advanced30 min studyLesson 401 in Physics

Unitary and antiunitary symmetry operators

Identify probability-preserving transformations without confusing antilinearity with a minus sign.

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What you’ll understand

  • Distinguish linear unitary maps from antilinear antiunitary maps.
  • Check transition-probability preservation by calculating overlaps.
  • Separate a symmetry of quantum kinematics from a symmetry of a chosen Hamiltonian.

A symmetry preserves the distinctions an experiment can make. In quantum mechanics those distinctions are encoded by transition probabilities between rays, where vectors differing only by an overall phase describe the same pure state. Preserving norms is part of the requirement, but preserving the probabilities between every pair of states is stronger. The surprising result is that complex conjugation can participate in a symmetry even though it reverses the complex structure.

Two ways to preserve a probability

Assume a bijection of the rays of a complex Hilbert space preserves all transition probabilities. Wigner's theorem says that it can be represented by a unitary or an antiunitary operator, up to irrelevant phases. A unitary operator is complex linear and preserves inner products. An antiunitary operator is antilinear and conjugates inner products. Both preserve their absolute squares. The theorem concerns the entire state space, not a rule that happens to preserve the norm of one selected preparation.

Antilinearity means that a scalar is conjugated when it passes through the map. In particular, A(i∣ψ⟩)=−iA∣ψ⟩A(i|\psi\rangle)=-iA|\psi\rangle. This is different from a linear operator with some negative matrix entries. In a chosen orthonormal basis, any antiunitary map can be written A=UKA=UK, where KK conjugates the components and UU is unitary. The matrix describing UU and the meaning of KK depend on that basis; the antilinear character of AA does not.

U†U=I,⟨Uϕ∣Uψ⟩=⟨ϕ∣ψ⟩,⟨Aϕ∣Aψ⟩=⟨ϕ∣ψ⟩∗U^\dagger U=I,\qquad \langle U\phi|U\psi\rangle=\langle\phi|\psi\rangle,\qquad \langle A\phi|A\psi\rangle=\langle\phi|\psi\rangle^*
States are normalized dimensionless vectors; U is linear and A is antilinear. The final equality establishes the same transition probability after either transformation.↔ Scroll wide equations horizontally.

Continuous symmetries and their generators

A continuously connected family starting at the identity stays in the unitary class. For a strongly continuous one-parameter unitary group, a self-adjoint generator describes the family by exponentiation. In finite dimensions this is an ordinary Hermitian matrix; in infinite dimensions its domain matters. The exponent must be dimensionless. A translation parameter measured in metres therefore pairs with momentum divided by Planck's reduced constant, whereas a rotation angle pairs with angular momentum divided by that constant.

The infinitesimal form gives a practical diagnostic. In finite dimensions, expanding the unitary condition to first order forces the generator to equal its adjoint. For unbounded operators the first-order condition instead gives symmetry on the relevant domain; self-adjointness follows from Stone’s theorem for the full strongly continuous group. This does not say that every unitary transformation is a physical symmetry of every Hamiltonian. A spin rotation is a legitimate probability-preserving operation, but a fixed external magnetic field selects a direction and usually prevents that rotation from being a symmetry of the same apparatus. Transforming both the apparatus and the state is a different comparison.

U(s)=e−isG/ℏ,U(s)∣ψ⟩=∣ψ⟩−isℏG∣ψ⟩+o(s),U(s)†HU(s)=HU(s)=e^{-isG/\hbar},\quad U(s)|\psi\rangle=|\psi\rangle-\frac{is}{\hbar}G|\psi\rangle+o(s),\quad U(s)^\dagger H U(s)=H
For a strongly continuous group, G is self-adjoint and sG has units of action. The expansion holds in Hilbert-space norm for a state in the domain of G; an O(s squared) remainder would require the domain of G squared. The final equality additionally requires a time-independent unitary symmetry of H.↔ Scroll wide equations horizontally.

Phases, composition, and physical implementation

Because states are rays, representative symmetry operators can compose with an extra phase. Such a projective representation is physically consistent: transition probabilities cannot detect the phase on an isolated transformed state. Relative phases between coherently compared operations can still matter. Spin one half supplies an important example, since a full spatial turn returns the ray while changing the sign of a representative spinor. Keeping the vector and ray statements separate avoids an apparent contradiction.

