Relativistic energy level shift from $s$ to $d$ orbitals

Question

I have read this question:

This question is not about the energy levels, negative and kinetic, why it is lower (more negative) close to the nucleus. I understand those. I understand electrons have kinetic and potential energy. I understand as you go closer to the nucleus, the electron will have more kinetic and less potential (more negative) energy. I understand that electrons do not orbit in a classical way, I understand they are existing around the nucleus at a certain energy level as per QM.

This question is about the energy shift from s to d because of relativistic effects (and because of that gold absorbs blue color).

How do electrons in a lower energy level have more kinetic energy?

Dependence of the energy of an electron on distance from the nucleus

https://physics.stackexchange.com/a/72412/132371

Where Johannes's answer says:

For gold (with atomic number 79 and hence a highly charged nucleus) this classical picture translates into relativistic speeds for electrons in s orbitals. As a result, a relativistic contraction applies to the s orbitals of gold, which causes their energy levels to shift closer to those of the d orbitals (which are localized away from the nucleus and classically speaking have lower speeds and therefore less affected by relativity).

I understand that relativistic effects come into play, because of the Heisenberg uncertainty principle, and not because at s orbitals electrons just move fast like classical planets. On an s energy level, electrons do have a higher probability of being closer to the nucleus, so they are constrained into a smaller space, thus we can tell their position with more precision, thus their momenta will be more uncertain, they will have more kinetic energy.

Electrons on the d orbitals (away from the nucleus) have higher energy (potential), which means (since electrons have negative EM energy) smaller kinetic energy in absolute term (less negative).

Electrons on the s orbital (closer to the nucleus) have lower (potential) energy, which means (since electrons have negative EM energy) higher kinetic energy in absolute term (more negative).

What this answer does not say, is how the s orbital's energy level will shift closer to the d orbital's energy level because of relativistic effects.

I do not understand if s orbitals have higher kinetic energy, and d orbitals lower kinetic energy, how can relativistic effects (Heisenberg uncertainty principle) shift the energy level of s orbital electrons closer to the d orbital electrons, so s orbital electron's energy level will be shifted to lower kinetic energy?

It should be the opposite, closer to the nucleus, the Heisenberg uncertainty principle should give (because of less uncertainty in the position, more uncertainty in the momentum) more kinetic energy to the electron.

I do not understand how can an s orbital electron act like a d orbital electron?

Question:

1. how can relativistic effects (Heisenberg uncertainty principle) shift the energy level to lower kinetic energy (from s to d) without the electron actually moving to the higher energy (d orbital) level?

2. How can an s orbital electron's energy level shift closer to a d orbital electron's energy level without moving to that higher energy level as per QM and change the absorption frequency?

Answer 1

I understand that electrons do not orbit in a classical way, I understand they are existing around the nucleus at a certain energy level as per QM.

It doesn't look like it.

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When you do the slow-speeds Taylor expansion of the relativistic energy in terms of momentum, $$ E_k = \sqrt{m^2c^4 + p^2c^2} = mc^2 + \frac{p^2}{2m} - \frac{p^4}{8m^3c^2} + \cdots, $$ you get a rest-energy offset (which it's safe to neglect), the newtonian kinetic energy, and bunch of relativistic corrections. When you move over to the quantum-mechanical domain, this translates into a kinetic energy operator $$ \hat E_k = \frac{\hat p^2}{2m} - \frac{\hat p^4}{8m^3c^2}, $$ correct to subleading order. Generally speaking, the relativistic corrections are too hard to solve directly, and we use the framework of time-independent perturbation theory, which tells us that if $|\psi⟩$ is an eigenstate of the non-relativistic hamiltonian, its energy shift from the relativistic corrections will be $$ \Delta E_\psi = -\left<\psi\middle|\frac{\hat p^4}{8m^3c^2}\middle|\psi\right>. $$ Why does the $s$ level get shifted more than the $d$ level? Because the matrix element is bigger. Anything else is added heuristics; those sometimes work, and sometimes fail, and they must always be referred back to the formal QM calculations.

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In this specific case, though, the heuristics are different to what you state in your question,

Electrons on the d orbitals (away from the nucleus) have higher energy (potential).

The difference between $s$ and $d$ electrons in this case has very little to do with the base non-relativistic approximation to their energy. Instead, it's to do with the centrifugal barrier experienced by the $d$ electrons and not by the $s$ electrons, as I explained in my answer to Why are the higher angular momentum states of a hydrogen atom closer to the nucleus?. Basically, the lack of a centrifugal barrier means that the $s$ electrons spend much more time than the $d$ electrons at very small radii (though, to be clear, they also spend more time at larger radii), and the added time at small radii where the kinetic energy is large (and therefore the local wavelength is very short) means that those states get a much larger contribution to their expectation value of $⟨\hat p^4⟩$, which determines the relativistic shift.

Answer 2

1. The mass of the particle increases which results in the fact that acceleration requires more work.
2. The level shifts, not the energy itself.