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Discrete Symmetries

Parity and time reversal impose different restrictions on quantum systems. Parity reverses spatial position and is unitary; time reversal reverses motion and is antiunitary, so it also conjugates complex amplitudes. That distinction determines whether a symmetry can force an energy doublet. This chapter develops the nonrelativistic examples, then introduces charge conjugation, CPT, and the matrix symmetries used in quantum matter.

Helpful background. Symmetry principles distinguishes a transformation from invariance of a Hamiltonian; spinors supplies the Pauli matrices used for spin-1/21/2 time reversal.

Question Read next
How does inversion act on states and vectors? Parity
Why must reversed motion conjugate amplitudes? Time reversal and antiunitary time reversal
How does the representation depend on spin? Spinless particles and spin-1/21/2 particles
When is a doublet protected? Kramers degeneracy
How do the rules restrict a concrete model? Discrete symmetries in Hamiltonians
What new assumptions enter relativistic theory? Charge conjugation and CPT
How are time-reversal and particle–hole constraints classified? Symmetry classification

Discrete operations and fixed external parameters

Section titled “Discrete operations and fixed external parameters”

Here “discrete” refers to the specified physical symmetry group. Inversion is not connected to the identity within the group of ordinary three-dimensional spatial rotations, and antiunitary time reversal cannot be continuously connected to the identity through unitaries. A finite subgroup can also consist of rotations that lie within a larger continuous group. The term does not claim that a unitary parity matrix has no mathematical logarithm; such a logarithm need not generate a continuous physical spatial symmetry.

For a time-independent Hamiltonian, test SHS−1=HSHS^{-1}=H, including its domain. Distinguish that from a relation between different Hamiltonians. A magnetic-field family may satisfy

ΘH(B)Θ−1=H(−B),\Theta H(\mathbf B)\Theta^{-1}=H(-\mathbf B),

while the fixed Hamiltonian at nonzero B\mathbf B generally breaks time reversal. The apparatus and external field cannot silently be reversed halfway through a fixed-field symmetry test.

For an ordinary scalar wavefunction, (Πψ)(x)=ψ(−x)(\Pi\psi)(\mathbf x)=\psi(-\mathbf x), and Π2=I\Pi^2=I gives eigenvalues ±1\pm1. An even potential with inversion-invariant domain and boundary conditions preserves parity. Energy states can then be chosen even or odd, but there is no requirement that an even state and an odd state have equal energy.

Position and momentum are polar vectors and reverse sign under inversion. Their cross product L=X×P\mathbf L=\mathbf X\times\mathbf P is axial and remains unchanged; spin is also parity even. Consequently, L⋅S\mathbf L\cdot\mathbf S can preserve parity. An operator’s rotational rank alone does not tell its parity. Parity selection rules applies this distinction to matrix elements.

Antiunitarity reverses the complex phase of evolution

Section titled “Antiunitarity reverses the complex phase of evolution”

An antiunitary operator obeys

Θ(a∣ψ⟩+b∣ϕ⟩)=a∗Θ∣ψ⟩+b∗Θ∣ϕ⟩,\Theta(a|\psi\rangle+b|\phi\rangle) =a^*\Theta|\psi\rangle+b^*\Theta|\phi\rangle,

and ⟨Θϕ∣Θψ⟩=⟨ϕ∣ψ⟩∗\langle\Theta\phi|\Theta\psi\rangle=\langle\phi|\psi\rangle^*. It preserves transition probabilities but sends ii to −i-i. This is required when position is unchanged and momentum reverses: [X,−P]=−iℏI[X,-P]=-i\hbar I, consistent with antiunitary conjugation of the canonical commutator.

If ΘHΘ−1=H\Theta H\Theta^{-1}=H, then

Θe−iHt/ℏΘ−1=e+iHt/ℏ.\Theta e^{-iHt/\hbar}\Theta^{-1}=e^{+iHt/\hbar}.

Thus Θ∣ψ(−t)⟩\Theta|\psi(-t)\rangle solves the same autonomous Schrödinger equation. Applying only t↦−tt\mapsto-t to a wavefunction omits the necessary conjugation.

