Title: ETH_Sem5 / AdvQC / 13_Relativistic_Effects_Chemistry

Q: Which two kinds of relativistic effects are often distinguished? How can you choose which one is implemented?
A: 1. scalar-relativistic or kinematic effects
2. spin-orbit (SO) effects
We can apply Dirac's relation, taking \(\bm{p}\bm{X}\) as one operator,
\[\begin{aligned}&\bm{\sigma}\cdot\bm{p}\bm{X}\bm{\sigma}\cdot\bm{p} = \underbrace{(\bm{p}\bm{X}) \bm{p}}_{\mathrm{sf}} + \underbrace{\mathrm{i}\bm{\sigma}\cdot[(\bm{p}\bm{X})\times\bm{p}]}_{\mathrm{sd}} \\ &\thus \bm{\sigma}\cdot\bm{p}V\bm{\sigma}\cdot\bm{p} = \underbrace{\bm{p}V\bm{p}}_{\mathrm{sf}} + \underbrace{\mathrm{i}\bm{\sigma}\cdot(\bm{p}V\times\bm{p})}_{\mathrm{sd}} \end{aligned}\]
to seperate the spin-free (sf) from the spin-dependent (sd) part! In the case of DKH the resulting even terms \(\mathcal{E}_k\) can always be decomposed to give, with an example for \(\mathcal{E}_{1+}(i,j)\),
\[\begin{aligned} H_{\mathrm{bd}} & =U H_D U^{\dagger} =\cdots U_4 U_3 U_2 U_1 H_1 U_1^{\dagger} U_2^{\dagger} U_3^{\dagger} U_4^{\dagger} \cdots \\ & =\sum_{k=0}^{\infty} \mathcal{E}_k =\sum_{k=0}^{\infty}\left(\begin{array}{cc} \\ \mathcal{E}_{k+} & 0 \\ 0 & \mathcal{E}_{k-} \end{array}\right) \\&=\sum_{k=0}^{\infty}\left(\begin{array}{cc}\\ \mathcal{E}_{k+}^{\mathrm{sf}}+\mathcal{E}_{k+}^{\mathrm{sd}} & 0 \\ 0 & \mathcal{E}_{k-}^{\mathrm{sf}}+\mathcal{E}_{k-}^{\mathrm{sd}} \end{array}\right) \end{aligned}\]
\[\begin{aligned}\mathcal{E}_{1+}&=A_i V_{i j} A_j+A_i (\bm{\sigma}\cdot\bm{P}_i) V_{i j} (\bm{\sigma}\cdot\bm{P}_j) A_j \\&=\underbrace{A_i\left(V_{i j}+\bm{P}_i V_{i j} \bm{P}_j\right) A_j}_{\mathcal{E}_{1+}^{\mathrm{sf}}(i, j)}+\underbrace{\mathrm{i} A_i \bm{\sigma} \cdot\left(\bm{P}_i \times V_{i j} \bm{P}_j\right) A_j}_{\mathcal{E}_{1+}^{\mathrm{sd}}(i, j)}\end{aligned}\]
For the elimination techniques we also always have the \((\bm{\sigma}\cdot\bm{p})\omega(\bm{\sigma}\cdot\bm{p})\) structure which can be separated using the Dirac relation above.

Q: When can a scalar-relavtivistic description be used? What is a typical scalar-relativistic effect?
A: When spin-orbit coupling can be neglected, there remains a 'contribution of relativity' to the kinetic energy operator. The deviation of the non-relativistic from a relativistic kinetic energy is then called a kinematic relativistic effect. As only spin-orbit coupling introduces a two-component structure into the wave function, its neglect implies that scalar functions can be used as eigenfunctions in a relativistic description neglecting spin-orbit coupling. Hence, a kinematic relativistic effect is often also called a scalar-relativistic effect. Inner shells — so-called core-penetrating shells (Tauchbahnen) — contract in a relativistic description. They then better shield the atomic nucleus such that valence shells expand with an effect on chemical bonding (longer/shorter bond lengths depending on the orbitals involved).
Radial functions of Kr-like Rn:

Q: When are spin-orbit (SO) effects (not) negligible? What is a typical SO effect?
A: Spin-orbit (SO) effects are negligible in closed-shell cases; in open-shell systems they can be significant for the electronic energy.
Experimental atomic spin-orbit splitting: (in cm\(^{-1}\))

Q: Where is the relativistic effect on the electron density the largest?
A: The relativistic effect is more pronounced for regions close to the nuclei and tends to increase for heavier elements. However, even very heavy atoms can have comparatively small relativistic effects in some cases (e.g. Fukui function).
Relativistic effect on the electron density in \(\ce{Pt(C2H2)}\): (Difference between rel. and nonrel. electron density)
\(\rho_\mathrm{DC}(\pos) - \rho_\mathrm{nr}(\pos)\): (correct HF) a - c
\(\rho_\mathrm{ZORA}(\pos) - \rho_\mathrm{nr}(\pos)\) (SO but order 0): d - f
\(\rho_\mathrm{DKH10}(\pos) - \rho_\mathrm{nr}(\pos)\) (order 10 but scalar): g - i
Relativistic effects most pronounced for Pt. Scalar-relativistic DKH covers most relativistic effects.

Q: Is spin a purely relativistic effect?
A: Dirac considered his equation as a way to introduce spin into quantum mechanics.
In the late 1960s, Lévy-Leblond showed that spin 1/2 also emerges in a nonrelativistic theory from irreducible representations of the Galilei group. The L\'{e}vy-Leblond equation is a nonrelativistic equation including spin
\[\left(\begin{array}{cc} V & c \bm{\sigma} \cdot \bm{p} \\ c \bm{\sigma} \cdot \bm{p} & -2 m_e c^2 \end{array}\right)\left(\begin{array}{c} \\ \tilde{\Psi}^L \\ \tilde{\Psi}^S \end{array}\right)=E^\mathrm{nr}\left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)\left(\begin{array}{c} \\ \tilde{\Psi}^L \\ \tilde{\Psi}^S \end{array}\right)\]
All difficulties with the elimination of the small component stem from the fact that the relation between large and small component contains the potential energy operator \(V\) as well as the energy. In the L\'{e}vy-Leblond equation this is avoided because the lower part of the matrix equation does not depend on either of them
\[\begin{aligned} V \tilde{\Psi}^L+c \bm{\sigma} \cdot \bm{p} \tilde{\Psi}^S & =E^\mathrm{nr} \tilde{\Psi}^L \\ c \bm{\sigma} \cdot \bm{p} \tilde{\Psi}^L-2 m_e c^2 \tilde{\Psi}^S & =0 \end{aligned}\]
From the second equation we obtain a relation between \( \tilde{\Psi}^S\) and \( \tilde{\Psi}^L\) which yields the Schrödinger equation once inserted in the first equation.
\[\begin{aligned}& \tilde{\Psi}^S =\frac{c \bm{\sigma} \cdot \bm{p}}{2 m_e c^2} \tilde{\Psi}^L \\ &\thus \left[V+c \bm{\sigma} \cdot \bm{p} \frac{c \bm{\sigma} \cdot \bm{p}}{2 m_e c^2}\right] \tilde{\Psi}^L=\left[V+\frac{\bm{p}^2}{2 m_e}\right] \tilde{\Psi}^L=E^\mathrm{nr} \tilde{\Psi}^L \end{aligned}\]