# Heisenbergs Uncertainty Principle

1. Jul 5, 2008

### Niles

Hi all

While watching the first stage of the Tour de France, I thought of Heisenberg and his Uncertainty Principle.

Does the Uncertainty Principle also work in the macroscopic world? I.e., can I tell with 100% accuracy how fast a car is going and where it is?

Regards,
Niles.

2. Jul 5, 2008

### CompuChip

We usually treat objects like cars classically, neglecting any quantum effects. So effectively, $\hbar = 0$ and there is no uncertainty. But if you want to write down the wave function of the combined system of electrons, protons and neutrons that make up the car, be my guest

3. Jul 5, 2008

### muppet

The theoretical guts of the HUP is based on the idea that prior to measurement, the properties of a system are fundamentally indeterminate; after you've measured it as having some particular values you can go on to measure even canonically conjugate variables to arbitrary accuracy.
Decoherence theory would say (I think!!) that any thermodynamically irreversible interaction is what constitutes a measurement. By virtue of its size, the car will be undergoing billions of such interactions (heat exchange with its surroundings, light bouncing off it, etc).
So the short answer would be no

4. Jul 5, 2008

### kahoomann

How can you tell with 100% accuracy how fast a car is going? You have to measure it first. But the measurement itself would change its speed at the same time. So the accuracy will be less than 100%

5. Jul 5, 2008

### CompuChip

Right, but as soon as you started measuring the conjugate variables, the original values you got would immediately become unreliable.

Suppose I set up two lasers at a very accurately measured distance, I measure when the light beams are blocked by the front of the car and I do the calculation (taking into account the processing speed of my circuit, the finite speed of the light and the signals in my wiring, etc). How does that not give me the velocity of the car to arbitrary accuracy? There is nothing in the HUP that prohibits me from measuring it as precisely as I please (and am able to set up my experiment).

6. Jul 5, 2008

### Crosson

I think this method would work, in principle, only until the measurement precision reached the wavelength of the laser light, since then we would have to take into account the exact nature of the reflection on that scale.

Your measurements of the reflected laser beams are subject to the energy-time uncertainty principle. In order for your method to work, you need to measure the exact time at which the reflected beam arrives. But this introduces a corresponding uncertainty in the wavelength which introduces all kinds of trouble.

7. Jul 5, 2008

### HallsofIvy

Even in the macroscopic world you have to distinguish between "100% accuracy" and "accuracy within the limits of my instruments". If you are going to measure the speec of a car with "100% accuracy" you will need to tell us how your instruments are going to give that accuracy.

8. Jul 6, 2008

### Niles

Hmm, I can see we are dealing with uncertainties from the instruments and from HUP. Let us just say that we can neglect the uncertanties from the instrument, so what they tell us is 100% accurate (it's just a thought-experiment).

Isn't the setup where I measure the speed of the car equivalent to "Heisenbergs Microscope" (http://en.wikipedia.org/wiki/Uncertainty_principle)? [Broken] So even in the macroscopic world, 100% accuracy doesn't exist because of HUP (still neglecting the uncertanties from instruments)?

Last edited by a moderator: May 3, 2017
9. Jul 6, 2008

### ZapperZ

Staff Emeritus
Have you encountered any situation where, the MORE accurate you measure the position of a classical object, the less you are able to predict its momentum? I haven't, and I haven't seen any report of such an observation.

So lacking that fact, we have no basis to claim that the HUP exists in a classical measurement. If it does, then it shouldn't have been that weird in the first place.

Zz.

Last edited by a moderator: May 3, 2017
10. Jul 6, 2008

### kahoomann

Can you tell me where is the boundary between a classical measurement and a quantum measurement if such thing exist? Cars are classical objects but are made of atoms, so quantum mechanics still apply.
That's why Schrodinger use Schrodinger's cat to show how absurd it is since cat is a classical object

Following quote is from http://en.wikipedia.org/wiki/Uncertainty_principle
In quantum mechanics, the position and velocity of particles do not have precise values, but have a probability distribution. There are no states in which a particle has both a definite position and a definite velocity. The narrower the probability distribution is in position, the wider it is in momentum.

