---
title: Parity, Time Reversal, and Discrete Symmetries
module: The Oscillator Algebraically, and Symmetry
moduleNumber: 5
lessonNumber: 4
order: 504
summary: >
  Parity and time reversal are symmetries no continuous generator can reach.
  Parity is a unitary involution whose eigenvalues label states even or odd, fixing
  the dipole selection rules. Time reversal is antiunitary: it conjugates i, flips
  momenta and spins, and for half-integer spin squares to minus one, which by
  Kramers' theorem makes every level of a time-reversal-invariant Hamiltonian at
  least doubly degenerate.
topics: [The Oscillator Algebraically, and Symmetry]
sources:
  - book: Sakurai & Napolitano
    ref: "Ch. 4 — Symmetry in Quantum Mechanics; §4.2 Parity, §4.4 The Time-Reversal Discrete Symmetry"
  - book: Griffiths & Schroeter
    ref: "Ch. 6 — Symmetries & Conservation Laws; parity and selection rules"
draft: false
---

The [continuous symmetries](/quantum-mechanics/oscillator-and-symmetry/symmetries-generators-and-conservation-laws)
of the previous lesson were all reached by exponentiating a Hermitian generator.
Two of the most important symmetries cannot be. Parity, the reflection of space
through the origin, and time reversal, the running of the film backward, are
**discrete**: there is no infinitesimal version to exponentiate. One is unitary and
supplies a two-valued quantum number and a set of selection rules; the other is
antiunitary and, for half-integer spin, guarantees a degeneracy nothing weaker can
remove.[^sn-pt]

## The parity operator

The **parity** operator $\hat\Pi$ inverts every position coordinate,

$$
\hat\Pi\,\psi(\vec r) = \psi(-\vec r).
$$

Applying it twice restores the original, so $\hat\Pi^{2} = 1$. It is both Hermitian
and unitary, and its eigenvalues, squaring to one, are

$$
\hat\Pi\,\psi = \pm\,\psi,
$$

with $+1$ for **even** states and $-1$ for **odd** states. Parity acts on the
canonical operators by flipping the polar vectors and leaving the axial ones fixed:

$$
\hat\Pi\,\hat{\vec r}\,\hat\Pi = -\hat{\vec r},
\qquad
\hat\Pi\,\hat{\vec p}\,\hat\Pi = -\hat{\vec p},
\qquad
\hat\Pi\,\hat{\vec L}\,\hat\Pi = +\hat{\vec L}.
$$

Position and momentum reverse because they are ordinary (polar) vectors; angular
momentum $\hat{\vec L} = \hat{\vec r}\times\hat{\vec p}$ is a cross product of two
polar vectors and so does not reverse — it is a **pseudovector**. Spin $\hat{\vec S}$,
sharing the transformation properties of angular momentum, is also even under parity.

$$
% caption: Under parity a polar vector such as position $\vec r$ reverses,
% $\vec r \to -\vec r$, while the axial vector $\vec L = \vec r\times\vec p$ (a sense
% of rotation) is unchanged. This is why $\hat{\vec r}$ and $\hat{\vec p}$ are
% parity-odd but $\hat{\vec L}$ and $\hat{\vec S}$ are parity-even.
\begin{tikzpicture}[>=stealth, font=\scriptsize]
  \definecolor{acc}{HTML}{4A6FA5}
  % polar vector panel
  \draw[black, dashed] (0,-1.6) -- (0,1.6);
  \draw[black, dashed] (-1.9,0) -- (1.9,0);
  \draw[->, very thick] (0,0) -- (1.4,0.9);
  \node[anchor=south west, font=\tiny] at (1.45,0.9) {position};
  \draw[->, acc, thick, dashed] (0,0) -- (-1.4,-0.9);
  \node[acc, anchor=north east, font=\tiny] at (-1.2,-0.85) {reversed};
  \node[black, anchor=north, font=\tiny] at (0,-1.75) {polar: flips};
  % axial vector panel
  \begin{scope}[xshift=5.4cm]
    \draw[black, dashed] (0,-1.6) -- (0,1.6);
    \draw[black, dashed] (-1.9,0) -- (1.9,0);
    \draw[->, very thick] (0.9,0) arc (0:290:0.9);
    \node[anchor=west, font=\tiny] at (0.55,0.95) {rotation};
    \node[anchor=north, font=\tiny] at (0,-1.75) {axial: unchanged};
  \end{scope}
\end{tikzpicture}
$$

