# Uncertainty Principle

1. Jun 29, 2006

### neophysique

I'm trying to understand the Uncertainty Principle,
for the case of momentum and position, but I'm stumped.
Suppose one can measure the position of a particle
with arbitrary precision. Suppose the measurement
of this particle's positions were made at time t= 0 and
then time t=1. The vector length of the particle path
divided by time then would yield the velocity of
the particle at time t=1, per definition. If the mass
of the particle was also constant then per definition
mv= momentum should yield the exact momentum
of the particle at t=1 , no?

If the above is accurate, how does one get to an
Uncertainty Principle? I'm sure the math leads
to this Principle but I'm having a hard time
relating it physically.

2. Jun 29, 2006

### Rach3

This leads to inconsistencies. For example, you'd expect the position at t=1/2 to be halfway between the position at t=0 and t=1, classicaly, right? Do all three measurements; this isn't the case. Did something accelerate? The more position measurements you do, the closer together you space them in time... the more wildly they flucutate and disagree with each other. Even in an extremely short time interval, they might be observed at two very remote positions, "implying" extreme velocity. Or they might stay in the same spot.

Of course each position measurement was arbitraily precise - except then the momentum measurements you're implicity performing get a large statistical distribution.

(This explanation was inspired by the very first chapter of Landau & Lif****z, authors whose names are inexplicably censored by this forum's software. You can actually read that part for free in the excerpt on Amazon's page (link)).

Last edited by a moderator: Jun 29, 2006
3. Jun 29, 2006

### Rach3

It comes from the formalism of Quantum Mechanics, which has its own postulates and cannot be derived from classical physics since it actively disagrees with classical physics.

4. Jun 29, 2006

### neophysique

Perhaps the results would depend on the type of experiments
performed. The results you describe seemed to come from
experiments in which the particle's starting and ending points
are not confined in space.

The forum software censors that name probably cause it contains
a curse word.

5. Jun 29, 2006

### HallsofIvy

How do you determine the position? Imagine you have a really large magnifying glass so that you can actually see the electron. You still have to shine a light on it so that you can see it. That light carries momentum, furthermore to get a very accurate position you will need to use small wavelength light (you can only get the postion to within a half wavelength of the light you use) which has higher momentum. Although you can use the two positions to accurately calculate the speed (and momentum) of the electron before you shown the light on it, the light itself changes the momentum.

6. Jun 29, 2006

### DrChinese

The HUP is actually independent of the type of experiment being performed. There is no question that you can observe the non-commuting attributes of a particle. The question is: what do you learn from such observations? If you cannot use the results of those observations to predict values at some future point in time, you didn't really learn as much as you thought. After measuring position and momentum, you will always find inconsistent results with follow-on measurements. This is true of any non-commuting observables. For example, measuring the y spin component of an electron completely invalidates any previous x or z spin component measurement, but NOT subsequent y spin measurements (or other observables that do commute). That is the HUP.

7. Jun 29, 2006

### Gokul43201

Staff Emeritus
No, that's not right. You have only determined an average velocity during the interval.

8. Jun 29, 2006

### RandallB

Simplifying is good to help understand, but you’re glossing over some significant points in your OP example. Restating the problem let your test particle follow a single straight line and we can give you both a start time of t=0 and a distance of d=0 precisely. Your measurements at “About” d=10 away at t=1 need to produce precise measurements relative to that starting coordinate.

Here’s the problem
1) You can actually measure the WHERE the particle is by detecting its exact crossing of the d=10.000 distance exactly --- BUT the more exact the location measurement is the less precise (uncertain) is your measurement of WHEN the particle arrived there! E.G. the time comes out between 0.99? to 1.00?
2) BUT wait you can revise your measurement to see exactly WHEN it arrives to see it as precisely t=1.000; BUT now this new measurement is unclear as to exactly WHERE it is with results like d = 9.99? to 10.00? so now you know WHEN but are uncertain about WHERE.

