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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

Mfi=⟨f∣O∣i⟩.M_{fi} = \langle f|O|i\rangle.

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

πf ηO πi=1.\pi_f\,\eta_O\,\pi_i=1.

The parity operator is introduced in Parity. Here we apply it to transition amplitudes, including the distinction between opposite-parity mixing and ordinary configuration mixing.

The algebra requires actual parity eigenstates and an operator of definite parity. A parity-invariant Hamiltonian allows stationary states to be chosen with this property; it does not force an arbitrary superposition in a degenerate space to have definite parity. External electric fields, asymmetric boundaries, and weak interactions can mix opposite parities.

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

Π∣i⟩=πi∣i⟩,Π∣f⟩=πf∣f⟩,\Pi|i\rangle = \pi_i|i\rangle, \qquad \Pi|f\rangle = \pi_f|f\rangle,

where

πi,πf=±1.\pi_i,\pi_f=\pm1.

An operator has definite parity ηO\eta_O if

ΠOΠ−1=ηOO,ηO=±1.\Pi O\Pi^{-1} = \eta_O O, \qquad \eta_O=\pm1.

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

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

⟨f∣O∣i⟩=⟨f∣Π−1(ΠOΠ−1)Π∣i⟩=πf ηO πi⟨f∣O∣i⟩.\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

(1−πfηOπi)⟨f∣O∣i⟩=0.\left( 1-\pi_f\eta_O\pi_i \right) \langle f|O|i\rangle=0.

A nonzero matrix element requires

πfηOπi=1.\pi_f\eta_O\pi_i=1.

Equivalently:

Operator parity Nonzero matrix element requires
even, ηO=+1\eta_O=+1 πf=πi\pi_f=\pi_i
odd, ηO=−1\eta_O=-1 πf=−πi\pi_f=-\pi_i

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

In one dimension, a parity eigenfunction obeys

ψn(−x)=πnψn(x).\psi_n(-x) = \pi_n\psi_n(x).

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

O(−x)=ηOO(x),O(-x) = \eta_OO(x),

then the integrand in

⟨f∣O∣i⟩=∫−∞∞ψf∗(x)O(x)ψi(x) dx\langle f|O|i\rangle = \int_{-\infty}^{\infty} \psi_f^*(x)O(x)\psi_i(x)\,dx

has parity πfηOπi\pi_f\eta_O\pi_i. If that product is −1-1, 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.

Under spatial inversion,

R↦−R,P↦−P.\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 R\mathbf R;
  • momentum P\mathbf P;
  • electric dipole moment d=∑aqara\mathbf d=\sum_a q_a\mathbf r_a.

Axial vectors are parity even. Examples include:

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

Some useful scalar and tensor examples are:

Operator Parity
R2\mathbf R^2, P2\mathbf P^2, L2\mathbf L^2 even
L⋅S\mathbf L\cdot\mathbf S even
electric dipole d\mathbf d odd
magnetic dipole μ\boldsymbol\mu even
electric quadrupole Qij∼3RiRj−δijR2Q_{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 R\mathbf R, or parity even, like L\mathbf L.

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

ΠXΠ−1=−X.\Pi X\Pi^{-1}=-X.

Therefore

⟨even∣X∣even⟩=0,⟨odd∣X∣odd⟩=0.\langle \text{even}|X|\text{even}\rangle=0, \qquad \langle \text{odd}|X|\text{odd}\rangle=0.

But XX can connect even and odd states. By contrast, X2X^2 is even, so

⟨even∣X2∣odd⟩=0.\langle \text{even}|X^2|\text{odd}\rangle=0.

For the centered harmonic oscillator,

πn=(−1)n.\pi_n=(-1)^n.

Thus XX connects only oscillator states with opposite parity, while X2X^2 connects only states with the same parity. The ladder-operator calculation further refines this to Δn=±1\Delta n=\pm1 for XX and Δn=0,±2\Delta n=0,\pm2 for X2X^2.

For orbital angular momentum eigenfunctions,

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

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

πℓ=(−1)ℓ.\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)ℓf=−(−1)ℓi.(-1)^{\ell_f} = - (-1)^{\ell_i}.

Equivalently, ℓf−ℓi\ell_f-\ell_i must be odd. If an operator is even, ℓf−ℓi\ell_f-\ell_i must be even.

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

The electric dipole operator is parity odd:

ΠdΠ−1=−d.\Pi\mathbf d\Pi^{-1} = -\mathbf d.

Therefore electric-dipole transitions require opposite parity:

πf=−πi.\pi_f=-\pi_i.

