# Trying to understand Heisenberg Uncertainty Principle in a physical sense

1. Sep 29, 2011

### Runner 1

I'm trying to understand the Heisenberg Uncertainty Principle, as it relates to experimental measurements, because it's kind of confusing me. We just learned the derivations for it in my QM class -- basically it's two standard deviations multiplied together (corresponding to measurements of incompatible observables).

Before I explain what I'm asking, I want to be clear that I'm not trying to understand it from a "philosophical" angle, i.e. Copenhagen interpretation. I'm just trying to understand the aspects of QM as they relate to physical measurements.

So the usual soundbite is "The Heisenberg Uncertainty Principle says that the more accurately momentum or position is know, the less accurately the other one may be known".

Another favorite is "Even with a perfect measuring device, there is still an inherent uncertainty in knowing both the position and momentum of a particle".

These statements are meaningless to me, and they sort of gloss over a good physical explanation. I view them as cop outs -- like saying "Well, I don't really understand HUP, but I'm just going to repeat something everyone else says that sounds fancy and scientific so I still appear as if I know what I'm talking about".

Let's consider a hypothetical situation in which humans have perfected particle detectors down to Planck scale. And let's assume these detectors record data to a trillion significant digits. The detector is a big slab of some material, and when a particle hits it, it registers the particle's position on the slab, and the momentum as it strikes.

So what exactly does the operator of this detector see on their computer screen? Will it show two numbers, each with a trillion digits, or will the computers just shut off to prevent HUP from being violated? (I kid).

In other words, assuming we lived in a universe without HUP, how would the results of an actual high-resolution experiment differ from those with HUP?

Thanks if you can shed any light on this!

Last edited: Sep 29, 2011
2. Sep 29, 2011

### HallsofIvy

Staff Emeritus
Here is one way of looking at it- you can't measure anything without changing the result. For example, to measure the temperature of water, you put a thermometer in it, the thermometer heats up (or cools down) to the temperature of the water- but that, of course, cause the water to cool down (or heat up) slightly. Or to measure the air pressure in your automobile tires, you put your pressure gauge on the tire and let air pass through it. Of course, that means you have let a slight amount of air out of the tire, so the pressure is no longer what you measured.

How do we measure the position of, say, an electron? Shine light on it so we can see it, of course! Well, not visible light, necessarily, because an electron is far smaller than the wave length of visible light and looking at something smaller than the wave length you just see a blur. We have to be talking about electro-magnetic waves of smaller wave length. But the energy of light is inversely proportional to the wave length. That is, as you make the wavelength smaller, the energy with which you are hitting the electron becomes greater- which changes the electron's momentum. That's why you have this trade off- the more accurately you measure the position by reducing the wavelength, the more you lose accuracy on the momentum.

3. Sep 29, 2011

### Runner 1

So essentially, you're measuring the movement of one moving Frisbee using another moving Frisbee? (Very loose analogy -- I know).

This is sort of how I envisioned it, but why do people say "there is an inherent uncertainty even with a perfect measurement device"? What does that even mean? Since there is no such thing as a perfect measuring device, how can they know there is an inherent uncertainty?

4. Sep 29, 2011

### DrChinese

Another way to look at it is that there is no way to prepare a system in 2 non-commuting eigenstates. However, you can prepare a particle in commuting bases (such as position and spin).

I personally think it is best to think of it that a particle does not have simultaneously well defined values for non-commuting observables. Please note that tests with entangled particle pairs follow the HUP too. Can't beat the HUP.

You may be right when you say something about "we don't really understand" but the HUP really says all there is to say.

5. Sep 29, 2011

### Runner 1

Okay, I think this helps a lot...

Is there a citeable experiment that has actually tested HUP? Or is it too small in terms of scale such that we only have the theory at the moment? (I'm just curious to see an actual experiment directly affected by the limit so I can understand this a bit more).

6. Sep 29, 2011

### DrChinese

Here is an example, you can google and find more:

http://www.news.cornell.edu/stories/sept06/schwab.quantum.html

7. Sep 29, 2011

### xts

You may make such experiment yourself in half an hour at no cost.

Take a piece of alufoil, make (with a pin) a tiny hole in it, illuminate it with toy laser and watch the Airy's pattern on the opposite wall. Now - make a bit bigger hole and watch how the pattern changes.

That is exactly the experiment you want: as you measure precisely the position (small hole) the pattern is wide (transverse momentum of the photons gets uncertain, so the direction gets also uncertain). As you measure the position with poor precision (large hole) - the pattern gets narrow, as the momentum uncertainity is much smaller now.

As you know the wavelength (thus longitudinal momentum of photons), size of the hole and angular size of the pattern (thus ratio of transverse to longitudinal momentum) - you may check they are consistent with Heisenberg's principle.

Last edited: Sep 29, 2011
8. Sep 29, 2011

### Staff: Mentor

Yes.

But if you fire another identically-prepared particle at it (from some kind of gun that is built as precisely as possible), you get a different set of two numbers. Fire another particle, you get yet another different set of two numbers. After you've fired a lot of particles and recorded the results, you calculate the standard deviations of both the position and the momentum. These are $\Delta x$ and $\Delta p$, and they satisfy the HUP.

Furthermore, from detailed knowledge of the particle gun's construction, you can (in principle) predict only statistical quantities relating to the particles' positions and momenta at the detector, such as the mean (a.k.a. expectation value) and standard deviation.