## Parity as a good quantum number

When the potential is symmetric, $V(-\vec r) = V(\vec r)$, the Hamiltonian commutes
with parity, $[\hat\Pi,\hat H] = 0$. Then energy and parity are compatible
observables, and a nondegenerate energy eigenstate is automatically an eigenstate of
$\hat\Pi$ — it has definite parity. The proof mirrors the
[oscillator argument](/quantum-mechanics/wave-mechanics-1d/operators-expectation-values-and-the-harmonic-oscillator):
if $\hat H\psi = E\psi$ with $E$ nondegenerate and $V$ even, then $\psi(-\vec r)$
solves the same equation at the same energy, so it can differ from $\psi(\vec r)$
only by a constant, and that constant squares to one.

The bound states of every symmetric potential organize this way. The infinite and
finite square wells centered at the origin, the harmonic oscillator, and any even
$V$ produce eigenstates alternating even, odd, even, ... as the number of nodes
grows. Parity is exact here in a way it is only approximate for the weak
interaction, where it is famously violated.

$$
% caption: In a symmetric potential every nondegenerate eigenstate has definite
% parity. An even state satisfies $\psi(-x) = \psi(x)$ (eigenvalue $+1$); an odd
% state satisfies $\psi(-x) = -\psi(x)$ (eigenvalue $-1$), passing through zero at
% the origin.
\begin{tikzpicture}[>=stealth, font=\scriptsize]
  \definecolor{acc}{HTML}{4A6FA5}
  % even panel
  \draw[black] (-2.2,0) -- (2.2,0) node[right, font=\tiny, text=black]{$x$};
  \draw[black, dashed] (0,-1.3) -- (0,1.6);
  \draw[very thick] (-2.0,0.05) .. controls (-0.9,0.1) and (-0.6,1.3) .. (0,1.3)
    .. controls (0.6,1.3) and (0.9,0.1) .. (2.0,0.05);
  \node[anchor=north, font=\tiny] at (0,-1.5) {even (parity $+1$)};
  % odd panel
  \begin{scope}[xshift=6.0cm]
    \draw[black] (-2.2,0) -- (2.2,0) node[right, font=\tiny, text=black]{$x$};
    \draw[black, dashed] (0,-1.5) -- (0,1.6);
    \draw[acc, very thick] (-2.0,-0.05) .. controls (-1.1,-0.15) and (-0.7,-1.2) .. (0,0)
      .. controls (0.7,1.2) and (1.1,0.15) .. (2.0,0.05);
    \node[acc, anchor=north, font=\tiny] at (0,-1.5) {odd (parity flips sign)};
  \end{scope}
\end{tikzpicture}
$$

## Parity selection rules

Parity constrains which transitions an atom can make. The rate of an electric-dipole
transition between states $\lvert i\rangle$ and $\lvert f\rangle$ is set by the
matrix element $\langle f\rvert\hat{\vec r}\lvert i\rangle$. The position operator is
parity-odd, so under $\hat\Pi$ the integrand picks up the product of the two states'
parities times $-1$. If $\lvert i\rangle$ and $\lvert f\rangle$ have the **same**
parity, the integrand is odd and the integral over all space vanishes:

$$
\langle f\rvert\hat{\vec r}\lvert i\rangle = 0
\quad\text{unless}\quad
\Pi_f\,\Pi_i = -1 .
$$

