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

or

\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.

 

 

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