Quantum Symmetries

A quantum symmetry compares states or experimental arrangements related by a transformation. There are two questions to distinguish: does the transformation preserve the transition probabilities between pure states, and is it a symmetry of the particular Hamiltonian being studied? The first is a statement about quantum kinematics; the second depends on the system and its external fields. Symmetries form groups when transformations can be composed and undone. For pure states, the kinematic condition is preservation of transition probabilities:

$$\lvert\langle\phi|\psi\rangle\rvert^2 \quad \text{is unchanged}.$$

In complex Hilbert-space quantum mechanics, Wigner’s theorem says that a bijection of pure-state rays preserving these probabilities is implemented on vectors by a unitary or antiunitary operator. Its overall phase is physically irrelevant. Every unitary operator satisfies this kinematic condition, but it need not be a symmetry of a given Hamiltonian.

Helpful background. Rays and global phase explains why different vectors can describe the same pure state; probability amplitudes explains the overlap in this formula.

Physical States Are Rays

Pure states are rays rather than individual vectors. If

$$\lvert\psi'\rangle=e^{i\alpha}\lvert\psi\rangle,$$

then $\lvert\psi'\rangle$ and $\lvert\psi\rangle$ represent the same physical pure state. A quantum symmetry therefore acts first on rays. A representative vector is chosen only after that.

This is why phases appear naturally in symmetry theory. A transformation can act exactly on physical rays while acting on chosen vectors only up to phases.

Preserving Transition Probabilities

For normalized pure states, the transition probability from $\lvert\psi\rangle$ to $\lvert\phi\rangle$ is

$$P_{\psi\to\phi} = \lvert\langle\phi|\psi\rangle\rvert^2.$$

A symmetry $S$ should preserve this number:

$$\lvert\langle S\phi|S\psi\rangle\rvert^2 = \lvert\langle\phi|\psi\rangle\rvert^2.$$

Unitary transformations preserve the inner product itself. Antiunitary transformations preserve its complex conjugate. Both preserve its absolute square.

Transformations of States

For a unitary symmetry $U$,

$$\lvert\psi\rangle \longmapsto U\lvert\psi\rangle.$$

For an antiunitary symmetry $T$, the same expression is often written formally,

$$\lvert\psi\rangle \longmapsto T\lvert\psi\rangle,$$

but $T$ is antilinear:

$$T(a\lvert\psi\rangle+b\lvert\phi\rangle) = a^*T\lvert\psi\rangle+b^*T\lvert\phi\rangle.$$

This distinction is essential for time reversal and for any operation involving complex phases.

Transformations of Observables

For a unitary transformation, observables transform as

$$A\longmapsto UAU^\dagger.$$

For a general unitary or antiunitary symmetry $S$, the compact expression is

$$A\longmapsto SAS^{-1}.$$

For a self-adjoint observable $A$ and a state with a finite expectation value, this preserves the expectation when both are transformed consistently:

$$\langle S\psi|SAS^{-1}|S\psi\rangle = \langle\psi|A|\psi\rangle,$$

For antiunitary $S$, the general matrix element is complex-conjugated. The displayed equality holds because the expectation of a self-adjoint observable is real. Transforming only the state can change the probabilities for a fixed measurement apparatus.

Transformations of Hamiltonians

A transformation is a symmetry of a time-independent Hamiltonian when

$$SHS^{-1}=H.$$

For a unitary $U$, this is equivalent to

$$[U,H]=0.$$

For a strongly continuous one-parameter unitary group $U(\alpha)=e^{-i\alpha G/\hbar}$ with self-adjoint generator $G$, the infinitesimal version is

$$[G,H]=0.$$

In finite dimensions these commutation tests are equivalent. For unbounded operators, the symmetry must preserve the domain of $H$; a formal commutator on a few vectors is not sufficient. Commuting spectral projections of $G$ and $H$ give the precise condition for a conserved generator. See Conservation Laws for the physical consequences.

An antiunitary symmetry has a different relation to time evolution. If $SHS^{-1}=H$, then

$$S e^{-iHt/\hbar}S^{-1}=e^{+iHt/\hbar}.$$

It sends forward evolution to backward evolution because it conjugates $i$. This is the characteristic relation for time reversal.

Symmetry Versus Change of Representation

Do not confuse a physical symmetry with a mere change of basis. A basis change rewrites the same state and operators in different coordinates. A physical symmetry compares a system with a transformed system and asks whether the physical predictions are preserved.

The formulas can look similar because both use unitary transformations. Keep track of whether the apparatus, the state, or merely its coordinates have changed; Active and Passive Transformations works through this distinction.

Examples

Parity in one dimension acts on a wavefunction by

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

It is a symmetry of

$$H=\frac{p^2}{2m}+V(x)$$

on the full line when $V(-x)=V(x)$ and the Hamiltonian’s domain is parity-invariant. Boundary conditions matter: reflecting a half-line generally changes the configuration space.

Spatial translations are represented by

$$U(a)=e^{-iaP/\hbar}.$$

They are symmetries of a free particle, but not of a particle in a generic position-dependent potential.

Spin rotations act on spin-$1/2$ states by

$$U(\hat{\mathbf n},\theta) = \exp\left( -\frac{i}{2}\theta\,\hat{\mathbf n}\cdot\boldsymbol\sigma \right).$$

They are symmetries only when the Hamiltonian respects the corresponding rotational invariance.

Common pitfalls

Preserving overlaps does not imply a symmetry of every experiment. Rotating a spin changes its probabilities relative to a fixed detector. Rotate the detector too to compare the same relative arrangement.

A symmetric Hamiltonian need not have a symmetric state. A wavepacket may sit on one side of an even potential. The reflected packet is a different state governed by the same Hamiltonian; see States, Observables, and Hamiltonians.

Exercises

1. Why is preserving $\lvert\langle\phi\rvert\psi\rangle\rvert^2$ more physically direct than preserving $\langle\phi\rvert\psi\rangle$ itself?

Solution

The absolute square is the transition probability. The inner product itself depends on phase choices for the two ray representatives. Since global phases of state vectors are not physical, the probability is the direct invariant.

2. For parity on the full line, show that $P^2=I$ when $(P\psi)(x)=\psi(-x)$. Is parity a symmetry of $H=p^2/(2m)+m\omega^2x^2/2+Fx$ for a fixed nonzero force parameter $F$?

Solution

Apply parity twice:

$$(P^2\psi)(x) = (P\psi)(-x) = \psi(x).$$

Thus $P^2$ acts as the identity on wavefunctions. The kinetic and quadratic terms are unchanged, but $Fx\mapsto-Fx$. Parity therefore relates Hamiltonians with opposite $F$; it is not a symmetry of the fixed nonzero-$F$ Hamiltonian.

References

• V. Bargmann, “Note on Wigner’s theorem on symmetry operations,” Journal of Mathematical Physics 5, 862–868 (1964). doi:10.1063/1.1704188.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chapter 4.
• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.