# 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
$\displaystyle p = h / \lambda$, 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:

- Use low-energy photons (long wavelength) for precise momentum.
- Use high-energy photons (short wavelength) for precise position.

The product of the absolute uncertainties of position and speed satisfies

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

where $\displaystyle \Delta{} x$ and
$\displaystyle \Delta{} v$ 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).

## Example

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

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

which evaluates to

$\displaystyle \Delta{} x \ge 1.2 \times 10^{- 4} \, \text{m}$.

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.”