Electric Dipole Transitions and Selection Rules

Derivation

Electric Dipole Transitions and Selection Rules

D-228 Home PU-301 Threads light · symmetry · matter Depends on Fermi's Golden Rule, minimal-coupling-em-hamiltonian, Spherical Harmonics as Angular Momentum Eigenstates

Statement

Starting from the minimal-coupling Hamiltonian and Fermi's golden rule, we derive the electric-dipole (E1) spontaneous-emission rate for an atomic transition \(|i\rangle \to |f\rangle\), \(\;\Gamma_{fi} = \dfrac{\omega_{fi}^3\,|\langle f|\,e\vec r\,|i\rangle|^2}{3\pi\varepsilon_0\hbar c^3}\), and show that its angular integral is nonzero only when \(\Delta \ell = \pm 1\), \(\Delta m = 0,\pm 1\) with a change of parity — the atomic selection rules.

Why it matters

Almost every spectral line you can see — the sodium doublet, hydrogen Lyman-\(\alpha\), the light from stars — is an electric-dipole transition. The selection rules explain which lines appear, why some levels are metastable (the \(2s\) state of hydrogen lives \(10^{8}\) times longer than \(2p\)), and why spectra come in orderly series rather than a structureless continuum.

The same rules encode conservation of angular momentum and parity between atom and photon. They are the microscopic reason a photon carries one unit of spin, and they underpin optical pumping, laser gain, and the interpretation of every atomic spectrum.

Assumptions

The atom couples to the field through minimal coupling \(\;\vec p \to \vec p - q\vec A\).Drop it and there is no principled interaction Hamiltonian; the \(\vec A\cdot\vec p\) coupling that produces radiation would be missing. The field is weak enough that first-order perturbation theory (Fermi's golden rule) applies.Drop it and one must resum to all orders — Rabi flopping, power broadening and multiphoton processes replace the single rate. The wavelength greatly exceeds the atomic size, \(k a_0 \sim \alpha \ll 1\), so \(e^{i\vec k\cdot\vec r}\approx 1\) (the dipole approximation).Drop it and higher multipoles (M1, E2) enter at order \((ka_0)^2\); "forbidden" lines reappear and the selection rules below are only the leading term. The states are simultaneous eigenstates of \(\hat H_0,\ \hat L^2,\ \hat L_z\) with definite parity (central potential, LS coupling).Drop it — e.g. an external field mixing parity, or strong configuration interaction — and \(\ell\) is no longer good; the rules soften into approximate ones.

Derivation

\[ \hat H = \frac{1}{2m}\big(\vec p - q\vec A\big)^2 + V(r) = \hat H_0 - \frac{q}{m}\,\vec A\cdot\vec p + \frac{q^2}{2m}\vec A^2 \]

Expand minimal coupling and adopt the Coulomb gauge \(\nabla\cdot\vec A=0\) so \(\vec A\cdot\vec p = \vec p\cdot\vec A\). For an electron \(q=-e\); drop the \(\vec A^2\) term (second order in the field). A

\[ \hat H'(t) = \frac{e}{m}\,\vec A\cdot\vec p, \qquad \vec A(\vec r,t) = A_0\,\hat{\boldsymbol\varepsilon}\,e^{i(\vec k\cdot\vec r-\omega t)} + \text{c.c.} \]

Identify the linear coupling as the perturbation and insert a monochromatic mode of polarization \(\hat{\boldsymbol\varepsilon}\perp\vec k\). The \(e^{+i\omega t}\) part drives emission, \(e^{-i\omega t}\) absorption. A

\[ \Gamma_{fi} = \frac{2\pi}{\hbar}\,\big|\langle f|\hat H'|i\rangle\big|^2\,\rho(E_f) \;\propto\; \Big|\big\langle f\big|\,e^{i\vec k\cdot\vec r}\,\hat{\boldsymbol\varepsilon}\cdot\vec p\,\big|i\big\rangle\Big|^2 \]

