Do I understand Heisenberg?

[drkitten]

You are wrong, actually. You can talk about the velocity of an object at a precise instant. It doesn't matter that the limit never actually reaches zero. It becomes so small that it is negligable, and drops out of the equation. Any term that is a constant multiplied by this limit also will drop out.

Um, no, you can't. Not in a physical system.

Heisenberg's uncertainty principle deals with uncertainty in measurement; any physical measurement you take of position will of necessity be an estimate of the "true" position, plus an error term. Similarly, any measurement of time that you take will inevitably be an estimate of time. Since the error term is not[\i] related to your limits, the errors will inevitably dominate your calculations, and you'll basically be taking the limit as estimated time goes to zero of a random error function. Hardly "negligible."

This is old stuff and applies even in a Newtonian universe -- with machines with finite precision, you cannot achieve infinitely accurate results. Heisenberg's accomplishment was to show that, under the assumptions of quantum mechanics, there is a limit beyond which the measurement precision cannot go due to the way in which the observed and the observer interact in the process of measuring.

 

[hammegk]

Heisenberg's accomplishment was to show that, under the assumptions of quantum mechanics, there is a limit beyond which the measurement precision cannot go due to the way in which the observed and the observer interact in the process of measuring.

Hmm. I'd contend the measurement problem can be described mathematically (in the limit) by H, but that isn't really the point of his principle.

 

[pgwenthold]

You are wrong, actually. You can talk about the velocity of an object at a precise instant. It doesn't matter that the limit never actually reaches zero. It becomes so small that it is negligable, and drops out of the equation.

Um, "negligible" is not the same as zero. Moreover, "negligible" is only meaningful in context of the available precision. When you say "negligible," all you are saying is that we lack the precision to look that close. The HUP says that it doesn't matter how close you look, you can't see that limit.

 

[rwald]

Renfield, while you can take limits as dx goes to zero in a mathematical, imaginary universe, in the real universe you can't. So even though calculus says you can find the instantaneous velocity and position, calculus doesn't fully model reality; quantum mechanics comes a lot closer (at least on small scales), and says that there is a fundamental limit to the accuracy that you can measure (and again, this limit isn't a function of the measurement equiptment; it really is in the system being measured).

 

[Dancing David]

I'm not so sure about that. Its by measuring (ie observing) the particle that it has to decide where it is, or what its momentum is. Until then it only exists in a sort of indeterminate state. We have to observe its momentum/position for its wave function to collapse.

It remains a wave function, it is a wave function, the wave does not collapse, it intersects another wave function and produces changes in both wave functions. They remain wave functions.

 

[Dancing David]

posted by Renfield

You really can find the exact velocity of something at a particular instant.

Yes, but on the quantum scale, the precision of the velocity is inversely proportional to the location of the vector. If you find the velocity with absolute precision then the ability to determine placement and direction of the vector decreases, by the HIP.

 

[TeaBag420]

Renfield is wrong, my original answer is right.

 

[Brian]

Thought of something else. I was thinking of a photograph. What do you call the unit of time that a photo of a moving object captures?

Let's see if I can explain what I think I mean.

Set up a camera with no shutter in a perfectly dark room, Set an object in motion at a known, measured speed (measured in k/s?)
Set of a flash that flashed precisely as long as it takes light to reach the object and bounce back to the camera.

Oh oh getting fuzzy now.

Scratch all that, I'll leave it so you'll know what I was thinking.

Simpler. Is the distance the object traveled in the time it takes light to hit the film at a certain distance the most precisely you can determine a moving objects location? Is that what the picture would be of?

Of course the closer the camera was to the object the shorter the distance the object traveled would be... Ah, but some one above mentioned a threshold of how close 2 objects can get... so if I get a real tiny camera...
oh, never mind

BTW- it was made clear to me that this has absolutely nothing to do with Heisenberg.

 

[drkitten]

Simpler. Is the distance the object traveled in the time it takes light to hit the film at a certain distance the most precisely you can determine a moving objects location? Is that what the picture would be of?

In relation to Heisenberg, you're actually not far off. In order to take "a picture" of an object, you need to shine light off of it and observe where the reflection(s) hit.

Unfortunately, light is made of photons, which have energy, and thus mass, and thus momentum. So when you shine light on an object, you're actually bouncing the equivalent of very small baseballs off of it, and the object will absorb some of the momentum from the baseballs, and thus change its velocity. So to measure its position (by bouncing photons off of it) will put uncertainty into its velocity.

