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 the fixed Hamiltonian $H(\mathbf B)$ is not time-reversal invariant unless $\mathbf B=0$ or another symmetry identifies $\mathbf B$ with $-\mathbf B$.

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.

Thus a real even potential has both parity and time-reversal symmetry. 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 assumes the external fields are fixed unless stated otherwise.

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
$\mathbf E\cdot(\mathbf S\times\mathbf P)$	depends on $\mathbf E$	often allowed with fixed $\mathbf E$	prototype spin–orbit-like structure in inversion-breaking systems

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.

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 flux or magnetic field is therefore a standard way to break time reversal.

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 square of the antiunitary symmetry is therefore a Hamiltonian-level input, not a decorative detail.

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.

Cross-Links

• Symmetry Constraints on Hamiltonians
• Parity
• Time Reversal
• Antiunitary Time Reversal
• Time Reversal for Spinless Particles
• Time Reversal for Spin-1/2 Particles
• Kramers Degeneracy
• Spin in Magnetic Fields
• Spin–Orbit Coupling
• Parity Selection Rules
• Symmetry Classification Preview

References

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

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 parity. The Hamiltonian is real in the position representation and contains only $P^2$, not $P$, so spinless time reversal is preserved.

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.