Physically, the uncertainty principle requires that when the position of an atom is measured with a photon, the reflected photon will change the momentum of the atom by an uncertain amount inversely proportional to the accuracy of the position measurement. The amount of uncertainty can never be reduced below the limit set by the principle, regardless of the experimental setup.

Last edited: Jul 6, 2008
11. Jul 6, 2008

### ZapperZ

Staff Emeritus
I've cited several papers already https://www.physicsforums.com/showthread.php?t=241396" trying to induce more and more decoherence onto a system. A paper that I had highlighted showed that even ONE interactions with another particle can destroy the single-particle state to behave more in a classical fashion.

Here's what we know: classical system and quantum system are very much different from each other. Where they meet, and how they meet, no one knows. We have seen "large" systems that can behave quantum mechanically, ranging from something as large as a buckyball, to a superfluid having 10^11 particles. However, these are quantum systems, not classical. To simply infer that there is a continuous crossover between classical and quantum systems is to trivialize the problem via an unfounded speculation. If it is that trivial, why are we continuing to pour money and effort into studying this boundary?

Zz.

Last edited by a moderator: Apr 23, 2017
12. Jul 6, 2008

### Crosson

ZZ, Phillip Ball and decoherence theory are addressing the transition between quantum superpositions and classical mixtures. The reason we need to address this is because the existence of quantum superpositions contradict normal classical experiments (I'm not talking about BEC, etc).

In contrast, the HUP does not contradict normal macroscopic experiments. There is no need to search for a HUP transition, because there is no contradiction, the effects of HUP become vanishingly small at macroscopic scales.

Similarly, we do not need to explore the transition between continuous and discrete energy states, because we can show that discrete states become arbitrarily close together as h -> 0.

Prior to decoherence theory, there was no way to make superpositions transform into mixtures by letting h get small, or m get large, etc, that is the reason for all the effort you mention.

13. Jul 6, 2008

### ZapperZ

Staff Emeritus
But that is my point. We cannot detect the effects of HUP for classical measurement even IF it is there (and that is a big IF). Thus, that is why I asked for signatures of the HUP in classical measurement. We know how the HUP rears itself in quantum measurement and that's who we know the HUP is valid.

It is still isn't obvious when and where a quantum system becomes totally classical where all the quantum behavior goes away. It is why this is continued to be studied and tested. Maybe the experimentalist in me is rearing its ugly head again, but considering all the myriad of tests and experiments being planned to study this, I'd say that the whole community of physics wants to know as well. This means that this is not a done deal where we can simply think that there is a smooth, continuous boundary between the two.

Zz.

14. Jul 6, 2008

### Crosson

The question was "Does HUP also work in the classical world?"

The answer should be "yes, because HUP is a fundamental property of matter and we have no reason to believe it should go away at a certain scale."

Compare this with superposition, which we know does go away. That's why we need to detect the boundary. There are no experiments that indicate the HUP goes away at some large scale, so why would we look for the transition?
What does your 'this' refer to? I don't know of any experiments being planned to test the HUP in the classical domain. If 'this' refers to 'the classical/quantum transition' then I think that statement is misleadingly vague, because the transition they are interested in is from superpositions to mixtures.

15. Jul 6, 2008

### ZapperZ

Staff Emeritus
I disagree. The HUP is actually isn't "fundamental", but rather a consequence of what is some time called the First Quantization. It is how we define what observables are and how 2 non-commuting observables influence each other. In fact, and I'm sure you know this, we seldom actually use the HUP in detailed calculations, and certainly not anything based on First Principles, even in QFT. While we invoke the HUP to explain various effects (zero-point energy, deBoer effect, etc), you seldom can simply get away with just using it without any kind of formal derivation to describe a phenomena.