A dipole transition connects only states of opposite parity. Written in terms of the
orbital quantum number, where parity is $(-1)^{\ell}$, this is the **Laporte rule**:
$\ell$ must change by an odd amount, and combined with the angular-momentum content
of $\hat{\vec r}$ it sharpens to $\Delta\ell = \pm 1$. The
[oscillator selection rule](/quantum-mechanics/wave-mechanics-1d/operators-expectation-values-and-the-harmonic-oscillator)
$\Delta n = \pm 1$ is the one-dimensional case of the same parity argument.

$$
% caption: A dipole integrand $\psi_f\,x\,\psi_i$ between same-parity states is an
% odd function: positive and negative lobes have equal area and cancel, so the
% matrix element vanishes. Only opposite-parity states give a nonzero (allowed)
% integral.
\begin{tikzpicture}[>=stealth, font=\scriptsize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (-2.6,0) -- (2.6,0) node[right, font=\tiny, text=black]{$x$};
  \draw[black, dashed] (0,-1.4) -- (0,0.7);
  % odd integrand: positive right lobe, negative left lobe
  \draw[very thick] (-2.3,0.03) .. controls (-1.7,-1.3) and (-0.6,-1.2) .. (0,0)
    .. controls (0.6,1.2) and (1.7,1.3) .. (2.3,0.03);
  \fill[acc!18] (0,0) .. controls (0.6,1.2) and (1.7,1.3) .. (2.3,0.03) -- (0,0);
  \fill[black] (-2.3,0.03) .. controls (-1.7,-1.3) and (-0.6,-1.2) .. (0,0) -- (-2.3,0.03);
  \node[acc, anchor=center, font=\tiny] at (1.4,0.55) {positive};
  \node[black, anchor=center, font=\tiny] at (-1.4,-0.55) {negative};
  \node[black, anchor=west, font=\tiny] at (-2.4,0.95) {odd: cancels};
\end{tikzpicture}
$$

## The time-reversal operator

Time reversal reverses the direction of motion: positions are untouched but
velocities and momenta flip. The Schrödinger equation
$i\hbar\,\partial_t\psi = \hat H\psi$ is not invariant under $t\to-t$ alone, because
the single time derivative changes sign while $\hat H$ does not. It becomes invariant
if the transformation also conjugates the wavefunction, $\psi(\vec r,t)\to\psi^{\ast}(\vec r,-t)$.
Complex conjugation is not a linear operation on the Hilbert space, and this is the
essential feature.

> **Definition (Antiunitary operator).** An operator $\hat\Theta$ is **antiunitary**
> if it is antilinear, $\hat\Theta(c_1\lvert\psi_1\rangle + c_2\lvert\psi_2\rangle)
> = c_1^{\ast}\hat\Theta\lvert\psi_1\rangle + c_2^{\ast}\hat\Theta\lvert\psi_2\rangle$,
> and preserves the magnitude of inner products,
> $|\langle\hat\Theta\phi\vert\hat\Theta\psi\rangle| = |\langle\phi\vert\psi\rangle|$.
> Time reversal is the antiunitary case allowed by Wigner's theorem.

The defining relation is $\hat\Theta\,i\,\hat\Theta^{-1} = -i$: time reversal
conjugates the imaginary unit. From this and the requirement that it flip momentum,
its action on the observables is

$$
\hat\Theta\,\hat{\vec r}\,\hat\Theta^{-1} = \hat{\vec r},
\qquad
\hat\Theta\,\hat{\vec p}\,\hat\Theta^{-1} = -\hat{\vec p},
\qquad
\hat\Theta\,\hat{\vec L}\,\hat\Theta^{-1} = -\hat{\vec L},
\qquad
\hat\Theta\,\hat{\vec S}\,\hat\Theta^{-1} = -\hat{\vec S}.
$$

Angular momentum and spin reverse because they are built from, or transform like,
$\vec r\times\vec p$ with one factor of $\vec p$ flipped. For a spinless particle the
operator is simply complex conjugation, $\hat\Theta = \hat K$, and $\hat\Theta^{2} = +1$.
A consequence is that the energy eigenfunctions of a time-reversal-invariant,
spinless Hamiltonian can always be chosen real.