(You can see this intuitively on the macro scale if you imagine trying to determine where a bowling pin is by rolling bowling balls towards it. Yes, you can hit it, and even detect the noise when you do, but as soon as you hit it, you've knocked it somewhere else entirely. For "bowling pin" read "subatomic particle," and for "bowling ball" read "high-energy photon.")

You could try to avoid this by using lighter and lighter baseballs, which really means using photons of less mass/energy, and hence lower frequency. However, photons of lower frequency (quantum wave effects), so you affect the momentum less, but you also get a less clear picture of the position. To get a good picture of the position, you need powerful photons (and lots of them), to get a good picture of velocity, you need to use as few photons, of as low energy, as possible. You can't do both at once --- hence the Uncertainty.

 

[garys_2k]

But again, it is not a measurement issue, it's that the position and momentum are fundamentally blurry at quantum scales. No need to bring photons into it, even in the dark those properties are linked and simultaneously mutually imprecise.

 

[drkitten]

But again, it is not a measurement issue, it's that the position and momentum are fundamentally blurry at quantum scales. No need to bring photons into it, even in the dark those properties are linked and simultaneously mutually imprecise.

That is a widely accepted interpretation of Heisenberg's mathematics, yes, and the one that Heisenberg himself believed. Einstein, Bethe, and many other influential physicists did (and do) not accept this interpretation.

The Copenhagen interpretation has never been proven or disproven. I therefore take an instrumentalist view that your statement above is to the best of our current knowledge neither true nor false.

 

[garys_2k]

So how do phonton to particle interactions influence a two-slit experiment, say where the particles being sent through the slits are quite heavy wrt the photons (buckeyballs, viruses)? AFAIK the results of these experiments do not show a dependence on ambient photon energy levels.

 

[drkitten]

So how do phonton to particle interactions influence a two-slit experiment, say where the particles being sent through the slits are quite heavy wrt the photons (buckeyballs, viruses)? AFAIK the results of these experiments do not show a dependence on ambient photon energy levels.

No, because in the two-slit experiment, ambient photon energy is not the measuring instrument.

The instrumentalist interpretation is that the result of the experiment depends on the measuring instrument. I'll leave it to your interpretation to determine how the result of the two-slit experiment hinges on how many slits you leave open.....

 

[garys_2k]

No, because in the two-slit experiment, ambient photon energy is not the measuring instrument.

The instrumentalist interpretation is that the result of the experiment depends on the measuring instrument. I'll leave it to your interpretation to determine how the result of the two-slit experiment hinges on how many slits you leave open.....

OK, fair enough, thanks.

I guess it comes down to differences in opinion as to what is going on at the most fundamental level. I'll just leave it to say that all evidence of property blurring is consistent with the Copenhagen interpretation (at least as I understand it).

Whether particles are measured with slits or photons, or even directly observed (as was done very recently), the amount of fuzziness is consistent with Heisenberg's original equation. I don't see how that consistency across diverse measurement platforms could happen if it was based on measurement precision.

 

[Skeptic]

Do I understand Heisenberg?

Maybe you do, I'm not certain.

OLD JOKE:

A cop stops Werner Heisenberg on the road. He asks him: "Do you have any idea how fast you were going???". Heisenberg says, "no, but I know exactly where I am!".

Speaking of which, I am "two degrees of seperation" from Heisenberg. I know someone who met Heisenberg and even shook his hand.

When I asked him why, he said that it is an honor to shake the hand of a great scientist and Nobel prize winner. I expressed the view that's its a bit less of an honor to shake the hand of the head of Hitler's nuclear bomb program.

 

[TeaBag420]

Do I understand Heisenberg?

Maybe you do, I'm not certain.

OLD JOKE:

A cop stops Werner Heisenberg on the road. He asks him: "Do you have any idea how fast you were going???". Heisenberg says, "no, but I know exactly where I am!".

Speaking of which, I am "two degrees of seperation" from Heisenberg. I know someone who met Heisenberg and even shook his hand.

When I asked him why, he said that it is an honor to shake the hand of a great scientist and Nobel prize winner. I expressed the view that's its a bit less of an honor to shake the hand of the head of Hitler's nuclear bomb program.

So is it safe to assume that you're not a big fan of the Apollo space program?

 

[TeaBag420]

In relation to Heisenberg, you're actually not far off. In order to take "a picture" of an object, you need to shine light off of it and observe where the reflection(s) hit.

Unfortunately, light is made of photons, which have energy, and thus mass, and thus momentum. So when you shine light on an object,

Nope. Photons are massless.

http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html