Unitary Representations

A unitary representation is a representation by unitary operators. It is the mathematical form taken by most ordinary quantum symmetries after a Hilbert space has been chosen.

The ordinary representation page explains how a group becomes a family of linear maps. This page adds the Hilbert-space requirement: the maps preserve the inner product, and therefore preserve norms, orthogonality, and transition probabilities.

Definition

Let $G$ be a group and let $\mathcal H$ be a complex Hilbert space. A unitary representation of $G$ on $\mathcal H$ is a homomorphism

$$U:G\to \mathcal U(\mathcal H),$$

where $\mathcal U(\mathcal H)$ is the group of unitary operators on $\mathcal H$. Thus

$$U(gh) = U(g)U(h), \qquad U(e)=I,$$

and each operator satisfies

$$U(g)^\dagger U(g) = U(g)U(g)^\dagger = I.$$

Equivalently, for all $\phi,\psi\in\mathcal H$,

$$\langle U(g)\phi\vert U(g)\psi\rangle = \langle\phi\vert\psi\rangle.$$

The Hilbert space $\mathcal H$ is the representation space. In finite dimension, choosing an orthonormal basis turns the operators $U(g)$ into unitary matrices. In infinite dimension, the basis-independent operator statement is usually the safer object.

Why Unitarity Matters

Quantum probabilities are built from inner products. If a symmetry is represented by $U(g)$, then

$$\lvert\langle U(g)\phi\vert U(g)\psi\rangle\rvert^2 = \lvert\langle\phi\vert\psi\rangle\rvert^2.$$

Therefore a unitary representation preserves transition probabilities. It also preserves normalization:

$$\lVert U(g)\psi\rVert = \lVert\psi\rVert.$$

This is why unitary operators are the default representatives of quantum symmetries. They implement transformations without changing the probabilistic content of state vectors.

Unitarity is stronger than invertibility. An arbitrary representation assigns invertible linear maps. A unitary representation assigns invertible linear maps that also preserve the Hilbert-space geometry.

Examples

The trivial unitary representation sends every group element to $I$.

A finite permutation representation is unitary once the basis vectors are declared orthonormal. If $\sigma\in S_n$ and

$$U(\sigma)e_i = e_{\sigma(i)},$$

then $U(\sigma)$ permutes an orthonormal basis and therefore preserves inner products.

Parity in one-dimensional wave mechanics is represented by

$$(P\psi)(x) = \psi(-x).$$

On $L^2(\mathbb R)$, this preserves the inner product by a change of variables. Since $P^2=I$, it is a unitary representation of a two-element group when parity is a symmetry operation under consideration.

Translations on the line are represented by

$$(U(a)\psi)(x) = \psi(x-a),$$

with group law

$$U(a)U(b) = U(a+b).$$

In generator form,

$$U(a) = \exp \left( -\frac{i}{\hbar}aP \right),$$

where $P$ is the self-adjoint momentum operator under the usual domain assumptions.

Spin rotations give finite-dimensional unitary representations. For spin-$1/2$,

$$U(\hat{\mathbf n},\theta) = \exp \left( -\frac{i}{2}\theta\,\hat{\mathbf n}\cdot\boldsymbol\sigma \right),$$

acting on $\mathbb C^2$.

Invariant Subspaces Split Orthogonally

Unitary representations have a useful structural property. If $W\subset\mathcal H$ is a closed invariant subspace, then its orthogonal complement $W^\perp$ is also invariant.

To see this, take $v\in W^\perp$ and $w\in W$. Since $W$ is invariant under every $U(g)$, it is also invariant under $U(g)^{-1}=U(g^{-1})$. Therefore $U(g)^{-1}w\in W$. Then

$$\langle w\vert U(g)v\rangle = \langle U(g)^{-1}w\vert v\rangle = 0.$$

So $U(g)v$ is orthogonal to every $w\in W$, hence $U(g)v\in W^\perp$.

This means a reducible unitary representation naturally decomposes as an orthogonal direct sum once an invariant subspace is found:

$$\mathcal H = W\oplus W^\perp.$$

That orthogonal splitting is one reason unitary representation theory is so effective in quantum mechanics. It lets symmetry sectors be separated without losing the inner-product structure that carries probabilities.

Finite and Compact Groups

Many representations can be made unitary by choosing a suitable inner product.

For a finite group $G$ and a finite-dimensional complex representation $\rho$, start with any positive-definite inner product $\langle\cdot\vert\cdot\rangle_0$ and average it over the group:

$$\langle v\vert w\rangle_G = \frac{1}{\lvert G\rvert} \sum_{g\in G} \langle \rho(g)v\vert\rho(g)w\rangle_0.$$

This averaged inner product is $G$-invariant, so $\rho(g)$ becomes unitary with respect to it.

For compact Lie groups, the same idea uses normalized Haar measure:

$$\langle v\vert w\rangle_G = \int_G \langle \rho(g)v\vert\rho(g)w\rangle_0\,dg.$$

This averaging principle is the source of many clean decomposition theorems for finite groups and compact groups. It is also why compact rotation groups and their covers have especially tractable angular-momentum representation theory.

The statement does not extend blindly to all groups or all infinite-dimensional settings. Noncompact groups, continuous spectra, and unbounded generators require more care.

Continuous Unitary Representations

For a Lie group, one usually wants a representation that varies continuously or smoothly with the group element. In infinite-dimensional Hilbert spaces, the standard condition is often strong continuity:

$$\lim_{g\to e} \lVert U(g)\psi-\psi\rVert = 0$$

for every $\psi\in\mathcal H$.

