# Heisenberg's Uncertainty Principle

1. Mar 5, 2014

### X_Art_X

Hi Guys,
Newbie question from a layperson so please don't beat me up! :D

I know that Heisenberg's Uncertainty Principle relates to measurement/observation of
particles with regard to quantum physics.

My question is whether or not it applies to measuring anything in the observable Universe. i.e..:

A foot race between friends where pushing the button on a stopwatch causes some air to brush past the racers,
The measurement of voltage/current/impedance of a circuit where the multimeter's leads adds
impedance to the circuit or a drain on the current being measured.

I can't relate it to the measurement of time with a clock.

Might sound like a silly question, but is there an accepted answer?
Cheers, Art.

2. Mar 5, 2014

### phoenixXL

No, it doesn't apply.

Heisenberg's Uncertainty Principle states that at the quantum level, the velocity and the position of the particle can't be known at the same time.
He gave a relation with the Uncertainty in position and momentum (not with position and momentum).
$$Δx.Δp\ >=\ \frac{h}{2\pi}$$
In the above relation the variables are the uncertainties.

3. Mar 5, 2014

### maajdl

Yes it applies to anything.
But there are precise mathematical rules behind it.
It is about "observables" and "compatible observables" and "incompatible observables".

For example, the positions of two particles are "compatible observables".
This means that you can measure both of them together with any precision.

Conversely, the poisition (x) and the velocity (vx) of a given particle are "incompatible observables".
Measuring one observable (x) with increasing precision, decreases the precision (knowledge, information) on the second (v).

"Incompatible" is not the right word.
One prefers to talk about non-commuting and commuting observables, in reference to the math behind that.

http://en.wikipedia.org/wiki/Uncertainty_principle

4. Mar 5, 2014

### X_Art_X

Remember I said layman!! :D I don't pretend to understand math

If it applies to say, timing the observed angle of the travel of the Sun across the sky,
could it be explained how measuring the time between Sunrise and Sunset could affect the daylight time?
Or is that ridiculous?

5. Mar 5, 2014

### CWatters

What do you mean by "I can't relate it to the measurement of time with a clock"?

The photons you bounce off the hands of the clock affect the accuracy of the clock.

6. Mar 5, 2014

### X_Art_X

but not the time being measured as I understand.

7. Mar 5, 2014

### maajdl

A mechanical clock show the time by the position of something.
There is actually no other way!
So the principle applies also to time.
But there is more to that ... with some maths!

8. Mar 5, 2014

### CWatters

Well yes. You are still bouncing photons off the earth to perform the measurement.

The measurement aspect of Heisenberg's Uncertainty Principle isn't the whole story. You might be interested in this..

http://www.livescience.com/18567-wacky-physics-heisenberg-uncertainty-principle.html

9. Mar 5, 2014

### Staff: Mentor

You've been misled by some of the popular "explanations" of the uncertainty principle, which try to explain it by saying that any measurement has to perturb the thing being measured. Although Heisenberg himself first explained it that way, as he and other quantum pioneers learned more and figured out what was really going on, they dropped that basically bogus explanation - and as so often happened in the history of QM, the popular press didn't notice and has kept on spreading misinformation that is now almost a century stale.

There are some real and interesting subtleties around how measurements disturb the system being measured (don't pass so quickly over CWatters's comment above - he's trying to get you to think about exactly what it means to "measure time"!), but these aren't part of the uniquely quantum nature of the Uncertainty Principle. So you can forget about them for a moment, they'll just get in the way.

Quantum mechanically, the uncertainty principle says that it is impossible for a quantum system to be in a state in which it simultaneously has a definite position and a definite momentum. It's not just that we can't measure both - it's that if the momentum is definite, whether known or not, then the position is not definite, and vice versa. This the fundamental quantum uncertainty, and it is inherent in the way that quantum mechanics describes the state of a system, whether measured or not.

10. Mar 5, 2014

### abitslow

There are two types of laymen...laypeople... those who are ilnumerate and those who aren't. If your mind "fuzzes" over when exposed to very small or very large numbers or simple algebraic equations, then you are ilnumerate. With practice and motivation, that is a problem you can fix, but we can't fix it for you, nor can we do much to dumb down our answers to overcome such disability.
The uncertainty principle is all about AxB = h/(2π). Where A and B are two measurements, or slightly more accurately, are the variation of two measurements. Now, if you understand that (in some tentative sort of way), then the FIRST thing you should be asking is:"What is the VALUE of h?". h÷(2π) is about
0.0000000000000000000000000000000001 (Jxs) (that's 1E-34). A Joule is the amount of energy required to accelerate a 1 kg object by 1 m/s² over a distance of 1 meter. Whether that is a lot compared to, say, the amount of force you would need to lift a hair from your shirt doesn't really matter. Why? Because of the 34 zeroes. A nanometer is 1E-9 meters. The point is, the scale where the above equation becomes significant are outside of your perception. An atom is say 0.1 to 0.01 nanometers; so h/2π is still a factor of 1E-24 below that. Still meaninglessly small. HTH. So since the observeable universe is quantum mechanical: yes it applies to the observeable universe, but no its not something that a macroscopic obsever (you or I) could ever notice directly.

11. Mar 5, 2014

### ZapperZ

Staff Emeritus
There are hints here that seem to indicate a common misconception about the HUP. It appears that you are equating the HUP with some sort of issue with measurement accuracy.

You might want to start with what I had written earlier and see if you might think differently afterwards:

Please also note that there are a number of reasons why "your friends" will not exhibit clear quantum effects and why applying the HUP to them is absurd.

Zz.

12. Mar 5, 2014

### maajdl

X_Art_X,

Have you noted how many words in this single sentence are mathematical?
How could we answer without more reference to maths?

13. Mar 5, 2014

### maajdl

14. Mar 5, 2014

### FactChecker

X_Art_X,
This may not be your OP question, but a point that Nugatory made deserves to be emphasized. It is almost existential. In physics, if there is no way to detect or measure something, than you can assume that it doesn't exist and see what the consequence is.

The statement that a precise position and momentum cannot exist is very profound. It explained some things that could not be understood before.