A paradox about Heisneberg Uncertainty ?

[lightarrow]

Have you ever done an experiment? Try one. Pick a ruler, and measure a length of something. Now, what is the uncertainty in that length? That is the uncertainty in that single measurement and it depends entirely on the accuracy of the instrument. This is not the HUP. I can make it more or less accurate by changing instrument, regardless of what was measured before, or what will be measured after.

I think you still have a strong misconception of the HUP. Maybe you might want to read what I wrote a while back on this.

Zz.

So, if I make only one measurement I can have more accuracy than with more measurements? This seems nonsense to me.

 

[Cthugha]

So, if I make only one measurement I can have more accuracy than with more measurements? This seems nonsense to me.

You are showing a general misconception of the HUP.
If you do one measurement, you get a result, which shows an accuracy corresponding to the accuracy of your measuring equipment.
If you are repeating these measurements under exactly the same circumstances (same initial state) you will find, that

a) for each of the measurements you get a result, which corresponds to the accuracy of your measuring equipment.
b) there is still a spread in your results, which is not due to the accuracy of your measuring equipment, although the initial state was the same at every measurement.

Case b) is what the HUP is about. The results are not reproducible exactly. Although the result of each measurement was exact, there is an intrinsic spread due to the quantum nature of what you are trying to measure.

 

[country boy]

When we talk about accuracy of measurements in this context, we essentially talk about standard deviations of the measurements, or they estimators s:

s = \sqrt{\frac{1}{(n-1)}\cdot\sum_{i=1}^n (X_i - X_m)^2}

where X_i is the i-esim measurement of the variable X and X_m is its aritmetical average.

For 1 measurement, n=1 so

s = \infty.

So it's not possible to talk about "arbitrary accuracy" in one single measure.

I have another question about how you seem to be applying the standard deviation. The calculation of the variance or standard deviation requires all of the measurements to have been completed so that a mean can be determined. Alternatively, I guess you could have some prior knowledge of the mean from a previous experiment or from a theoretical prediction. Either way, without some prediction of the outcome you cannot use the standard deviation on a single measurement, can you?

By the way, since the mean of a single measurement is the measurement itself, the numerator and denominator are both zero, so the standard deviation is indeterminate (see ZapperZ's comments). That means formula still works, though, because in fact you can't determine s from one measurement.

It's interesting that you could start with the prediction of a mean derived from the wavefunction and then calculate something like the standard deviation (with n instead of n-1) as you accumulate measurements. The standard deviation should then be greater than or equal to the spread predicted from the wavefunction. This probably only has value as a mental exercise.

 

[ZapperZ]

So, if I make only one measurement I can have more accuracy than with more measurements? This seems nonsense to me.

It's nonsense because you didn't understand what I just said.

The accuracy of a single measurement depends on the accuracy of your instrument. Now do you dispute this? Let's get that out of the way first.

Now, if you make repeated measurement of the SAME thing (i.e. of the width of the slit, for example), then you will have an average value, and a standard deviation of the width of the slit based on the repeated measurement. However, each of the measurement in your data set is accompanied by its own uncertainty based on the uncertainty of a single measurement. This is the uncertainty that I'm talking about, and it has nothing to do with the HUP. That is the uncertainty of the instrumentation. If I was using a ruler to measure the width of the slit, I get a certain amount of uncertainty. If I instead switch to a micrometer caliper, then I have a smaller uncertainty in that measurement because I just switched to a more accurate derive that measures a length. So now I've just reduced the uncertainty of my length measurement with no regards to what is going on before and after that measurement. What is nonsense is to confuse that with the HUP.

Zz.

 

[lightarrow]

It's nonsense because you didn't understand what I just said.

The accuracy of a single measurement depends on the accuracy of your instrument. Now do you dispute this? Let's get that out of the way first.

Now, if you make repeated measurement of the SAME thing (i.e. of the width of the slit, for example), then you will have an average value, and a standard deviation of the width of the slit based on the repeated measurement. However, each of the measurement in your data set is accompanied by its own uncertainty based on the uncertainty of a single measurement. This is the uncertainty that I'm talking about, and it has nothing to do with the HUP. That is the uncertainty of the instrumentation. If I was using a ruler to measure the width of the slit, I get a certain amount of uncertainty. If I instead switch to a micrometer caliper, then I have a smaller uncertainty in that measurement because I just switched to a more accurate derive that measures a length. So now I've just reduced the uncertainty of my length measurement with no regards to what is going on before and after that measurement. What is nonsense is to confuse that with the HUP.

