Heisenberg's Uncertainty Principle — Question 2

Question 2

Localization sets a lower bound on kinetic energy even with zero mean momentum

A nonrelativistic electron has $\sigma_x=.100\,\mathrm{nm}$ and mean momentum $p_0$. Use $\hbar^2/(2m_e)=.0381\,\mathrm{eV\,nm^2}$. All uncertainties here are standard deviations, with $\sigma_x\sigma_p\ge\hbar/2$.

1. Derive a lower bound on its mean kinetic energy in terms of $p_0$ and $\sigma_x$.

2. Evaluate the minimum possible mean kinetic energy for $p_0=0$.

Question 2 - Solution

Localization sets a lower bound on kinetic energy even with zero mean momentum

(a) Method and calculation

$$\langle K\rangle=\frac{\langle p^2\rangle}{2m_e}=\frac{p_0^2+\sigma_p^2}{2m_e}.$$$$\boxed{\langle K\rangle\ge\frac{p_0^2}{2m_e}+\frac{\hbar^2}{8m_e\sigma_x^2}}.$$

(b) Interpretation and further calculation

$$p_0=0:\quad K_{\min}=\frac{.0381\,\mathrm{eV\,nm^2}}{4(.100\,\mathrm{nm})^2}.$$$$\boxed{K_{\min}=.9525\,\mathrm{eV}}.$$

Physical check

Zero mean momentum is compatible with nonzero kinetic energy: positive and negative momenta contribute equally to the mean while their squares remain positive.