Last edited: Sep 29, 2011
9. Sep 29, 2011

### Runner 1

Huh... not what I was expecting. So you actually do get numbers, but in a sense these numbers aren't "correct" (meaning you can't pin the nature of the particle down to two numbers) because they vary each time an identical experiment is performed?

10. Sep 29, 2011

### DrChinese

You can prepare particles in identical states. Their non-commuting bases will respect the HUP. With any individual particle, you can get numbers, but they don't really mean anything if they cannot be used to predict the result of another experiment.

The classical example might be a red striped sock. Next time I look at it, it is red AND striped. Quantum particles don't work that way. And yet they do have attributes that remain the same UNTIL you look at the other non-commuting attribute.

1. So for a classical particle: Red, red. Striped, striped. Red, red. Striped, striped. I.e. it is always red and striped.

2. So for a quantum particle: Red, red. Striped, striped. Green, green. Striped. Green. Plaid. I.e. what is it at any time? Who knows? And yet it will remain plaid until you check something else. Of course this is the ideal case, in practice you may disturb it additionally and get slightly different results. This is what can cause additional confusion.

11. Sep 29, 2011

### Runner 1

I have one more question. What you are saying is that identically prepared experiments give different results, and that it is only the statistics of these results that can be calculated. Is this correct?

And if so, is it possible that the state of a particle is dependent on the time elapsed since some universal reference time? In other words, the experiments are identical in all respects, except that they are performed at different points in time. (Of course I know this isn't possible -- this sort of question is simply to understand why not).

12. Sep 29, 2011

### 1mmorta1

The machine you describe will only measure position. The result will be an extremely precise measurement of WHERE the particle WAS. HOWEVER, the data will be useless, because taking such a precise reading on position will make the velocity infinitely immeasurable, and you will have no way of even pondering where the particle has ended up.

13. Sep 29, 2011

### 1mmorta1

Runner1, your thinking too much of particles as, well, particles. They are not. And they are not wave functions either....they are a duality. All the time. It is the choice of the observer to perform experiments which demonstrate one property or the other. The really tricky thing is that the wave/particles seem to be transcendent in that, whichever property you demonstrate, going back in time will reveal that property continuosly for the wave/particles tested, as if they knew which experiment you were going to perform. Very difficult to grasp conceptually, but proven mathematically.

14. Sep 29, 2011

### Runner 1

I'm not thinking of them as anything really. My question is about turning vague statements such as "They are not [particles]. And they are not wave functions either....they are a duality" into something experimentally demonstrable. And yes, for that particular sentence, the double-slit experiment demonstrates the wave-particle duality. My question concerns an analogous experiment for the Uncertainty Principle ...which DrChinese provided a link to, so my question is mostly resolved.

EDIT: Thought I'd explain with a table for you:

PHP:
Vague statement:         Electrons are waves & particles       An electron\'s position & momentum are inherently uncertain
Clarifying experiment:   Double-slit experiment                ???

Last edited: Sep 29, 2011
15. Sep 30, 2011

### DrChinese

Actually, a particle in an eigenstate remains in that state until something changes it. So basically, no, time is not the factor.

For example: imagine you have 2 clone particles (this is feasible). You can measure ANY idenctical observable on these and get the same result - always. And yet... if you measure different observables, say X and Y (whatever those happen to be, but they are non-commuting), you will find that neither happens to have Y and X as their paired value. Instead, the difference will be consistent with the HUP.

16. Sep 30, 2011

### haael

All problems with understanding Uncertainty Principle arise from thinking of particles as tiny hard balls. It is sufficient to imagine them as clouds to grasp the idea of fundamentally uncertain position.

To get why position and momentum measurements are incompatible, we need to employ a bit more maths. First, we must accept that "position" and "momentum" are only macroscopic concepts and do not need to actually exist.
Imagine any function. Now try to check its support (domain subset where it is nonzero) and period. It's easy to see that not all functions have any period (only periodic functions for that matter) and support is often bigger than a single point.

Actually, assigning a quantum particle position and momentum is just like trying to describe function's support and period with just one number. You cannot have both at the same time. Periodic functions have infinite support, functions with narrow support are not periodic.

Particles are not tiny balls. They are more complex concept.

17. Sep 30, 2011

### Runner 1

I think you replied to the wrong thread. But good explanation for particle/wave duality!

18. Sep 30, 2011

### 1mmorta1

Ah I see what you were asking. Do you feel satisfied with the answer you have to the question you were asking?

19. Oct 2, 2011

### Runner 1

In addition to the experiment DrChinese provided, I also found one on my own that is a GREAT explanation to the question I asked:

http://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_condensation" [Broken]

For Bose-Einstein condensate, Wikipedia says:

So in a world without HUP, the peak would be infinitely narrow. With HUP, it forms a mound. I think this example should be mentioned in QM classes as a way that something as abstract as the Uncertainty Principle relates to real data. I believe it would clarify a lot of things for many students, much like the double-slit experiment does for wave-particle duality.

Last edited by a moderator: May 5, 2017
20. Oct 2, 2011

### ThomasT

Hi Runner 1, I'm just curious why you didn't expect the reply that jtbell gave in post #8. You said in the OP that you wanted to understand aspects of QM (such as the uncertainty relations) as they relate to physical measurements. The basic inequality (Δx)(Δp) ≥ h seems to me to communicate a pretty clear physical meaning.

Last edited: Oct 2, 2011