Relativistic many-electron Hamiltonian

The multi-electron Hamiltonian including the relativistic spin-orbit coupling term (but excluding other relativistic corrections) is

$\hat{H}_R=\hat{H}_T+\hat{H}_{so}\; \; \; \; \; \; \; \; 283$

where $\hat{H}_T$ and $\hat{H}_{so}$ are given by eq240 and eq261a respectively.We shall now show that $\hat{J}_z$ and $\hat{J}^2$ commute with $\sum_{i=1}^n\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i$ and therefore with $\hat{H}_R$ . Assuming $n=2$ ,

$\left[\hat{J}_z,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=\left [\hat{L}_z^T+\hat{S}_z^T ,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right ]=\left [\hat{L}_{1z}+\hat{L}_{2z}+\hat{S}_{1z}+\hat{S}_{2z} ,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right ]$

Expanding the RHS of the second equality of the above equation and using eq99, eq100, eq101, eq165, eq166 and eq167, $\left[\hat{J}_z,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=0$ . In general,

$\left[\hat{J}_z,\sum_{i=1}^n\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=0\; \; \; \; \; \; \; \; 284$

Similarly,

$\left[\hat{J}_x,\sum_{i=1}^n\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=0\; \; \; \; \; and\; \; \; \; \;\left[\hat{J}_y,\sum_{i=1}^n\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=0\; \; \; \; \; \; \; \; 285$

Furthermore, $[\hat{J}_z,\hat{H}_T]=[\hat{L}_z^T,\hat{H}_T]+[\hat{S}_z^T,\hat{H}_T]$ . Using eq243, $[\hat{J}_z,\hat{H}_T]=0$ . Combining this with eq284,

$\left[\hat{J}_z,\hat{H}_R\right]=0\; \; \; \; \; \; \; \; 286$

Next,

$\left[\hat{J}^2,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=\left[\hat{J}_x^2,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]+\left[\hat{J}_y^2,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]+\left[\hat{J}_z^2,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]$

Using the identity $[\hat{A}\hat{B},\hat{C}]=[\hat{A},\hat{C}]\hat{B}+\hat{A}[\hat{B},\hat{C}]$ and substituting eq284 and eq285 in the above equation, $\left[\hat{J}^2,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=0$ . In general,

$\left[\hat{J}^2,\sum_{i=1}^n\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=0\; \; \; \; \; \; \; \; 287$

Using eq244, eq245, $\left[\hat{J}^2,\hat{H}_T\right]=\left[\hat{{L}^2}^T+\hat{{S}^2}^T+2\hat{\boldsymbol{\mathit{S}}}^T\cdot\hat{\boldsymbol{\mathit{L}}}^T,\hat{H}_T\right]=0$ . So,

$\left[\hat{J}^2,\hat{H}_R\right]=0\; \; \; \; \; \; \; \; 288$

Therefore, $\hat{J}^2$ and $\hat{J}_z$ commute with $\hat{H}_R$ . We have also shown in an earlier article that $\hat{P}_{ij}$ commutes with $\hat{J}^2$ , $\hat{J}_z$ and $\hat{H}_R$ . This implies that we can select a common complete set of eigenfunctions for $\hat{J}^2$ , $\hat{J}_z$ , $\hat{P}_{ij}$ and $\hat{H}_R$ . . Unlike the case of the hydrogen atom, $\hat{{L}^2}^T$ and $\hat{{S}^2}^T$ no longer commute with $\hat{H}_{so}$ for a multi-electron atom (see eq182 and eq182a). Despite that, the quantum numbers $J$ , $L$ and $S$ are used to characterise the states of elements in the first three periods of the periodic table, because spin-orbit coupling is weak for these atoms.

Similar to hydrogen, the eigenvalue equation of $\hat{H}\psi=E\psi$ for a multi-electron atom is solved by treating $\hat{H}_{so}$ as a perturbation if spin-orbit coupling is weak, with the unperturbed part of the eigenvalue equation solved using the Hartree-Fock method.

Question

Evaluate $E_{so}$ using the first order perturbation theory.

Answer

For a multi-electron system, $\hat{J}^2=\hat{L}^2+\hat{S}^2+2\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}$ . So, $\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}=\frac{1}{2}\left ( \hat{J}^2-\hat{L}^2-\hat{S}^2\right )$ , which when substituted in $E_{so}=\langle\psi\vert\xi(r)\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}\vert\psi\rangle$ and using eq133, eq160 and eq205a, gives

$E_{so}=A\frac{\hbar^2}{2}[J(J+1)-L(L+1)-S(S+1)]\; \; \; \; \; \; \; \; 289$

where $A=\langle\psi\vert\xi(r)\vert\psi\rangle=\biggl\langle\psi\left | \sum_{i=1}^n\frac{1}{2m_e^{\;2}c^2r_i}\frac{dV_i}{dr_i}\right |\psi\biggr\rangle$ .

$A$ is evaluated empirically to be a constant for a given term.

Relativistic many-electron Hamiltonian

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