Parity Selection Rules

A parity selection rule is a statement that a matrix element vanishes because spatial inversion assigns incompatible signs to the initial state, final state, and operator.

The basic matrix element is

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

If the states have definite parity and the operator has definite parity, then $M_{fi}$ can be nonzero only when the product of the three parity signs is even:

$$\pi_f\,\eta_O\,\pi_i=1.$$

This page is the focused home for parity selection rules. The parity operator itself is introduced in Parity, while the broader symmetry logic is collected in Selection Rules.

Assumptions

The rule needs three assumptions:

• the Hamiltonian or approximation being used is invariant under parity;
• the initial and final states can be chosen as parity eigenstates;
• the operator $O$ has a definite parity.

The first assumption is often the one that fails in real systems. External electric fields, asymmetric boundary conditions, chiral environments, weak interactions, and parity-mixing perturbations can all weaken or remove a parity rule.

State and Operator Parity

Let $\Pi$ denote the parity operator. A parity eigenstate satisfies

$$\Pi|i\rangle = \pi_i|i\rangle, \qquad \Pi|f\rangle = \pi_f|f\rangle,$$

where

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

An operator has definite parity $\eta_O$ if

$$\Pi O\Pi^{-1} = \eta_O O, \qquad \eta_O=\pm1.$$

The operator is parity even when $\eta_O=+1$ and parity odd when $\eta_O=-1$.

Derivation

Insert the identity $\Pi^{-1}\Pi$ on both sides of the operator:

$$\begin{aligned} \langle f|O|i\rangle &= \langle f|\Pi^{-1} \left( \Pi O\Pi^{-1} \right) \Pi|i\rangle \\ &= \pi_f\,\eta_O\,\pi_i \langle f|O|i\rangle. \end{aligned}$$

Therefore

$$\left( 1-\pi_f\eta_O\pi_i \right) \langle f|O|i\rangle=0.$$

A nonzero matrix element requires

$$\pi_f\eta_O\pi_i=1.$$

Equivalently:

Operator parity	Nonzero matrix element requires
even, $\eta_O=+1$	$\pi_f=\pi_i$
odd, $\eta_O=-1$	$\pi_f=-\pi_i$

In words: even operators connect states of the same parity, while odd operators connect states of opposite parity.

Integral Version

In one dimension, a parity eigenfunction obeys

$$\psi_n(-x) = \pi_n\psi_n(x).$$

If $O(x)$ is multiplication by a function with parity $\eta_O$,

$$O(-x) = \eta_OO(x),$$

then the integrand in

$$\langle f|O|i\rangle = \int_{-\infty}^{\infty} \psi_f^*(x)O(x)\psi_i(x)\,dx$$

has parity $\pi_f\eta_O\pi_i$. If that product is $-1$, the integrand is odd and the integral over a symmetric domain vanishes.

The operator derivation is more general. It still works when the Hilbert space has spin, degeneracies, internal labels, or nontrivial coordinate systems.

Common Operator Parities

Under spatial inversion,

$$\mathbf R\mapsto-\mathbf R, \qquad \mathbf P\mapsto-\mathbf P.$$

Thus polar vectors built from position or momentum are parity odd. Examples include:

• position $\mathbf R$;
• momentum $\mathbf P$;
• electric dipole moment $\mathbf d=\sum_a q_a\mathbf r_a$.

Axial vectors are parity even. Examples include:

• orbital angular momentum $\mathbf L=\mathbf R\times\mathbf P$;
• spin angular momentum $\mathbf S$ in the usual nonrelativistic setting;
• magnetic moment operators proportional to angular momentum.

Some useful scalar and tensor examples are:

Operator	Parity
$\mathbf R^2$, $\mathbf P^2$, $\mathbf L^2$	even
$\mathbf L\cdot\mathbf S$	even
electric dipole $\mathbf d$	odd
magnetic dipole $\boldsymbol\mu$	even
electric quadrupole $Q_{ij}\sim 3R_iR_j-\delta_{ij}\mathbf R^2$	even

The parity label is independent of the rotational rank. A vector under rotations can be parity odd, like $\mathbf R$, or parity even, like $\mathbf L$.

One-Dimensional Example

In an inversion-symmetric one-dimensional potential, stationary states can be chosen even or odd. The position operator $X$ is odd:

$$\Pi X\Pi^{-1}=-X.$$

Therefore

$$\langle \text{even}|X|\text{even}\rangle=0, \qquad \langle \text{odd}|X|\text{odd}\rangle=0.$$

But $X$ can connect even and odd states. By contrast, $X^2$ is even, so

$$\langle \text{even}|X^2|\text{odd}\rangle=0.$$

For the centered harmonic oscillator,

$$\pi_n=(-1)^n.$$

Thus $X$ connects only oscillator states with opposite parity, while $X^2$ connects only states with the same parity. The ladder-operator calculation further refines this to $\Delta n=\pm1$ for $X$ and $\Delta n=0,\pm2$ for $X^2$.

Orbital Parity

For orbital angular momentum eigenfunctions,

$$Y_\ell^m(-\hat{\mathbf r}) = (-1)^\ell Y_\ell^m(\hat{\mathbf r}).$$

Thus a scalar central-potential state with orbital angular momentum $\ell$ has orbital parity

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

This immediately gives the parity part of many atomic selection rules. If an operator is odd, it connects orbital states with opposite parity:

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

Equivalently, $\ell_f-\ell_i$ must be odd. If an operator is even, $\ell_f-\ell_i$ must be even.