$$
% caption: Time reversal leaves position fixed but reverses momentum and spin,
% $\vec p \to -\vec p$ and $\vec S \to -\vec S$, sending a trajectory into the one
% that retraces it. Position, a polar vector, is untouched.
\begin{tikzpicture}[>=stealth, font=\scriptsize]
  \definecolor{acc}{HTML}{4A6FA5}
  % original
  \fill[black] (0,0) circle (2.2pt);
  \draw[->, acc, very thick] (0,0) -- (1.8,0);
  \node[acc, anchor=south, font=\tiny] at (1.4,0.05) {momentum};
  \draw[->, black, very thick] (0,0) -- (0,1.3);
  \node[black, anchor=west, font=\tiny] at (0.05,1.1) {spin};
  \node[black, anchor=north, font=\tiny] at (0.9,-0.4) {before};
  % time-reversed
  \begin{scope}[xshift=5.4cm]
    \fill[black] (0,0) circle (2.2pt);
    \draw[->, acc, very thick] (0,0) -- (-1.8,0);
    \node[acc, anchor=south, font=\tiny] at (-1.4,0.05) {reversed};
    \draw[->, black, very thick] (0,0) -- (0,-1.3);
    \node[black, anchor=east, font=\tiny] at (-0.05,-1.1) {reversed};
    \node[black, anchor=north, font=\tiny] at (0,-1.6) {after time reversal};
  \end{scope}
\end{tikzpicture}
$$

## Kramers degeneracy

The spinless case squares to $+1$, but spin changes the sign. Building time reversal
for a spin-$\tfrac12$ particle requires a rotation in spin space alongside the
conjugation, $\hat\Theta = -i\hat\sigma_y\hat K$, and this operator squares to
$\hat\Theta^{2} = -1$. More generally $\hat\Theta^{2} = (-1)^{2j}$: integer spin gives
$+1$, half-integer spin gives $-1$. The negative sign has a sharp physical
consequence.

> **Theorem (Kramers degeneracy).** If a Hamiltonian is time-reversal invariant,
> $[\hat\Theta,\hat H] = 0$, and the system has half-integer total angular momentum
> so that $\hat\Theta^{2} = -1$, then every energy level is at least doubly
> degenerate.

> **Proof.** Let $\hat H\lvert\psi\rangle = E\lvert\psi\rangle$. Since $\hat\Theta$
> commutes with $\hat H$, the state $\hat\Theta\lvert\psi\rangle$ has the same energy
> $E$. Suppose it were not a new state, $\hat\Theta\lvert\psi\rangle = c\lvert\psi\rangle$
> for some scalar $c$. Applying $\hat\Theta$ again and using its antilinearity,
> $\hat\Theta^{2}\lvert\psi\rangle = \hat\Theta(c\lvert\psi\rangle) = c^{\ast}\hat\Theta\lvert\psi\rangle
> = |c|^{2}\lvert\psi\rangle$. But $\hat\Theta^{2} = -1$ requires
> $|c|^{2} = -1$, impossible for a scalar. Therefore $\hat\Theta\lvert\psi\rangle$ is
> orthogonal to $\lvert\psi\rangle$ and independent of it, giving a second state at
> energy $E$. The level is degenerate.

