Heisenberg's Uncertainty Principle — Question 3

Question 3

A Gaussian image width must be converted before applying the uncertainty inequality

An electron position density is Gaussian, $\rho(x)=\rho_0e^{-x^2/(2\sigma_x^2)}$, with measured full width at half maximum $w=.200\,\mathrm{nm}$. Use $\hbar c=197.3\,\mathrm{eV\,nm}$. All uncertainties here are standard deviations, with $\sigma_x\sigma_p\ge\hbar/2$.

1. Derive the relation between w and $\sigma_x$.

2. Calculate the lower bound on $\sigma_pc$ and compare with treating w incorrectly as $\sigma_x$.

Question 3 - Solution

A Gaussian image width must be converted before applying the uncertainty inequality

(a) Method and calculation

$$\rho(w/2)=\rho_0/2\quad\Rightarrow\quad\frac{w^2}{8\sigma_x^2}=\ln2.$$$$\boxed{\sigma_x=\frac{w}{2\sqrt{2\ln2}}=.0849322\,\mathrm{nm}}.$$

(b) Interpretation and further calculation

$$\boxed{\sigma_pc\ge\frac{\hbar c}{2\sigma_x}=1161.51\,\mathrm{eV}}.$$$$\frac{\text{correct lower bound}}{\hbar c/(2w)}=\frac w{\sigma_x}=\boxed{2\sqrt{2\ln2}=2.35482}.$$

Physical check

FWHM and standard deviation describe different features of a distribution. Their conversion is shape-dependent; the Gaussian conversion cannot be applied to every profile.