Spin does not by itself fix parity. When spin and orbital angular momentum are coupled, the total $j$ label is not enough to determine parity; one must still know the orbital or intrinsic parity content.

Electric Dipole Rule

The electric dipole operator is parity odd:

$$\Pi\mathbf d\Pi^{-1} = -\mathbf d.$$

Therefore electric-dipole transitions require opposite parity:

$$\pi_f=-\pi_i.$$

For central-potential orbital states, this means $\ell_f-\ell_i$ is odd. Combining this with the rotational rank-$1$ rule gives the familiar orbital electric-dipole result

$$\Delta\ell=\pm1,$$

where the angular-momentum labels are valid and the usual boundary cases are respected. The electric-dipole specialization, including polarization and magnetic quantum numbers, is treated in Dipole Transitions.

Magnetic Dipole and Electric Quadrupole

Magnetic dipole operators are axial vectors and are parity even:

$$\eta_{M1}=+1.$$

Thus magnetic-dipole matrix elements connect states of the same parity, subject to their rotational rank-$1$ angular rules. The $E\lambda/M\lambda$ rank-and-parity table is collected in Multipole Operators.

Electric quadrupole operators are built from two powers of position, with the trace removed. They are parity even:

$$\eta_{E2}=+1.$$

Thus electric-quadrupole matrix elements also connect states of the same parity, subject to rank-$2$ angular rules. For scalar central-potential orbital states, parity permits only even changes in $\ell$. Combining with the rank-$2$ triangle condition gives the usual possibilities

$$\Delta\ell=0,\pm2,$$

with additional exclusions when the angular coefficient itself vanishes.

Broken and Approximate Parity

If parity is not an exact symmetry, states need not have definite parity. A weakly mixed state may have the schematic form

$$|\widetilde i\rangle = |i,+\rangle + \epsilon |i,-\rangle, \qquad |\epsilon|\ll1.$$

An otherwise forbidden matrix element can then become nonzero at order $\epsilon$. For an odd operator $O_-$ between two mostly even states,

$$\langle \widetilde f|O_-|\widetilde i\rangle = \epsilon_i \langle f,+|O_-|i,-\rangle + \epsilon_f^* \langle f,-|O_-|i,+\rangle + O(\epsilon^2).$$

This is the symmetry reason that external fields, asymmetric environments, weak interactions, or configuration mixing can make a parity-forbidden transition weakly allowed. The word “forbidden” always refers to a specified Hamiltonian, operator, and approximation.

Common Mistakes

• Applying a parity selection rule when the Hamiltonian is not parity symmetric.
• Knowing the state parity but forgetting to determine the operator parity.
• Assuming every vector operator is parity odd; axial vectors are parity even.
• Assuming same $j$ means same parity. Total angular momentum does not determine parity by itself.
• Calling a transition impossible rather than forbidden at a specified order or by a specified operator.
• Combining parity and angular-momentum rules without checking both separately.

References

• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon Press, 1977.

Exercises

1. Two states have the same parity. Can an odd operator connect them?

Solution

No. If $\pi_f=\pi_i$, then for an odd operator $\eta_O=-1$,

$$\pi_f\eta_O\pi_i = - \pi_f\pi_i = -1.$$

The nonzero condition $\pi_f\eta_O\pi_i=1$ fails, so the matrix element vanishes.

2. In a centered harmonic oscillator, which matrix elements can be nonzero by parity: $\langle n'|X|n\rangle$ and $\langle n'|X^2|n\rangle$?

Solution

The oscillator parity is $\pi_n=(-1)^n$. Since $X$ is odd, $X$ requires opposite parity:

$$(-1)^{n'}=-(-1)^n.$$

Thus $n'-n$ must be odd. Since $X^2$ is even, $X^2$ requires the same parity:

$$(-1)^{n'}=(-1)^n.$$

Thus $n'-n$ must be even. The actual oscillator ladder calculation gives the stronger rules $\Delta n=\pm1$ for $X$ and $\Delta n=0,\pm2$ for $X^2$.

3. Combine parity with the rank-$1$ orbital rule for an electric dipole operator in a central potential.

Solution

The rank-$1$ rotational rule permits

$$\ell_f=\ell_i,\ell_i\pm1$$

where valid. The electric dipole operator is parity odd, so it requires opposite parity:

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

Therefore $\ell_f-\ell_i$ must be odd. The $\ell_f=\ell_i$ option is removed, leaving

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

4. Suppose two mostly even states contain small odd admixtures:

$$|\widetilde a\rangle=|a,+\rangle+\epsilon_a|a,-\rangle, \qquad |\widetilde b\rangle=|b,+\rangle+\epsilon_b|b,-\rangle.$$

Estimate the leading parity-allowed contribution to $\langle\widetilde b|O_-|\widetilde a\rangle$ for an odd operator $O_-$.

Solution

Expand to first order:

$$\begin{aligned} \langle\widetilde b|O_-|\widetilde a\rangle &= \langle b,+|O_-|a,+\rangle + \epsilon_a \langle b,+|O_-|a,-\rangle \\ &\quad + \epsilon_b^* \langle b,-|O_-|a,+\rangle + O(\epsilon^2). \end{aligned}$$

The first term vanishes because an odd operator cannot connect two even states. The leading contribution is therefore

$$\epsilon_a \langle b,+|O_-|a,-\rangle + \epsilon_b^* \langle b,-|O_-|a,+\rangle.$$