For central-potential orbital states, this means ℓf−ℓi\ell_f-\ell_i is odd. Combining this with the rotational rank-11 rule gives the familiar orbital electric-dipole result

Δℓ=±1,\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 operators are axial vectors and are parity even:

ηM1=+1.\eta_{M1}=+1.

Thus magnetic-dipole matrix elements connect states of the same parity, subject to their rotational rank-11 angular rules. The Eλ/MλE\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:

ηE2=+1.\eta_{E2}=+1.

Thus electric-quadrupole matrix elements also connect states of the same parity, subject to rank-22 angular rules. For scalar central-potential orbital states, parity permits only even changes in ℓ\ell. Combining with the rank-22 triangle condition gives the usual possibilities

Δℓ=0,±2,\Delta\ell=0,\pm2,

with the full triangle condition still essential: an initial ℓi=0\ell_i=0 requires ℓf=2\ell_f=2, not 00. A particular magnetic component can impose additional zeros.

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

∣i~⟩=∣i,+⟩+ϵ∣i,−⟩,∣ϵ∣≪1.|\widetilde i\rangle = |i,+\rangle + \epsilon |i,-\rangle, \qquad |\epsilon|\ll1.

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

⟨f~∣O−∣i~⟩=ϵi⟨f,+∣O−∣i,−⟩+ϵf∗⟨f,−∣O−∣i,+⟩+O(ϵ2).\langle \widetilde f|O_-|\widetilde i\rangle = \epsilon_i \langle f,+|O_-|i,-\rangle + \epsilon_f^* \langle f,-|O_-|i,+\rangle + O(\epsilon^2).

A parity-odd perturbation, such as a static electric field, can supply these opposite-parity admixtures. By contrast, ordinary electrostatic configuration interaction and spin–orbit coupling preserve parity: they mix only configurations with the same total parity and cannot by themselves remove this zero. The corresponding rate is generally quadratic in the small admixture, unless a separate nonzero amplitude interferes with it.

  • Assigning parity labels to states that are actually mixtures of opposite parity. The matrix-element proof needs parity eigenstates, whether or not they are energy eigenstates.
  • 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 jj 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.
  1. Two states have the same parity. Can an odd operator connect them?
Solution

No. If πf=πi\pi_f=\pi_i, then for an odd operator ηO=−1\eta_O=-1,

πfηOπi=−πfπi=−1.\pi_f\eta_O\pi_i = - \pi_f\pi_i = -1.

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

  1. In a centered harmonic oscillator, which matrix elements can be nonzero by parity: ⟨n′∣X∣n⟩\langle n'|X|n\rangle and ⟨n′∣X2∣n⟩\langle n'|X^2|n\rangle?
Solution

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

(−1)n′=−(−1)n.(-1)^{n'}=-(-1)^n.

Thus n′−nn'-n must be odd. Since X2X^2 is even, X2X^2 requires the same parity:

(−1)n′=(−1)n.(-1)^{n'}=(-1)^n.

Thus n′−nn'-n must be even. The actual oscillator ladder calculation gives the stronger rules Δn=±1\Delta n=\pm1 for XX and Δn=0,±2\Delta n=0,\pm2 for X2X^2.

  1. Combine parity with the rank-11 orbital rule for an electric dipole operator in a central potential.
Solution

The rank-11 rotational rule permits

ℓf=ℓi,ℓi±1\ell_f=\ell_i,\ell_i\pm1

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

(−1)ℓf=−(−1)ℓi.(-1)^{\ell_f} = - (-1)^{\ell_i}.

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

Δℓ=±1.\Delta\ell=\pm1.
  1. Suppose two mostly even states contain small odd admixtures:
∣a~⟩=∣a,+⟩+ϵa∣a,−⟩,∣b~⟩=∣b,+⟩+ϵb∣b,−⟩.|\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 ⟨b~∣O−∣a~⟩\langle\widetilde b|O_-|\widetilde a\rangle for an odd operator O−O_-.

Solution

Expand to first order:

⟨b~∣O−∣a~⟩=⟨b,+∣O−∣a,+⟩+ϵa⟨b,+∣O−∣a,−⟩+ϵb∗⟨b,−∣O−∣a,+⟩+O(ϵ2).\langle\widetilde b|O_-|\widetilde a\rangle = \langle b,+|O_-|a,+\rangle + \epsilon_a \langle b,+|O_-|a,-\rangle + \epsilon_b^* \langle b,-|O_-|a,+\rangle + O(\epsilon^2).

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

ϵa⟨b,+∣O−∣a,−⟩+ϵb∗⟨b,−∣O−∣a,+⟩.\epsilon_a \langle b,+|O_-|a,-\rangle + \epsilon_b^* \langle b,-|O_-|a,+\rangle.
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