Symmetry Constraints on Hamiltonians

Symmetry is a filter on Hamiltonians. Instead of writing every operator allowed by dimension or convenience, one writes only terms compatible with the stated symmetries.

The basic criterion is:

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

For a unitary symmetry $U$, this is $UHU^\dagger=H$. For a continuous unitary symmetry generated by $G$, the infinitesimal condition is

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

Constraining the operator terms

To constrain a Hamiltonian by symmetry:

1. Choose the Hilbert space and degrees of freedom.
2. State how the symmetry acts on states and operators.
3. Write a general Hamiltonian in the relevant operator basis.
4. Impose $SHS^{-1}=H$ for each symmetry.
5. Remove terms whose coefficients must vanish or combine terms whose coefficients must be equal.
6. Interpret the remaining parameters physically.

The method constrains the spectrum before it is solved. Include boundary conditions and operator domains in the specification of $H$: a symmetric differential expression need not define a symmetric physical problem. For unbounded generators, use invariance under the unitary group rather than relying only on a formal commutator.

Spin-1/2 Example

Every Hermitian Hamiltonian for a two-level system can be written

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

with real $a$ and real vector $\mathbf b$. If there is no external vector selecting a direction and the Hamiltonian must be invariant under all spin rotations, then the vector term is forbidden:

$$H=aI.$$

If an external magnetic field $\mathbf B$ is present, a term

$$H=-\gamma\,\mathbf B\cdot\mathbf S$$

is covariant if spin and field are transformed together. It is not invariant under every spin rotation at a fixed field. If $\mathbf B$ is fixed along $z$, the Hamiltonian no longer has full rotational symmetry; it retains rotations about the $z$ axis.

Parity in One Dimension

For

$$H=\frac{p^2}{2m}+V(x),$$

parity acts as

$$PxP^{-1}=-x, \qquad PpP^{-1}=-p.$$

The kinetic term is invariant. The potential is invariant only when

$$V(-x)=V(x).$$

On the full line with a parity-invariant domain, parity symmetry therefore constrains the potential to be even. Walls or boundary conditions must be reflected as well when testing a different geometry.

Translation Invariance

Spatial translations are generated by momentum:

$$U(a)=e^{-iaP/\hbar}.$$

For a Hamiltonian

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

continuous translation invariance on the full line requires

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

Since

$$[P,V(X)]=-i\hbar V'(X),$$

a differentiable potential must be constant on the full line. Discrete translation symmetry is weaker: $V(x+a_0)=V(x)$ permits a nonconstant periodic potential. This is why the free particle is translation invariant and a generic potential is not. The focused version of this test is Translation-Invariant Hamiltonians.

Central Potentials

For a particle in three dimensions,

$$H=\frac{\mathbf p^2}{2m}+V(r), \qquad r=\lvert\mathbf r\rvert,$$

the Hamiltonian is rotationally invariant if its domain has the same symmetry. Therefore

$$[H,L_i]=0$$

for each component of orbital angular momentum. Energy eigenstates can be organized using angular momentum quantum numbers, even when the radial equation is still nontrivial.

Symmetry Breaking Terms

Adding a small term can break a symmetry and split degeneracies. For example,

$$H=aI+b_z\sigma_z$$

with $b_z\ne0$ selects the $z$ direction. It commutes with $\sigma_z$ but not with $\sigma_x$ or $\sigma_y$. The symmetry has been reduced from full spin rotation to rotations about the $z$ axis.

This is not a failure of symmetry reasoning. It is one of its main uses: identifying what a perturbation breaks and which quantum numbers remain good.

An antiunitary constraint depends on the physical action

For the same two-dimensional matrix space, $K$ and $\Theta=-i\sigma_yK$ impose different constraints. In the $\sigma_z$ basis, $K$ fixes $\sigma_x,\sigma_z$ and reverses $\sigma_y$, so $KHK^{-1}=H$ requires only $b_y=0$. Physical spin-$1/2$ time reversal reverses all three Pauli matrices, so $\Theta H\Theta^{-1}=H$ requires $\mathbf b=0$. Merely saying “time reversal” without specifying how the degrees of freedom transform is insufficient.

Exercises

1. Which terms in $H=aI+b_x\sigma_x+b_y\sigma_y+b_z\sigma_z$ commute with $\sigma_z$?

Solution

The terms $aI$ and $b_z\sigma_z$ commute with $\sigma_z$. The Pauli commutators give

$$[\sigma_z,\sigma_x]=2i\sigma_y, \qquad [\sigma_z,\sigma_y]=-2i\sigma_x,$$

so nonzero $b_x$ or $b_y$ breaks the symmetry generated by $\sigma_z$.

2. For a particle on the full line, let $H=p^2/(2m)+V(x)$ with differentiable $V$ and suppose all spatial translations are symmetries. What does this imply about $V$? Why does a constant potential inside a finite box not imply the same symmetry?

Solution

Since $[P,p^2/(2m)]=0$ and $[P,V(X)]=-i\hbar V'(X)$, the condition $[P,H]=0$ implies $V'(X)=0$ as an operator on the interval. Thus $V(x)$ is constant there. In a finite box, arbitrary translations move the walls and do not preserve the Hamiltonian domain. The local commutator calculation inside the box does not establish a symmetry of that boundary-value problem.

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, chapter 4.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• M. Tinkham, Group Theory and Quantum Mechanics, Dover, 2003 reprint of the 1964 edition.