Discrete Symmetries in Hamiltonians

Discrete symmetries constrain a Hamiltonian by ruling out terms that transform with the wrong sign or the wrong complex structure. The basic test is always the same:

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

but the details depend on whether $S$ is unitary or antiunitary, and on whether external fields are transformed as part of the physical comparison.

Parity is unitary. Time reversal is antiunitary. That single difference is responsible for many common sign mistakes.

Fixed Hamiltonian Versus Covariant Family

There are two related but different questions:

• Is one fixed Hamiltonian invariant?
• Is a family of Hamiltonians covariant when external parameters are transformed?

For a Hamiltonian $H(\lambda)$ depending on external parameters $\lambda$, the covariant statement is

$$S H(\lambda) S^{-1} = H(S\lambda).$$

The fixed-Hamiltonian symmetry statement is stronger:

$$S H(\lambda) S^{-1} = H(\lambda).$$

For example, a magnetic field is time-reversal odd. A spin Hamiltonian in a field may satisfy

$$\Theta H(\mathbf B)\Theta^{-1} = H(-\mathbf B),$$

but a nonzero fixed Zeeman coupling is not invariant under ordinary time reversal. A combined operation $U\Theta$ can sometimes preserve it if a spatial or internal $U$ reverses the transformed coupling again. That combined symmetry is distinct from $\Theta$.

Transformation Table

For the standard nonrelativistic variables:

Quantity	Parity	Time reversal
$\mathbf X$	$-\mathbf X$	$\mathbf X$
$\mathbf P$	$-\mathbf P$	$-\mathbf P$
$\mathbf L=\mathbf X\times\mathbf P$	$\mathbf L$	$-\mathbf L$
$\mathbf S$	$\mathbf S$	$-\mathbf S$
$i$	$i$	$-i$
$\mathbf E$ as an external field	$-\mathbf E$	$\mathbf E$
$\mathbf B$ as an external field	$\mathbf B$	$-\mathbf B$

The rows for $\mathbf E$ and $\mathbf B$ are about transforming the physical external field. If the field is held fixed, a Hamiltonian can fail to be invariant even when the family is covariant.

Spinless Scalar Hamiltonians

For

$$H = \frac{P^2}{2m}+V(X)$$

in one dimension:

• parity requires $V(-x)=V(x)$;
• spinless time reversal requires a real Hamiltonian in the position representation.

Provided the self-adjoint domain also preserves inversion and conjugation, a real even potential has both symmetries. An asymmetric boundary or a complex boundary twist can invalidate the corresponding conclusion even when the differential expression looks symmetric. A real but asymmetric potential preserves spinless time reversal but breaks parity. A complex absorbing potential is not a closed self-adjoint Hamiltonian and should not be treated as an ordinary time-reversal-invariant quantum Hamiltonian.

The spinless time-reversal representation is Time Reversal for Spinless Particles.

Finite-Dimensional Matrix Test

If time reversal is represented as

$$\Theta=UK,$$

where $K$ complex-conjugates a chosen basis, then the Hamiltonian condition is

$$U H^* U^\dagger = H.$$

For spinless time reversal, $U=I$, so

$$H^*=H.$$

Together with Hermiticity, this means $H$ is real symmetric in that basis.

For a two-state Hamiltonian with spinless time reversal,

$$H = aI+b_x\sigma_x+b_y\sigma_y+b_z\sigma_z,$$

the $\sigma_y$ term is forbidden because $\sigma_y$ is imaginary:

$$K\sigma_yK^{-1}=-\sigma_y.$$

Thus

$$b_y=0.$$

This is a basis-dependent matrix form of the invariant statement that the Hamiltonian commutes with the chosen antiunitary symmetry.