Apply Fermi's golden rule; energy conservation fixes \(\hbar\omega = E_i-E_f\equiv\hbar\omega_{fi}\). All the atomic physics sits in the matrix element. B

\[ e^{i\vec k\cdot\vec r} = 1 + i\vec k\cdot\vec r + \cdots \;\approx\; 1, \qquad k a_0 \sim \frac{2\pi a_0}{\lambda} \sim \alpha \approx 7\times10^{-3} \]

Dipole approximation: over the atomic scale \(a_0\) the field phase barely varies, so keep the leading term. Corrections are the higher multipoles. B t3

\[ [\hat{\vec r},\hat H_0] = \frac{i\hbar}{m}\,\vec p \;\Longrightarrow\; \langle f|\vec p|i\rangle = \frac{m}{i\hbar}\langle f|[\hat{\vec r},\hat H_0]|i\rangle = i m\,\omega_{fi}\,\langle f|\vec r|i\rangle \]

Since \(\hat H_0=\hat p^2/2m + V\), only the kinetic term fails to commute with \(\vec r\), giving \([\vec r,\hat H_0]=i\hbar\vec p/m\). Sandwiching between eigenstates turns the momentum matrix element into a position (dipole) one — the "length gauge" form. This is exact, not an approximation. C t3

\[ \vec d \equiv -e\,\vec r, \qquad \Gamma_{fi} = \frac{\omega_{fi}^3}{3\pi\varepsilon_0\hbar c^3}\,\big|\langle f|\vec d|i\rangle\big|^2 = \frac{e^2\omega_{fi}^3}{3\pi\varepsilon_0\hbar c^3}\,\big|\langle f|\vec r|i\rangle\big|^2 \]

Sum over the two transverse polarizations and integrate the emitted photon over solid angle (each contributes \(\tfrac{1}{3}|\vec d_{fi}|^2\) after averaging \(|\hat{\boldsymbol\varepsilon}\cdot\vec d|^2\)); with the free-space photon density of states this is the Einstein \(A\) coefficient. The matrix element of \(\vec r\) is what remains. C

\[ z = r\sqrt{\tfrac{4\pi}{3}}\,Y_1^0, \qquad x\pm iy = \mp r\sqrt{\tfrac{8\pi}{3}}\,Y_1^{\pm1} \]

Write the Cartesian components of \(\vec r\) in the spherical basis: each is \(r\) times a \(Y_1^{q}\). The dipole operator is a rank-1 spherical tensor \(T^{(1)}_q\), \(q=0,\pm1\). B

\[ \langle n'\ell' m'|\,r\,Y_1^{q}\,|n\ell m\rangle \propto \int Y_{\ell'}^{m'*}\,Y_1^{q}\,Y_\ell^{m}\,d\Omega \propto \begin{pmatrix} \ell' & 1 & \ell \\ -m' & q & m \end{pmatrix}\begin{pmatrix} \ell' & 1 & \ell \\ 0 & 0 & 0 \end{pmatrix} \]

The angular part is a Gaunt integral of three spherical harmonics; the Wigner–Eckart theorem expresses it through two 3-\(j\) symbols. Nonvanishing requires the projections and triangle conditions of these symbols. C t3

\[ m' = m + q \;\Rightarrow\; \boxed{\Delta m = 0,\pm1}, \qquad |\ell-1|\le \ell' \le \ell+1 \;\Rightarrow\; \Delta\ell \in \{-1,0,+1\} \]

The first 3-\(j\) symbol vanishes unless \(-m'+q+m=0\); the triangle rule of both symbols limits \(\Delta\ell\) to \(0,\pm1\). Angular momentum \(\vec L_{\rm atom}+\vec L_{\rm photon}\) is conserved, the photon carrying one unit. B

\[ \hat P\,\vec r\,\hat P^{-1} = -\vec r, \qquad \hat P\,Y_\ell^m = (-1)^\ell Y_\ell^m \;\Rightarrow\; (-1)^{\ell'}(-1)^{1}(-1)^{\ell} = +1 \]

\(\vec r\) is parity-odd, so the integrand's total parity \((-1)^{\ell'+\ell+1}\) must be even for a nonzero integral: \(\ell'+\ell\) is odd. This kills \(\Delta\ell=0\), and the second 3-\(j\) symbol \(\binom{\ell'\,1\,\ell}{0\,0\,0}\) vanishes for the same reason. B

\[ \Delta\ell = \pm1 \quad(\text{Laporte's rule: parity must flip}) \]

