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Suppose we have a hydrogen atom, and measure the position of the electron; we must not be able to predict exactly where the electron will be, or the momentum spread will then turn out to be infinite. Every time we look at the electron, it is somewhere, but it has an amplitude to be in different places so there is a probability of it being found in different places. These places cannot all be at the nucleus; we shall suppose there is a spread in position of order $a$. That is, the distance of the electron from the nucleus is usually about a. We shall determine a by minimizing the total energy of the atom.

The spread in momentum is roughly $ℏ/a$ because of the uncertainty relation, so that if we try to measure the momentum of the electron in some manner, such as by scattering x-rays off it and looking for the Doppler effect from a moving scatterer, we would expect not to get zero every time—the electron is not standing still—but the momenta must be of the order $p≈ℏ/a$. Then the kinetic energy is roughly $\frac{1}{2}mv^2= p^2/2m= ℏ^2/2ma^2$. (In a sense, this is a kind of dimensional analysis to find out in what way the kinetic energy depends upon the reduced Planck constant, upon $m$, and upon the size of the atom. We need not trust our answer to within factors like $2, π,$ etc. We have not even defined a very precisely.) Now the potential energy is minus $e^2$ over the distance from the center, say $−e^2/a$, where, as defined in Volume I, $e^2$ is the charge of an electron squared, divided by $4πϵ_0$. Now the point is that the potential energy is reduced if $a$ gets smaller, but the smaller $a$ is, the higher the momentum required, because of the uncertainty principle, and therefore the higher the kinetic energy. The total energy is $E=ℏ^2/2ma^2−e^2/a.\tag{2.10}$ We do not know what $a$ is, but we know that the atom is going to arrange itself to make some kind of compromise so that the energy is as little as possible. In order to minimize $E$, we differentiate with respect to $a$, set the derivative equal to zero, and solve for $a$. The derivative of $E$ is $dE/da=−ℏ^2/ma^3+e^2/a^2,\tag{2.11}$ and setting $dE/da=0$ gives for a the value $a_0=ℏ^2/me^2=0.528~\text{angstrom},=0.528×10^{−10} ~\text{ meter}.\tag{2.12}$ This particular distance is called the Bohr radius, and we have thus learned that atomic dimensions are of the order of angstroms, which is right. This is pretty good—in fact, it is amazing, since until now we have had no basis for understanding the size of atoms! Atoms are completely impossible from the classical point of view, since the electrons would spiral into the nucleus. ....

This is the excerpt from Feynman's Lectures on The size of an atom. While reading this I couldn't conceive one thing; how he wrote $p\approx \hbar/a$. $a$ is $\Delta y$ that is the uncertainty of position: we can find the electron within $\pm a$ from the nucleus. So, that means $\Delta p \approx \hbar / a$ & not $p\approx \hbar/a$.

Feynman concluded

...the potential energy is reduced if $a$ gets smaller, but the smaller $a$ is, the higher the momentum required, because of the uncertainty principle...

Momentum is not higher but rather the uncertainty becomes higher 'because of Uncertainty Principle'.

Also, as it approaches the nucleus, kinetic energy increases. So, why does the atom need to compromise to decrease $E$ when it approaches the nucleus? After all, it is inevitable that when the electron is very close to the nucleus, it has high KE, isn't it?

So, my questions are:

  • How/Why did Feynman write $p \approx \hbar/ a $ instead of $\Delta p \approx \hbar/a$?

  • Why would the energy $E$ decrease when the electron approaches the nucleus? After all, the KE would become high at the proximity of the nucleus, isn't it?

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    $\begingroup$ Feynman is making a semiclassical argument by combining concepts of classical mechanics like potential and kinetic energy with a concept from quantum mechanics (uncertainty). The result of that combination is not supposed to be a relevant theoretical result but it is just a handwaving argument of what order of magnitude one would expect the actual result to be. So what is the actual result? It's the solution to the Schroedinger equation for the 1/r potential which happens to come out on a similar scale. Within that theory, of course, both potential and kinetic energy are meaningless terms. $\endgroup$ – CuriousOne Sep 11 '15 at 4:40
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The average vector momentum of an electron bound to an atom is exactly zero. (Otherwise, the electron would leave the atom!)

The average magnitude of the momentum can't be zero, because of the uncertainty principle. So Feynman is using the approximation $\vec p = \vec 0 + \Delta p \hat p$, where the magnitude $\Delta p$ comes from the uncertainty principle and the unit vector $\hat p$ points in a completely random direction.

As for your second question, you're almost there. The kinetic energy does become larger for an electron nearer the nucleus — and, thanks to the uncertainty principle, so does the momentum! It has to be this way because the kinetic energy is approximately $T=p^2/2m.$

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  • $\begingroup$ Hi, sir; thanks for the answer. Still not cleared because Uncertainty Principle not speaks of any value but of standard deviations or uncertainties. Momentum doesn't become higher but the uncertainty in the value of momentum becomes here. I'm not understanding that:( $\endgroup$ – user36790 Sep 24 '15 at 15:31
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    $\begingroup$ If you repeatedly sample the momentum you will get momentum values from a distribution whose width is $\Delta p$. Some of those momenta will be smaller than $\Delta p$, but some will be larger. It works out that $\Delta p$ is a good average. Contrariwise, if you repeatedly sampled a distribution of momenta and always found some value with magnitude smaller than $\Delta p$, then you would say that $\Delta p$ is a poor estimator of the uncertainty in the momentum. Standard deviations aren't the same as typical values, but they are closely related. $\endgroup$ – rob Sep 24 '15 at 17:45
  • $\begingroup$ So, standard deviations are close to actual values, is it so? Does this always happen? $\endgroup$ – user36790 Sep 24 '15 at 18:11
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    $\begingroup$ It's always true for variables where the absolute magnitude of the mean value is smaller than the standard deviation. $\endgroup$ – rob Sep 25 '15 at 0:55
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In QM that electron would need more and more kinetic energy when it closes to nucleus, so at some point Coulombic force becomes too weak. The result is an equilibrium position with nucleus, i.e. hydrogen atom.

Heisenberg's uncertainty principle is useless when it comes to hydrogenic $1/r$ potential, but for harmonic oscillator $r^2$ potential we can use Heisenberg's uncertainty principle to easily calculate the ground state energy.

What you need is the following "uncertainty principle" that solves $1/r$ potential and the ground state energy for hydrogen atom, when $\psi$ is the wave for electron. You can prove it as an easy homework exercise: $$\int\frac{|\psi(x,t)|^2}{|x|}dx \le \sqrt{\int |\nabla\psi(x,t)|^2dx}.$$

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