Antiunitarity also has an operational boundary. Ordinary closed-system time evolution generated by a Hamiltonian is unitary, so complex conjugation of an arbitrary unknown input is not another universal Hamiltonian gate. Antiunitary symmetries relate descriptions or possible histories; they need not be directly implementable as deterministic preparation devices. Time reversal is the central physical example. Whether it preserves a particular Hamiltonian requires examining magnetic fields, angular momenta, and all other time-odd quantities rather than only conjugating a displayed column.

U(g1)U(g2)=eiω(g1,g2)U(g1g2),∣⟨ϕ∣ψ⟩∣2∈[0,1]U(g_1)U(g_2)=e^{i\omega(g_1,g_2)}U(g_1g_2),\qquad |\langle\phi|\psi\rangle|^2\in[0,1]
The dimensionless phase omega expresses a projective composition law. This displayed law is stated for unitary representatives; antiunitary composition also conjugates scalar phases.↔ Scroll wide equations horizontally.

Conjugate, swap, and check the norm

In an orthonormal two-state basis let A=σxKA=\sigma_xK, with σx\sigma_x exchanging the two components. Apply it to ∣ψ⟩=(3/2,i/2)T|\psi\rangle=(\sqrt3/2,i/2)^T, calculate A2∣ψ⟩A^2|\psi\rangle, and test its action on i∣ψ⟩i|\psi\rangle.

  1. Conjugation first sends the column to (3/2,−i/2)T(\sqrt3/2,-i/2)^T. Swapping afterward gives A∣ψ⟩=(−i/2,3/2)TA|\psi\rangle=(-i/2,\sqrt3/2)^T. Its norm squared is 1/4+3/4=11/4+3/4=1, so normalization is preserved.

  2. Apply the same operations again: conjugate to (i/2,3/2)T(i/2,\sqrt3/2)^T and exchange to obtain (3/2,i/2)T(\sqrt3/2,i/2)^T. Thus this particular antiunitary has square II; antiunitarity alone does not require square −I-I.

  3. The scalar ii must also be conjugated. Consequently A(i∣ψ⟩)=−iA∣ψ⟩A(i|\psi\rangle)=-iA|\psi\rangle, rather than iA∣ψ⟩iA|\psi\rangle. Checking this scalar distinguishes the map from an ordinary linear swap.

The result

The transformed state is normalized and A squared is the identity, while its action on a complex scalar proves that A is antilinear.

Explore the model

Interactive lab

Interactive lab · quantum

Conjugation reflects a pure-state coordinate

For each run Reset and set both controls. Compare (θ,ϕ)=(60∘,90∘)(\theta,\phi)=(60^\circ,90^\circ) with (60∘,−90∘)(60^\circ,-90^\circ) while paused. Predict equal basis probabilities, unchanged Bloch x and z, and opposite y. These states illustrate K in this basis. The pure-state instrument does not implement an antiunitary gate or a measurement collapse.

Conjugation reflects a pure-state coordinateA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forProbability preservation can coexist with a reflected complex phase.

Check your understanding

Which relation identifies an antiunitary map A?

Try it yourself

Let D=diag⁡(1,i)D=\operatorname{diag}(1,i) and ∣+⟩=(1,1)T/2|+\rangle=(1,1)^T/\sqrt2. Find D∣+⟩D|+\rangle and KD∣+⟩KD|+\rangle. Are their probabilities in the original basis different, and what are their Pauli-y expectations?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

For a normalized column (a,b), the Pauli-y expectation is twice the imaginary part of the product of the conjugate of a and b.

Compare with the solution

The columns are (1,i)T/2(1,i)^T/\sqrt2 and (1,−i)T/2(1,-i)^T/\sqrt2. Both assign probability 1/21/2 to each basis outcome. Their Pauli-y expectations are respectively +1+1 and −1-1, since the relevant products are i/2i/2 and −i/2-i/2. A fixed-basis probability list therefore misses the distinction, whereas a y-basis measurement detects it.

Take it with you

  • A unitary map is linear; an antiunitary map conjugates complex scalars.
  • Both preserve transition probabilities, but Hamiltonian invariance is an additional requirement.
  • Continuous unitary groups have self-adjoint generators with dimensionless exponents.

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