Parity

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

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

On an inversion-invariant configuration space, the standard scalar action of parity is represented by a unitary operator usually denoted $\Pi$. Its action on position and momentum operators is

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

Applying spatial inversion twice returns the original scalar wavefunction. Together with unitarity, this gives

$$\Pi^2=I, \qquad \Pi^{-1}=\Pi^\dagger=\Pi.$$

Thus scalar orbital parity is also self-adjoint. Whether it is a symmetry of a particular Hamiltonian is a separate question.

In three dimensions inversion has determinant $\det(-I_3)=(-1)^3=-1$ and cannot be produced by a proper rotation, whose determinant is $+1$. A plane reflection also has determinant $-1$, but flips only the normal coordinate; the two differ by a proper rotation. In one dimension reflection about the chosen origin is parity. In two dimensions, full inversion $(x,y)\mapsto(-x,-y)$ is a rotation by $\pi$, with determinant $+1$; an orientation-reversing reflection instead flips one coordinate.

Helpful background. Unitary symmetries supplies operator conjugation, Momentum Operator as Generator gives the derivative representation, and Orbital Angular Momentum defines the axial vector used below.

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 the operator domain and boundary conditions are also inversion invariant and

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

An even differential expression on a half-line is not enough: inversion must map the physical domain back to itself. 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 a nonzero matrix element of $X$ between parity eigenstates requires opposite parities. This condition does not guarantee a nonzero value; other selection rules or the specific states can make the matrix element vanish. The detailed proof is in Parity Selection Rules, and the broader logic is 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.

For the centered harmonic oscillator, the ground-state Gaussian is even. Since $a^\dagger$ is a linear combination of $X$ and $P$, parity gives $\Pi a^\dagger\Pi^{-1}=-a^\dagger$. Repeated raising therefore gives

$$\Pi|n\rangle=(-1)^n|n\rangle.$$

Commutation with $H$ permits definite parity; the raising-operator argument explains the alternating pattern. Symmetry alone does not tell which sign a particular energy level has.

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.
• Assigning definite parity to arbitrary superpositions. The matrix-element proof needs parity eigenstates; Hamiltonian symmetry is a way of finding them, not an extra algebraic step in the proof.
• Confusing parity with time reversal; parity does not reverse time.

Exercises

1. Show directly that both position and momentum change sign under parity in one dimension. Work with smooth rapidly decreasing wavefunctions on the full line.

Solution

Use $(\Pi\psi)(x)=\psi(-x)$ and $\Pi^{-1}=\Pi$. Position acts by multiplication, so

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

For momentum, $P=-i\hbar\,d/dx$ and the chain rule give

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

and applying $\Pi$ gives

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

Thus $\Pi X\Pi^{-1}=-X$ and $\Pi P\Pi^{-1}=-P$ on this core.

2. If $\lvert a\rangle$ and $\lvert b\rangle$ have the same parity and the matrix element exists, show that $\langle a\vert X\vert b\rangle=0$. Deduce the position expectation value in each energy eigenstate of the centered harmonic oscillator.

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.

The centered oscillator has $\Pi|n\rangle=(-1)^n|n\rangle$. Taking both states to be $|n\rangle$ gives $\langle n|X|n\rangle=0$ for every $n$.

3. A normalized oscillator state is $|\psi\rangle=(|0\rangle+|1\rangle)/\sqrt2$. Is it a parity eigenstate? Find the two parity-measurement probabilities.

Solution

Parity sends it to $(|0\rangle-|1\rangle)/\sqrt2$, which is orthogonal to the original state. The projectors are $P_\pm=(I\pm\Pi)/2$, so $P_+|\psi\rangle=|0\rangle/\sqrt2$ and $P_-|\psi\rangle=|1\rangle/\sqrt2$. Each outcome has probability $1/2$. The oscillator Hamiltonian has parity symmetry, but this prepared state does not have definite parity.

4. On the full line, a displaced oscillator has $V(x)=m\omega^2(x-a)^2/2$ with $a\ne0$. Does $V(-x)\ne V(x)$ rule out every inversion symmetry? Find an inversion operator that leaves its Hamiltonian invariant.

Solution

The potential is not invariant under reflection about the old origin. Reflection about the oscillator center acts as

$$(\Pi_a\psi)(x)=\psi(2a-x).$$

It reverses displacement from the center and reverses momentum:

$$\Pi_a(X-aI)\Pi_a^{-1}=-(X-aI), \qquad \Pi_a P\Pi_a^{-1}=-P.$$

Consequently both $P^2/(2m)$ and $m\omega^2(X-aI)^2/2$ are invariant. The full-line oscillator domain is also preserved. Parity must specify its center; an asymmetric expression relative to an arbitrary origin does not establish the absence of inversion symmetry.

References

• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon Press, 1977.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chapter 4.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.