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The attractive Coulomb potential energy is $V(r)=-e^2/(4\pi {ϵ}_{0}r)$, soUse precisely the radial normalization stated in the PDF: the radial probability measure is $4\pi r^2|\psi {|}^{2}dr$. The supplied function hasIts integral is one, since ${\int }_0^{\infty }{r}^4{e}^{-r/a_0}dr=24a_0^5$. In particular,This calculation uses the supplied convention, rather than multiplying a separately normalized standard radial function by an extra $4\pi$.

The addition of angular momentum bracket is $j(j+1)-l(l+1)-s(s+1)=3/4-2-3/4=-2$. Substitution in the spin-orbit energy shift in a central potential therefore givesHere $a_0=5.29\times 10^{-11}\mathrm{m}$ is the supplied Bohr radius. The negative sign means the $j=1/2$ member of the $l=1$ pair is shifted downward by this spin-orbit interaction; it is not the total hydrogen binding energy.

For the isotropic three-dimensional quantum harmonic oscillator, $V(r)=m{\omega }^2{r}^2/2$, hence $r^{-1}dV/dr=m{\omega }^2$ is constant. No radial integral remains, so the spin-orbit energy shift in a central potential reduces toFor $l=0$, the only possibility is $j=1/2$, and the bracket vanishes. For $l=1$, coupling to spin one-half permits $j=1/2$ and $j=3/2$, whose brackets are $-2$ and $+1$. ThusThese spin-orbit shifts of an isotropic harmonic oscillator are a zero line, a downward parabola, and an upward parabola of half the downward magnitude. They are independent of the radial excitation number, and their magnetic multiplicities are respectively two, two and four. Their weighted mean within the $l=1$ manifold is zero: $2\Delta E_{1,1/2}+4\Delta E_{1,3/2}=0$.

The Rashba effect, named for Emmanuel Rashba, is a momentum-dependent spin-orbit interaction associated with broken spatial inversion symmetry. In an isotropic two-dimensional model, choose $\lambda >0$ and write the Hamiltonian $H={\hslash }^2{k}^{2}I/(2m)+\lambda (\hat{\mathbf{z}}\times \mathbf{k})\cdot \mathbf{\sigma }$, where $I$ is the identity matrix and $\mathbf{\sigma }$ is the vector of Pauli matrices. The eigenvalues of the spin term give $E_{\pm }(k)={\hslash }^2{k}^2/(2m)\pm \lambda k$. Define $k_R=m\lambda /{\hslash }^2$ and $E_R={\hslash }^2{k}_R^2/(2m)$. The lower branch is $E_{-}(k)={\hslash }^2(k-k_R{)}^2/(2m)-E_R$ and has a ring of minima at $k=k_R$. A negative $\lambda$ reverses the spin directions; ordering the two energies still uses $|\lambda |$. The alternative parameter $q_R=2k_R$ must not be mistaken for the minimum radius.

Neglect hyperfine structure and use a magnetic field weak compared with the spin-orbit interaction. The sodium D doublet has transitions $3p{}^2{P}_{1/2}\to 3s{}^2{S}_{1/2}$ and $3p{}^2{P}_{3/2}\to 3s{}^2{S}_{1/2}$. Their Landé g-factors are $2/3$, $4/3$ and $2$ respectively. With $m_u,m_l$ denoting upper and lower magnetic labels, the photon shift isThe first transition has four allowed shifts ${\mu }_{B}B\{\pm 2/3,\pm 4/3\}$; the second has six, ${\mu }_{B}B\{\pm 1/3,\pm 1,\pm 5/3\}$. The most blue-shifted component of the second is $m_u=1/2\to m_l=-1/2$, with shift $5{\mu }_{B}B/3$. The apparently larger shift $m_u=3/2\to m_l=-1/2$ is excluded by the electric-dipole selection rule. The $3p{}^2{P}_{3/2}$ level lies above $3p{}^2{P}_{1/2}$, so in this weak-field model the $5/3$ component also has the shortest wavelength of the whole doublet. ^[ref]

For a central-potential Hamiltonian operator with the spin-orbit interaction $H_{\mathrm{so}}=[r^{-1}{V}^{\prime }(r)/(2m^2{c}^2)]\mathbf{L}\cdot \mathbf{S}$, use $\mathbf{J}=\mathbf{L}+\mathbf{S}$ andA coupled angular momentum eigenstate makes this angular factor a number. The radial operator commutes with all angular factors, giving the energy shift in the title. It is independent of the magnetic projection $m_j$, so a fixed-$j$ multiplet retains its $2j+1$ degeneracy.

For spin one-half, the angular bracket is $l$ when $j=l+1/2$ and $-(l+1)$ when $j=l-1/2$. Only the former exists at $l=0$, where it is zero. For the Coulomb potential energy, $r^{-1}{V}^{\prime }(r)=e^2/(4\pi {ϵ}_0{r}^3)$; the radial normalization convention must be respected when computing its expectation value.

At large but finite magnetic field, opposite-spin admixtures in approximate product eigenstates let nominally spin-forbidden electric-dipole channels borrow intensity. Their amplitudes scale as spin-orbit interaction energy divided by Zeeman energy; their intensities therefore vanish quadratically in that ratio.

An equal incoherent mixture within a fixed total-$J$ multiplet has density operator $\rho =P_J/(2J+1)$, where $P_J$ is its orthogonal projection. If the applied-field Hamiltonian operator is proportional to $L_z+2S_z$ and no spin-orbit interaction is retained, its energy eigenstates are the uncoupled states $|m_L,m_S⟩$. The Born rule gives their populations by the formula above, or equivalently by squared Clebsch-Gordan decomposition coefficients. Equal populations of coupled states need not give equal populations of magnetic energy eigenvalues.

For $L=1,S=1/2,J=3/2$, the nonextremal coupled states have weights $2/3$ and $1/3$ in their two uncoupled components. The five dimensionless magnetic energies $-2,-1,0,1,2$ then have populations $1/4,1/6,1/6,1/6,1/4$; the zero energy contains two independent states, each with population $1/12$. In a different, weak-field approximation with a dominant spin-orbit interaction, projecting into the isolated multiplet gives the Landé g-factor spectrum instead. The full operator and its projection must not be interchanged.