# A question about the uncertainty principle.

1. Mar 3, 2008

### kenewbie

My understanding of it: You cannot measure both the location and momentum (as in vector) of a particle and have both be correct to a given percentage, where that percentage is the uncertainty.

Lets say I build a long tube. Theoretically this tube is so small that you can send a single photon down, nothing bigger. Now, I pick at spot in this tube, to measure when a photon travels past. If this is possible, I I know where the photon is at a given time.

And, a photon always travels at the speed of light, right? So I know a large amount of the momentum as well, and given the tube there are very few directions for the vector to point, as far as I can tell, only 2.

What am I missing?

k

2. Mar 3, 2008

### malawi_glenn

a photon have no size, neither an electron or any elementary particle. So there is not even in principle to construct that device of yours.

3. Mar 3, 2008

### Fredrik

Staff Emeritus
For starters, you won't know its momentum unless you know its wavelength.

But suppose that we put in some filters before the detector, filters that only let a narrow range of wavelengths pass through (e.g. only "blue")...

I started writing an explanation, but I realized that it's going to take too long to write a good one, so I'll leave it for someone who isn't as tired as I am right now.

The bottom line is that the uncertainty principle must hold, and this tells us that if the detector really "squeezes" the photon into a short region of space (so that we can say that we know where it is), then it can only do so by "unsqueezing" the photon's wavelenth into a much wider range (so that we can't say that we know its wavelength).

4. Mar 3, 2008

### Fredrik

Staff Emeritus
That's technically correct of course, but the wave function is often peaked around a certain position, and the peak has a certain width...well, you know the rest.

5. Mar 3, 2008

### malawi_glenn

yes, but my arguing was that you cant even in principle construct that device he proposed ;)

6. Mar 3, 2008

### Gigi

Well, quantum mechanics is all about measurement and observers.

Heisenberg's Uncertainty Principle states that you cannot measure simultaneously the precise position and momentum of a particle.

In your case, lets assume that you make a measurement of the position of the photon, thus you know it without any uncertaintly.

What you say after that is that from theory you know that photons have a speed of c, thus you can work out the momentum associated with them.

Yes, you can theoretically, but you cannot measure them at the same time. Thus in this case you are not talking about simultaneous measurements- which is what Heisenberg's uncertainly principle prohibits, but you are talking about 1 measurement and 1 theoretical calculation.

That is where your assumption fails.

7. Mar 3, 2008

### Valery

Is Heisenberg principle is a proven fact, or it is just a principle, current state of the scientific knowledge? I see that some really and religiously believe that it is absolutely true. What if Einstein was right? Perhaps, that one day it will be disproved.

Keep trying!

8. Mar 3, 2008

### malawi_glenn

Proofs regarding its form $$\Delta x \Delta p$$ exists, and also on the form $$\Delta E \Delta t$$

See for example natural spectral broadening, zero point energy and degenerate (gas)pressure.

9. Mar 3, 2008

### Valery

I alway wonder, what if there is some undiscovered formula or law of nature that connects the momentum with the position. That is, if we know the rule and one of the parameters and then plug it into that formula and, bingo, we know the angular momentum. What would it do to the uncertainty principle?

10. Mar 3, 2008

### malawi_glenn

where is angular momentum?.. p is linear momentum.

but now there exists no things as hidden variables, there are millions posts here about hidden variable theories and Bells theorem etc.

The uncertanty relation just follows very very simple from the position and momentum operator commutator :)

11. Mar 3, 2008

### Valery

O.K., momentum. My mistake.

12. Mar 4, 2008

### kenewbie

Glenn: Meh, ok. I should probably get more basic knowledge about particle behavior in general before I start wondering about things like that :) It sort of just hit me that we always know the speed of a photon, and was wondering if that makes for a special case with the uncertainty principle, since we have one variable for an equation figured out ahead of the actual measurement.

Gigi: That did not make any sense to me. Are you saying that if you can measure the momentum of a particle then take a calculator and use 10 minutes to calculate the position it had at the time when you did you measurement, that does NOT violate the uncertainty principle?

My understanding is that it is not a principle against measurement, it is a principle against knowledge.

On another note, which I am sure stems from my gross ignorance on the subject, I cannot shake the feeling that the whole measurement problem is related to us trying to measure a cue ball by throwing other cue balls at it. That the uncertainty is not "physical", it is merely our inability to do better measurements at the moment.

This is all extremely fascinating and I need to read up on a whole lot of physics in a very short time :)

k

13. Mar 4, 2008

### Gigi

Hi:) No, this is not what I meant.

Let me put it simply.

The uncertainty principle says that you cannot 'measure' the exact momentum and position of a particle at the same time. The idea of measurement implies that there must be an observer that will use some set up to measure.

In your case you measure experimentally-observe - the position thus there is a collapse of the wavefunction and then you proceed to make a mere calculation of the momentum based on theory.
You do not suggest simultaneous measurements, but rather one observation-measurement and one theoretical calculation.

