Experimental test of the Uncertainty principle

[IRobot]

Sorry if this question has already been asked many times, but I found no answer after a quick search. Have already made an experimental test of the Heisenberg's inequalities:
\Delta p \Delta x \geq \hbar

I'm quite sure our instruments' imprecision are larger than Planck's constant, but I prefer be sure.

 

[ansgar]

you mean that if we have such experiments where the experimental uncertainty is smaller than the calculated HUP?

 

[Doc Al]

Those deltas refer to distributions of measured values, not to the accuracy of anyone particular measurement.

 

[ansgar]

Those deltas refer to distributions of measured values, not to the accuracy of anyone particular measurement.

yes, I forgot to point that out

 

[IRobot]

Sorry, my english is bad so it's hard for me to explain myself.

Those deltas (Heisenberg's ones) refer to the spatial extension and the extension in Fourier Space (momentum space), like the width of a Gaussian, but what i am saying is, could we make a measurement of a electron, for example, where the measurement uncertainty of his position and his momentum are violating the Heisenberg's inequalities. Or let's say it in another way: Is actual instrument precision enough to dispute the uncertainty principle

 

[ansgar]

You can not speak of HUP for one measurment

 

[Doc Al]

Those deltas (Heisenberg's ones) refer to the spatial extension and the extension in Fourier Space (momentum space), like the width of a Gaussian, but what i am saying is, could we make a measurement of a electron, for example, where the measurement uncertainty of his position and his momentum are violating the Heisenberg's inequalities. Or let's say it in another way: Is actual instrument precision enough to dispute the uncertainty principle

Can the "uncertainty" of a particular measurement be less than the delta in Heisenberg's inequality? Absolutely! Does that dispute the uncertainty principle? No.

 

[ZapperZ]

Sorry, my english is bad so it's hard for me to explain myself.

Those deltas (Heisenberg's ones) refer to the spatial extension and the extension in Fourier Space (momentum space), like the width of a Gaussian, but what i am saying is, could we make a measurement of a electron, for example, where the measurement uncertainty of his position and his momentum are violating the Heisenberg's inequalities. Or let's say it in another way: Is actual instrument precision enough to dispute the uncertainty principle

You need to learn a little bit more of the thing that you want to test. If not, you are going to be testing the wrong thing.

Look carefully at the definition of those "delta's". It is a SPREAD in measurements, i.e. MANY measurements.

I've written in another thread the misconception of the HUP using the single slit as an illustration. Look for it.

Zz.

 

[ajw1]

There seems to be evidence that falsifies this limit, for example: http://www.springerlink.com/content/r581155315556545/".
(text can also be found http://cfcul.fc.ul.pt/equipa/3_cfcul_elegiveis/croca/berkeley-paper.pdf")

If this is correct the answer to IRobot should simply be: yes we can test the uncertainty principle, and the result is sometimes negative.

 

[IRobot]

Thank you for the reference Zapperz, I've found your note and read it. So now I understood why my example was absurd. But I've a new one which I think more relevant:

Let's take your one slit montage and send this time a Gaussian wave packet in it. We fix aperture of the split, to fix the \Delta x[\tex]. We then make a series of measure of the momentum p_x[\tex] on the screen. We should have a Gaussian repartition of the momentum. But since it's a Gaussian repartition even really high values of p_x[\tex] will be measured (with really low probabilities ofc) and the HUP only deals with the width of a Gaussian at half amplitude (sorry I think I am not clear at all). So we could violate the HUP, except if the inequality is define by the width at middle amplitude for a Gaussian without excluding firmly an "exotic value" of p_x[\tex]; but then could it still be violated for an non-Gaussian wave packet?

 

[IRobot]

There seems to be evidence that falsifies this limit, for example: http://www.springerlink.com/content/r581155315556545/".
(text can also be found http://cfcul.fc.ul.pt/equipa/3_cfcul_elegiveis/croca/berkeley-paper.pdf")

If this is correct the answer to IRobot should simply be: yes we can test the uncertainty principle, and the result is sometimes negative.

