Heisenberg's Uncertainty Principle — Question 8

Question 8

A freely evolving minimum-uncertainty packet spreads through momentum dispersion

Initially a free particle has an unchirped Gaussian with $\sigma_x(0)=a$, $\sigma_p=\hbar/(2a)$ and $C_{xp}(0)=0$. In free evolution $x(t)=x(0)+p(0)t/m$ and p is constant.

1. Derive the position variance at time t.

2. Find the positive time when the width doubles and the uncertainty product then.

Question 8 - Solution

A freely evolving minimum-uncertainty packet spreads through momentum dispersion

(a) Method and calculation

$$\sigma_x^2(t)=\sigma_x^2(0)+\frac{2t}{m}C_{xp}(0)+\frac{t^2}{m^2}\sigma_p^2.$$$$\boxed{\sigma_x(t)=a\sqrt{1+(t/t_0)^2}},\quad t_0=2ma^2/\hbar.$$

(b) Interpretation and further calculation

$$\sigma_x(t_d)=2a\quad\Rightarrow\quad\boxed{t_d=\sqrt3\,t_0=2\sqrt3\,ma^2/\hbar}.$$$$\boxed{\sigma_x(t_d)\sigma_p=(2a)\frac\hbar{2a}=\hbar}.$$

Physical check

Free evolution preserves the momentum distribution but builds position-momentum correlation. The ordinary uncertainty product can increase even though the state remains pure.