Parity as Spatial Inversion

Parity is spatial inversion. In ordinary three-dimensional quantum mechanics it sends every position vector to its negative:

$$\mathbf r\mapsto-\mathbf r.$$

This page treats parity as a spatial operation, alongside translations and rotations. The full discrete-symmetry treatment, including broader selection rules and internal parity labels, lives in Parity.

The Inversion Operation

For a spinless scalar wavefunction, the parity operator $\Pi$ acts by

$$(\Pi\psi)(\mathbf r) = \psi(-\mathbf r).$$

Applying inversion twice gives the original point, so

$$\Pi^2=I.$$

For ordinary scalar wavefunctions, $\Pi$ is unitary and self-adjoint:

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

The eigenvalues are therefore

$$\pi=\pm1.$$

A parity eigenstate satisfies

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

In one spatial dimension this becomes

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

Thus $\pi=+1$ means an even wavefunction and $\pi=-1$ means an odd wavefunction.

Position and Momentum

Parity reverses ordinary polar vectors. For the position operator,

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

In one dimension this statement follows directly from the wavefunction action:

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

Momentum also changes sign:

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

In position representation, this is the chain rule. Since $P=-i\hbar\,d/dx$,

$$\frac{d}{dx}\psi(-x) = -\psi'(-x),$$

so a derivative changes sign under inversion.

By contrast, orbital angular momentum is unchanged:

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

Both $\mathbf X$ and $\mathbf P$ reverse, so their cross product does not. This is why angular momentum is an axial vector rather than a polar vector.

Not a Rotation in Three Dimensions

Rotations in three dimensions preserve orientation and have determinant $+1$. Full inversion is represented on ordinary vectors by $-\mathbf{1}_3$, whose determinant is

$$\det(-\mathbf{1}_3)=-1.$$

Therefore parity is not a rotation in three dimensions. It belongs to the larger orthogonal group $O(3)$, not to the rotation group $SO(3)$.

This is the key distinction from the rotation story in Rotations Preview: rotations are continuous transformations connected to the identity, while parity is a discrete transformation.

Lower-dimensional language needs care. In one dimension, parity looks like reflection through the origin. In two dimensions, the map $(x,y)\mapsto(-x,-y)$ is a rotation by $\pi$. In the usual three-dimensional quantum-mechanics convention, parity means full spatial inversion.

When Parity Is a Symmetry

Parity is available as a transformation even when it is not a symmetry of the Hamiltonian. It is a symmetry only if

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

or equivalently

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

For a one-dimensional Hamiltonian

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

parity is a symmetry when

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

and the domain or boundary conditions are also invariant under $x\mapsto -x$.

If parity is a symmetry, stationary states can often be chosen with definite parity. If parity is broken by the potential, by boundary conditions, or by external fields, even and odd labels are no longer protected.

Even and Odd Examples

The simplest examples are one-dimensional functions:

$$\cos kx \quad\text{is even}, \qquad \sin kx \quad\text{is odd}.$$

For a centered harmonic oscillator,

$$H=\frac{P^2}{2m}+\frac12m\omega^2X^2,$$

the potential is even. The energy eigenfunctions satisfy

$$\psi_n(-x)=(-1)^n\psi_n(x).$$

Thus the oscillator ground state is even, the first excited state is odd, the second is even, and so on. The oscillator page records the explicit wavefunctions, while Parity and Nodes explains why parity and node counting are so useful in symmetric one-dimensional wells.

Selection Rule Preview

Parity gives the cleanest first selection rule. Suppose $\lvert i\rangle$ and $\lvert f\rangle$ are parity eigenstates:

$$\Pi\lvert i\rangle=\pi_i\lvert i\rangle, \qquad \Pi\lvert f\rangle=\pi_f\lvert f\rangle.$$

Suppose an operator $O$ has definite parity $\eta_O$:

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

Then a nonzero matrix element requires

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

For example, $X$ is odd under parity. In a parity-symmetric problem, $X$ connects states of opposite parity and has zero matrix element between states of the same parity:

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

This argument predicts symmetry-forced zeros before doing an integral. The fuller parity rule is developed in Parity Selection Rules, with the broader selection-rule machinery in Symmetry and Selection Rules Preview and Selection Rules.

Common Mistakes

• Assuming parity is a symmetry whenever the symbol $\Pi$ can be defined.
• Forgetting that both position and momentum reverse under spatial inversion.
• Treating angular momentum as a polar vector under parity.
• Calling parity a rotation in three-dimensional space.
• Applying even and odd labels when the potential or the domain is not inversion symmetric.
• Forgetting possible internal parity actions for spinor, molecular, nuclear, or field-theoretic states.

Cross-Links

• Rotations Preview
• Parity
• Parity Operator
• Symmetry Constraints on Hamiltonians
• Symmetry and Selection Rules Preview
• Selection Rules
• Parity Selection Rules
• Parity and Nodes
• Quantum Harmonic Oscillator
• Orbital Angular Momentum

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.
• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.

Exercises

1. Starting from $(\Pi\psi)(x)=\psi(-x)$, show that $\Pi X\Pi^{-1}=-X$.

Solution

Since $\Pi^{-1}=\Pi$,

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

First apply $\Pi$:

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

Then apply $X$:

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

Finally apply $\Pi$ again:

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

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

2. Use parity to show that $\langle n|X|n\rangle=0$ for any harmonic-oscillator energy eigenstate $\lvert n\rangle$.

Solution

The oscillator eigenstate has parity

$$\Pi\lvert n\rangle=(-1)^n\lvert n\rangle,$$

and $X$ is parity odd:

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

Insert $\Pi^{-1}\Pi=I$ around $X$:

$$\begin{aligned} \langle n|X|n\rangle &= \langle n|\Pi^{-1}\Pi X\Pi^{-1}\Pi|n\rangle \\ &= (-1)^n(-1)(-1)^n \langle n|X|n\rangle \\ &= -\langle n|X|n\rangle. \end{aligned}$$

Thus the expectation value equals its negative, so it must vanish.

3. Why is full spatial inversion not a rotation in three dimensions?

Solution

A proper rotation in three dimensions is an element of $SO(3)$ and has determinant $+1$. Full inversion is represented by $-\mathbf{1}_3$. Its determinant is

$$\det(-\mathbf{1}_3)=(-1)^3=-1.$$

Since the determinant is $-1$, inversion reverses orientation and is not connected continuously to the identity inside $SO(3)$. It is a discrete spatial symmetry, not a rotation.