In a chosen basis, write Θ=UK\Theta=UK with UU unitary and KK componentwise complex conjugation. Then Θ2=UU∗\Theta^2=UU^*, not U2U^2. Rephasing an antiunitary does not change its square because the second phase is conjugated.

Spin determines which time-reversal square applies

Section titled “Spin determines which time-reversal square applies”

For a spinless particle without magnetic coupling, the usual position representation has Θ=K\Theta=K and Θ2=I\Theta^2=I. A real potential with a compatible time-reversal-invariant domain admits this symmetry. A nondegenerate eigenfunction can be rephased to be real; no doublet follows.

In the spin-1/21/2 SzS_z basis, choose

Θ=−iσyK,Θ(αβ)=(−β∗α∗).\Theta=-i\sigma_yK,\qquad \Theta\begin{pmatrix}\alpha\\\beta\end{pmatrix} =\begin{pmatrix}-\beta^*\\\alpha^*\end{pmatrix}.

This reverses every spin component and has Θ2=−I\Theta^2=-I. In particular, a single-spin Hamiltonian H=aI+b⋅σH=aI+\mathbf b\cdot\boldsymbol\sigma with real coefficients is time-reversal invariant at fixed parameters only if b=0\mathbf b=0. The two energies must coincide.

The square must be computed on the complete sector in question. Two independent spin-1/21/2 factors have product time reversal with square +I+I. Their rotationally and time-reversal-invariant exchange Hamiltonian AS1⋅S2A\mathbf S_1\cdot\mathbf S_2, with A>0A>0, has a nondegenerate singlet ground state. There is no conflict with the single-spin conclusion.

Kramers pairing requires a negative square

Section titled “Kramers pairing requires a negative square”

Let HH be self-adjoint and Θ\Theta antiunitary. Suppose

ΘHΘ−1=H\Theta H\Theta^{-1}=H

and, on the sector under consideration,

Θ2=−I.\Theta^2=-I.

If

H∣ψ⟩=E∣ψ⟩,H|\psi\rangle=E|\psi\rangle,

then

HΘ∣ψ⟩=EΘ∣ψ⟩.H\Theta|\psi\rangle = E\Theta|\psi\rangle.

The energy is real because HH is self-adjoint. Antiunitarity and the negative square also imply

⟨ψ∣Θψ⟩=0.\langle\psi|\Theta\psi\rangle=0.

Thus ∣ψ⟩|\psi\rangle and Θ∣ψ⟩\Theta|\psi\rangle are distinct orthogonal states with the same energy. Every finite-dimensional normalizable energy eigenspace in that sector has even dimension. The theorem pairs normalizable eigenvectors; continuum scattering states require their own spectral treatment.

The theorem needs every hypothesis:

  • Θ\Theta is antiunitary;
  • Θ2=−I\Theta^2=-I on the relevant sector;
  • the fixed Hamiltonian is invariant;
  • the transformed state remains in the same physical sector.

Spinless time reversal with Θ2=+I\Theta^2=+I does not force a doublet. A fixed magnetic field breaks the hypothesis ΘHΘ−1=H\Theta H\Theta^{-1}=H. A random twofold degeneracy need not be a Kramers degeneracy.

Spin–orbit coupling does not automatically break time reversal. Since

ΘLΘ−1=−L,ΘSΘ−1=−S,\Theta\mathbf L\Theta^{-1} = -\mathbf L, \qquad \Theta\mathbf S\Theta^{-1} = -\mathbf S,

the scalar product is even:

Θ(L⋅S)Θ−1=L⋅S.\Theta \left( \mathbf L\cdot\mathbf S \right) \Theta^{-1} = \mathbf L\cdot\mathbf S.

Spin–orbit interactions can reorganize levels while preserving Kramers pairs in a negative-square sector.