That cannot violate Heisenberg's principle. It would only be violated if you made 2 simultaneous measurements, yielding the exact value of the position and momentum.

Last edited: Mar 4, 2008
14. Mar 4, 2008

### kenewbie

Gigi: Einstein's box of light:

(from wikipedia)
This is an after-the-fact measurement, and if it could be done it would violate the principle. How is this different from my after the fact calculation?

k

15. Mar 4, 2008

### Gigi

Well they are proposing a measurement using experiment-an observation - and you are proposing a theoretical calculation. These are entirely different routes, as in the first case you have a collapse of the wavefunction, in the second you dont.

Now in the case they are describing, see what they say in the end: that it is 'naively possible' to violate Heisenberg's uncertaintly principle.

Why naively? First of all the measurements suggested are not simultaneous.

If one makes a measurement on a wavefunction that represents the state of the particle, then the wavefunction collapses and the state changes. Thus any subsequent measurement would be made on a different state.

Even when we define the expectation value, i.e. talk about the average position and so on, we say in the definition that it is the average of repeated measurements performed on an ensemble of particles all in the same state and not the average of repeated measurements on one and the same syste.

Heisenberg's Uncertainty Principle goes hand in hand with a number of concepts in quantum mechanics.

Last edited: Mar 4, 2008
16. Mar 5, 2008

### kenewbie

I would think the naively refers to the fact that no such scale exists and that he implores you to ignore that for the sake of argument.

And the box of light is even less of an direct observation than my "idea" was. The box never touches a photon at all, it uses only second hand data to perform a calculation. My set up does at least use direct measurement to get the position.

I am sorry, but it seems to me that you are just talking and don't really know much about this and just make it up as you go. This feeling that I have could also just be me wailing hopelessly around in an area where I know absolutely nothing, and so I fail to understand your arguments.

For the sake of understanding, I would really like a second opinion on Gigis statements. I want to understand this. Please dont take this the wrong way Gigi, but you have 16 posts total and could be just a random person trying to confuse me for fun.

Anyone feel like commenting?

17. Mar 5, 2008

### Gigi

No, I am not trying to confuse you at all- there is no point-I think my life is busy enough as it is.
I have 20 posts as I only registered 4 days ago!

As I said before, a calculation is not an observation that would collapse the wavefunction. You can read about measurement in QM in a number of books and the effect it has on the wavefunction.
You can also read the definition of the expectation value and its relationship with Heisenberg's Uncedrtainty Principle.

All this can be found in Introduction to Quantum Mechanics by David Griffiths, as well as another book that is much simpler and descriptive by Jim Al Khalili called A guide for the perplexed.

Last edited: Mar 5, 2008
18. Mar 6, 2008

### Fredrik

Staff Emeritus
I don't think the idea of measuring position and momentum simultaneously makes sense. The reason is what you said here:

Would you collapse the wavefunction in two different ways at the same time? Then you'd have two wavefunctions.

This is how I think about the uncertainty relation:

One of the things that QM tells us is that neither the position nor the momentum of a particle is well-defined. The uncertainty of an observable (like position or momentum) is a measure of how ill-defined that observable is when the particle is in is current quantum state. So what the uncertainty principle tells us is that by making the position more well-defined we're making the momentum less well-defined, and vice versa.

Not exactly. The measurement would only tell us that the photon is localized in a small region of space. But you can make that region very small if you'd like, so the uncertainty can also be made very small.

That's not correct. Knowing the speed isn't enough. You need to know the wavelength.

I don't agree with what you're saying about theoretical calculations vs. measurements. If he could measure the position with a small uncertainty and calculate the momentum (correctly) with a small uncertainty, then this would violate the uncertainty principle. The reason he can't calculate the momentum (with a small uncertainty) is that the position measurement has changed the quantum state of the photon (into a state with a large uncertainty of the wavelength).

You probably just meant that his calculation would be irrelevant if he calculates the wavelength uncertainty of the state the particle is in before the position measurement, when he should be calculating the wavelength uncertainty of the state the particle is in after the position measurement. You probably understand these things well enough, but you explained them in a strange way.

Last edited: Mar 6, 2008
19. Mar 6, 2008

### kenewbie

Thank you, that is how I have always thought it was. That the principle states that you CANNOT pinpoint both momentum and position to a certain degree, regardless of HOW you go about doing it.

I am sorry, I did not mean to disrespect you in any way. Put yourself in my position, I know very little about QP, and what you said goes against one of the few things I thought I knew. Then we start discussing it and I cannot understand how some of the statements you make avoid contradicting each other. So I wanted a second opinion of sorts, someone to confirm or elaborate on what you said so I can understand it. Again I apologize, I'm just not quite sure how to go about it differently.

k

20. Mar 6, 2008

### reilly

All you need to do is to say that what Gigi, or anyone else, said disagrees with your understanding and, perhaps why. A personal attack says a great deal more about the attacker than the attackee.
Regards,
Reilly Atkinson