omega t)} + \frac{\omega^2}{\Omega^2} \sin^2{(\omega t)} \right)\right),</math>
where we have used the notation <math>\mathcal{N}(\mu, \sigma^2)</math> to denote a normal distribution of mean ''μ'' and variance ''σ''<sup>2</sup>. Copying the variances above and applying [[list of trigonometric identities|trigonometric identities]], we can write the product of the standard deviations as
<math display="block">\begin{align}
\sigma_x \sigma_p&=\frac{\hbar}{2}\sqrt{\left( \cos^2{(\omega t)} + \frac{\Omega^2}{\omega^2} \sin^2{(\omega t)} \right)\left( \cos^2{(\omega t)} + \frac{\omega^2}{\Omega^2} \sin^2{(\omega t)} \right)} \\
&= \frac{\hbar}{4}\sqrt{3+\frac{1}{2}\left(\frac{\Omega^2}{\omega^2}+\frac{\omega^2}{\Omega^2}\right)-\left(\frac{1}{2}\left(\frac{\Omega^2}{\omega^2}+\frac{\omega^2}{\Omega^2}\right)-1\right) \cos{(4 \omega t)}}
\end{align}</math>

From the relations
<math display="block">\frac{\Omega^2}{\omega^2}+\frac{\omega^2}{\Omega^2} \ge 2, \quad |\cos(4 \omega t)| \le 1,</math>
we can conclude the following (the right most equality holds only when {{nowrap|1=Ω = ''ω''}}):
<math display="block">\sigma_x \sigma_p \ge \frac{\hbar}{4}\sqrt{3+\frac{1}{2} \left(\frac{\Omega^2}{\omega^2}+\frac{\omega^2}{\Omega^2}\right)-\left(\frac{1}{2} \left(\frac{\Omega^2}{\omega^2}+\frac{\omega^2}{\Omega^2}\right)-1\right)} = \frac{\hbar}{2}. </math>