Charge conjugation, CPT, and particle–hole structure

Section titled “Charge conjugation, CPT, and particle–hole structure”

Charge conjugation maps charge-qq sectors to charge-−q-q sectors. An electron-only nonrelativistic space is not closed under the physical particle–antiparticle operation because it contains no positron sector. Charge conservation, [H,Q]=0[H,Q]=0, is a different statement from invariance under charge conjugation. Complex conjugating a one-particle wavefunction is likewise not, by itself, the full field-theoretic operation.

The CPT theorem belongs to local Lorentz-covariant quantum field theory with its usual spectrum and Hilbert-space assumptions. It does not follow by multiplying three symbols in a finite-dimensional model, and it does not imply separate CC, PP, or TT invariance. The CPT introduction and QFT bridge identify the required framework; Streater and Wightman provide the axiomatic development.

In noninteracting band and Bogoliubov–de Gennes models, time-reversal, particle–hole, and chiral constraints instead organize the ten Altland–Zirnbauer classes. Particle–hole structure in Nambu space is not simply relativistic charge conjugation. A class alone does not specify a topological phase: the spatial dimension, gap, locality, and allowed perturbations also matter. Continue with symmetry classification for explicit matrices and examples.

  1. Determine the parity and time-reversal character of the Hermitian dilation operator
D=12(X⋅P+P⋅X).D = \frac12 \left( \mathbf X\cdot\mathbf P +\mathbf P\cdot\mathbf X \right).
Solution

Under parity, both X\mathbf X and P\mathbf P reverse. Each product therefore remains unchanged:

ΠDΠ−1=D.\Pi D\Pi^{-1}=D.

The operator is parity even.

Under time reversal, X\mathbf X is even and P\mathbf P is odd. The coefficient 1/21/2 is real, so antiunitarity introduces no additional sign. Both ordered products reverse:

ΘDΘ−1=−D.\Theta D\Theta^{-1}=-D.

Thus DD is time-reversal odd. A real term proportional to DD can preserve parity while breaking time reversal.

  1. Let a spinless two-state Hamiltonian be
H=aI+bxσx+byσy+bzσz,H = aI +b_x\sigma_x +b_y\sigma_y +b_z\sigma_z,

with real coefficients and time reversal represented by KK. Which coefficients are constrained by time-reversal invariance?

Solution

In the standard Pauli basis, II, σx\sigma_x, and σz\sigma_z are real, while σy\sigma_y is imaginary. Therefore

KHK−1=aI+bxσx−byσy+bzσz.KHK^{-1} = aI +b_x\sigma_x - b_y\sigma_y +b_z\sigma_z.

The condition KHK−1=HKHK^{-1}=H requires

by=0.b_y=0.

The real coefficients aa, bxb_x, and bzb_z remain unconstrained by this symmetry. No Kramers degeneracy follows because K2=+IK^2=+I.

  1. Let Θ\Theta be antiunitary with Θ2=−I\Theta^2=-I, and suppose
ΘHΘ−1=H.\Theta H\Theta^{-1}=H.

Show that a normalizable energy eigenstate and its time-reversed partner have the same energy and are orthogonal.

Solution

If

H∣ψ⟩=E∣ψ⟩,H|\psi\rangle=E|\psi\rangle,

then

HΘ∣ψ⟩=ΘH∣ψ⟩=Θ(E∣ψ⟩)=E∗Θ∣ψ⟩.\begin{aligned} H\Theta|\psi\rangle &= \Theta H|\psi\rangle \\ &= \Theta\left( E|\psi\rangle \right) \\ &= E^*\Theta|\psi\rangle. \end{aligned}

Self-adjointness makes EE real, so the partner has the same energy.

Let

c=⟨ψ∣Θψ⟩.c=\langle\psi|\Theta\psi\rangle.

Antiunitarity gives

⟨Θψ∣Θ2ψ⟩=c∗.\langle\Theta\psi|\Theta^2\psi\rangle = c^*.

Using Θ2=−I\Theta^2=-I, the left side is also

−⟨Θψ∣ψ⟩=−c∗.-\langle\Theta\psi|\psi\rangle = -c^*.

Hence c∗=−c∗c^*=-c^* and c=0c=0. The two states are orthogonal and form a Kramers pair.

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