Selection Rules

A selection rule is a symmetry statement that makes a matrix element vanish. In transition problems the relevant matrix element often has the form

$$M_{fi} = \langle f|O|i\rangle,$$

where $O$ is the perturbing operator, coupling operator, or observable being tested. If symmetry forces $M_{fi}=0$, the corresponding transition, coupling, or mixing is forbidden at that order and under those assumptions.

An exact zero is stronger than a small rate. To decide which is meant, identify the states, operator, and approximation used in the calculation.

The argument needs definite transformation laws for the states and operator. A symmetry of the Hamiltonian often supplies suitable energy eigenstates, but the matrix-element identity also holds for deliberately prepared symmetry eigenstates. Mixing those states or changing the interaction can change the rule.

Symmetry Zeros of a Matrix Element

Let $U$ be a unitary symmetry. Suppose

$$U|i\rangle = u_i|i\rangle, \qquad U|f\rangle = u_f|f\rangle,$$

and suppose the operator transforms as

$$UOU^\dagger = u_O O.$$

Then

$$\begin{aligned} \langle f|O|i\rangle &= \langle f|U^\dagger UOU^\dagger U|i\rangle \\ &= u_f^*u_Ou_i\, \langle f|O|i\rangle. \end{aligned}$$

Therefore a nonzero matrix element requires

$$u_f^*u_Ou_i=1.$$

If this condition fails, the matrix element must vanish. Parity rules, charge-conservation rules, and many discrete symmetry rules are versions of this simple argument.

Parity Rules

The focused derivation and common parity examples are collected in Parity Selection Rules. The compact version is as follows.

For parity, the eigenvalues are

$$\pi_i,\pi_f=\pm1.$$

If the operator has parity $\pi_O$, then

$$\Pi O\Pi^{-1} = \pi_O O.$$

The nonzero condition becomes

$$\pi_f\pi_O\pi_i=1.$$

Thus:

• an even operator connects states of the same parity;
• an odd operator connects states of opposite parity.

For example, the electric dipole operator is odd under parity. In a parity-symmetric atom or molecule, an electric-dipole matrix element between two states of the same parity vanishes.

Charge-Like Rules

For a continuous $U(1)$ symmetry with conserved generator $Q$, states may be labeled by charge:

$$Q|i\rangle = q_i|i\rangle, \qquad Q|f\rangle = q_f|f\rangle.$$

An operator carries charge $q_O$ when $[Q,O]=q_OO$. Taking this commutator between charge eigenstates gives

$$(q_f-q_i)\langle f|O|i\rangle=q_O\langle f|O|i\rangle.$$

A nonzero matrix element therefore requires

$$q_f=q_i+q_O.$$

If $O$ is neutral, then $q_f=q_i$. This is the selection-rule form of charge conservation. In nonrelativistic quantum mechanics the same logic appears for particle number, magnetic quantum numbers, oscillator number under special ladder operators, and other additive labels when the operator has a definite charge under the symmetry.

Angular Momentum Rules

Angular momentum selection rules use the fact that operators can transform as irreducible spherical tensors. A rank-$k$ tensor component $T_q^{(k)}$ carries magnetic label $q$.

The Wigner–Eckart theorem gives the rotational selection rules

$$m_f=m_i+q,$$

and

$$|j_i-k| \le j_f \le j_i+k.$$

The first rule is a magnetic quantum-number rule. The second is a triangle rule, with $j_f$ advancing in integer steps from $|j_i-k|$. It expresses whether $j_i$ and the operator rank $k$ can couple to $j_f$. A rank-$1$ operator excludes $0\to0$, even though the shorthand $\Delta j=0,\pm1$ might seem to include it.

For a vector operator, $k=1$, so

$$\Delta m=q=0,\pm1,$$

and

$$j_f=j_i-1,\ j_i,\ j_i+1$$

where the values are allowed by nonnegative angular momentum and the triangle condition.

Parity Plus Rotation

Many familiar rules combine more than one symmetry. Electric dipole transitions in central-potential orbital states are the standard example.

The position operator is a vector, so angular momentum allows

$$\ell_f=\ell_i-1,\ell_i,\ell_i+1.$$

But position is odd under parity, and orbital parity is

$$\pi_\ell=(-1)^\ell.$$

The parity rule requires opposite parity:

$$(-1)^{\ell_f} = -(-1)^{\ell_i}.$$

Combining both statements gives

$$\Delta\ell=\pm1.$$

The $\Delta\ell=0$ option is allowed by rotations for a vector operator but forbidden by parity in this setting.

Spin Rules

Spin selection rules depend on the operator. If an operator acts only on spatial variables and commutes with total spin, then spin labels are preserved:

$$\Delta s=0, \qquad \Delta m_s=0$$

in the basis where those labels are meaningful.

