# Heisenberg Uncertainty Principal

1. Jan 22, 2010

### Wannabeagenius

Hi All,

I understand that it is impossible for man to know both the position and momentum of a particle with arbitrary accuracy. This is a result of the physics of photons which are necessary to observe an object.

Now although man cannot know the exact position and momentum simultaneously, that doesn't mean that a particle does not have an exact position and momentum. The position and momentum is exact but man cannot know what it is.

I believe that there is something fundamentally wrong with my logic but I'm not sure what it is?

Thank you,

Bob

2. Jan 22, 2010

### f95toli

No, it is an intrinsic property of non-commuting variables. It is always true regardless if we are observing the system or not, and it has nothing to do with photons.

3. Jan 22, 2010

### Wannabeagenius

f95toli,

Should I be able to understand this from an elementary treatment of the subject or should I be patient and wait until I have a better foundation?

Thank you,
Bob

4. Jan 23, 2010

### bobquantum

First, we must understand what the Heisenberg Uncertainity Principal dictates
The postion and momentum of a particle cannot both be know exactly because particles can both act as waves or point particles. Therefore, the Heisenberg Uncertainity Principal means that there is actually no single point where the particle is at any point in time because the wave is a continuous function. There is a speed however, to as how fast the wave us travelling.

5. Jan 23, 2010

### Nuno Amiar

At the most elementary interpretation, Heisenberg's uncertainty principal states that for every two variables that have a special relation, such as position and momentum, the act of measuring one variable, lets say, position, is such a physical procedure that it affects the momentum of the particle. Therefore it is impossbile to know with an infinite precision both position and momentum at the same time.
To fully understand the principal and all its implications you must first know amongst other things what a state of a system is and its dynamics. That shouldn't be too hard to learn but it requires some study on the topic.

6. Jan 23, 2010

### SpectraCat

I thought that when quantum physicists misbehaved, they got sent to the Heisenberg Uncertainty Principal's office ...

7. Jan 23, 2010

### dlgoff

Last edited by a moderator: Apr 24, 2017
8. Jan 23, 2010

### Nuno Amiar

That put a smile on my face. I'm new to this physics jokes :p
I have a question, so please do correct my reasoning. Is ZapperZ suggesting that I could in theory arrange a system where I could know both position and momentum with infinite precision at the same time? If not, how should we interpret this limit? Isn't it that the act of observing changes the state of the system? Is there no contradiction between my reasoning and ZapperZ's?

Last edited by a moderator: Apr 24, 2017
9. Jan 23, 2010

### ZapperZ

Staff Emeritus
If you look at the HUP and the uncertainty in, say, position and momentum, you will notice that these are statistical spread in the values. It is similar to the variance of the measured value around some average value. It says nothing about one single measurement of the position, followed by one single measurement of the momentum. You can make each measurement as accurately as technologically possible, and the uncertainty in the measured value has nothing to do with the HUP - these are instrument's uncertainty/errors.

dlgoff: thanks for the citation. I didn't think anyone paid any attention to what I've written way back when. :)

Zz.

10. Jan 23, 2010

### Nuno Amiar

But isn't there a colapse of the wave fuction upon passing the single slit (measurement) which changes the shape and distribution of momentum?

11. Jan 23, 2010

### f95toli

The idea of a "collapsing wavefunction" is not really physical and most people (including me) don't like the concept, there is after all no "collapse operation" in QM.
The idea of a collapse goes back to the early days of QM where many people though that there was some basic difference between the microscopic and the macroscopic word, the "transition" from one to the other would be due to this collapse.
However, now we know that this is not correct, there is no sharp dividing line between these worlds. This is where things like decoherence etc comes into play.

Nowadays the idea of a "collapsing wavefunction" is really only used in pop-sci, in reality it is much more complicated than that.

12. Jan 23, 2010

### Nuno Amiar

Thanks for the fast response. This makes the whole topic a lot more interesting. I haven't grasped the slightest yet about QM. I still have to figure out why I'm wrong though. I thought the whole concept of the wave collapse was if you have a system made of superpositioned states that by performing a measurement on it it will stay in the state that was observed with a given probability, evolving in time with the dynamics of the Schrodinger equation. Is this not backed up by experience? Or is there any simpler way to account for this phenomena?

