Simple question?

1. May 1, 2006

johnnybgood

I often think about the physics that I have read. I am currently woundering about the uncertainty principal. As I have read, we can only determine the point of a particle or its velocity but not both. My question is; if we were to determine the point of a particle twice, would we not know its velocity on the second observation given the distance traveled in the time?

2. May 1, 2006

neutrino

Even for classical point particles, what you say holds true only if the particle is not accelarating. In the general case, it givens you the average velocity. But even if you say that an electron is not accelarating, the uncertainity principle still holds good at the quantum level.

The usual term is 'position'.

3. May 2, 2006

pivoxa15

At the atomic level, whenever you make a measurement hence disturb the particle, its wavefunction collapses. When you make the second measurement, that electron acts according to a new wavefunction and the second measurement corresponds to the second wavefunction. Therefore the two measurements are unrelated as they were dependent on two different wavefunctions. You cannot link the results from the two measurements and calculate its speed and so on.

4. May 2, 2006

ZapperZ

Staff Emeritus
I really, really, did not want to get into this because I foresee a very long and tedious discussion that has occured already many times on here. But I simply have to correct the wrong idea about the HUP here.

https://www.physicsforums.com/journal.php?do=showentry&e=79&enum=24

The HUP is NOT about the "uncertainty" in a single measurement. Let's get that out right away. I can make as accurate of a determination of the position AND momentum of a particle in a single measurement as I wish, limited only by the technology I have on hand.

For example, in the CCD screen at the end of an electron analyzer (see figure), an electron can make a DOT when it hits it. When that occurs, I know that at that instant, it was in that position. But I also can tell its lateral momentum due to the electron optics in the system, and that how far it has drifted laterally tells me the lateral momentum of THAT electron when it hits the detector. The ONLY uncertainty in the values are DETECTOR UNCERTAINTY, i.e. how many pixels per sq. inch on the detector, etc... This is NOT the HUP!

The HUP is based on the statistics of many repeated measurement under the identical situation. In classical mechanics, if you have such a situation, you ALWAYS get the identical answer repeatedly, allowing for a statistical variation due to inherent instrument errors/uncertainty. But you'd get a gaussian distribution of the values that you are measuring. In QM, you do not get that. As you try to confine the particle to an even greater certainty in position by narrowing the slit, for example, your ability to PREDICT where the particle will land will become fuzzier. This is reflected if you try to do this many time and look at the spread in where the particles land on the detector, even when they were prepared identically.

So no, the HUP isn't the uncertainty in a single measurement that you have made. It is a fallacy to think that "oh, I've made a position measurement, and so, my momentum measurement will produce huge uncertainty, or even undefined." That is wrong. You can make as an accurate of a measurement of those two as you wish. There's nothing in the HUP that prevents you from doing that. In fact, most experiments rely on such ability.

Zz.

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5. May 3, 2006

pivoxa15

So the HUP arises out of statistics and does not apply on a case by case basis? It becomes more relavant as you pile the molecules? So with just one measurement, I am able to get delxdelp much smaller than hbar/2 if my equipment is good enough?

6. May 3, 2006

ZapperZ

Staff Emeritus
How do you think $$\Delta x$$ is defined in the first place? Look at its definition and see if it makes sense when you have just ONE single measurement.

Zz.

7. May 3, 2006

Rach3

pivoxa -

There's a crucial distinction here. What you cannot do is make a statement about what are the x and p of an electron, and then expect your experiment to validate that. A measurement that measures x and p simulatenously is allowed; but it is neither predictable, nor even repeatable! You can even make your experiment measure both numebers to arbitrarily high precision, as with Zz's CCD's. The data you get however, is,

(i) not repeatable - repeating the measurements will get you totally different numbers

(ii) not predictable - complete knowledge of the physical state of the system is not enough (!) to let us calculate a measurement outcome*, unless we're in an eigenstate of our observable; of course x and p have no simulataneous eigenstates, so there's no predicting what will happen!