So as you put it nailing down the WHEN would give you the momentum your looking for to go with your WHERE measurement. The trick is getting both WHEN and WHERE at the same time.

Now here is the part your need to get, this not because of HUP quantum uncertainty -- these are experimental observations made I the 1920’s, and EXPLAINED by HUP uncertainty.
What was claimed by that explanation – that HUP could predict the amount of that uncertainty and no measurement could ever be better than that prediction.

But hold on, you’ve got better rulers and better stop watches than they ever had 80 years ago, and two hands to use both at once, surly you can now do better than they did back then.
NOPE – HUP so far has predicted correctly; so well in fact that the theoretical principles associated with that are largely responsible for most of the high technology created in the 20th century.

Exactly why is it this way? – plenty of competing theories working on being the one that explains that in detail, -- including some that claim there is no real improvement to be made over what we already know – no winner yet.

9. Jun 29, 2006

### Rach3

This is quite wrong. There's no non-commutivity between position operators X and time t, which isn't even an operator but a parameter. HUP places no restriction on simultaenous measurements of position and time; even if some alien formalism had such a thing, it wouldn't have units of action (hbar) but rather of distance*time = distance^2, so it would clearly be incompatible with the uncertainty principle of QM.

10. Jun 29, 2006

### Eye_in_the_Sky

The Uncertainty Principle is about "what will be", not "what was".

Or, if you want to fix your eye only on the past, it is about "what could have been", not just "what was".

11. Jun 29, 2006

### RandallB

Nonsense: If QM allowed that at one observation you could measure the exact location and time at that location; and also be able to measure with precision both time and location at second observation how would you not be able to define a precise Momentum and Location for the particle?
It is the essence of the OP ? by neophysique.
He deserves a response that addresses the context of his question; not some theory detail like position operators in units of hbar, or some philosophy like – what is, will be, what was, to what will be.

12. Jun 29, 2006

### Rach3

You might want to be more specific, some of us do not follow you.

13. Jun 29, 2006

### Rach3

No. There's nothing in HUP against measuring both position and momentum, simulatenously, to arbitrary precision - that's a misconception. The uncertainty is not the "measurement uncerainty", the precision of the measuring apparatus, it is a statistical uncertainty over ensembles of identical systems. Read my above explanation or the linked excerpt from the opening of Landau's book, for an illuminating example.

Better yet, fall back on the formalism. What happens to a wavefunction when you measure it's position? How does it evolve after that? And how much does a later position measurement tell you?

Remember that position and time aren't on equal footing in non-relativistic QM formalism. Time is a parameter, which is always known to arbitrary precision.

14. Jun 29, 2006

### Eye_in_the_Sky

Suppose that at time t1 the system is found to be at position x1 with spread equal to ∆x. More precisely, at time t1 we perform a filtering-type measurement for which the output wavefunction is some φ(x) centered at x1 with standard deviation ∆x.

The Uncertainty Principle then tells us:

Any subsequent measurement of position will give us a result x2 which lies in some interval of x-values consistent with a momentum uncertainty on the order of h/∆x.

This refers to "what will be".

Specifically, suppose that at time t2 the system is found to be at position x2. The Uncertainty Principle merely informs us that the value of x2 is quite likely to fall in the above mentioned associated interval.

Of course, having made that measurement, we can then infer a momentum for the system over the duration of the time interval (t1,t2). This refers to "what was".

Next, if we then look back on the whole affair, we realize that the result x2 which we obtained did not have to be what is was. Rather, as required by the Uncertainty Principle, it could have been anything consistent with the associated interval of x-values. And this, of course, refers to "what could have been".

I repeat:
Is what I meant clear enough now?

15. Jun 29, 2006

### RandallB

Better than that, fall back on why they needed to create that “formalism” and the view used by QM.
QM gives a better solution to describing WHEN and WHERE (location & momentum) at atomic level measurements, than classical. Making it more useful to follow the QM explanation, even as it turned out to be classically non-local.

