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Cyclotron resonance is resonant absorption by charged particles in a magnetic field when an applied oscillating electric field drives their cyclotron motion. For a parabolic band, the resonance frequency is ${\omega }_c=|q|B/m^{\ast }$, so measuring it determines the cyclotron effective mass. In the quantum description, transverse electric dipole transitions connect adjacent Landau levels, whose separation is $\hslash {\omega }_c$. For an electron and a field along $z$, only the circular polarization rotating in the electron's orbital sense drives the positive-frequency resonance. Scattering broadens the absorption peak; resolving it requires an orbital period short compared with the relaxation time.

An electric dipole transition is a transition driven, in the long-wavelength approximation, by the electric-dipole interaction $H_{\mathrm{int}}=-\mathbf{d}\cdot \mathbf{E}(t)$. Its leading amplitude is proportional to a matrix element of the electric dipole moment between the initial and final states. A vanishing element gives an electric-dipole selection rule; a nonzero element alone does not ensure a large probability, since the drive frequency, duration and strength also matter. For a charged quantum harmonic oscillator driven along its axis, $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$ gives the single-quantum rule $\Delta n=\pm 1$.

A spherical tensor operator of rank $k$ transforms under rotation like an angular momentum $k$. The Wigner-Eckart theorem writes its matrix elements asThe reduced matrix element contains the radial and other dynamical information; the Clebsch-Gordan coefficients determine the dependence on the projections. Ratios within a fixed pair of multiplets therefore follow without a radial integral. The angular selection rules include $m^{\prime }=m+q$ and $|j-k|\le j^{\prime }\le j+k$.

For the position vector, choose the rank-one componentsLet $R=⟨{\lambda }^{\prime },3/2\Vert T^{(1)}\Vert \lambda ,1/2⟩$. The stretched Clebsch-Gordan coefficients and one lowering step giveFor example, apply the total angular momentum ladder operator to the stretched state. The result has components with coefficients $1/\sqrt{3}$ and $\sqrt{2/3}$, corresponding to lowering the initial angular momentum and the vector component, respectively.

Only the $q=+1$ component of $x$ contributes to $M_1$. The $q=0$ operator $z$ cannot connect the projections in $M_2$, so that matrix element vanishes. With the convention above,Thus the requested relative magnitudes and their squared values areThese ratios presume a common nonzero reduced matrix element. The position operator has odd parity, so an electric dipole transition requires opposite-parity multiplets; for one Electron the electric-dipole selection rule also gives $\Delta \ell =\pm 1$. The displayed $j,m$ labels alone do not establish that condition. If the common reduced matrix element vanishes, all three matrix elements vanish and no nonzero ratio can be inferred.

The Zeeman effect comes from $-\mathbf{\mu }\cdot \mathbf{B}$. For an Electron, the orbital and spin magnetic moments are approximately $-{\mu }_B\mathbf{L}/\hslash$ and $-2{\mu }_B\mathbf{S}/\hslash$, respectively. This explains both the sign and the relative factor two in the term linear in $B$. The second term is the spin-orbit interaction: the Electron's motion through the atomic electric field produces a magnetic interaction in its rest frame. The relativistic correction, including Thomas precession, and the radial expectation value are incorporated into $K$. This effective Hamiltonian retains the linear Zeeman splitting and spin-orbit coupling, while omitting diamagnetic and hyperfine structure corrections.

Define the angular momentum ladder operators by $L_{\pm }=L_x\pm i{L}_y$ and similarly for $S_{\pm }$. Direct expansion givesso adding $L_z{S}_z$ gives the required scalar product. The off-diagonal terms change $m_{\ell }$ and $m_s$ in opposite directions. Consequently $J_z=L_z+S_z$ commutes with $H$, and $m_j$ remains a good quantum number at every magnetic field.

In the given product eigenbasis, the diagonal matrix elements areFor $\ell =1$ and $s=1/2$, the only off-diagonal couplings are between $(1,-1/2)$ and $(0,1/2)$, and between $(0,-1/2)$ and $(-1,1/2)$. Each relevant orbital ladder factor is $\hslash \sqrt{2}$ and each spin ladder factor is $\hslash$. Multiplying by $K/{\hslash }^2$ gives $\sqrt{2}K$. ThereforeAll other off-diagonal entries vanish either by the ladder endpoints or by conservation of $m_j$. This derives every entry of the displayed matrix.