Single Spin-One-Half

For a single spin-$1/2$ degree of freedom,

$$\Theta=-i\sigma_yK, \qquad \Theta^2=-I.$$

A general Hermitian two-level Hamiltonian is

$$H = aI+\mathbf b\cdot\boldsymbol\sigma.$$

Since

$$\Theta\boldsymbol\sigma\Theta^{-1} = -\boldsymbol\sigma,$$

time-reversal invariance with no transformed external time-odd parameter requires

$$\mathbf b=0.$$

Therefore a truly isolated single spin-$1/2$ cannot have a preferred Zeeman axis while preserving time reversal. If the vector $\mathbf b$ is supplied by a magnetic field, the covariant statement is instead

$$\Theta H(\mathbf B)\Theta^{-1} = H(-\mathbf B).$$

The spinor convention is developed in Time Reversal for Spin-1/2 Particles, and the degeneracy consequence is Kramers Degeneracy.

Common Hamiltonian Terms

The following table uses real couplings, Hermitian operators, and spatially uniform external fields held fixed. Position-dependent coefficients may require symmetrizing products with momentum.

Term	Parity	Time reversal	Comment
$P^2/(2m)$	allowed	allowed	kinetic energy
$V(X)$ with $V(-x)=V(x)$	allowed	allowed if real	even scalar potential
$\lambda X$ with fixed $\lambda$	breaks parity	allowed if real	uniform electric-force model with fixed direction
$-\mathbf d\cdot\mathbf E$	covariant if $\mathbf E\mapsto-\mathbf E$	allowed if $\mathbf E$ fixed	fixed electric field selects a parity-breaking direction
$-\gamma\mathbf S\cdot\mathbf B$	allowed for axial $\mathbf B$	breaks time reversal if $\mathbf B$ fixed	covariant if $\mathbf B\mapsto-\mathbf B$
$\mathbf L\cdot\mathbf S$	allowed	allowed	both $\mathbf L$ and $\mathbf S$ are time-reversal odd
$\mathbf E\cdot(\mathbf S\times\mathbf P)$	odd for fixed nonzero $\mathbf E$	even	spin and momentum both reverse under time reversal

The table is a symmetry filter, not a derivation of coefficients. Dynamics, approximations, and microscopic modeling decide which allowed terms actually appear and how large their coefficients are.

A Spin-Dependent Term That Preserves Time Reversal

For a particle moving in the $xy$ plane, consider

$$H(\mathbf p)=\frac{p_x^2+p_y^2}{2m}I +\alpha(\sigma_xp_y-\sigma_yp_x),$$

with real $\alpha$ in velocity units. Time reversal flips both spin and momentum, leaving the second term invariant. Inversion flips momentum but not spin, so it reverses that term when the inversion-breaking environment fixing $\alpha$ is held fixed. The energies are

$$E_\pm(\mathbf p)=\frac{p^2}{2m}\pm|\alpha|\,|\mathbf p|.$$

At nonzero momentum the two branches split, but time reversal relates a state at $\mathbf p$ to a partner at $-\mathbf p$ with the same energy. The two spin states are degenerate at $\mathbf p=0$. Thus time-reversal invariance does not prohibit spin-dependent dispersion; it specifies which states are paired.

Central Spin–Orbit Coupling

The standard scalar spin–orbit term

$$H_{LS} = \xi(r)\,\mathbf L\cdot\mathbf S$$

is parity even because both $\mathbf L$ and $\mathbf S$ are axial vectors. It is time-reversal even because both factors change sign under time reversal:

$$\Theta(\mathbf L\cdot\mathbf S)\Theta^{-1} = (-\mathbf L)\cdot(-\mathbf S) = \mathbf L\cdot\mathbf S.$$

Thus spin–orbit coupling by itself is not a time-reversal-breaking term. It can split angular-momentum multiplets, but in a $\Theta^2=-I$ sector it still leaves Kramers pairs when the Hamiltonian is otherwise time-reversal invariant.

The angular-momentum treatment is Spin–Orbit Coupling.

