Feynman Lectures and Uncertainty Principle

[forcefield]

I read the Quantum Physics section of the online version of Feynman lectures http://feynmanlectures.caltech.edu/I_02.html#Ch2-S3 and I don't understand how he can deduce electron momentum from the Uncertainty Principle. I agree that the momentum is uncertain but how can he deduce that it is very large ?

This is the relevant content:
"If they were in the nucleus, we would know their position precisely, and the uncertainty principle would then require that they have a very large (but uncertain) momentum"

 

[dauto]

Ask yourself: How can the expected value of the magnitude of the momentum be smaller than its uncertainty? Think of a set of numbers with large standard deviation. This set could have zero average. $\langle x\rangle=0$, but it canot have small average of magnitude $\langle |x|\rangle >> 0$.

 

[Avodyne]

It would have been better if Feynman had said "the uncertainty principle would then make it very likely that they have a very large (but uncertain) momentum"

 

[San K]

I read the Quantum Physics section of the online version of Feynman lectures http://feynmanlectures.caltech.edu/I_02.html#Ch2-S3 and I don't understand how he can deduce electron momentum from the Uncertainty Principle. I agree that the momentum is uncertain but how can he deduce that it is very large ?

This is the relevant content:
"If they were in the nucleus, we would know their position precisely, and the uncertainty principle would then require that they have a very large (but uncertain) momentum"

What Mr. Fenyman, must have, meant was -- A very large spread in the probability distribution of Momentum.

 

[forcefield]

What Mr. Fenyman, must have, meant was -- A very large spread in the probability distribution of Momentum.

That is how I understand the uncertainty principle. But does that really imply that we know something about the probability of individual momentum values ? Does that really imply that a large momentum is more likely than a small momentum ?

 

[kith]

Picture a Gaussian distribution. The probability to obtain a momentum in a certain interval is the area under the curve. Even though the most probable value may be zero, the area for a small interval around zero is much smaller than the remaining area if the spread is large wrt to this intervall.

 

[nikol]

Sometimes I am getting confused about this too and then this is my line of reasoning that clears my mind: we would like to probe very small distances e.g. is the electron at position $x$ or at position $(x+ 10^{-9})m$ or in other words our $\Delta x$ is of the order of $nm$. But how do we do that - in particle physics what we measure actually is energy and momentum(we cannot take a ruler and measure the distance between particles). Now by the Heisenberg principle $\Delta E$ is very big, of the order of $GeV$. Which energy $E$ is this $\Delta E$ uncertainty of - well the energy of our electron (the one we want to determine the position of). So if we try to measure this energy we will get numbers spread from $(E - \Delta E)$ to$(E + \Delta E)$. Now what if $E$is very small, like only $eV$ - we did not manage to measure nothing here. The only way to get meaningful results will be $E$ to be bigger that its own error (so at least of order $GeV$), otherwise we did not measured anything. I hope that this is helping.