Symmetry Principles

A quantum symmetry preserves transition probabilities, but it need not preserve a particular Hamiltonian. This chapter separates transformations of rays, transformations of observables, and invariance of the dynamics. It then explains how a specified symmetry restricts Hamiltonians and organizes energy levels.

Helpful background. Rays and global phase explain why a state vector’s overall phase is irrelevant; groups explain how transformations compose.

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Question	Read next
What does a quantum symmetry preserve?	Quantum symmetries and Wigner’s theorem
Is the state changing, or its description?	Active and passive transformations
Which object is invariant?	States, observables, and Hamiltonians
Why does complex conjugation matter?	Unitary and antiunitary symmetries
Why can composition include a phase?	Projective representations
How do groups act on states and spectra?	Groups and representations and multiplets
Which Hamiltonian terms are allowed?	Symmetry constraints
When is coherence between sectors unobservable?	Superselection sectors

Symmetries act on quantum rays

Pure physical states are rays rather than individual normalized vectors. For Wigner’s theorem, take a bijective transformation of the pure-state rays that preserves transition probabilities. If rays represented by $|\psi\rangle$ and $|\phi\rangle$ are mapped to rays represented by $|\psi'\rangle$ and $|\phi'\rangle$, then

$$|\langle\phi'|\psi'\rangle|^2 = |\langle\phi|\psi\rangle|^2.$$

Wigner’s theorem states, under its standard assumptions, that such a ray transformation can be represented on Hilbert-space vectors by either a unitary or an antiunitary operator. The representative is not unique: multiplying it by an overall phase leaves its action on rays unchanged.

This theorem answers how prediction-preserving transformations can act. It does not say that every such transformation is a symmetry of a particular Hamiltonian. Dynamics must still be tested.

Hamiltonian symmetry differs from state symmetry

Let $S$ represent a candidate transformation. An active state transformation is

$$|\psi\rangle \longmapsto S|\psi\rangle.$$

For a unitary transformation, an observable transforms by conjugation:

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

A time-independent Hamiltonian is invariant when

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

For unitary $S$, this is equivalent to

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

For an unbounded $H$, invariance includes its domain. For antiunitary $S=\Theta$, the same Hamiltonian equation has a different implication for the propagator:

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

The sign changes because $\Theta i\Theta^{-1}=-i$. Time reversal relates forward and backward evolution rather than commuting with evolution at fixed $t$.

Three statements must remain distinct:

1. $S$ is a valid transformation of quantum states.
2. $S$ is a symmetry of the Hamiltonian.
3. A particular state ray is invariant under $S$.

A parity-symmetric Hamiltonian, for example, can evolve a state that is not itself a parity eigenstate. Symmetry of the laws is not invariance of every allowed state.

A symmetry can constrain a model before diagonalization

For a Hermitian two-level Hamiltonian,

$$H=a_0I+a_x\sigma_x+a_y\sigma_y+a_z\sigma_z,$$

the coefficients are real. Impose the unitary symmetry $P=\sigma_z$. Conjugation reverses $\sigma_x$ and $\sigma_y$ but leaves $I$ and $\sigma_z$ unchanged, so $PHP^{-1}=H$ requires $a_x=a_y=0$. The resulting energies are $a_0\pm a_z$. The symmetry provides two invariant sectors, but it does not force their energies to agree.

This example also separates conservation from superselection. A coherent state $(|+z\rangle+|-z\rangle)/\sqrt2$ is still allowed, and its relative phase is visible through an observable such as $\sigma_x$. A superselection rule would additionally restrict the physical observable algebra so that coherence between the sectors cannot be detected. Commutation of $P$ with this particular Hamiltonian alone does not impose that restriction.

Multiplets depend on the representation

For a finite-dimensional unitary representation, group the Hilbert space into irreducible representation spaces $V_\lambda$ and their multiplicity spaces $M_\lambda$:

$$\mathcal H=\bigoplus_\lambda V_\lambda\otimes M_\lambda.$$

An invariant Hamiltonian has the form $H=\bigoplus_\lambda I_{V_\lambda}\otimes h_\lambda$. Thus it acts identically on the components of an irreducible multiplet but may mix different copies of that representation. This is why full rotational symmetry enforces the magnetic degeneracy within a spin-$j$ multiplet while allowing distinct radial or other multiplicity levels to have different energies. The multiplet article explains the construction; Wigner and the modern textbooks listed below develop the representation-theoretic argument.

Projective action and spinors

Quantum states are rays, so composition may be exact on rays while vector representatives compose only up to phase:

$$U(g_1)U(g_2) = e^{i\omega(g_1,g_2)} U(g_1g_2).$$

This is a projective representation. The phase does not alter an individual ray, but its consistency encodes real structure. Spin-$1/2$ is the central example: physical rotations form $SO(3)$, while spinors carry ordinary representations of its double cover $SU(2)$.

For a spin-$1/2$ rotation,

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

so

$$U(\hat{\mathbf n},2\pi)=-I.$$

The vector changes sign, but the ray does not. Projective Representations explains the composition law; later spin pages develop the concrete rotations.

Exercises

Symmetric Hamiltonian, nonsymmetric state

Let $[H,P]=0$ with $P^2=I$, and let $|e\rangle$ and $|o\rangle$ have parity $+1$ and $-1$. Is the state

$$|\psi\rangle = \frac{|e\rangle+|o\rangle}{\sqrt2}$$

parity invariant?

Solution

Applying parity gives

$$P|\psi\rangle = \frac{|e\rangle-|o\rangle}{\sqrt2}.$$

This is not the original state times one overall phase, so the ray is not parity invariant. The Hamiltonian can nevertheless be parity symmetric. Because $[H,P]=0$, the even and odd components evolve within their respective sectors.

Classify spinless time reversal

Why can the spinless operation $\Theta=K$ preserve transition probabilities even though it is not unitary?

Solution

Complex conjugation is antiunitary. It reverses scalar multiplication,

$$K(c|\psi\rangle) = c^*K|\psi\rangle,$$

and conjugates inner products:

$$\langle K\phi|K\psi\rangle = \langle\phi|\psi\rangle^*.$$

Taking the absolute square removes that conjugation, so transition probabilities are preserved. Wigner’s theorem allows both unitary and antiunitary representatives for precisely this reason.

Filter a two-level Hamiltonian

For

$$H=a_0I+a_x\sigma_x+a_y\sigma_y+a_z\sigma_z,$$

impose invariance under $P=\sigma_x$. Which coefficients must vanish?

Solution

Conjugation gives

$$\begin{aligned} \sigma_x\sigma_x\sigma_x&=\sigma_x, \\ \sigma_x\sigma_y\sigma_x&=-\sigma_y, \\ \sigma_x\sigma_z\sigma_x&=-\sigma_z. \end{aligned}$$

Thus

$$PHP^{-1} = a_0I+a_x\sigma_x-a_y\sigma_y-a_z\sigma_z.$$

Equality with $H$ requires $a_y=a_z=0$. The coefficients $a_0$ and $a_x$ remain allowed.

References

• H. F. Jones, Groups, Representations and Physics, 2nd ed., Institute of Physics Publishing, 1998.
• 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.
• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, translated by J. J. Griffin, Academic Press, 1959.