# Relativistic Treatment of Core Electrons (DFT)

1. Jul 19, 2013

### citw

I realize this is something I should probably know intuitively, but why are core states (generally) treated relativistically in density functional calculations? What exactly makes these relativistic bound states rather than non-relativistic? I think this is some basic physics I'm forgetting.

2. Jul 19, 2013

### hilbert2

For example, the fact that mercury is liquid unlike most metals and that gold is not silver-colored are results of relativistic effects.

The closer the orbiting electron is to the nucleus, the higher the expectation value of its speed (result of position-momentum uncertainty principle), and relativistic effects are therefore more important in the case of core electrons.

3. Jul 19, 2013

### citw

Could you explain this consequence of the uncertainty principle?

4. Jul 19, 2013

### hilbert2

When the electron is more localized near the nucleus (like in a core orbital), we know its position more precisely, and therefore its momentum is more uncertain according to the equation $\Delta x \Delta p \geq \frac{\hbar}{2}$ . When the momentum is more uncertain, the expectation value of speed (norm of velocity vector) must be higher. At high speeds, relativistic effects become important.

Last edited: Jul 19, 2013
5. Jul 19, 2013

### citw

Sorry, I thinking I'm missing something obvious. If

$$\Delta x\Delta p_x\geq \frac{\hbar}{2}$$

and Δx is small near the nucleus, Δp is large, meaning

$$\Delta p_x=\sqrt{\big<p^2\big>-\big<p\big>^2}$$

is large... the uncertainty in momentum should be large, but wouldn't this be more suggestive of a smaller momentum expectation value? E.g.,

$$\text{if }\sqrt{\big<p^2\big>-\big<p\big>^2}\text{ is large, } \big<p\big>^2\text{ could be small, or }\big<p^2\big>\text{ could be large.}$$

6. Jul 19, 2013

### hilbert2

If the electron is in a bound state, $\big<p\big>=0$, because in the center-of-mass frame the electron is not moving to any specific direction on average. A higher uncertainty of momentum implies a higher expectation value for its absolute value.

7. Jul 19, 2013

The higher speed near the core is related to <KE> not dp, and can be understood via the virial theorem (http://en.wikipedia.org/wiki/Virial_theorem), which applies to both QM and CM. Briefly, there is a definite mathematical relationship between <KE> and the mean potential for a Coulombic potential.

Note that in a closed shell atom, the potential at a distance R from the nucleus, V(R), depends mainly on the charge that is located in the region between 0 and R (from Gauss' Law). So, deep core electrons in heavy atoms experience a largely unshielded nuclear potential.

I believe that hilbert2 is essentially answering the question of how core electrons come about, which is more closely related to the uncertainty principle.

Last edited: Jul 19, 2013
8. Jul 20, 2013

### hilbert2

^ Yes, we don't even need quantum mechanics to see that a particle orbiting in a $1/r$ potential has to have high orbital speed if $r$ is small. The centripetal acceleration of the particle in circular orbit is $a = \frac{v^{2}}{r}$, and as $a\propto \frac{1}{r^{2}}$ we get $v \propto \sqrt{\frac{1}{r}}$ .

In the solar system the planet Mercury, which is closest to the sun, has highest orbital speed and its orbital motion does not agree fully with newtonian mechanics but can be explained with special relativity.

9. Jul 20, 2013

### DrDu

Both their potential and kinetic energy are higher than that of the valence electrons. The high kinetic energy means that they have speeds near the speed of light so that relativistical corrections become important.

10. Jul 20, 2013

### citw

What I've been trying to figure out (and what everyone is trying to explain to me) is why this is the case.

11. Jul 20, 2013

### citw

Ok, this is what I was thinking (stronger attraction near the nucleus), the the virial theorem is the perfect explanation. This is great, thanks.