advanced30 min studyLesson 406 in Physics

Parity and electric-dipole transitions

Use inversion symmetry to prove matrix-element zeros and combine parity with angular selection rules.

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What you’ll understand

  • Classify polar and axial quantities under spatial inversion.
  • Derive the parity condition on an electric-dipole matrix element.
  • Apply the condition within its long-wavelength and definite-parity assumptions.

An integral can vanish because every contribution has an equal opposite partner. Parity turns that observation into an operator argument applicable to atoms, oscillators, and other inversion-symmetric systems. Electric-dipole transitions provide a particularly useful example: the dipole operator reverses under inversion, so it connects opposite parity sectors. This statement complements angular-momentum selection rules and has a defined approximation regime; it is not a prohibition on every possible interaction with light.

Spatial inversion is a discrete unitary symmetry

For a spinless spatial wavefunction take (Pψ)(r)=ψ(−r)(P\psi)(\mathbf r)=\psi(-\mathbf r). With P2=IP^2=I, parity eigenvalues are +1+1 and −1-1. An inversion-symmetric Hamiltonian commutes with parity, so energy eigenstates can be organized into parity sectors. A nondegenerate eigenstate has definite parity. Within a degenerate eigenspace one may choose a parity basis, but an arbitrary superposition in that space need not itself have definite parity. Commutation does not make every written vector a parity eigenvector.

Position and momentum are polar vectors and reverse under inversion. Their cross product, orbital angular momentum, is axial and remains unchanged. Spin transforms as an axial vector too; it is not reversed by parity. This is a common point of confusion because both parity and time reversal change familiar motions, but their operator actions differ. In central potentials the spatial spherical harmonic contributes parity (−1)ℓ(-1)^\ell, whereas the radial part stays unchanged because radius is nonnegative and invariant.

P2=I,P†rP=−r,P†LP=L,P∣ℓm⟩=(−1)ℓ∣ℓm⟩P^2=I,\qquad P^\dagger\mathbf rP=-\mathbf r,\qquad P^\dagger\mathbf LP=\mathbf L,\qquad P|\ell m\rangle=(-1)^\ell|\ell m\rangle
P is dimensionless and unitary. The final expression concerns the orbital angular wavefunction; intrinsic parities or additional particles require their own factors.↔ Scroll wide equations horizontally.

An odd operator links opposite sectors

In the electric-dipole approximation the interaction is minus the dipole moment dotted into the applied electric field. The approximation treats the field as nearly uniform across the bound state, requiring the state size to be small compared with the radiation wavelength. For a single charge measured relative to a fixed origin, the dipole is charge times position and is odd under inversion. Insert a pair of parity operators into its matrix element between definite-parity states to expose a sign constraint.

If the two state parities are equal, the matrix element equals its own negative and must vanish. If they are opposite, parity places no further restriction on that matrix element. The surviving integral can still vanish for angular or radial reasons. In a one-electron central-field model, parity requires an odd change of orbital label; combining that with the rank-one triangle condition gives a change of exactly one. Neither ingredient alone yields the full familiar electric-dipole rule.

Mfi=⟨f∣d∣i⟩=−ηfηiMfi,Mfi≠0 ⟹ ηfηi=−1M_{fi}=\langle f|\mathbf d|i\rangle=-\eta_f\eta_iM_{fi},\qquad M_{fi}\ne0\ \Longrightarrow\ \eta_f\eta_i=-1
The parity eigenvalues eta are real signs and d has units coulomb metres in SI. This necessary condition assumes definite-parity states and an odd electric-dipole operator.↔ Scroll wide equations horizontally.

Scope of a forbidden transition

A static external electric field can mix opposite-parity states, because its dipole interaction breaks the inversion symmetry of the fixed-system Hamiltonian. A previously vanishing amplitude can then acquire a small value. Weak interactions also violate parity, but ordinary electromagnetic spectroscopy often starts from a parity-conserving approximation and estimates such corrections separately. Calling a line forbidden therefore requires stating the operator and the symmetry assumptions, rather than assigning an absolute label to a pair of levels.