For a strongly continuous one-parameter unitary group $U(t)$, Stone’s theorem says that there is a self-adjoint generator $A$ such that

$$U(t) = e^{-itA}.$$

With physical units, time evolution is written

$$U(t) = e^{-itH/\hbar},$$

where $H$ is the Hamiltonian. Spatial translations and rotations have analogous self-adjoint generators: momentum and angular momentum.

Thus continuous unitary representation theory is the mathematical home of the slogan:

Continuous quantum symmetries have self-adjoint generators.

Wigner Theorem Preview

Wigner’s theorem connects unitary representations to the physical meaning of symmetry. A symmetry of pure quantum states is a transformation of rays preserving transition probabilities. Under standard assumptions, such a ray transformation is implemented on Hilbert-space vectors by either a unitary or an antiunitary operator, unique up to phase.

Therefore ordinary continuous symmetries are usually represented by unitary operators, while time reversal and related symmetries may be antiunitary. Because pure states are rays, a group action on physical states may lift to a projective representation on vectors:

$$U(g)U(h) = e^{i\alpha(g,h)}U(gh).$$

For the conceptual quantum statement, see Quantum Symmetries. For the phase ambiguity and spinor consequences, see Projective Representations. For active physical transformations of Hamiltonians and observables, see Unitary Symmetries.

Common Mistakes

• Treating every invertible representation as unitary without specifying an inner product.
• Forgetting that the basis must be orthonormal before matrices can be called unitary.
• Assuming finite-dimensional compact-group intuition applies unchanged to noncompact groups.
• Confusing a unitary representation of a symmetry group with a passive change of basis.
• Ignoring the projective phase freedom of quantum ray representations.
• Treating antiunitary symmetries as ordinary unitary representations.
• Forgetting domain questions for generators in infinite-dimensional Hilbert spaces.

Cross-Links

• Representations
• Unitary Operators
• Inner Products
• Hilbert Spaces
• Lie Groups
• Lie Algebras
• SO(3)
• SU(2)
• SU(2) versus SO(3)
• Heisenberg Group
• Antiunitary Symmetries, First Look
• Tensor Product Representations
• Symmetry Groups and Representations
• Quantum Symmetries
• Unitary Symmetries
• Projective Representations
• Unitary Time Evolution

References

• B. C. Hall, Lie Groups, Lie Algebras, and Representations, 2nd ed., Springer, 2015.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• J.-P. Serre, Linear Representations of Finite Groups, Springer, 1977.
• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.
• V. Bargmann, “On unitary ray representations of continuous groups,” Annals of Mathematics 59, 1–46, 1954.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, Academic Press, 1980.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

Exercises

1. Show that a unitary representation preserves transition probabilities.

Solution

For a unitary representation,

$$\langle U(g)\phi\vert U(g)\psi\rangle = \langle\phi\vert U(g)^\dagger U(g)\psi\rangle = \langle\phi\vert\psi\rangle.$$

Taking absolute squares gives

$$\lvert\langle U(g)\phi\vert U(g)\psi\rangle\rvert^2 = \lvert\langle\phi\vert\psi\rangle\rvert^2.$$

2. Prove that a permutation representation is unitary in the orthonormal basis it permutes.

Solution

Let $\{e_1,\ldots,e_n\}$ be orthonormal and let $U(\sigma)e_i=e_{\sigma(i)}$. Then

$$\langle U(\sigma)e_i\vert U(\sigma)e_j\rangle = \langle e_{\sigma(i)}\vert e_{\sigma(j)}\rangle = \delta_{\sigma(i),\sigma(j)} = \delta_{ij}.$$

Since inner products of basis vectors are preserved, the whole inner product is preserved by linearity.

3. Let $W$ be invariant under a unitary representation. Show that $W^\perp$ is invariant.

Solution

Take $v\in W^\perp$ and $w\in W$. Since $W$ is invariant under $U(g^{-1})$, the vector $U(g^{-1})w$ lies in $W$. Then

$$\langle w\vert U(g)v\rangle = \langle U(g)^{-1}w\vert v\rangle = \langle U(g^{-1})w\vert v\rangle = 0.$$

Thus $U(g)v$ is orthogonal to every vector in $W$, so $U(g)v\in W^\perp$.

4. For a finite group, verify that the averaged inner product is invariant.

Solution

Use

$$\langle v\vert w\rangle_G = \frac{1}{\lvert G\rvert} \sum_{g\in G} \langle \rho(g)v\vert\rho(g)w\rangle_0.$$

For $h\in G$,

$$\langle \rho(h)v\vert \rho(h)w\rangle_G = \frac{1}{\lvert G\rvert} \sum_{g\in G} \langle \rho(g)\rho(h)v\vert\rho(g)\rho(h)w\rangle_0.$$

Since $\rho(g)\rho(h)=\rho(gh)$ and the map $g\mapsto gh$ permutes the elements of $G$, the sum is the same as the original sum. Hence the averaged inner product is invariant.

5. If $U(t)=e^{-itA}$ with $A$ self-adjoint, show formally that $U(t+s)=U(t)U(s)$.

Solution

Because $A$ commutes with itself, exponentials of scalar multiples of $A$ combine:

$$U(t)U(s) = e^{-itA}e^{-isA} = e^{-i(t+s)A} = U(t+s).$$

In finite dimension this follows directly from the power series. In infinite dimension, the statement is made precise through the spectral theorem for self-adjoint operators.