Minimal Coupling and Magnetic Fields

For a charged spinless particle,

$$H(\mathbf A,\Phi) = \frac{1}{2m} \left(\mathbf P-q\mathbf A(\mathbf X)\right)^2 +q\Phi(\mathbf X).$$

Under time reversal,

$$\Theta H(\mathbf A,\Phi)\Theta^{-1} = H(-\mathbf A,\Phi),$$

up to gauge convention. Since

$$\mathbf B=\nabla\times\mathbf A,$$

this corresponds to $\mathbf B\mapsto-\mathbf B$. A fixed magnetic field generally breaks ordinary time reversal. For a multiply connected region, flux must be compared modulo gauge-equivalent flux quanta: integer and half-integer flux quanta on a ring are special cases where the reversed problem is equivalent. See the boundary-condition example.

Degeneracy and Labels

If parity is a symmetry,

$$[H,\Pi]=0,$$

then energy eigenstates can be organized into even and odd sectors, at least within each degenerate eigenspace. Parity symmetry by itself does not force degeneracy.

If time reversal is a symmetry and $\Theta^2=+I$, degeneracy is not generally forced either. Spinless real Hamiltonians can have nondegenerate eigenstates.

If time reversal is a symmetry and $\Theta^2=-I$ on the relevant sector, Kramers degeneracy applies: normalizable energy levels occur in orthogonal pairs. The sign of the antiunitary square decides whether this pairing is compulsory.

Common Mistakes

• Testing a Hamiltonian with a fixed external field while using transformation rules for the whole field family.
• Treating time reversal as unitary when applying it to matrix entries or factors of $i$.
• Saying a term is forbidden because it is odd under one variable while forgetting that an external parameter may also transform.
• Treating spin–orbit coupling as time-reversal breaking by itself.
• Assuming parity symmetry forces degeneracy.
• Assuming time reversal always gives Kramers degeneracy; the sign of $\Theta^2$ matters.
• Calling a non-Hermitian effective potential an ordinary Hamiltonian symmetry problem without stating the open-system approximation.

Exercises

1. Let

$$H = \frac{P^2}{2m} +\frac{1}{2}m\omega^2X^2 +\lambda X.$$

Assume $\lambda$ is a fixed real parameter. Which of parity and spinless time reversal are symmetries?

Solution

The linear term changes sign under parity:

$$\Pi X\Pi^{-1}=-X,$$

so fixed nonzero $\lambda$ breaks inversion about $x=0$. Spinless time reversal is preserved on the usual full-line domain. Completing the square gives

$$H=\frac{P^2}{2m}+\frac12m\omega^2(X-x_0)^2- \frac{\lambda^2}{2m\omega^2},\qquad x_0=-\frac{\lambda}{m\omega^2}.$$

It still has reflection symmetry about the shifted center $x_0$. Thus specifying the inversion center is part of specifying parity.

2. For

$$H=aI+b_x\sigma_x+b_y\sigma_y+b_z\sigma_z,$$

impose spinless time reversal $KHK^{-1}=H$. Which coefficient is forbidden?

Solution

$I$, $\sigma_x$, and $\sigma_z$ are real matrices, while $\sigma_y$ is imaginary. Therefore

$$K\sigma_yK^{-1}=-\sigma_y.$$

The condition $KHK^{-1}=H$ forces $b_y=0$.

3. Explain why $\mathbf L\cdot\mathbf S$ is time-reversal even.

Solution

Time reversal sends both orbital and spin angular momentum to their negatives:

$$\Theta\mathbf L\Theta^{-1}=-\mathbf L, \qquad \Theta\mathbf S\Theta^{-1}=-\mathbf S.$$

Therefore

$$\Theta(\mathbf L\cdot\mathbf S)\Theta^{-1} = (-\mathbf L)\cdot(-\mathbf S) = \mathbf L\cdot\mathbf S.$$

So the scalar spin–orbit term does not break time reversal by itself.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• M. S. Dresselhaus, G. Dresselhaus, and A. Jorio, Group Theory: Application to the Physics of Condensed Matter, Springer, 2008.
• 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.
• M. Tinkham, Group Theory and Quantum Mechanics, Dover, 2003.