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I am wondering about this. I have often encountered a claim that the electrons in an atom have an "orbital speed", and in some cases that even so much as a number is assigned to this supposed "speed", like that the electron in hydrogen at the 1s ground state "orbits" at a very specific number of $\approx \frac{1}{137} c$. But what exactly does that mean?

Apparently, it seems to go back to the Bohr atomic model, where the electrons are still conceived of as classical (Newtonian)-mechanical particles, following Newton's laws of motion except that they are restricted to only certain discrete orbits. And if you go through the calculation you find the ground state speed is this amount.

But as we all know, Bohr's model is deader than dead from quantum mechanics. But in this real quantum mechanical context, there is now, it seems, no way to make sense of this. There is no definite velocity for the electron, because its momentum is uncertain; but even taking this into account you can still take the expected value, but the expected value of momentum is zero. You can't even use the angular momentum to "coax" something analogous to an orbital speed or angular velocity out even though one might be led into thinking that would be a potentially useful route - after all, if the regular momentum is zero perhaps there should be some angular momentum though, no? Yet at ground state $1\mathrm{s}$, the angular momentum is zero and thus the orbital speed should be zero. It's only the excited states with the letter after the number being p, d, f, etc. (or ground state electrons in those levels in multi-electron atoms) that have nonzero orbital angular momentum. So what on earth does this number $\frac{1}{137} c$ even mean in this context, such that it's still legitimate to quote it (don't have cites but I've seen posts even here to this sort)? How can you extract that from a quantum wave function in any way that has anything at all to do with motion, or orbits, or anything like that? Especially when it's clearly said "electrons DO NOT ORBIT".

Which of course makes one want to perhaps just dismiss this as flat out wrong. BBBBUUTTTTTTT... it seems that that's a little too easy, and actually there is something going on after all, because in heavy atoms, "relativistic" effects occur that change the structure, suggesting high "speeds" - in particular the "orbitals" contract as though they were somehow relativistic orbits at a speed, due to relativity! Thus suggesting this number was meaningful after all! But then what exactly is its meaning in a fully quantum-mechanical treatment with no Bohr-like semi-classical models?

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You are quite correct that electrons don't orbit, however we can calculate an expectation value for the kinetic energy. That calculation is described in Calculating the (expected) kinetic energy of an electron in the ground state of a Coulomb potential? if you're interested in the gory details.

The expectation value of the kinetic energy turns out to be simply the energy of the ground state i.e. 13.6eV. This shouldn't be unexpected as we expect the virial theorem to apply to the hydrogen atom, in which case we have:

$$ KE = -\tfrac{1}{2}PE $$

and the total energy is then simply:

$$ E = KE + PE = KE - 2KE = -KE $$

Anyhow, if you naively use the equation $KE = \tfrac{1}{2}m_ev^2$ you get $v \approx c/137$. There is no simple physical interpretation of this velocity but it is a useful indication that we don't don't need to consider relativistic corrections to get a useful description of the hydrogen atom.

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  • $\begingroup$ So I guess the more "correct" metric (that doesn't invoke a pseudo-Newtonian velocity) of whether or not relativity would be necessary is whether or not the expected KE is within range of the electron's rest energy, right? And since that's not the case for hydrogen (13.6 eV << 511 000 eV) then it is not very relativistic. $\endgroup$ – The_Sympathizer Oct 28 '17 at 9:41
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    $\begingroup$ Yes. As a general rule we'd simply compare the total energy to the rest mass energy. $\endgroup$ – John Rennie Oct 28 '17 at 9:45

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