You don’t have start with the chicken and demand they understand QM theory, you can start with the egg and how a solution to real problems was raised from that egg.

16. Jun 29, 2006

### prochatz

I don't think that the problem is the method we use. The Uncertainty Principle does not depend on the method. It's a part of our world. The wrong must be somewhere else. (I totally agree with the procedure and the results you mentioned).

In my opinion, neophysique' s thought is wrong on the derivation. I don't even know if we can use the length vector in such problems. It seems too classical.

17. Jun 29, 2006

### masudr

In quantum mechanics, observables are not functions of co-ordinate space.

The transition from CM to QM is easily made from the Hamiltonian formulation. In this approach, our state is defined on a state space, separate from the phase space.

Observables are operators defined on the state space, where the form of the operator comes from replacing x by the $\hat{x}$ operator and p by the $\hat{p}.$ In the x-basis, these become $\hat{x}=x$ and $\hat{p}=-i\hbar d/dx$.

Therefore, per definition, possible values of momentum is given by the eigenvalues of the momentum operator, not mv.

18. Jun 29, 2006

### Eye_in_the_Sky

The prescription offered by neophysique is a valid one. It is correct to say that at time t=1 the particle is at the place in which it has been found to be, and moreover, that it must have had a particular momentum in order to get there.

Here are Feynman's words on this sort of question:

19. Jun 30, 2006

### Eye_in_the_Sky

I must clarify

This requires clarification.

The said "validity" and "correctness" is understood to be in a context where 'permission' is granted to engage in a semi-classical type of analysis. Beginner and intermediate textbooks on Quantum Mechanics are replete with examples of such analyses, the purpose of which is to give the reader a deepened understanding, albeit from an informal, and perhaps it is even fair to say "ill-defined", point of view.

So, strictly speaking (i.e. when 'permission' is not granted) there is no room for neophysique's analysis. It is as prochatz says, the "thought is wrong on the derivation ... It seems too classical", and as masudr explains, "possible values of momentum is given by the eigenvalues of the momentum operator, not" m(x2-x1)/∆t.

Nonetheless, if we accept that 'permission' has been granted and we want to push our classical way of thinking as far as it can go, then a rule like ∆x∙∆p~h (now stripped away from its purely formal -- but unambiguous -- meaning in terms of operators and state vectors) needs to be clarified. What I have indicated in my earlier posts, that the Uncertainty Principle must then apply only to prediction, but not to retrodiction, gets the job done.

The above having been said, it is now a good idea to go back and look at the problem again, this time around from a strictly "orthodox" perspective. So, here we go:

A measurement of position is performed at time t=0, and then again at time t=1. During the intermediate times 0<t<1, we cannot say that the particle must have been somewhere, and that therefore it must have described some kind of path. No, on the contrary, all we can say is that over the course of those intermediate times the particle was in a state of "superposition of possibilities" described by a wavefunction evolving according to Schrödinger's equation, and then, at time t=1, a particular possibility became actualized.

If we ask the question, "What path did the particle take to get from its initial position to its final position?", then we must answer that the particle had some amplitude to go along each and every path, and when we compute these amplitudes and combine these paths according to the appropriate rules, then we are able to compute the probability for the particle to reach its final position, given its initial position. In short, since we did not measure the path, the particle did not have a (particular) path. Therefore, the particle did not have a well-defined momentum over the course of the intermediate times 0<t<1.

That having been said, perhaps now someone else would like to explain it all over again, this time around from the perspective of MWI, and then again, from the perspective of Bohm.

20. Jun 30, 2006

### HallsofIvy

My point was not that it depended on the particular method of observation but rather to show how, for a particular method, the observation necessarily changes the situation. Some other method of observation would entail a different mechanism but still change the situation. By "accurately" observing the position of a particle twice you can calculate the momentum just before the second observation but that will not be the momentum after the observation.