So "the position" of the electron, and "the momentum", are not compatible numbers in any physically meaningful way. How incompatible are they? Well, as it turns out, contrary to what I just said: you can make predictions, and you can get repeatable measurements! ...At a small cost, you have to allow some uncertainty. Take the slit experiment: you make only an approximate measurement of position (finite slit width), and get only an approximate momentum (on a finite area of the target). Within these constraints though, you get the same numbers every time! When you look at the statistics of thousands of photons and their images on the target, they've all hit the same place (to within an uncertainty)! It's these uncertainties which are in HUP.

A slightly different way to realize this, would be that Zz's experiment makes very precise measurements of the electron (both x and p), but what that doesn't tell you is where the electron "had been" and how fast it "had been moving". That information never existed, and those questions are not in any way answerable.

footnote
*the crux of the probabilistic nature of QM - at most you can only know the probability of whether something can happen, not defitively whether it will

Well, writing this post put me to sleep. Good night all! I'll chase down the typos in the morning.

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8. May 3, 2006

pivoxa15

Isn't $$\Delta x$$ the uncertainty in position of a particle in any measurement?

If the particle was stationary than $$\Delta x$$ is 0. But if it was moving than $$\Delta x$$ must be non zero if I also want to know its velocity with good accuracy. This is because I need to know at least two positions of the particle in order to get a velocity. Hence del(x)del(p)>0. All this is intutitive or classical so far. I have always thought that what I have just said did not make sense at the atomic level for any measurement and that del(x)del(p)>hbar/2. But you seem to point out that this is a misconception. Is it because the HUP works better the most particles in your system?

9. May 3, 2006

Rach3

pivoxa -

Delta X is effectively just the standard deviation of the statistics of X.

In either operator notation of QM, or notation of ordinary statistics, it's the positive square root of

<x-<x>>

where <> refers to the expectation value (~average). In short, the "average distance of X from the average value of X". Of course with QM expectation values, this "average" is an exact number.

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10. May 3, 2006

ZapperZ

Staff Emeritus
Please read what Rach said. You have obviously missed the definition of these "uncertainties" in QM. It requires the knowledge of the AVERAGE value of the observable for the uncertainty to make ANY sense. You still have wrong impression that most people have, which is that it is the INSTRUMENTATION uncertainty.

Zz.

11. May 3, 2006

DrChinese

Just to echo what ZapperZ is saying...

We know from the EPR paradox that when you have a pair (A and B) of entangled particles, you cannot measure momentum on A and position on B as a way to beat the limits of the HUP. That is because the HUP does not limit due to INSTRUMENTATION uncertainty, just as Zz said. If it were due to problems on the measuring side, then there would be no EPR paradox. Clearly, the issue goes much deeper than that.

Further, you can always perform measurements on any particle that may cause you to mistakenly think you have beat the limits of the HUP. But as mentioned, this is taken care of when you put together the results of an ensemble of measurements - because the standard deviation of the sample will still show the underlying uncertainty.

In my opinion, it is easiest to follow when you consider spin measurements rather than position or momentum measurements. Once you know any spin component (say in the direction arbitrarily labeled 0 degrees), you can measure it over and over and continue to get the same answer (even on a *single* photon, and not an ensemble). But if you subsequently measure a different spin component (say at 45 degrees) and then try to return to the original one you measured (at 0 degrees), you will not necessarily get the original answer (as you might otherwise expect) ! You can easily deduce from this example that INSTRUMENTATION (i.e. the measurement apparatus) has absolutely NOTHING to do with these results.

When you know one value with certainty, non-commuting values have complete uncertainty and will thus appear to be random as to value - exactly as happens in spin measurements.

12. May 3, 2006

nrqed

Really? I don't see how this could even be defined in QM. If one measures X and then P on gets a different result in gneral than measuring P and then X (I mean P along X of course). So I don't know how a simultaneous measurement could be defined in QM. If we take a very simple system (let's say the 1D infinite harmonic oscillator) an dwe specifify the quantum state (let's say the ground state) what does QM predict for the simulatenous measurement of momentum and position?? How would one go about that??

13. May 3, 2006

nrqed

But what Rach seems to be saying that one can make a *simultaneous* measurement of two non-commuting observables, let's say S_x and S_y. I am confused by that.