# How can I calculate Heisenberg uncertainty?

Dec 14, 2014

Heisenberg's uncertainty principle states that the momentum and precision of a particle cannot be simultaneously measured with arbitrarily high precision.

This is not something can that be put on the innacuracy of the measurement instruments, nor on the quality of the experimental methods; the uncertainty comes from the wave properties inherent in the quantum world.

Heisenberg's uncertainty principle is defined as

$\Delta p \cdot \Delta x \ge \frac{h}{4 \pi}$,

and shows how the more accurately a particle's position is known ( the smaller the $\Delta x$ is), the less accurately the momentum of the particle ($\Delta p$) is known (and vice-versa).

Here's a kind of silly example on this:

You are pulled over by a police officer for speeding; since you're familiar with Heisenberg's uncertainty principle, you try to convince the officer that if he knew were you were, he could not possibly know how fast you were going. The officer replies: "Well, I knew where you were by 0.5m, and your car's weight is 1300kg, so I know your velocity by..."

Start by using the equation $\Delta x \cdot \Delta p \ge \frac{h}{4 \pi}$. We know $\Delta x$ = $0.5 m$ and that $\Delta p = m \cdot \Delta v$, therefore

$\Delta x \cdot m \cdot \Delta v \ge \frac{h}{4 \pi} \to \Delta v = \frac{h}{4 \pi} \cdot \frac{1}{\Delta x \cdot m}$

Since $h$ represents Planck's constant and $m$ is the car's mass, we get

$\Delta v \ge \frac{6.626 \cdot {10}^{- 31} J \cdot s}{4 \pi \cdot 1300 k g \cdot 0.5 m} = 8.112 \cdot {10}^{- 35} \frac{m}{s}$

So the officer had an uncertainty of $8.112 \cdot {10}^{- 35} \frac{m}{s}$ on your speed...