Beyond the long-wavelength dipole approximation, magnetic-dipole and electric-quadrupole operators have different parity and angular transformation properties. A line absent at electric-dipole order may remain visible through these weaker mechanisms or through multiphoton processes. Even in the ideal dipole model, population and frequency matching still matter: an allowed matrix element does not supply occupied initial states or a resonant field. Symmetry saves an integral calculation, but it does not replace the dynamical and experimental conditions for observing a transition.

x=ℏ2mω(a+a†),⟨n+1∣x∣n⟩=ℏ2mωn+1x=\sqrt{\frac{\hbar}{2m\omega}}(a+a^\dagger),\qquad \langle n+1|x|n\rangle=\sqrt{\frac{\hbar}{2m\omega}}\sqrt{n+1}
This is the one-dimensional harmonic oscillator with positive mass m and angular frequency omega. The position matrix element has length units and links opposite oscillator parities.↔ Scroll wide equations horizontally.

Dipole strengths in a symmetric oscillator

Let x0=ℏ/(2mω)=2 nmx_0=\sqrt{\hbar/(2m\omega)}=2\,\mathrm{nm} for a one-dimensional harmonic oscillator. Starting in n=1n=1, find the position matrix elements to n=0n=0, n=2n=2, and n=3n=3, and compare the two nonzero squared strengths.

  1. Oscillator parity is (−1)n(-1)^n, so the initial state is odd. The final state n=3n=3 is also odd, and its position matrix element vanishes immediately because position is odd.

  2. The lowering term gives ⟨0∣x∣1⟩=x01=2 nm\langle0|x|1\rangle=x_0\sqrt1=2\,\mathrm{nm}. The raising term gives ⟨2∣x∣1⟩=x02=22 nm\langle2|x|1\rangle=x_0\sqrt2=2\sqrt2\,\mathrm{nm}. No other ladder terms reach these states.

  3. Squaring yields 4 nm24\,\mathrm{nm}^2 and 8 nm28\,\mathrm{nm}^2, a ratio of one to two. Multiplying by charge squared would give squared dipole moments; actual absorption and emission rates also involve the radiation conditions.

The result

The n=3 channel vanishes, while the two nonzero position strengths are 4 and 8 square nanometres.

Explore the model

Interactive lab

Interactive lab · function

Even potential and odd slope

For each paused run Reset and set both controls. Compare coefficient a=0.5a=0.5 at positions x=1x=1 and x=−1x=-1. Predict the same quadratic value 0.5 and slopes +1 and −1. This illustrates an even potential and its odd derivative. The plotted quadratic does not calculate wavefunctions, dipole matrix elements, or spectral lines.

Even potential and odd slopeThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forInversion symmetry relates values at opposite positions and can force an odd integrand to cancel.

Check your understanding

Which one-electron central-field transition passes both the orbital parity and rank-one electric-dipole tests?

Try it yourself

On [−L,L][-L,L], take normalized states ψe=1/2L\psi_e=1/\sqrt{2L} and ψo=3/(2L3) x\psi_o=\sqrt{3/(2L^3)}\,x. Compute ⟨e∣x∣e⟩\langle e|x|e\rangle and ⟨o∣x∣e⟩\langle o|x|e\rangle. These are illustrative interval states, not asserted energy eigenfunctions.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The first integrand is odd. In the second use the integral of x squared over the symmetric interval, which is two L cubed divided by three.

Compare with the solution

The first integral vanishes. The product of normalization constants in the second is 3/(2L2)\sqrt3/(2L^2), so multiplying by 2L3/32L^3/3 gives L/3L/\sqrt3. The result has length units and connects the odd and even states. Their normalizations are verified from the integrals of one and x squared on the stated interval.

Take it with you

  • Parity is a unitary inversion; it leaves axial angular momentum unchanged.
  • An electric-dipole matrix element between equal-parity states vanishes under the stated approximation.
  • A dipole-forbidden line can appear through symmetry mixing or higher electromagnetic multipoles.

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