The pair $\lvert\psi\rangle$ and $\hat\Theta\lvert\psi\rangle$ is a **Kramers
doublet**. An electron bound in any electrostatic environment, however irregular,
keeps this twofold degeneracy: no purely electric field can split it, because
electric fields are time-reversal invariant. Only a magnetic field, which is
time-reversal odd and breaks $[\hat\Theta,\hat H] = 0$, lifts the doublet — the
[Zeeman splitting](/quantum-mechanics/approximation-methods/the-zeeman-and-stark-effects).

$$
% caption: Kramers' theorem: with half-integer spin and time-reversal symmetry each
% level is a doublet $\{\lvert\psi\rangle,\ \hat\Theta\lvert\psi\rangle\}$, split only
% by a magnetic field. Integer-spin levels carry no such protection and can be
% nondegenerate.
\begin{tikzpicture}[>=stealth, font=\scriptsize]
  \definecolor{acc}{HTML}{4A6FA5}
  % half-integer: doublet
  \node[black, anchor=south, font=\tiny] at (1.6,2.4) {half-integer spin};
  \draw[acc, thick] (0.4,1.6) -- (2.8,1.6);
  \node[acc, anchor=south, font=\tiny] at (1.6,1.7) {doublet (2 states)};
  % split by B field
  \draw[thick, dashed] (4.6,2.0) -- (6.0,2.0);
  \draw[thick, dashed] (4.6,1.2) -- (6.0,1.2);
  \node[black, anchor=west, font=\tiny] at (6.1,1.6) {magnetic field splits};
  \draw[->, black] (3.2,1.6) -- (4.4,1.6);
  % integer: single
  \node[black, anchor=south, font=\tiny] at (1.6,0.4) {integer spin};
  \draw[black, thick] (0.4,0.0) -- (2.8,0.0);
  \node[black, anchor=west, font=\tiny] at (2.9,0.0) {may be single};
\end{tikzpicture}
$$

## Parity and time reversal together

The two discrete symmetries jointly forbid one striking possibility: a permanent
electric dipole moment aligned with a particle's spin. Such a moment
$\vec d = d\,\hat{\vec S}/|\vec S|$ would need $d\,\hat{\vec S}$ to transform like the
polar, time-even vector $\hat{\vec r}$. But $\hat{\vec S}$ is parity-even and
time-odd, the exact opposite of $\hat{\vec r}$ on both counts. A nonzero $d$ therefore
requires violating **both** parity and time reversal. Searches for a neutron or
electron electric dipole moment are precision tests of combined $P$ and $T$
violation, and their null results to date bound physics beyond the standard model.[^edm]

The four symmetries of these two lessons complete the basic catalog:

| Symmetry | Type | Operator | Key consequence |
| --- | --- | --- | --- |
| Translation | continuous | unitary $e^{-i a\hat p/\hbar}$ | momentum conserved |
| Rotation | continuous | unitary $e^{-i\phi\hat n\cdot\hat{\vec L}/\hbar}$ | angular momentum conserved |
| Parity | discrete | unitary $\hat\Pi$, $\hat\Pi^{2}=1$ | even/odd states, dipole selection rules |
| Time reversal | discrete | antiunitary $\hat\Theta$ | real eigenfunctions; Kramers degeneracy |

Parity and time reversal round out the symmetry structure that
[angular momentum](/quantum-mechanics/angular-momentum/orbital-angular-momentum-and-spherical-harmonics)
and [spin](/quantum-mechanics/spin/spin-half-pauli-matrices-and-stern-gerlach) build
on directly.

[^sn-pt]: **Sakurai & Napolitano**, _Modern Quantum Mechanics_ (3rd ed., Cambridge, 2021), §4.2 (parity: the operator, even/odd eigenstates, selection rules) and §4.4 (time reversal: antiunitarity, $\hat\Theta^{2}=(-1)^{2j}$, and Kramers degeneracy). Cambridge listing: https://www.cambridge.org/highereducation/books/modern-quantum-mechanics/6C8BB37F5B120694E9AB9DF6EBFD6D48. Parity and selection rules also in **Griffiths & Schroeter**, _Introduction to Quantum Mechanics_ (3rd ed., Cambridge, 2018), Ch. 6.
[^edm]: A spin-aligned permanent electric dipole moment violates both $P$ and $T$; combined with the $CPT$ theorem this makes it a probe of $CP$ violation. Current experimental bounds on the neutron and electron electric dipole moments are compiled by the Particle Data Group, _Review of Particle Physics_: https://pdg.lbl.gov.