Relativistic one-electron Hamiltonian

The one-electron Hamiltonian including the relativistic spin-orbit coupling term (but excluding other relativistic corrections) adds the term in eq261 to eq45 to give:

$\hat{H}=-\frac{\hbar^2}{2m_e}\nabla^2-\frac{Ze^2}{4\pi\varepsilon_0r}+\frac{1}{2m_e^{\;2}c^2r}\frac{dV}{dr}\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}\; \; \; \; \; \; \; \; 282$

$\hat{H}=\frac{\hat{p}^2}{2m_e}-\frac{Ze^2}{4\pi\varepsilon_0r}+\frac{1}{2m_e^{\;2}c^2r}\frac{dV}{dr}\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}$

Since $V=-\frac{Ze^2}{4\pi\varepsilon_0r}$

$\hat{H}=\hat{H}_0+\frac{1}{2m_e^{\;2}c^2}\frac{Ze^2}{4\pi\varepsilon_0r^3}\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}$

where $\hat{H}_0=\frac{\hat{p}^2}{2m_e}-\frac{Ze^2}{4\pi\varepsilon_0r}$ .

From eq106 and eq107, we know that $[\hat{L^2},\hat{H}_0]=0$ . We have also shown in an earlier article (Pauli matrices) that $[\hat{L^2},\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}]=0$ and $[\hat{S^2},\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}]=0$ . Consequently, $\left\[\hat{L^2},\frac{\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}}{r^3}\right\]=0$ and $\left\[\hat{S^2},\frac{\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}}{r^3}\right\]=0$ .

Question

Show that $[\hat{J}^2,\hat{p}^2]=0$ , $[\hat{J}^2,\frac{1}{r}]=0$ , $[\hat{J^2},\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}]=0$ and $[\hat{J}_z,\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}]=0$ , where $\hat{\boldsymbol{\mathit{J}}}=\hat{\boldsymbol{\mathit{L}}}+\hat{\boldsymbol{\mathit{S}}}$ .

Answer

Substituting $\hat{J}^2=\boldsymbol{\mathit{J}}\cdot\boldsymbol{\mathit{J}}=\hat{L}^2+\hat{S}^2+2\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}$ in $[\hat{J}^2,\hat{p}^2]$ , expanding the expression and using eq104, we have $[\hat{J}^2,\hat{p}^2]=0$ . Similarly, $[\hat{J}^2,\frac{1}{r}]=0$ . Clearly, $[\hat{J}^2,\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}] =0$ . Finally, expanding the RHS of $[\hat{J}_z,\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}] =[\hat{J}_z,\hat{L}_x\hat{S}_x]+[\hat{J}_z,\hat{L}_y\hat{S}_y]+[\hat{J}_z,\hat{L}_z\hat{S}_z]$ , noting that $\hat{L}_i$ and $\hat{S}_i$ act on different vector spaces and using eq100, eq101, eq166 and eq167, $[\hat{J}_z,\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}] =0$ .

Since $\hat{L}^2$ , $\hat{S}^2$ , $\hat{J}^2$ and $\hat{J}_z$ all commute with $\hat{H}$ , we can select a complete set of eigenfunctions in the form of $\vert\hat{J}^2,\hat{J}_z,\hat{L}^2,\hat{S}^2\rangle$ for all operators. The eigenvalue equation of $\hat{H}\psi=E\psi$ is solved by treating $\hat{H}_{so}=\frac{1}{2m_e^{\;2}c^2}\frac{Ze^2}{4\pi\varepsilon_0r^3}\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}$ as a perturbation.

Question

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

Answer

For a one-electron system, $\hat{J}^2=\hat{L}^2+\hat{S}^2+2\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}$ . So, $\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}=\frac{1}{2}(\hat{J}^2-\hat{L}^2-\hat{S}^2)$ . Using this equation and eq133, eq160 and eq205a,

$E_{so}=\langle\psi\vert\xi(r)\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}\vert\psi\rangle=\langle\xi(r)\rangle\frac{\hbar^2}{2}[J(J+1)-L(L+1)-S(S+1)]$

where $\langle\xi(r)\rangle=\biggl\langle\psi\left | \frac{1}{2m_e^{\;2}c^2}\frac{ze^2}{4\pi\varepsilon_0r^3} \right |\psi\biggr\rangle$ .

Relativistic one-electron Hamiltonian

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