Thank you for the link. ;)

 

[SpectraCat]

There seems to be evidence that falsifies this limit, for example: http://www.springerlink.com/content/r581155315556545/".
(text can also be found http://cfcul.fc.ul.pt/equipa/3_cfcul_elegiveis/croca/berkeley-paper.pdf")

If this is correct the answer to IRobot should simply be: yes we can test the uncertainty principle, and the result is sometimes negative.

That paper looks highly suspect to me .. I have not read it carefully yet, but I am not sure the authors really understand the HUP. I found the following statement on the second page:

Analytically, the usual uncertainty relations are a mathematical consequence of the nonlocal
Fourier analysis [2];

This is very different from my own understanding of the HUP, which is that it is connected at a very deep level to the fact that quantum states are defined as vectors in a Hilbert space. It is then fairly simple to prove that an HUP relation emerges from simple vector algebra for any pair of non-commuting operators acting on kets in that space. Perhaps this is somehow equivalent to the author's definition above, but I cannot see how ... his definition looks much more restrictive.

I will have to read the rest of the paper at another time, but I suspect that in the end, what they are presenting is not true violations of the HUP, but rather (at most) apparent violations of a more narrowly defined special case.

 

[ZapperZ]

Thank you for the reference Zapperz, I've found your note and read it. So now I understood why my example was absurd. But I've a new one which I think more relevant:

Let's take your one slit montage and send this time a Gaussian wave packet in it.

Er.. what?

If you send a gaussian wave packet, it means that you're sending a large amount of photons that are not monochromatic! The pattern you get on the screen would be VERY difficult to decipher, much less, to apply the HUP to it.

The HUP does NOT deal with that gaussian wave packet. The HUP is a fundamental consequence of the non-commutation of observables, i.e. it is a fundamental property of QM!

I still think you do not realize that \Delta A = \sqrt{<A>^2 - <A^2>}.

Zz.

 

[IRobot]

By a Gaussian wave packet, I mean that we have prepared an electron in a way that his distribution of probablity is like a Gaussian, for example:
\psi = Ae^{i\omega t}e^{-(x-x_0(t))^2/2\sigma^2}<br /> \psi \psi^* = AA^*e^{-(x-x_0)²/\sigma^2}<br />

 

[DrChinese]

Let's take your one slit montage and send this time a Gaussian wave packet in it. We fix aperture of the split, to fix the \Delta x[\tex]. We then make a series of measure of the momentum p_x[\tex] on the screen. We should have a Gaussian repartition of the momentum. But since it&amp;#039;s a Gaussian repartition even really high values of p_x[\tex] will be measured (with really low probabilities ofc) and the HUP only deals with the width of a Gaussian at half amplitude (sorry I think I am not clear at all). ...

&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; I think there are a couple of things that help to highlight the HUP:&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; a) You can measure commuting observables without limit. The HUP only applies to non-commuting observables. This should be a good indicator that observational issues are not relevant.&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; b) With a pair of entangled particles, you can do the following: Alice measures an observable to high precision. Bob measures the non-commuting complementary observable, also to high precision. You now know the values of a non-commuting pair to higher precision than the HUP allows. Or do you? No, the HUP still applies. A subsequent investigation will show that the particles no longer have the values you thought they did, but they will follow the HUP.&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; In other words: ZapperZ is saying that measuring the values in and of itself tells you nothing more about the particle than the HUP allows. You can &amp;amp;quot;think&amp;amp;quot; you have beaten it, but you will not have.

 

[ZapperZ]

By a Gaussian wave packet, I mean that we have prepared an electron in a way that his distribution of probablity is like a Gaussian, for example:
\psi = Ae^{i\omega t}e^{-(x-x_0(t))^2/2\sigma^2}<br /> \psi \psi^* = AA^*e^{-(x-x_0)²/\sigma^2}<br />

This still doesn't tell you what you get AFTER you pass through the slit. The uncertainty of the position of the electron AT the slit is still the width of the slit! The uncertainty in the lateral momentum after the slit will still be equivalent to the transverse distance at the screen.

BTW, where you have seen such a gaussian description of a single, free electron going through a slit?

Zz.