Combining "\(\Delta\ell\) odd" (parity) with "\(|\Delta\ell|\le1\)" (angular momentum) leaves exactly \(\Delta\ell=\pm1\). No E1 transition connects states of equal parity. A

Result

\[ \Gamma_{fi} = \frac{e^2\,\omega_{fi}^3}{3\pi\varepsilon_0\hbar c^3}\,\big|\langle f|\vec r|i\rangle\big|^2, \qquad \boxed{\;\Delta\ell = \pm1,\quad \Delta m = 0,\pm1,\quad \text{parity flips}\;} \]

Reading. An excited atom radiates at a rate set by the cube of the transition frequency times the squared dipole matrix element. The transition is dead unless the orbital angular momentum changes by exactly one unit and the parity reverses; the magnetic quantum number may change by at most one, the deficit carried off as the photon's spin projection. In full \(jj\)/fine-structure notation the companion rules are \(\Delta J = 0,\pm1\) (but \(J=0\nrightarrow J=0\)), \(\Delta m_J = 0,\pm1\), and \(\Delta S = 0\) in LS coupling.

Units check. \([\omega^3]=\mathrm{s^{-3}}\); \([e^2/\varepsilon_0]=\mathrm{J\cdot m}\); \([|\vec r|^2]=\mathrm{m^2}\); \([\hbar c^3]=\mathrm{J\,s\cdot m^3 s^{-3}}=\mathrm{J\,m^3 s^{-2}}\). Then \(\dfrac{\mathrm{s^{-3}\cdot J\,m\cdot m^2}}{\mathrm{J\,m^3 s^{-2}}}=\mathrm{s^{-1}}\) — a rate, as required.

Limiting cases

Hydrogenic, low \(Z\): \(\omega_{fi}\propto Z^2\) and \(\langle r\rangle\propto Z^{-1}\), so \(\Gamma\propto Z^4\) — inner-shell transitions are enormously faster.
\(\Delta\ell=0\) or \(\Delta\ell=\pm2\): E1 amplitude vanishes exactly; the transition proceeds only via M1/E2 or two-photon paths, suppressed by \((ka_0)^2\sim\alpha^2\) — a "forbidden" line.
Long wavelength \(\omega\to0\): \(\Gamma\propto\omega^3\to0\); low-frequency transitions are intrinsically slow, which is why fine- and hyperfine-structure levels are long-lived.
\(J=0\to J=0\): strictly forbidden to all single-photon multipoles — a single photon cannot carry zero angular momentum.

Breaks when

Short wavelength / hard X-rays: \(ka_0\) is no longer small, \(e^{i\vec k\cdot\vec r}\neq1\), and higher multipoles become comparable — the dipole rate and its selection rules cease to dominate.
Strong external or laser fields: perturbation theory fails (Rabi cycling, AC-Stark shifts, ionization); the single golden-rule rate is meaningless and field-dressed states must be used.
Broken parity / mixed \(\ell\): in a static electric field (Stark) or in non-central potentials, \(\ell\) and parity are not conserved, so parity-forbidden lines appear (e.g. Stark-induced \(2s\!-\!1s\)).
Relativistic / high-\(Z\) atoms: spin–orbit mixing makes \(S\) and \(\ell\) approximate, weakening \(\Delta S=0\) and enabling intercombination lines.

Failure modes

Forgetting the parity rule and allowing \(\Delta\ell=0\) (e.g. calling \(2s\to1s\) an allowed E1 line).
Using \(\Delta m=\pm1\) but ignoring \(\Delta m=0\), or thinking \(\Delta m\) refers to the photon's helicity rather than the atomic projection change.
Writing \(\Gamma\propto\omega\) or \(\omega^2\) instead of \(\omega^3\) — dropping factors from the photon density of states.
Confusing the one-electron rule \(\Delta\ell=\pm1\) with the many-electron rule \(\Delta L=0,\pm1\); for the atom as a whole \(\Delta L=0\) is allowed while for the jumping electron \(\Delta\ell=0\) never is.
Applying \(J=0\nrightarrow J=0\) to \(\ell\) as well; \(\ell=0\to\ell=1\) is perfectly allowed.
Using the momentum form \(\langle f|\vec p|i\rangle\) with an approximate wavefunction and expecting it to equal the length form — they differ unless \(|i\rangle,|f\rangle\) are exact eigenstates.

