Uncertainty Principle: Misconceptions & Justifications

[tommy01]

Hi together ...

In many textbooks on particle physics i encounter - at least in my mind - a misuse of the Heisenberg uncertainty principle.
For completeness we talk about
\Delta p \Delta x \geq \hbar/2

For example they state that the size of an atom is of the order of a few Angstroms. Therefore \Delta x of an Electron is \approx 10^{-10}m then the conclude that the momentum is \approx \hbar/\Delta x.

But what justifies this? The only thing you get is \Delta p this means the momentum is about p \pm \Delta p how one can conclude from the uncertainty of an observable its mean value?

 

[Sourabh N]

You choose a frame where mean value is zero.

 

[tommy01]

They conclude that the momentum of the electron is \approx \hbar/\Delta x, so it surely isn't zero. They want to calculate the typical momentum of an electron in an atom, it would not be very helpful to chooce such a reference frame, because the answer would be that the typical momentum is zero.
It is cited in many textbooks. I just read in in aitchison, hey - gauge theories in particle physics. vol. 1 p.8.

 

[tommy01]

Sorry for bumping, but i can't believe that nobody who did some lectures on particle physics ever encountered this (or a similar misuse of the uncertainty principle) problem. I would be very glad for an answer.

Greetings.

 

[msumm21]

Do you have a link to such an argument in an online reference? I don't recall seeing this argument.

 

[Fredrik]

Sorry for bumping, but i can't believe that nobody who did some lectures on particle physics ever encountered this (or a similar misuse of the uncertainty principle) problem. I would be very glad for an answer.

Are they actually saying that the expected value of the momentum is of that order of magnitude? It seems valid to say e.g. that a typical momentum is of that order, or that the expected magnitude of the momentum is of that order.

 

[tommy01]

They say:

"The uncertainty principle gives an estimate for the typical electron momenta when they are confined to such a linear domain." Then they give the above formula.

But all they get from the uncertainty principle is the uncertainty \Delta p which is not the typical momentum. In my mind the typical momentum should be the same as the expected momentum or at least the same order of magnitude.

Greetings.

 

[A. Neumaier]

They say:

"The uncertainty principle gives an estimate for the typical electron momenta when they are confined to such a linear domain." Then they give the above formula.

But all they get from the uncertainty principle is the uncertainty \Delta p which is not the typical momentum. In my mind the typical momentum should be the same as the expected momentum or at least the same order of magnitude.

The uncertainty is a lower bound on the standard deviation of multiple observations of
a quantity. So different observations typically differ by about this much. Even when the expected value is zero, one can conclude that many observations will have this magnitude. if the expectation value is nozero, then even more.

Thus it is appropriate to regard the uncertainty as a lower bound for the order of magnitude of a typical momentum.

 

[tommy01]

Thanks, but i think it's meaningless to give a lower bound and speak of typical momenta.
The uncertainty can be a very small fraction of the actual momentum or vice versa, so i can't believe they meant it this way.

 

[A. Neumaier]

Thanks, but i think it's meaningless to give a lower bound and speak of typical momenta.
The uncertainty can be a very small fraction of the actual momentum or vice versa, so i can't believe they meant it this way.

If it is a very small fraction of the actual momentum, it is a lower bound, though a poor one. On the other hand, ''vice versa'' is very unlikely -- Look at a Gaussian random number generator and observe how long it takes until a realization is much smaller in magnitude than the standard deviation!

 

[tommy01]

Thanks again. But that's what i meant.
It is unlikely but not impossible. And a very poor lower bound doesn't justify to call \Delta p the typical momentum.
It seems as if one could conclude the mean value (which i assume is the typical value) from the standard deviation. But there's no justification in statistics for this. Isn't it?

 

[f95toli]

Note that they are talking about the momentum for a confined electron. Hence, classically the mean momentum would be zero (since it is not moving) but here the momentum is -of the order of magnitude- h/dx.
This is obviously not a rigorous derivation; but using the UP in this way is not wrong as such (I do it all the time) and is very often useful as a first approximation (and occasionally it will even be exact).

 

[Fredrik]

Thanks again. But that's what i meant.
It is unlikely but not impossible. And a very poor lower bound doesn't justify to call \Delta p the typical momentum. It seems as if one could conclude the mean value (which i assume is the typical value) from the standard deviation. But there's no justification in statistics for this. Isn't it?

Look at the definition of "standard deviation". It's a measure of how much a measurement result will typically deviate from the mean, which in this case is obviously zero.

 

[Deric Boyle]

A simple suggestion to you tommy01.
Use H. A. Robertsons method to calculate the uncertainity product for electron in H-atom compare it with lower bower uncertainity product.
If still unsatisfied consult an expert like A. Neumaier (he will surely help)