Uncertainty principle

According to the Heisenberg uncertainty principle, it is impossible to know the exact position and momentum of a particle at the same time. The more precisely you know one, the less precisely you know the other. Since a photon has momentum equal to p=h/λ\displaystyle p = h / \lambda, it can alter the path of an electron when it collides with it. But to measure the position or momentum of an electron, you have to shoot photons at it. The very act of looking at the particle in this way changes the results. We are left with two options:

The product of the absolute uncertainties of position and speed satisfies

ΔxΔvh2πm\displaystyle \Delta{} x \Delta{} v \ge \frac{h}{2 \pi m},

where Δx\displaystyle \Delta{} x and Δv\displaystyle \Delta{} v are the absolute uncertainties of the position, measured in metres (m), and of the speed, measured in metres per second (m/s), respectively, and where h is Planck’s constant and m represents the mass of the particle, measured in kilograms (kg).


If we know the speed of an electron to within 1.0 m/s, how precisely can we know its position?

Solving the inequality for the uncertainty of position, we have

Δxh2πmΔv=6.63×1034J/s(2(3.14)(9.11×1031kg)(1.0m/s)\displaystyle \Delta{} x \ge \frac{h}{2 \pi m \Delta{} v} = \frac{6.63 \times 10^{- 34} \, \text{J/s}}{\left ( 2 \left ( 3.14 \right ) \left ( 9.11 \times 10^{- 31} \, \text{kg} \right ) \left ( 1.0 \, \text{m/s} \right ) \right.},

which evaluates to

Δx1.2×104m\displaystyle \Delta{} x \ge 1.2 \times 10^{- 4} \, \text{m}.

We are therefore off by at least 1.2 × 10−4 m for the position, which is quite a lot considering that this number is about 100 billion times larger than the radius of the electron—it’s sort of like saying that you know the position of your friend: “he’s somewhere in the solar system.”