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In https://en.wikipedia.org/wiki/Lamb_shift about the lamb shift, it's mentioned that the change in the electron's frequency due to QED effects (vacuum polarization and self-energy correction) is about 1 GHz, which would translate to an energy change of hf = 6.63E-25 J. This is 3E-7 times of the hydrogen electron's kinetic energy (13.6 eV). It might seem small, but given the huge velocity of the electron in an atom (2E6 m/s), the change in the velocity for a factor of 3E-7 in energy would be about 0.3 m/s (dv = dE/mv).

Now, given the very small size of the atom (1E-10 m), even in one nanosecond, a 0.3 m/s uncertainty in the velocity of the electron would make the uncertainty in its position 3 times more than the radius of the atom, which means a complete uncertainty of position inside the atom.

Is this argument correct? So why are QED effects assumed to be small?

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  • $\begingroup$ your second paragraph doesn't make much sense. You should compare uncertainty due to velocity correction to the original velocity, which is on the order of (fine structure constant)*(speed of light). The "orbit" (semiclassically speaking) of the electron is hardly modified. $\endgroup$ – wcc May 2 at 4:25
  • $\begingroup$ @AmIAStudent The correction is 0.3 m/s, while the original velocity is 2E6 m/s. So the correction is %1.5E-5. This is small of course, but an electron moves 2mm every nanosecond (20 million times of the hydrogen atom's radius). So this small uncertainty would spread dramatically. $\endgroup$ – Ali Lavasani May 2 at 4:32
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I'm assuming all your numbers are correct, and in a certain sense I agree with your conclusion. But here's a different, more precise I think, way of phrasing it. If you had two hydrogen atoms, one which experienced QED and one which didn't, the energy difference of the two electrons would be ~1GHz, and so the two quantum wavefunctions would coherently go out of phase after only 1 nanosecond, and that's pretty fast.

However, it sounds like you're thinking too literally about the atom as having a trajectory around the nucleus, and it's unclear what you mean by a "huge change" to the position of the electrons, and what you mean by an increased positional uncertainty. It's quantum mechanics - everything's waves. I haven't calculated it, but I'm sure the ground state of the hydrogen atom (which tells you where the atom is) with QED is nearly identical to the ground state without QED. Similarly, 1 GHz is a tiny amount of energy compared to the ionization energy of hydrogen. Of course, small is relative, and a 1GHz shift is also easily measurable in modern atomic physics experiments. Hope that helps.

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