# Heisenberg's principle in classical world [duplicate]

I have a ball with me and I have been observing it's position which remains constant at a place only, hence it must be largely localised, further as it remains on the same place from so long, it's velocity must be largely localised to 0 only. Now with both momentum and position know to any degree, Where is the uncertainty?? With the understanding of quantum mechanics, we must expect the "fixed" ball to have any arbitrary momentum, but that is not the case. Where am I going wrong??

• You yourself used the words “largely localised”. That’s the resolution. The position is not exactly zero. Only if it was, the momentum would be completely uncertain. – Superfast Jellyfish Feb 9 '20 at 18:55
• The position of the ball is only an expectation value with standard deviation $\sigma_x$. The same is true for the velocity (or equivalently the momentum) which has standard deviation $\sigma_p$. According to Heisenberg, the product of these standard deviations can not be smaller than the constant $\hbar/2$. – user56224 Feb 9 '20 at 19:05
• I also suggest learning more about "uncertainty" in QM. It is just the standard deviation of repeated measurements of similarly prepared quantum states. It doesn't have to do with how "certain" we are about the "true" position or the "true" momentum. – BioPhysicist Feb 10 '20 at 4:41

## 3 Answers

The Heisenburg uncertainty principle is as follows:

$$\Delta p \Delta x \ge \frac{\hbar}{2}$$

where $$\hbar$$ is the reduced Planck constant, equal to about $$10^{-34}$$ Js. This is very small in terms of every day measurements. For example, if there is an uncertainty in position of only 1 micrometre, the uncertainty in the momentum need only be around $$10^{-28}$$ kg m/s. This is negligable on a classical scale, so it appears that you can know both position and momentum with absolute certainty, even though you obviously cannot.

• But, don't we know the position exactly. I mean, for infinite time span, we have only found that particle to be located at that point, thus it must be having a probability of 1 of being their and thus must have been with known position completely. What is wrong with this? – Yashkalp Sharma Feb 9 '20 at 20:11
• @YashkalpSharma you've only been measuring the position to the accuracy of say a photon wavelength ($\sim 500\text{nm}$). Or probably worse because the ball is in fact in thermal motion and vibrating ever so slightly. – jacob1729 Feb 9 '20 at 20:27

What we call uncertainty is the expected value of the standard deviation of position and momentum.

From statistical theory, if the standard deviation of a random variable with normal distribution is $$\sigma$$, the standard deviation of the mean is: $$\sigma_{\mu} = \frac{\sigma}{\sqrt{n}}$$, where $$n$$ is the number of elements of the sample.

So, if by position we understand position of the center of mass, and by momentum also CM momentum, we have no lower bound for the standard deviation (uncertainty). It is only necessary to increase the sample, that means the number of atoms of the object.

An object with enough number of atoms (each of them subject to the uncertanty principle) can have: $$\Delta x_{CM}\Delta p_{CM} < \frac{\hbar}{2}$$, where the $$\Delta$$'s are the standard deviation of the average position and average momentum of the center of mass.

First, while the amount of uncertainty required by the Principle is significant at quantum mechanical scales, it's tiny on macroscopic scales. Are you certain where the ball is, down to an atom's radius?

Second, the uncertainty is regarding the object's momentum, not its velocity. If you have an object that's 10,000 times the mass of an electron, then it can have 1/10,000 the uncertainty in its velocity, while having the same uncertainty in its momentum. A ball is larger than an electron by a factor larger than Avogadro's Number, so even a tiny amount of uncertainty in its velocity results in an uncertainty in it momentum that is massive by quantum mechanical standards.