Unitary Symmetries

A unitary symmetry is a symmetry represented on Hilbert space by a unitary operator $U$. The defining operator identity is

$$U^\dagger U=UU^\dagger=I.$$

Translations, rotations, parity, and internal phase transformations have unitary implementations. Whether any one is a symmetry of a specific Hamiltonian requires a further test. Assigning operators $U(g)$ that obey group multiplication gives a unitary representation; multiplication only up to phase gives a projective representation.

Preservation of Inner Products

A unitary operator preserves inner products:

$$\langle U\phi|U\psi\rangle = \langle\phi|\psi\rangle.$$

It therefore preserves norms and transition probabilities:

$$\lvert\langle U\phi|U\psi\rangle\rvert^2 = \lvert\langle\phi|\psi\rangle\rvert^2.$$

This is why unitary operators are natural representatives of quantum symmetries on state vectors.

State and Operator Transformations

Under an active unitary transformation,

$$\lvert\psi\rangle\longmapsto U\lvert\psi\rangle.$$

Operators transform as

$$A\longmapsto UAU^\dagger.$$

If both states and operators are transformed consistently, expectation values are unchanged:

$$\langle U\psi|UAU^\dagger|U\psi\rangle = \langle\psi|A|\psi\rangle.$$

This invariance is a statement about representing the same physical relation after the transformation.

Symmetry of a Hamiltonian

A unitary transformation is a symmetry of a Hamiltonian when

$$UHU^\dagger=H.$$

Equivalently,

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

For a time-independent $H$ and a time-independent $U$ preserving its domain, this gives $Ue^{-iHt/\hbar}=e^{-iHt/\hbar}U$. Thus transforming a state and evolving it can be done in either order. If $\lvert E\rangle$ is an energy eigenstate, then $U\lvert E\rangle$ is also an energy eigenstate with the same energy.

Continuous Unitary Symmetries

A one-parameter unitary group that is strongly continuous has the form

$$U(\alpha)=e^{-i\alpha G/\hbar},$$

where $G$ is a self-adjoint generator. In finite dimensions this means a Hermitian matrix; for an unbounded operator it also specifies the domain. If $U(\alpha)$ is a symmetry for all $\alpha$, then

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

The commutator statement holds on appropriate common domains. In the unbounded case, commuting spectral projections give the full symmetry condition; a formal commutator alone is insufficient. For a time-independent generator this is the quantum symmetry-conservation link. Examples include:

• translations generated by momentum,
• rotations generated by angular momentum,
• time translations generated by the Hamiltonian.

Examples

Translations in one dimension:

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

A free-particle Hamiltonian

$$H=\frac{P^2}{2m}$$

commutes with $U(a)$ for all $a$.

Rotations, with their spatial group described by $SO(3)$, are represented by

$$U(\hat{\mathbf n},\theta) = \exp\left( -\frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf J \right).$$

A rotationally invariant Hamiltonian commutes with all components of $\mathbf J$.

Parity in one dimension is unitary and satisfies

$$P^2=I.$$

It is a symmetry when the Hamiltonian is invariant under $x\mapsto -x$.

Exercises

1. Show that a unitary operator preserves transition probabilities.

Solution

Using $U^\dagger U=I$,

$$\langle U\phi|U\psi\rangle = \langle\phi|U^\dagger U|\psi\rangle = \langle\phi|\psi\rangle.$$

Taking the absolute square gives preservation of transition probabilities.

2. If $[U,H]=0$ and $H\lvert E\rangle=E\lvert E\rangle$, show that $U\lvert E\rangle$ is also an energy eigenstate with energy $E$.

Solution

Since $HU=UH$,

$$H(U\lvert E\rangle) = UH\lvert E\rangle = E(U\lvert E\rangle).$$

Thus $U\lvert E\rangle$ is either zero or an eigenstate with the same energy. A unitary operator cannot map a nonzero vector to zero.

3. For a qubit with $H=\Delta\sigma_z/2$ and $\Delta\ne0$, compare $U_z(\theta)=e^{-i\theta\sigma_z/2}$ with $U_x(\pi)=-i\sigma_x$. Which is a symmetry?

Solution

$U_z$ commutes with $\sigma_z$, so $U_zHU_z^\dagger=H$. By contrast, $\sigma_x\sigma_z\sigma_x=-\sigma_z$, so $U_x(\pi)HU_x(\pi)^\dagger=-H$. Both operators preserve inner products, but only the first preserves this Hamiltonian. The second exchanges its two energy levels.

References

• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013, chapters 10 and 16. doi:10.1007/978-1-4614-7116-5.
• 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.