Last edited: Jan 23, 2010
13. Jan 24, 2010

### ansgar

ZapperZ's post is entirely correct, HUP is a statistical feature of many measurements - not a single one.

14. Jan 24, 2010

### Nuno Amiar

So, it is possible with a single experiment to measure both position and momentum arbitarly? Can anyone point to me such apparatus or thought experiment.

15. Jan 24, 2010

### ZapperZ

Staff Emeritus
You have seen those numerous experiments where photons or electrons with very low intensity pass through a double-slit, haven't you? Well then, how well you know the momentum depends on the "size" of the spot on the screen. The centroid of the spot gives you the ability to calculate the lateral momentum, and the width of the spot corresponds to the uncertainty in momentum. If I use a simple photographic screen, the spot is bigger. If I use a CCD, the spot will be smaller. If I use a CCD with better resolution and lower noise, it will be even smaller. All of this reduces the uncertainty in the momentum without ever changing the uncertainty in the position, meaning these values of uncertainty isn't part of the HUP at all. It is more of the function of the technology and methodology.

Zz.

16. Jan 24, 2010

### SpectraCat

Hmmm ... I guess I see what you are saying here ... since a single particle must necessarily give single-valued answers, the measurement uncertainties are not relevant to the HUP. Instead, it is the predictability of the result that comes in to question. In classical physics, if we knew the momentum and position, we could predict the future trajectory with arbitrary precision. However, in Q.M., since the particles can only be measured once (meaningfully anyway), the only way to test predictability is with multiple experiments. Even so, is it correct to say that the HUP doesn't come into play for the single measurement, because doesn't it define the range of possible values for the measurement?

17. Jan 24, 2010

### ansgar

That is a different issue, that issue is also applicable to classical systems as well. The main thing is that if even if we in principal we could measure the x and p with infinite precision there would still be a HUP when measuring several particles of the same system.

18. Jan 24, 2010

### Nuno Amiar

Thanks for the clarification. It makes sense that the principal should be viewed as a statistical implication and the problem I was thinking, to be both valid in classical and quantum physics.
Is there any known connection? I thought one of the postulates was to relate the eigenvalues of operators in order to define observables. Consequently the commutation relations (HUP) would mean that it is not indiferent to measure x then px or px and after x. Thus the connection between the statistical spreads and the measurement apparatus.

Last edited: Jan 24, 2010
19. Jan 24, 2010

### ansgar

have you derived HUP for your self?

20. Jan 24, 2010

### Nuno Amiar

Yes. Let $$\hat{A}$$,$$\hat{B}$$ two operators representing two different observables and let $$\sigma$$A,$$\sigma$$B be the respective standart deviations, then the HUP states that IF $$\hat{A}$$ and $$\hat{B}$$ do not commute then the product $$\sigma$$A$$\sigma$$B is always greater than the square of the (minus halve) commutator.
The standart deviations are the statistical spread ZapperZ was arguing about and the commutator is a result of the fact that measuring AB is different than BA. As a consequence, you can't know both A and B to arbitrary precision.

21. Jan 24, 2010

### ansgar

but "measuring AB is different than BA" is not the same as knowing both A and B with arbitrary precision.....

22. Jan 24, 2010

### Nuno Amiar

It's not the same, it is quite the opposite. Because A,B do not commute, you cannot know both to arbitrary precision.

23. Jan 24, 2010

### ansgar

why? where in the formalism and definitions of QM can you use to draw that conclusion?

the wavefunction is a statstical object, please dont forget that

24. Jan 24, 2010

### Nuno Amiar

That would have something to do with the basis of the space spawned by an operator A sharing the same eigenvectors of another operator B. If they share a common vector base then the state of the system is fully described in both basis at the same time. Otherwise this cannot be true. But that is a little more technical for me yet to be more precise on this issue.

Last edited: Jan 24, 2010
25. Jan 24, 2010

### ansgar

this issue of "can not be measured with infinite precision simoultaneously" is only found in popular explanations of QM, can you point to me a text book with full derivation etc which can tell that since A and B does not commute the observables a and b can not be known with finite precision simoultanously......? I think ZapperZ wants to know this too