# Do QED effects make a huge change to the position of the electrons?

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?

• 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.
– wcc
May 2 '19 at 4:25
• @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. May 2 '19 at 4:32