# HUP and Conservation of momentum

1. Mar 9, 2013

### g.lemaitre

The HUP says that I can not know the position and the momentum of a particle simultaneously, therefore, from time T1 to time T2 I cannot predict its position. How does this not violate the law of conservation of momentum? If you cannot know its position and momentum exactly that how do we know the law of conservation of momentum is true at the quantum level?

2. Mar 9, 2013

### Staff: Mentor

As far as I know, that isn't how the HUP works. You can measure the position and measure the momentum and do each measurement to an arbitrary degree of precision. What you cannot do is prepare a state where all of the particles in that state have a well defined position and a well defined momentum.

I am sure one of the real experts here will correct me if I am wrong on this.

3. Mar 9, 2013

### Staff: Mentor

That's it exactly. For a good discussion of this point, consult Ballentine's text: Quantum Mechanics, A Modern Development.

4. Mar 9, 2013

### wuliheron

You don't know because to paraphrase Socrates the only thing we know about indeterminacy is that we don't know. Accepting this brings enlightenment, rejecting it misery.

5. Mar 9, 2013

### Staff: Mentor

We do know that conservation of momentum is "true" at the quantum level by Noether's theorem. The Lagrangians are all invariant under translations, therefore momentum is conserved. The HUP doesn't change that.

6. Mar 9, 2013

### g.lemaitre

I find it very hard to believe that we know momentum is conserved at the quantum level. Let P = position and M = momentum.

10am: P = (7,4) M = ?
11am: P = ? M = 4

(7,4) is an x,y coordinate.

How am I supposed to be able to conclude that momentum of the particle above was conserved from 10 am to 11 am if I don't even know what it was to begin with?

7. Mar 9, 2013

### Staff: Mentor

Again, you are misunderstanding how the HUP works. You can measure both P and M to arbitrary precision at the same time.

The form of the Lagrangian guarantees it.

If you don't know what it was to begin with then what would make you think it wasn't conserved?

8. Mar 9, 2013

### g.lemaitre

Arbitrary precision isn't good enough. Say that momentum is between 3 and 5 at time t1, then between 3 and 5 at time t2 - how do you know it's conserved? It could be 3 at time t1 and 5 at time t2?

My point exactly. If physicists can't know what x then how do they know x is conserved?

9. Mar 9, 2013

### Jorriss

That's all we have. We can only bound our uncertainty. It's very likely momentum is conserved because the bound on it is very low.

There are also many indirect ways to check conservation laws. For example conservation of angular momentum can be checked by looking at particle decay patterns, spectroscopic transitions, etc.

No one experiment shows conservation of momentum but there is plenty to suggest it and nothing that has violated it.

10. Mar 9, 2013

### Staff: Mentor

This is silly. Arbitrary precision means there is not a limit to the precision. That isn't good enough? How could arbitrary precision possibly not be good enough? If you want infinite precision then you have come to the wrong universe.

No, it is the opposite of your point.

We have a theory, that theory predicts many things, including that momentum is conserved. We have more than 100 years worth of historical measurements. Those measurements all agree with the theory, including experiments on the conservation of momentum.

You have a new piece of data which does not contradict the theory. So what would make you think momentum wasn't conserved?

11. Mar 10, 2013

### g.lemaitre

I'm not talking about infinite precision. The Planck Length is 10^-35 m. I'm pretty sure quantum effects occur between 10^-15 and 10^-35 m since as I recall a proton's position and momentum can be known simultaneously but an electron cannot and a proton is 10^-15 m if I have my numbers correct. So particles smaller than 10^-15 m how do you know that momentum is conserved?

To use the analogy of fireflies: we see fireflies when they light up but we do not know where they are between light-ups. How do you prove that their momentum is conserved between light-ups, imagining that the fireflies lengths is smaller than 10^-15 m?

I don't even see how you can input the values into Noether's theorem.

No, it is the opposite of your point.

I'm talking about experiments on the quantum level.

12. Mar 10, 2013

### atyy

Let's talk about conservation of energy. In general, it is the average value of energy that is conserved, ie. does not change with time. So at t=0. we prepare an infinite number of systems identically in a particular state.

At t=1, we measure the energy of half of them. In general we will get a bunch of different results, but if we average across all measurements, there will be some average value <E1>.

At t=2, we measure the energy of the other half of them. Again, we get a bunch of different results, but if we average across all measurements, there will be some average value <E2>.

We will find that <E1>=<E2>, ie. the average energy does not change with time. This is what we mean by conservation of energy.

13. Mar 10, 2013

### g.lemaitre

Actually I got this quote from Kurdt posted 7 years ago in this forum at this thread:

I remember the analogy that you can borrow a loan from a bank on Friday so long as you return it on Monday when no one is looking.

I'm still not satisfied with the explanations for conservation of momentum.

If a particle is at 0,0 at time t1 with uncertainty of a radius of 2*10^-18. Then we find it at time t2 at 1,0 where the units are 10^-18 m and its uncertainty is still the same radius, and we know that it's momentum is east between 30 degrees and 330 degrees, then I don't see how we know that it went on a straight line from 0,0 to 1,0

14. Mar 10, 2013

### Jorriss

You don't seem to get what 'to know' means in science. That's the problem.

15. Mar 10, 2013

Staff Emeritus
+1

And you should read some of the responses he got in that thread, rather than implicitly asking us to type them in again special just for you.

16. Mar 10, 2013

### g.lemaitre

Your analogy is flawed. You're committing the fallacy of division

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

You're assuming that because a group obeys a law that its parts obeys a law. Imagine that you only had the ability to measure the strength of 3 billion humans. Given the crudity of your measurements you would assume that there is a conservation of strength among humans. When you're measuring devices finally reach the level where they can measure the strength of individual humans you'll find that not all humans have equal strength, nor is it conserved.

17. Mar 10, 2013

### DennisN

18. Mar 10, 2013

### g.lemaitre

19. Mar 10, 2013

### Jano L.

If you accept HUP as you stated it, it refers to knowledge of some human. But the conservation laws refer to the actual state of the particles. There is thus no direct contradiction.

Why do we think that the law of conservation of momentum is valid for particles in situations when we can't measure their momentum? Because it worked very well so far, and thus it became one of the cornerstones of the theory. There is no reason to abandon it. So we assume that conservation of momentum holds always, even if we do not check.

20. Mar 10, 2013

### Staff: Mentor

Then arbitrary precision must be good enough.

You recall incorrectly. The HUP does not prevent that, as I explained in post 2.

By Noether's theorem, as I said in post 5.

g.lemaitre, this is going around in circles. You are posting the same wrong statements. Please try actually responding to the substance of my post. You mention that you don't know how to plug numbers into Noether's theorem. What do you know about Noether's theorem?

Last edited: Mar 10, 2013