In atomic spectroscopy, electric dipole transitions often satisfy an approximate $\Delta S=0$ rule in an $LS$-coupling description because the leading electric dipole operator does not act on spin. But this is not a universal law. Spin-dependent interactions can mix different $S$ sectors or introduce an operator with a different spin transformation law. Spin-independent electrostatic configuration mixing preserves $S$ and cannot alone open a $\Delta S\ne0$ line. Likewise, the leading atomic M1 operator proportional to $\mathbf L+g_s\mathbf S$ preserves $S$ in pure $LS$ coupling.

Check whether the particular interaction commutes with $S^2$ before attributing a weak line to spin mixing.

Forbidden Versus Weakly Allowed

“Forbidden” usually means forbidden in a specified approximation. A transition can become weakly allowed when:

• a different multipole operator contributes;
• a higher-order perturbative process is included;
• spin–orbit or hyperfine mixing changes the state labels;
• an external field breaks the symmetry;
• the environment or boundary conditions no longer respect the symmetry;
• the initial or final state is only approximately a symmetry eigenstate.

For example, an electric-dipole forbidden transition may occur through magnetic dipole or electric quadrupole multipoles, two-photon processes, or symmetry-breaking mechanisms. The selection rule still matters: it tells which leading mechanism vanishes and why the observed process is weaker.

Rates Versus Rules

Selection rules constrain amplitudes. Rates require additional physics. In a golden-rule calculation,

$$\Gamma_{i\to f} = \frac{2\pi}{\hbar} |M_{fi}|^2 \rho(E_f).$$

Here $M_{fi}$ is the relevant interaction matrix element and $\rho$ is the density per unit energy at the energy allowed by the process, including photon energy for absorption or emission. For isolated discrete levels, finite-time amplitudes or a line profile replace a naive continuum density. If symmetry gives $M_{fi}=0$, the leading rate vanishes. If $M_{fi}$ is allowed, the actual rate still depends on the reduced matrix element, density of final states, linewidths, populations, polarization, and experimental geometry.

The transition-rate machinery belongs to Selection Rules in Transition Rates. The electric-dipole specialization is derived in Dipole Transitions, while Atomic Selection Rules applies the hierarchy to fine and hyperfine labels, polarization, metastability, and perturbative mixing. This page explains the general symmetry zeros.

Common Mistakes

• Calling a transition forbidden without naming the symmetry and operator.
• Treating “allowed” as meaning “large.”
• Applying parity rules when the system is not parity symmetric.
• Forgetting that angular momentum rules and parity rules are separate constraints.
• Using weak-field quantum numbers in a strong-field regime where the coupling scheme has changed.
• Treating spin selection rules as exact when spin-dependent interactions mix the states.
• Forgetting that a higher-order or different-multipole process can bypass a leading selection rule.

Exercises

1. Two states have parity $\pi_i=+1$ and $\pi_f=+1$. Can an odd operator connect them?

Solution

For an odd operator, $\pi_O=-1$. The nonzero condition is

$$\pi_f\pi_O\pi_i=1.$$

Here

$$(+1)(-1)(+1)=-1,$$

so the matrix element must vanish.

2. Let $Q$ count excitations and let $[Q,O]=2O$. Can $O$ connect a state with $q_i=3$ to one with $q_f=4$? What about $q_f=5$?

Solution

The commutator requires $(q_f-q_i-2)\langle f|O|i\rangle=0$. Thus $3\to4$ vanishes, while $3\to5$ is not excluded by this rule. The latter can still vanish for another symmetry or because of the detailed matrix element.

3. Why does the electric dipole orbital rule exclude $\Delta\ell=0$ even though a vector operator can satisfy the triangle rule with $\ell_f=\ell_i$?

Solution

The vector nature of the dipole operator gives the triangle rule

$$\ell_f=\ell_i-1,\ell_i,\ell_i+1$$

where allowed. But the electric dipole operator is odd under parity. Orbital states have parity $(-1)^\ell$, so a dipole matrix element requires opposite parity between the initial and final orbital states. The case $\ell_f=\ell_i$ has the same parity and is therefore forbidden. The remaining options are

$$\Delta\ell=\pm1.$$

4. Give one mechanism that can make a forbidden transition weakly allowed.

Solution

For an opposite-parity pair with an E1 spin-forbidden amplitude, spin–orbit mixing can admix the required spin character. If the admixture is $\epsilon$ and the borrowed allowed amplitude is $M_0$, then $M\simeq\epsilon M_0$ and the rate is of order $|\epsilon|^2|M_0|^2$, apart from phase-space factors. Spin–orbit coupling preserves parity, so it cannot by itself open an E1 line between two same-parity states.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• C. J. Foot, Atomic Physics, Oxford University Press, 2005.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.