So one simply can't say the HUP works but superposition doesn't, since they are ALL related and came out of the same description. You can't simply unravel one without affecting the other. To me, the HUP is one of only a suite of consequences and properties (such as superposition, Hermiticity, pure/mixed states, etc... ) that comes along with the whole baggage.

Zz.

16. Jul 6, 2008

### muppet

It's worthwhile reiterating that the physical origin of the uncertainty principle has nothing to do with the act of observation. If you don't believe that, have a look at a proof of it. You won't find any reference to the wavelength of an incident photon from a gamma-ray microscope or anything like that. Remember that Heisenberg thought that a trajectory did not exist until it was measured; the HUP quantified an indeterminacy inherent in nature.
The reason so many people misunderstand the HUP is because Heisenberg, immediately after its formulation, carried out his famous thought experiment (which he infamously got wrong first time around...) about a gamma ay microscope. The reason he did this was to satisfy himself that you couldn't carry out an experiment which would immediately measure something to greater precision than was allowed by the HUP.

17. Jul 6, 2008

### Crosson

Yes, 'fundamental' was a poor word choice for the HUP. I should have said 'is derived under very general assumptions' which as Zz points out are not totally valid anyway.

Interesting, I thought the explanation would be along the lines of 'the decoherence times are short', 'the uncertainty is small', but you bring up the more interesting possibility of a total cutoff for quantum effects, rather then an asymptotic vanishing.

18. Jul 6, 2008

### ZapperZ

Staff Emeritus
Actually, I don't think I did, because I have no idea if it is simply a smooth crossover or an abrupt phase-transition type either. So I certainly can't insist on a particular viewpoint.

Now, having said that, and having cited the paper where even just ONE interaction with another particle is sufficient to destroy the single-particle quantum state and cause the system to start to resemble a classical description, one is tempted to think that maybe the transition is abrupt. I find this to be very tantalizing, but still quite far from being convincing.

Zz.

19. Jul 8, 2008

### kahoomann

The gamma-ray microscope is just an example or application of HUP. No one is saying it's the proof of HUP itself.

Following quote is from http://en.wikipedia.org/wiki/Uncertainty_principle
Physically, the uncertainty principle requires that when the position of an atom is measured with a photon, the reflected photon will change the momentum of the atom by an uncertain amount inversely proportional to the accuracy of the position measurement. The amount of uncertainty can never be reduced below the limit set by the principle, regardless of the experimental setup.

20. Jul 8, 2008

### ZapperZ

Staff Emeritus
But I could come in and change that, and replace it with this that is equally valid:

One can also determine the "instantaneous" position of an atom by shooting a stream of atom through a "slit". The ones that passes through the slit had that position the moment it passed through. Here, I used NO PHOTON to measure its position. The uncertainty it the position corresponds to the width of the slit.

I could also located an atom using something ranging from an STM to AFM. No photons there either!

So it is not a "requirement" that one must use a photon to measure a position. That's a fallacy.

Zz.

21. Jul 8, 2008

### Count Iblis

Yes, the uncertainty principle poses a fundamental limit on the accuracy of any measurement. A good example of a macroscopic manifestation of this effect that is commonly encountered in the lab is in electric circuits. Even if there is no EMF between two points, a voltmeter will still show fluctuations in the voltage. At room temperature, this is mainly due to Nyquist noise (also inside the voltmeter itself). But if you cool everything, inclluding your voltmeter down to absolute zero, there are still fluctuations in the voltage.

The fluctuations are in fact fluctuations in the power (voltage times current). In a quantum mechanical description, voltage and current are operators that do not commute. They satisfy commutation relations similar to position and momentum.

You can also consider a supercondicting LC circuit. Suppose the angular resonance frequency is omega. Then the circuit has energy levels of:

(n+1/2) hbar omega

The lowest possible energy the LC circuit can have is thus
1/2 hbar omega. So, you always have a fluctuating voltage and current and this is readily detected in the lab.

22. Jul 8, 2008

### reilly

First, the HUP. Of course it holds in all circumstances which are subject to Quantum Mechanics. Uncertainty relations occur for all linear wave equations. There are two issues: first, what does it mean physically?,2. under what circumstances can it be observed in action?