Discussion

The selection rules are bookkeeping for two conserved quantities. Angular momentum: a dipole photon carries spin 1, so the atom must shed exactly one unit — hence \(|\Delta\ell|\le1\) and \(\Delta m=0,\pm1\), with \(\Delta m\) fixed by the photon's polarization (\(\sigma^\pm\) light drives \(\Delta m=\pm1\), \(\pi\) light drives \(\Delta m=0\)). Parity: the photon field \(\vec A\) is odd, so absorbing or emitting one flips the atom's parity, forbidding \(\Delta\ell=0\). Together they leave the single clean rule \(\Delta\ell=\pm1\).

The \(\omega^3\) scaling is the deep reason spectroscopy is possible at all. Because rates grow as the cube of frequency, optical and UV transitions are fast (nanoseconds) and bright, while microwave transitions between fine- or hyperfine-structure levels are slow (seconds to megayears) and correspondingly narrow — the basis of atomic clocks and the 21 cm line of interstellar hydrogen.

Metastability follows directly. When every E1 channel to lower states is blocked by the rules, an atom is stranded: hydrogen \(2s\) (\(\ell=0\)) cannot reach \(1s\) (\(\ell=0\)) by E1 and survives \(\sim0.12\ \mathrm{s}\) — decaying by two-photon emission — versus \(1.6\ \mathrm{ns}\) for \(2p\). Astrophysical "nebular" and "coronal" lines are exactly these forbidden transitions, visible only because interstellar densities are too low for collisions to de-excite the trapped atoms first.

Rigorously, the rules are the Wigner–Eckart theorem specialized to a rank-1 tensor: the matrix element factorizes into a geometric 3-\(j\) symbol (which alone carries the selection rules) and a reduced matrix element (all the radial dynamics). This makes clear that the rules are pure group theory — properties of \(SO(3)\) and parity — independent of the specific potential. The same machinery, with rank \(k\), generates the E2 rules (\(\Delta\ell=0,\pm2\), no parity change) and M1 rules, and the vanishing of \(\binom{\ell'\,k\,\ell}{0\,0\,0}\) for \(\ell'+\ell+k\) odd is the general Laporte statement.

Common misconceptions. "Forbidden" does not mean impossible — it means E1-forbidden; the line still appears, just weaker by \(\sim\alpha^2\) or via multiphoton paths. And the rules constrain the transition, not the states: a stationary atom has no permanent dipole (\(\langle n\ell m|\vec r|n\ell m\rangle=0\) by parity), yet transition dipoles between opposite-parity states are large.

Worked examples

Hydrogen \(2p\to1s\) (Lyman-\(\alpha\)): find the spontaneous-emission lifetime.

Check the rules first: \(2p\ (\ell=1)\to1s\ (\ell=0)\) has \(\Delta\ell=-1\), parity flips — E1-allowed. A

\[ \hbar\omega = 13.6\,\mathrm{eV}\big(1-\tfrac14\big)=10.2\,\mathrm{eV} \;\Rightarrow\; \omega = \frac{10.2\times1.602\times10^{-19}}{1.055\times10^{-34}} = 1.55\times10^{16}\ \mathrm{s^{-1}} \]

Transition frequency from the Rydberg energies. A

\[ \int_0^\infty R_{10}\,R_{21}\,r^3\,dr = \frac{2}{\sqrt{24}}\,a_0^{-4}\!\int_0^\infty r^4 e^{-3r/2a_0}dr = \frac{2}{\sqrt{24}}\cdot\frac{24\,a_0}{(3/2)^5} = 1.29\,a_0 \]

