Parity

Parity is the spatial inversion transformation. It sends position to minus position while leaving time unchanged:

$$\mathbf x\mapsto-\mathbf x.$$

In quantum mechanics parity is represented, when it is a symmetry or a useful transformation, by a unitary operator usually denoted $\Pi$. Its defining action on position and momentum operators is

$$\Pi\,\mathbf X\,\Pi^{-1}=-\mathbf X, \qquad \Pi\,\mathbf P\,\Pi^{-1}=-\mathbf P.$$

Since applying spatial inversion twice returns the original coordinates,

$$\Pi^2=I$$

for ordinary scalar wavefunctions.

The chapter guide compares parity with antiunitary time reversal, Kramers degeneracy, Hamiltonian constraints, and the scoped bridges to charge conjugation and CPT.

Wavefunction Action

In one spatial dimension, parity acts on a wavefunction by

$$(\Pi\psi)(x)=\psi(-x).$$

In three dimensions,

$$(\Pi\psi)(\mathbf x)=\psi(-\mathbf x)$$

for a spinless scalar wavefunction. If the state has spin or other internal indices, parity may also act on those internal components depending on the system and representation. This page focuses on the standard nonrelativistic scalar action.

Even and Odd States

Because $\Pi^2=I$, parity eigenvalues are

$$\pi=\pm1.$$

A parity eigenstate satisfies

$$\Pi\lvert\psi\rangle = \pi\lvert\psi\rangle.$$

In position space this becomes

$$\psi(-x)=\pi\psi(x).$$

Thus $\pi=+1$ states are even and $\pi=-1$ states are odd:

$$\psi(-x)=\psi(x) \qquad\text{or}\qquad \psi(-x)=-\psi(x).$$

Parity is often the cleanest way to organize bound states in symmetric one-dimensional potentials.

The one-dimensional solving consequences are developed in Parity and Nodes.

When Parity Is a Symmetry

Parity is a symmetry of a Hamiltonian $H$ if

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

Equivalently,

$$[H,\Pi]=0.$$

For a one-dimensional Hamiltonian

$$H=\frac{P^2}{2m}+V(X),$$

parity is a symmetry when

$$V(-x)=V(x).$$

If the potential is not inversion symmetric, parity may still be a useful transformation, but parity eigenvalue is not generally conserved and energy eigenstates need not have definite parity.

Angular Momentum Under Parity

Orbital angular momentum is

$$\mathbf L=\mathbf X\times\mathbf P.$$

Since both $\mathbf X$ and $\mathbf P$ change sign under parity,

$$\Pi\,\mathbf L\,\Pi^{-1} = \mathbf L.$$

This is why angular momentum is called an axial vector or pseudovector. It behaves differently from ordinary polar vectors under spatial inversion.

Spin angular momentum is also an axial vector in the nonrelativistic setting:

$$\Pi\,\mathbf S\,\Pi^{-1} = \mathbf S$$

when parity does not act nontrivially on extra internal labels.

Parity Selection Rule

Suppose $\lvert a\rangle$ and $\lvert b\rangle$ are parity eigenstates:

$$\Pi\lvert a\rangle=\pi_a\lvert a\rangle, \qquad \Pi\lvert b\rangle=\pi_b\lvert b\rangle.$$

Let $O$ be an operator with parity $\pi_O$, meaning

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

Then the matrix element $\langle a\vert O\vert b\rangle$ can be nonzero only if

$$\pi_a\pi_O\pi_b=1.$$

For example, the position operator is odd under parity:

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

Therefore $X$ connects states of opposite parity, not states of the same parity, in a parity-symmetric problem. The detailed parity selection-rule proof is collected in Parity Selection Rules, and the broader selection-rule logic is collected in Selection Rules.

Example: Symmetric Potential

For an even potential $V(x)=V(-x)$, $H$ and $\Pi$ commute. If an energy level is nondegenerate, its eigenstate must also be a parity eigenstate. The wavefunction can then be chosen even or odd.

This is why the harmonic oscillator eigenfunctions alternate parity. The ground state is even, the first excited state is odd, the second is even, and so on.

Degenerate subspaces require more care: one can choose parity eigenstates inside the degenerate subspace, but arbitrary linear combinations may not themselves have definite parity.

Common Mistakes

• Assuming parity is a symmetry of every Hamiltonian.
• Forgetting that both position and momentum reverse sign.
• Treating angular momentum like an ordinary vector under parity.
• Applying parity selection rules when the Hamiltonian does not have parity symmetry.
• Confusing parity with time reversal; parity does not reverse time.

Cross-Links

• Unitary Symmetries

• States, Observables, and Hamiltonians

• Symmetry Constraints on Hamiltonians

• Exact Symmetry

• Explicit Symmetry Breaking

• Translations and Momentum

• Parity as Spatial Inversion

• Orbital Angular Momentum

• Time Reversal

• Charge Conjugation Preview

• Discrete Symmetries in Hamiltonians

• From Discrete Symmetries to CPT

• Selection Rules

• Parity Selection Rules

• Selection Rule Problems

• Parity and Nodes

• Symmetry Sectors in Many-Body Numerics — parity projectors, finite-lattice reflection maps, and the compatibility of reflection with translation momentum.

• Tests of Fundamental Symmetries for atomic parity violation, EDM signatures, reversal analysis, and experimental constraints.

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Butterworth-Heinemann, 1977.

Exercises

1. Show that the momentum operator changes sign under parity in one dimension.

Solution

Use $(\Pi\psi)(x)=\psi(-x)$ and $P=-i\hbar\,d/dx$. Since $\Pi^{-1}=\Pi$,

$$(\Pi P\Pi^{-1}\psi)(x) = \Pi P\psi(-x).$$

More directly,

$$(P\Pi\psi)(x) = -i\hbar\frac{d}{dx}\psi(-x) = i\hbar\psi'(-x),$$

and applying $\Pi$ gives

$$(\Pi P\Pi^{-1}\psi)(x) = i\hbar\psi'(x) = -P\psi(x).$$

Thus $\Pi P\Pi^{-1}=-P$.

2. If $\lvert a\rangle$ and $\lvert b\rangle$ have the same parity, show that $\langle a\vert X\vert b\rangle=0$ in a parity-symmetric system.

Solution

Since $X$ is parity odd,

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

Insert $I=\Pi^{-1}\Pi$ and use the parity eigenvalue equations:

$$\langle a\vert X\vert b\rangle = \pi_a\pi_b \langle a\vert\Pi X\Pi^{-1}\vert b\rangle = -\pi_a\pi_b \langle a\vert X\vert b\rangle.$$

If $\pi_a=\pi_b$, then $\pi_a\pi_b=1$, so the matrix element equals its negative and must vanish.