# The Heisenberg Uncertainty Principle and velocity

Hello all

I have a question concerning The Heisenberg Uncertainty Principle. The principle mathematically looks like this-

$$\Delta x\Delta p \geq \hbar/2$$

The principle states that you can not measure more than two quantities simultaneously. If you know a particle's position very precisely, then you won't know its momentum very precisely and vise versa. I want to know how it is used, so lets set up an example.

I have a caliper that can measure to an accuracy of ±.05mm. I confine an electron which has a mass of 9.11*10^-31 kilograms in a space of 10mm. Predict the error of velocity. So lets start with.

$$\Delta x\Delta p \geq \hbar/2$$
Lets solve for
$$\Delta v$$

$$\Delta x\Delta p \geq \hbar/2$$

$$\Delta x\Delta m\Delta v \geq \hbar/2$$

$$\Delta v \geq \hbar/(2 \Delta x\Delta m)$$

Ok now I just plug in values. If I do something wrong or stupid, don't be afraid to say something. I want to see how this works.

$$\Delta v \geq \hbar/(2*.05*9.11*10^-31)$$
For an answer I get 1.15759767 m / s. Now I have a question about this answer. Does it mean that the electron will have a velocity of v±1.15759767? I don't know what the final answer is suppose to represent.

Now, what the uncertainty principle says is that if you multiply the standard deviation of the positions by the standard deviation of the momenta, the product will be no less than $$\hbar/2$$. And the reason for this is not because your measurements were not good enough; it's because of the intrinsic properties of the quantum state. No matter how carefully you perform this experiment, you will never wind up with a product of standard deviations that is less than $$\hbar/2$$.