Zz.

ZapperZ, it was very clear what you intended with accuracy of a single measurement (= accuracy of the instrument) and the fact HUP concerns repeated measurement of the same thing.

What I wanted to discuss are 2 different things:

1. Independently on HUP, but talking about measurements in general, I don't understand how is possible to define the accuracy of a single measurement as the accuracy of the instrument - let's say 0.01 mm of something - if, repeating many times that measure with the same instrument, we have a standard deviation of 10 mm, for example. It was just a statistical consideration. Certainly, with only one measurement we have nothing else than the instrument's accuracy, but, IMO, we should never estimate the experiment accuracy from one only measurement, we should make at least two of them if we want to give a reasonable value of it (reasonable in the sense I wrote).

2. It's not clear to me if it's possible or not (as you say) to measure x and p of, let's say, a photon, simultaneously. This is because, infact, I'm now more confused on what we mean with "measure of position" and "measure of momentum".

Let's set, e.g., the following situation:

A plane, monocromatic electromagnetic wave traveling along the +z direction hits a screen with a slit, which width is d along the x direction (and D>>d in the y direction) and then hits a final screen parallel to the first and at distance L from it (so, a very standard arrangement).

According to what is written in some books (see, for example, A.Messiah - ch. 4, par.13), the light passing through the slit correspond to a position measurement of photons, with \Delta x = d and to a momentum measurement with \Delta p_x = p\cdot\sin\alpha where \alpha is the angle of diffraction. Now I ask, since this is not stated: does it mean that the position measurement is performed before the momentum measurement?

Thank you for your patience.

 

[lightarrow]

I have another question about how you seem to be applying the standard deviation. The calculation of the variance or standard deviation requires all of the measurements to have been completed so that a mean can be determined. Alternatively, I guess you could have some prior knowledge of the mean from a previous experiment or from a theoretical prediction. Either way, without some prediction of the outcome you cannot use the standard deviation on a single measurement, can you?

By the way, since the mean of a single measurement is the measurement itself, the numerator and denominator are both zero, so the standard deviation is indeterminate (see ZapperZ's comments). That means formula still works, though, because in fact you can't determine s from one measurement.

Yes, I agree.

 

[ZapperZ]

ZapperZ, it was very clear what you intended with accuracy of a single measurement (= accuracy of the instrument) and the fact HUP concerns repeated measurement of the same thing.

What I wanted to discuss are 2 different things:

1. Independently on HUP, but talking about measurements in general, I don't understand how is possible to define the accuracy of a single measurement as the accuracy of the instrument - let's say 0.01 mm of something - if, repeating many times that measure with the same instrument, we have a standard deviation of 10 mm, for example. It was just a statistical consideration. Certainly, with only one measurement we have nothing else than the instrument's accuracy, but, IMO, we should never estimate the experiment accuracy from one only measurement, we should make at least two of them if we want to give a reasonable value of it (reasonable in the sense I wrote).

I don't quite understand this. Take a ruler that has, as the smallest scale, 1 mm tick marks. You can accurately measure by eye, a length of as close to -0.5 mm in accuracy, but with an uncertainty of +/- 0.5. So you can read a length of 10.0 mm with +/- 0.5. That is the accuracy of that length measurement. You can make 100,000 measurement, but that is as accurately as you can make each of the single measurement.

Now, there could be a spread in that measurement, since in this case, it is done by eye. Someone else can come in and instead, would say it is 10.5 mm. Another person says it is 10.0 mm, another says it is 11.0 mm. etc... etc. NOW you have a statistical spread in the measurement. The spread would correspond to the standard deviation of all the measurement. If this spread follows a normal distribution, you'd expect that the average value would be closest to the 'actual' value and the standard deviation to get smaller and smaller as the sampling data gets larger. But the uncertainty in each of the measurement remains the same!

In high energy physics experiments, a form of the standard deviation is often cited as the "confidence level" whenever the statistics of a spectrum is given. However, in other experiments such as in condensed matter (i.e. photoemission, tunneling, etc.), we often cite the RESOLUTION as the uncertainty in the measurement. This means that this is the uncertainty that is present in one single measurement, and it is the best resolution of energy, momentum, etc. that the instrument has. You can take a gazillion measurement (and often we do when we're dealing with a stream of electrons hitting a CCD), but you cannot get any better than the instrument's resolution.

This is not the HUP.

Zz.