What does it mean -- the Fourier transform of a delta function is a constant, 1 in fact. The Fourier transform of a Gaussian is another Gaussian. For practical purposes, one can think of coordinate-momentum or time-energy (E&M) relations. The upshot of this math is that uncertainty relations. associated with linear wave equations, are responsible for the phenomena of diffraction - speaking somewhat figuratively. So, the HUP is a natural consequence of the Schrodinger Eq.

Let's sneak in superposition, another property of linear wave equations -- one that is linear in the wave function and its derivatives. We know that for virtually any F(x), that
F(X-cT) is a solution of Maxwell's free field wave equation. Suppose that F is a Gaussian, with unit variance. Where's the superposition in F? Well, we can expand F(Z) is a power series, with Z =X-cT, and clearly get a superimposed solution. Then with
Fourier transforms we can get another superimposed solutions as we can also with Hermite polynomials.

What does superposition mean? In and of itself, it does not mean much of anything. Its just a useful mathematical procedure. But, just as spherical coordinates facilitate the description of central potential problems, the frequency-wavelength approach is particularly suitable for diffraction problems. Here's where the superposition comes in, via Hugyen's Principle, which says that any slit through which light -- or water, or... -- passes acts as a distributed source of radiation -- hence superposition of sources is in play. Plug in the notion of optical path-length, and the diffraction distribution will emerge; just as velocity of a rowboat in a stream can be found by a superposition of appropriate vectors.

On the other hand, the Rutherford cross section is identical whether computed classically or quantum mechanically, and shows no signs of superposition, at least at first glance.

Superposition-- just a common garden mathematical technique, sometimes useful, sometimes not. In optics, in particular, it proves to be a very powerful tool, which makes the explanation of lots of phenomena very simple.

When is all this stuff detectable? Macroscopically, generally not. The fluctuations, ultimately induced by the HUP, are too small to be detected, and are averaged out anyway. But phenomena like superconductivity and other cooperative phenomena certainly occur -- in a sense what happens here is that lots of states combine, superimpose if you will, so that the components of the system are in lock step.

In the quantum world, "slits" must be no more than a few hundred angstroms , a few wavelengths-- if I remember correctly -- to generate a diffraction pattern. That's pretty small. But, you can see lots of wave phenomena with water waves.

Regards,
Reilly Atkinson

23. Jul 8, 2008

### Crazy Tosser

Heisenberg's uncertainty principle applies to the macroscopic world. Here is a problem from an old QM textbook:

A snooker ball of mass 0.1 kg rests on top of an identical ball and is stabilized by a dent $$10^{-4}$$m (1 millimeter, pretty damn big) wide on the surface of the lower ball. Calculate how long the system takes to topple, neglecting all but quantum disturbances.

Now, that's a realistic question, and an experiment like that could be carried out in theory. The answer is $$10^{27}$$ seconds... And the question is macroscopic...

24. Jul 9, 2008

### ZapperZ

Staff Emeritus
"... an experiment like that could be carried out in theory...."

The same can be said about any String Theory predictions.

A question in a textbook tends to IGNORE a whole lot of other non-negligible factor, because the pedagogical idea in presenting it is to test the knowledge of a principle, NOT to present any realistic case. A "ball" isn't a coherent object. That is the most crucial aspect of this. It is why in all the treads asking for the probability of a ball "tunneling" through a wall (it could be "carried out in theory") is rather meaningless unless one simply ignores a whole set of realistic conditions and make it into an unrealistic problem. That is what is going on here. It is why it takes a superconductor to produce a coherent, "macroscopic" object with 10^11 particles where quantum effects are now evident. It isn't easy!

Zz.

25. Jul 9, 2008

### Count Iblis

The fact that a pencil when balanced on its tip cannot remain in that state for longer than a minute due to quantum fluctuations alone is better example. The fact that the pencil isn't a coherent quantum object is not relevant in this estimate.