Uncertainty of a projectile

Problem

A $1\;\text{g}$ projectile travels at $300\;\text{m/s}$, with an uncertainty in velocity of $0{,}1\;\text{mm/s}$. Calculate the quantum limitation on the precision of its position. Is it significant? What about an electron with the same relative velocity uncertainty?

Solution

Step 1 — Uncertainty in the projectile’s momentum. $m = 10^{-3}\;\text{kg}$, $\Delta v = 0{,}1\;\text{mm/s} = 10^{-4}\;\text{m/s}$:

$\Delta p = m\,\Delta v = 10^{-3}\cdot 10^{-4} = 10^{-7}\;\text{kg·m/s}$

Step 2 — Position limit.

$\ev{\Delta x \geq \frac{\hbar}{2\,\Delta p} = \frac{1{,}055\cdot 10^{-34}}{2\cdot 10^{-7}} \approx 5{,}3\cdot 10^{-28}\;\text{m}}$

Is it significant? No. $5{,}3\cdot 10^{-28}\;\text{m}$ is a trillion trillion times smaller than a nucleus ($\sim 10^{-15}\;\text{m}$): utterly negligible. For a macroscopic object, quantum uncertainty has no measurable effect whatsoever.

Step 3 — The electron. Equal relative uncertainty: $\Delta v/v = 10^{-4}/300 \approx 3{,}3\cdot 10^{-7}$. For the same speed ($v = 300\;\text{m/s}$), $\Delta v = 10^{-4}\;\text{m/s}$ stays the same, but the mass is $\sim 10^{27}$ times smaller:

$\Delta p = m_e\,\Delta v = 9{,}11\cdot 10^{-31}\cdot 10^{-4} \approx 9{,}1\cdot 10^{-35}\;\text{kg·m/s}$

$\Delta x \geq \frac{\hbar}{2\,\Delta p} = \frac{1{,}055\cdot 10^{-34}}{2\cdot 9{,}1\cdot 10^{-35}} \approx 0{,}6\;\text{m}$

Hugely significant! For the electron the position limit is macroscopic ($\sim 0{,}6\;\text{m}$): it makes no sense to speak of a well-defined trajectory. The same relative uncertainty gives negligible effects for the projectile and dramatic ones for the electron, because $\Delta x \propto 1/m$: it is the mass that separates the classical world from the quantum one.

Links

Topics: Quantum physics Concepts: Uncertainty principle Objects: Projectile