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In Chapter 2-3,Vol I of the Feynman lectures, Feynman talks about a rule in quantum mechanics that says that one cannot know both where something is and how fast it is moving. That the uncertainty of the momentum and the uncertainty of the position are complementary,and the product of the two is bounded by a small constant. And that the law can be written as such: ΔxΔp≥ℏ/2 He then goes on to state: "This rule is the explanation of a very mysterious paradox: if the atoms are made out of plus and minus charges, why don’t the minus charges simply sit on top of the plus charges (they attract each other) and get so close as to completely cancel them out? Why are atoms so big? Why is the nucleus at the center with the electrons around it? It was first thought that this was because the nucleus was so big; but no, the nucleus is very small. An atom has a diameter of about 10^−8 cm. The nucleus has a diameter of about 10^−13 cm. If we had an atom and wished to see the nucleus, we would have to magnify it until the whole atom was the size of a large room, and then the nucleus would be a bare speck which you could just about make out with the eye, but very nearly all the weight of the atom is in that infinitesimal nucleus. What keeps the electrons from simply falling in? This principle: 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, i.e., a very large kinetic energy. With this energy they would break away from the nucleus. They make a compromise: they leave themselves a little room for this uncertainty and then jiggle with a certain amount of minimum motion in accordance with this rule."

-Why is it that the momentum has to be very large? -Isn’t is the uncertainty of the momentum that has to be large not the momentum itself? Since he stated earlier that in the rule,it is the uncertainty of the momentum,not the momentum itself that is complementary with the uncertainty of the position,and the product..by a small constant. -Is it because a larger momentum would mean a more uncertain momentum. If it is,why is that so?

-On the last line of making a compromise,whereby they leave themselves a little room for uncertainty —“they” refer to the electrons? —So,the way that the electrons leave a little room for uncertainty (increase the uncertainty of their position) is to surround at a distance around the atom’s nucleus,instead of being in the nucleus? —At the same time,the jiggling with a certain amount of minimum motion is to increase the uncertainty of their speed?

Qmechanic
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Charles
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  • The uncertainty only appears if we try to measure positions and momenta. You are trying to apply all classical terms to a non-classical system, the result is a hodgepodge of misconceptions about what a quantum system is. Uncertainty doesn't mean that there is a lot of jiggling around in atoms. It just means that what we see when we do measurements is limited by the uncertainty relation. Indeed, of you look at the actual wave function of quantum systems, the ground states are stationary except for a simple periodic phase term. – CuriousOne Jan 02 '15 at 12:15
  • @Charles : you ask "Isn’t is the uncertainty of the momentum that has to be large not the momentum itself?" There is no sharp value of the linear momentum (not even of the radial momentum - i.e. along the radius). So, to speak of the $\Delta p_r$ it's quite O.K. But, for economizing your worries, you can do a simple thing: calculate the mean radial momentum, and the $\Delta p_r$. The calculus for the ground state $n=1$ will give you the order of magnitude. The wave-function you can find in the Wikipedia. No need to bother with an analytical integration, do you have an EXCEL on your computer? – Sofia Jan 03 '15 at 02:30

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