Radial dipole integral with \(R_{10}=2a_0^{-3/2}e^{-r/a_0}\), \(R_{21}=\tfrac{1}{2\sqrt6}a_0^{-3/2}(r/a_0)e^{-r/2a_0}\); \(1.29\,a_0=6.83\times10^{-11}\,\mathrm{m}\). B

\[ |\langle 1s|\vec r|2p\rangle|^2 = \tfrac13\,(1.29\,a_0)^2 = 1.55\times10^{-21}\ \mathrm{m^2} \]

The angular part contributes a factor \(\tfrac13\) (summed over the three degenerate \(2p\) sublevels the rate is the same for each). B

\[ \Gamma = \frac{(1.602\times10^{-19})^2(1.55\times10^{16})^3}{3\pi(8.854\times10^{-12})(1.055\times10^{-34})(3\times10^8)^3}\,(1.55\times10^{-21}) \]

Insert numbers into the boxed rate. A

\[ \Gamma \approx 6.3\times10^{8}\ \mathrm{s^{-1}} \;\Rightarrow\; \tau = \frac{1}{\Gamma} \approx 1.6\ \mathrm{ns} \]

Reading. The measured \(2p\) lifetime is \(1.6\,\mathrm{ns}\) — our first-principles E1 rate reproduces it. Units: \(\mathrm{s^{-1}}\), a decay rate.

Why is hydrogen \(2s\) metastable? Compare \(2s\to1s\) with \(2p\to1s\).

Both would emit the same \(10.2\,\mathrm{eV}\) photon, so the frequency factor is identical — only the selection rules differ. A

\[ 2s:\ \ell=0,\ P=+1;\qquad 1s:\ \ell=0,\ P=+1;\qquad \Delta\ell = 0,\ \ \Delta P = 0 \]

Both states have even parity and \(\ell=0\): the E1 rule \(\Delta\ell=\pm1\) with a parity flip is doubly violated. A

\[ \langle 1s|\vec r|2s\rangle \propto \int Y_0^{0*}\,Y_1^{q}\,Y_0^{0}\,d\Omega = 0 \]

The angular integral vanishes identically (one \(Y_1\) between two \(\ell=0\) states), so the E1 amplitude is exactly zero — not merely small. B

\[ \tau_{2s}^{\text{(2-photon)}} \approx 0.12\ \mathrm{s}, \qquad \frac{\tau_{2s}}{\tau_{2p}} \approx \frac{0.12}{1.6\times10^{-9}} \approx 7.5\times10^{7} \]

With E1 forbidden, \(2s\) decays only by two-photon emission; its lifetime is tens of millions of times longer. B

\[ \boxed{\;\Gamma^{E1}_{2s\to1s}=0\;}\quad\Rightarrow\quad \tau_{2s}\approx0.12\ \mathrm{s}\ \gg\ \tau_{2p}\approx1.6\ \mathrm{ns} \]

Reading. Metastability is a direct consequence of the parity/\(\Delta\ell\) rule: block every E1 channel and the state is stranded for \(\sim10^{8}\) times longer. Units: \(\Gamma\) in \(\mathrm{s^{-1}}\), \(\tau\) in \(\mathrm{s}\).

