# Heisenberg's Uncertainty Principle

1. May 26, 2014

### BOAS

Hello,

i'm solving some quite simple problems using the uncertainty principle, but I don't have access to the solutions and I really don't have a feel for what a 'sensible' answer is... When finding the minimum uncertainty in velocity, I end up with things greater than the speed of light, so I want to check i'm not making an error somewhere.

1. The problem statement, all variables and given/known data

Suppose that an electron is trapped in a small region and the uncertainty in it's position is $10^{-15}$m. What is the minimum uncertainty in the particles momentum? What uncertainty in the electrons velocity does this correspond to?

2. Relevant equations

3. The attempt at a solution

$(\Delta y)(\Delta p_{y}) \geq \frac{h}{4 \pi}$

Since we're dealing the minimum uncertainty, we can equate the two.

$(\Delta p_{y}) = \frac{h}{4 \pi (\Delta y)}$ = 5.27 x 10-20 kgms-1

$p = mv$

$\Delta p_{y} = m (\Delta v_{y})$

$\Delta v_{y} = \frac{\Delta p_{y}}{m}$ = 5.79 x 1010 ms-1

This answer just seems absurd, maybe it's a consequence of the accuracy we know the electrons position to, but like I said I have no way of checking what a sensible answer is.

So, have I gone wrong somewhere?

Thanks for any help you can give.

2. May 26, 2014

### dauto

Your mistake is to use the formula p = mv which is the non-relativistic formula. Try using p = γmv, where γ = (1 - v2/c2)-1/2.

3. May 26, 2014

### BOAS

Aha! Thanks, I was looking at an old example from my notes but evidently it was intended to show why the heisenberg uncertainty principle is negligible at the macroscopic level of a ping pong ball.