Another confusion: How many measurements is many ?

[kof9595995]

Another confusion: How many measurements is "many"?

Since quantum mechanics gives only statistical prediction, we actually need numerous measurements to verify the theory. But how many measurements will make it statistically meaningful?
Let's say we want to verify the uncertainty principle, we make two measurements of the momentum and position. If the results of two measurements just happen to be the same, then the standard deviations are 0, it'll make delta x*deltap=0.
Of course we can say only two measurements won't make the result statistical, but how many is "many"? 100, 1000, 10000? What's the criterion?

 

[vanesch]

Since quantum mechanics gives only statistical prediction, we actually need numerous measurements to verify the theory. But how many measurements will make it statistically meaningful?

That's not a specificity of quantum theory, but of any statistical theory. Quantum theory calculates probability distributions for results (like any other model, say, in medicine, would). Next, it is a matter of statistical techniques to verify whether a time series of actual results is compatible with the theoretical distributions or not. And this result itself is (except in some trivial cases), itself a statistical prediction

In other words, you might have a probability distribution F, and a series of results g = {f1,f2,...fN} from measurement, and there is then a way to calculate what is the probability, P, that the series of results is actually drawn from distribution F.
Which means that if you have a series of such series, {g1, g2, ...,gM}, that you can calculate what is the probability P', that the different series g1, g2... had a probability P to be drawn from F.
And in order to verify this probability P', you could have series of series, {h1, h2, h3,...}...

and so on.

But that's not a difficulty of quantum mechanics per se. It is a difficulty of any theory that gives you probability distributions.

 

[alxm]

Let's just ignore QM and consider the classical case. You're measuring a property that has a definite value. In reality, you will always have measurement error, so you get a distribution of measured values. Given this, you can calculate your measured value, and the deviation. Deviations come with a confidence interval.
For instance, you say a measurement might be 1.0 meters ± 0.01 with a 95% confidence interval and 1.0 m ± 0.02 with a 99% confidence interval. (of course, even that requires knowledge of the nature of the distribution, which in itself is an assumption based on measurement...)

You can't get to 100% confidence.

Now, with a quantum-mechanical value, what you're measuring isn't a definite value, but a distribution in itself. So, to begin with, your measurement accuracy has to be smaller than the quantum-mechanical distribution, otherwise it'll just get lost in the noise. Which is the case with macroscopic measurements.

What you have with QM is an analagous situation, just a bit more complicated. Basically you'll have a situation where it's a matter of to what degree your measured distribution matches the actual quantum-mechanical distribution, rather than how sharply the distribution is concentrated around the 'actual' value.

But in either case, if you know the measurement distribution, or the quantum-mechanical one (whichever is most significant) you can calculate/estimate how many measurements you will need to get the expectation value with a certain degree of confidence.

 

[Meir Achuz]

As a simple rule, if you are measuring a number of events and find N, the standard deviation is +/-\sqrt{N}.

 

[kof9595995]

Em, I still don't get it. Maybe let me think and I'll ask you guys later.

 

[Bob_for_short]

Consider instead of "many" the notion of "sufficient". The latter depends on your requirements, i.e. it is subjective. If, say, accuracy of 5% is sufficient to you, than the sufficient number is quite determined and N_many ≥ N_suff.(5%).

 

[kof9595995]

Ok,guys, I have to admit I got no clue at all.
So let's start with the case I proposed:

Let's say we want to verify the uncertainty principle, we make two measurements of the momentum and position. If the results of two measurements just happen to be the same, then the standard deviations are 0, it'll make delta x*deltap=0.

So can we say there's a chance that the uncertainty principle can be violated?

 

[Meir Achuz]

No......
You can't apply statistics to one measurement.

 

[kof9595995]

No......
You can't apply statistics to one measurement.

But there are two.

 

[kote]

Let's say we want to verify the uncertainty principle, we make two measurements of the momentum and position. If the results of two measurements just happen to be the same, then the standard deviations are 0, it'll make delta x*deltap=0.

How can a measurement of position be the same as a measurement of momentum when they don't even have the same units?

 

[kof9595995]

How can a measurement of position be the same as a measurement of momentum when they don't even have the same units?

I mean we measure position twice and momentum twice, and the results happen to be the same respectively.

 

[vanesch]

Ok,guys, I have to admit I got no clue at all.
So let's start with the case I proposed:So can we say there's a chance that the uncertainty principle can be violated?

The standard deviation of the sample is not the standard deviation of the distribution (but will, as every other statistical property, TEND to the distribution value with many measurements).

Look at it this way:
you throw heads. You throw heads again. (one chance out of 4 to do this).

Does that mean that the standard deviation of the distribution of throws (50% heads, 50% tails) is 0 ? No, it means that you've a particular sample of which the sample moment of 2nd order is 0.

This is similar to thinking that if you make one trial, you have the average. You have the average of your sample (of course). But not of the distribution from which it was drawn.

In the same way as the single value of a trial (which is of course equal to the "average" of that sample) is not the average of the distribution (but rather a bad-quality estimator of that value), we also have that the single value of the second moment of a (small) sample is not the standard deviation of the distribution (but rather a bad-quality estimator of that value).

 

[kof9595995]

The standard deviation of the sample is not the standard deviation of the distribution (but will, as every other statistical property, TEND to the distribution value with many measurements).

Look at it this way:
you throw heads. You throw heads again. (one chance out of 4 to do this).

Does that mean that the standard deviation of the distribution of throws (50% heads, 50% tails) is 0 ? No, it means that you've a particular sample of which the sample moment of 2nd order is 0.

This is similar to thinking that if you make one trial, you have the average. You have the average of your sample (of course). But not of the distribution from which it was drawn.

In the same way as the single value of a trial (which is of course equal to the "average" of that sample) is not the average of the distribution (but rather a bad-quality estimator of that value), we also have that the single value of the second moment of a (small) sample is not the standard deviation of the distribution (but rather a bad-quality estimator of that value).

So the uncertainty principle is about the standard deviation of the distribution? I see, my understanding of uncertainty principle was totally wrong until now. Thanks for clarifying.

 

[kof9595995]

And by the way, do I have to learn a lot statistical math to get a deep understanding about QM? I never indeed paid too much attention to statistical math.

 

[Fredrik]

Depends on how deep you want to go, and what specific details you want to focus on. In my opinion you can get very far without "a lot of statistical math".

 

[kof9595995]

Depends on how deep you want to go, and what specific details you want to focus on. In my opinion you can get very far without "a lot of statistical math".

That's what I thought. Before my QM course started I took a glimpse of some QM books just to have a look at the fancy math notations, and didn't find too much statistical math inside. It's just recently the statistical aspects of QM really bothered me a lot.