 

[country boy]

...it was very clear what you intended with accuracy of a single measurement (= accuracy of the instrument) and the fact HUP concerns repeated measurement of the same thing.

There seem to be two kinds of measurement going on here. First, the measurement on the screen of the location of a single event. This position can be measured many times and the statistics will produce a progressively more accurate value for the position. In this case, the uncertainty approaches zero; as the number of measurements increases, the knowledge of the location improves without limit. The standard deviation among these measurements gives the measurement accuracy, presumably set by the measuring device or technique. So there is a "probability distribution" associated with the measurement.

Second, a large number of arrival events are measured. In this case we are determining, for a given slit width, the distribution of arrival positions on the screen. Again, the statistics of a large number of measurements will yield a progressively more accurate determination of the average arrival location (the center of the distribution on the screen). We can also calculate the standard deviation of the set of measurements. But now we notice that a smaller slit has the effect of making the standard deviation larger, exceeding our measurement accuracy for a single event. The standard deviation now gives the QM uncertainty, determined by the QM probability distribution. The standard deviation will be inversely proportional to the slit width.

The same statistical tools are used in both regimes, but they are measuring different things.

 

[lightarrow]

I don't quite understand this. Take a ruler that has, as the smallest scale, 1 mm tick marks. You can accurately measure by eye, a length of as close to -0.5 mm in accuracy, but with an uncertainty of +/- 0.5. So you can read a length of 10.0 mm with +/- 0.5. That is the accuracy of that length measurement. You can make 100,000 measurement, but that is as accurately as you can make each of the single measurement.

Now, there could be a spread in that measurement, since in this case, it is done by eye. Someone else can come in and instead, would say it is 10.5 mm. Another person says it is 10.0 mm, another says it is 11.0 mm. etc... etc. NOW you have a statistical spread in the measurement. The spread would correspond to the standard deviation of all the measurement. If this spread follows a normal distribution, you'd expect that the average value would be closest to the 'actual' value and the standard deviation to get smaller and smaller as the sampling data gets larger. But the uncertainty in each of the measurement remains the same!

A machinery makes steel balls; they should all be of the same diameter, but no machinery is perfect: the first ball is 10mm +/- 0.01 mm, the second is 10.5 mm +/- 0.01 mm, then 9.2 mm +/- 0.01 mm,...ecc.

The instrument accuracy (the instrument used to measure the ball's diameter) is 0.01 mm, the standard deviation of the balls dimension is, let's say, 1mm (computed on the entire population of balls made by the machinery).

If I perform a single measure on a single ball, I should say the accuracy is 0.01mm; if I perform more measures on more balls I discover that the accuracy is much less (= 1 mm).

So, should I perform 1 only measurement (on a single ball), to be sure to have the maximum accuracy? Certainly not.

 

[ueit]

2. It's not clear to me if it's possible or not (as you say) to measure x and p of, let's say, a photon, simultaneously. This is because, infact, I'm now more confused on what we mean with "measure of position" and "measure of momentum".

In the example I've given you, the detection is a measurement of both position and momentum. The position is the spot on the screen, the momentum is calculated from the two position measurements (the position of the source and that of the spot).

 

[ZapperZ]

A machinery makes steel balls; they should all be of the same diameter, but no machinery is perfect: the first ball is 10mm +/- 0.01 mm, the second is 10.5 mm +/- 0.01 mm, then 9.2 mm +/- 0.01 mm,...ecc.

The instrument accuracy (the instrument used to measure the ball's diameter) is 0.01 mm, the standard deviation of the balls dimension is, let's say, 1mm (computed on the entire population of balls made by the machinery).

If I perform a single measure on a single ball, I should say the accuracy is 0.01mm; if I perform more measures on more balls I discover that the accuracy is much less (= 1 mm).

So, should I perform 1 only measurement (on a single ball), to be sure to have the maximum accuracy? Certainly not.

And who said you should? In the example I gave you, BOTH types were used under difference circumstances. Did you miss that? No one said you should use one over the other. But my main point still stands, which is the uncertainty in a SINGLE measurement remains the same and independent of what was measured earlier or later.

And besides, what does this have anything to do with the HUP? That is my whole point - they don't!

Zz.

 

[lightarrow]

And who said you should? In the example I gave you, BOTH types were used under difference circumstances. Did you miss that? No one said you should use one over the other. But my main point still stands, which is the uncertainty in a SINGLE measurement remains the same and independent of what was measured earlier or later.

And besides, what does this have anything to do with the HUP? That is my whole point - they don't!