Problems

Circularly polarized light propagating along \(z\) drives an atomic transition. Which value of \(\Delta m\) does \(\sigma^{+}\) (right-circular) light produce, and why can it not drive \(\Delta m=0\)?
Solution
\(\sigma^{+}\) light corresponds to the spherical component \(q=+1\) of \(\vec A\) (it carries \(+\hbar\) of angular momentum along \(z\)). The matrix element \(\propto\langle\ell'm'|Y_1^{+1}|\ell m\rangle\) is nonzero only for \(m'=m+1\), i.e. \(\boxed{\Delta m=+1}\). It cannot drive \(\Delta m=0\) because that would require the \(q=0\) (\(\pi\), linear-along-\(z\)) component, which is absent for light travelling along \(z\); a photon moving along \(z\) has no \(z\)-polarization.
Classify each hydrogen transition as E1-allowed or forbidden, giving the reason: (a) \(3s\to2s\), (b) \(3p\to2s\), (c) \(3d\to2p\), (d) \(4f\to2s\), (e) \(3d\to1s\).
Solution
(a) \(\Delta\ell=0\) — forbidden (no parity flip). (b) \(\Delta\ell=+1\!\to\!-1\), i.e. \(\ell:1\to0\), \(\Delta\ell=-1\) — allowed. (c) \(\ell:2\to1\), \(\Delta\ell=-1\) — allowed. (d) \(\ell:3\to0\), \(\Delta\ell=-3\) — forbidden (\(|\Delta\ell|>1\)). (e) \(\ell:2\to0\), \(\Delta\ell=-2\) — forbidden. Only (b) and (c) produce single-photon E1 lines.
Verify the dipole approximation for Lyman-\(\alpha\) (\(\hbar\omega=10.2\,\mathrm{eV}\)): compute the photon wavelength, wavenumber \(k\), and the parameter \(ka_0\). At what order do neglected multipole corrections enter?
Solution
\(\lambda=\dfrac{2\pi c}{\omega}=\dfrac{2\pi(3\times10^8)}{1.55\times10^{16}}=1.22\times10^{-7}\,\mathrm{m}=122\,\mathrm{nm}\). \(k=2\pi/\lambda=5.17\times10^{7}\,\mathrm{m^{-1}}\). With \(a_0=5.29\times10^{-11}\,\mathrm{m}\), \(ka_0=2.7\times10^{-3}\approx\alpha/2.6\ll1\). Corrections enter at \(\mathcal O((ka_0)^2)\sim7\times10^{-6}\), i.e. the E2/M1 intensities are suppressed by \(\sim10^{-6}\) relative to E1 — the dipole approximation is excellent.
The \(3p\to1s\) radial integral in hydrogen is \(\int R_{10}R_{31}r^3dr = 0.52\,a_0\), and \(\hbar\omega=13.6(1-\tfrac19)=12.1\,\mathrm{eV}\). Estimate the \(3p\to1s\) partial rate and compare its order of magnitude with \(2p\to1s\).
Solution
\(\omega=\dfrac{12.1\times1.602\times10^{-19}}{1.055\times10^{-34}}=1.84\times10^{16}\,\mathrm{s^{-1}}\). \(|\langle1s|\vec r|3p\rangle|^2=\tfrac13(0.52\,a_0)^2=\tfrac13(2.75\times10^{-11})^2=2.52\times10^{-22}\,\mathrm{m^2}\). Then \(\Gamma=\dfrac{e^2\omega^3}{3\pi\varepsilon_0\hbar c^3}|\vec r_{fi}|^2\). Relative to Example 1, \(\Gamma\) scales as \(\omega^3|\vec r|^2\): \((1.84/1.55)^3\times(2.52\times10^{-22}/1.55\times10^{-21})=1.67\times0.163=0.27\). So \(\Gamma_{3p\to1s}\approx0.27\times6.3\times10^{8}\approx1.7\times10^{8}\,\mathrm{s^{-1}}\) — a few times smaller than \(2p\to1s\), as expected since the \(3p\) also decays to \(2s\).
Using parity alone, prove that a stationary atomic state \(|n\ell m\rangle\) of a central potential has no permanent electric dipole moment, \(\langle n\ell m|\vec r|n\ell m\rangle=0\). Why does this not contradict the large transition dipole between \(2p\) and \(1s\)?
Solution
Under parity \(\hat P|n\ell m\rangle=(-1)^\ell|n\ell m\rangle\) and \(\hat P\vec r\hat P^{-1}=-\vec r\). Then \(\langle n\ell m|\vec r|n\ell m\rangle=\langle n\ell m|\hat P^{-1}(\hat P\vec r\hat P^{-1})\hat P|n\ell m\rangle=(-1)^{\ell}(-1)^{\ell}(-\langle n\ell m|\vec r|n\ell m\rangle)=-\langle n\ell m|\vec r|n\ell m\rangle\). A quantity equal to its own negative is zero, so \(\boxed{\langle\vec r\rangle=0}\). No contradiction: the transition dipole \(\langle 1s|\vec r|2p\rangle\) connects states of opposite parity (\((-1)^0\) and \((-1)^1\)), for which the same argument gives \(+\langle\vec r\rangle=+\langle\vec r\rangle\) — no constraint — and the integral is nonzero (\(=1.29a_0/\sqrt3\) per component).