The stretched states have no mixing and yieldThe $m_j=1/2$ block is $\left(\begin{array}{cc}-K& \sqrt{2}K\\ \sqrt{2}K& a\end{array}\right)$, while the $m_j=-1/2$ block is $\left(\begin{array}{cc}-a& \sqrt{2}K\\ \sqrt{2}K& -K\end{array}\right)$. Taking the trace and determinant of each $2\times 2$ block gives the remaining four energy eigenvalues:The subscripts indicate $m_j=\pm 1/2$ and the superscripts indicate the upper or lower root. These formulas are the Zeeman spectrum for orbital angular momentum one and remain valid for either sign of $K$.

For the usual ordering $K>0$, including sodium, the zero-field states follow fromThe $j=3/2$ quartet lies at $K$ and the $j=1/2$ doublet at $-2K$, separated by $3K$. In a weak magnetic field, the Landé g-factors giveThese label the curves near $B=0$. In a strong magnetic field, the Paschen-Back effect makes $(m_{\ell },m_s)$ the appropriate approximate labels:for the mixed states, with the stretched-state formulas already exact. The branch connections for $B>0$ areWithin a fixed $m_j$, the nonzero coupling prevents a crossing; this is an avoided crossing. Curves with different $m_j$ may cross because the Hamiltonian does not couple them.

For the sodium estimate, the zero-field frequency separation fixesThe lower $3s$ doublet has shifts $\pm a$. The Sodium D-line Zeeman pattern gives photon-frequency shifts $a/h$ times $\{\pm 2/3,\pm 4/3\}$ for D${}_1$ and $\{\pm 1/3,\pm 1,\pm 5/3\}$ for D${}_2$. Thus the highest-frequency D${}_1$ component and lowest-frequency D${}_2$ component approach each other by $3a/h$. Extrapolating the weak-field Zeeman splitting givesso the crossover occurs at a field of order $10\mathrm{T}$, where $a\sim K$ and the weak-field approximation is becoming marginal.

The exact eigenvalues refine the estimate within this Hamiltonian. The relevant photon energies, with a common zero-field reference omitted, are $E_{+}^{(-)}+a$ and $E_{-}^{(+)}-a$. The corresponding electric dipole transitions have $\Delta m_j=+1$ and $-1$, respectively. Equating them givesIts positive solution is $a/K=\sqrt{10}/3$, hencein the retained model. The crossing is between optical transition frequencies; the participating upper states themselves need not be degenerate.

The infinite square well has energy levels $E_n=n^2{h}^2/(8m_e{L}^2)$, with $n=1,2,\dots$. The longest absorption wavelength corresponds to the smallest available energy difference from the ground state, namely $n=1\to 2$. Its position matrix element is nonzero because the two states have opposite parity about the center of the well, so it is an allowed electric dipole transition.

Equating the photon energy to this gap givesHere the photon gap is approximately $1.55\mathrm{eV}$, not the ground-state energy itself.

The nuclear beta decay is fast compared with the electronic orbital timescale. In the sudden approximation, the electronic wavefunction does not have time to change during the increase from $Z_i=1$ to $Z_f=2$. The subsequent probability of a particular daughter energy eigenstate is therefore its squared overlap with the initial state.

For the final $1s$ state, write ${\beta }_i=1/a_0$ and ${\beta }_f=2/a_0$. The sudden hydrogenic charge change gives the amplitudeThusThe final $2p$ states have spherical harmonics $Y_{1m}$, whereas the initial state is proportional to $Y_{00}$. Their angular overlap $\int Y_{1m}^{\ast }{Y}_{00}d\Omega$ vanishes by orthogonality. Accordingly $\boxed{P_{2p}=0}$ in the ideal spherically symmetric sudden-change model. Nuclear recoil or interactions with the emitted particles can produce small corrections. This is an overlap selection rule for a sudden change, not a photon-induced electric dipole transition.

Inverting the definitions of the creation operator and annihilation operator gives $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$. Their actions and the orthogonality of the oscillator energy eigenstates giveThus the electric dipole transition selection rule iswith the lower alternative absent for $n_1=0$. The field exchanges one oscillator quantum in first-order dipole coupling. For a long weak monochromatic drive, its frequency must also be resonant with the level separation, here $\omega$, to give a substantial transition probability. Higher-order multiphoton processes are beyond the stated single-position-matrix element approximation.