Zz.

Ok.
What about question n. 2 in my post n.35?

 

[lightarrow]

Which is what you consider the better way to express HUP?

 

[ZapperZ]

Ok.
What about question n. 2 in my post n.35?

But I thought I already answered that in the link that I gave to my blog entry? Besides, you never explained what you mean by "simultaneous". I had to clearly define what *I* meant by that, but no one else here who used that word seems to do that.

OK, one more time, and then I'm done with this one.

You measure the momentum AFTER the light passed through the slit. You do this by look at where the photon hit the screen/detector, and then deducing the transverse momentum it has gained in the x-direction. This means that in terms of the sequence of operators, the position operator came first, and the momentum operator came next. That's all we care about. So now you have x, p, and the uncertainty in x from the slit width. You still have no uncertainty in p that corresponds to the one referred to in the HUP. All you have in terms of the uncertainty in p as this point is the instrumentation uncertainty, i.e. how big was the spot on the detector that gives you the single-measurement uncertainty in p. But as I have said before, this is not part of the HUP. The spread that enters in the HUP isn't known yet from experimental observations.

So how do we get the HUP-related uncertainty in the p? You do this experiment many, many times. You'll see that if the slit width is small enough, the position where the photon hit the screen will have a very large spread. If you bin the results, you'll have roughly a gaussian spread centered directly behind the slit. The smaller the slit, the larger the spread. It is this spread that gives you the HUP uncertainty for p. It fits exactly in what the HUP predicted. The smaller the slit (i.e. the more I know about the position of the photon that got through), the less I can predict what x-component of the momentum will be of that photon that got through.

Zz.

 

[lightarrow]

But I thought I already answered that in the link that I gave to my blog entry? Besides, you never explained what you mean by "simultaneous". I had to clearly define what *I* meant by that, but no one else here who used that word seems to do that.

I thought it was clear: to me, it means "Exactly at the same time". I know you had already answered to it, I wanted to be sure of it.
Thank you to have clarified this, because it's not so clear in the books.

 

[ZapperZ]

I thought it was clear: to me, it means "Exactly at the same time".

But why is this so important and so relevant here? That has never been explained to me when I asked the first time. I have already indicated that this is a non-issue as far as the HUP is concerned, simply from the fact that the operators' order of measurement makes a whole lot of difference. So insisting on something being measured "exactly as the same time" introduces something that isn't contained in the HUP. So why are you requiring it?

Zz.

 

[reilly]

If you work it out, you'll find that x(t+dt) does not commute with x(t) -- formally, use the "time" unitary boost, exp(-i H t). So, this means given an, an x(0), any x(t+dt) is possible. Thus it will take many observations to get a mean value for x(t+dt). And the best one can do for v, or p, is an average value. No problem with the HUP;.
Regards,
Reilly Atkinson

 

[lightarrow]

If you work it out, you'll find that x(t+dt) does not commute with x(t) -- formally, use the "time" unitary boost, exp(-i H t). So, this means given an, an x(0), any x(t+dt) is possible. Thus it will take many observations to get a mean value for x(t+dt). And the best one can do for v, or p, is an average value. No problem with the HUP;.
Regards,
Reilly Atkinson

Ok, but, as also nrqed wrote in post n.7, what does the operator x(t)x(t+dt) mean, if AB means: "the operator B acts first, in the time, and then acts A"?

 

[jVincent]

So here a simple view of it:
classical view: Momentum is carried in little matter-lumps called particles which at any given time are located at _one_ place in space, and have _one_ definite momentum.

Now in QT momentum is carried in waves, which are not localized in space.
one wave is _everywhere_ and has _one_ definite momentum.

So we see a problem since in our world, things aren't everywhere, but only "one" place, so we seek out a method of creating a localized particle. We do this by adding up these waves, thus adding up different discrete momenta. So a particle is suddenly described as a wave-package.
This makes up the _one_ location particle, but this particle has infinite many momenta, otherwise it would not be at only one location.

This "uncertainty" principle doesn't have anything to do with measurements, it's simple a fact of the composition of our world. Thus no measurement can be made that breaks this "uncertainty"

In the classical world we can't measure the position of a car to be two different places at the same time, simply because if we do then we come to the conclusion that it wasn't the same car, since things can't be at to places at once. Even if we find a apparatus that measures position with uncertainty down to the nm scale, we still can't perform this feat, because it has nothing to do with uncertainty, but is related to the nature of what we are measuring.