Skip to main content

Section 2.3 Unitary representations

One type of representations is particularly important in physics: unitary representations. Those are widely used in quantum mechanics. So studying unitary representations in general allows physicists to learn lots of interesting things about quantum systems. You can look at these slides by Peter Woit if you want to know more about this, but we will come back to this later on in this course.

Mathematically, unitary representations are also generally easier to handle than general representations: they satisfy nice properties not shared by general representations. It turns out that for finite groups, all representations are equivalent to unitary representations. This is not so simple for infinite groups though.

Subsection 2.3.1 Review of linear algebra

Let us start by reviewing a bit of linear algebra.

Definition 2.3.1. Adjoint of a matrix.

Let \(A \in GL(n,\mathbb{C})\) be an \(n \times n\) matrix with complex-valued entries. We write \(A^*\) for its complex conjugate, obtained by complex conjugating the entries. We write \(A^\dagger = (A^*)^T\) for the transpose of the complex conjugated matrix. \(A^\dagger\) is called the adjoint (or Hermitian conjugate).

Remark 2.3.2.

Note that in the literature, particularly in the mathematics literature, the notation \(A^*\) is often use to denote the adjoint \(A^\dagger\text{.}\)

Definition 2.3.3. Hermitian matrices.

A matrix \(A \in GL(n,\mathbb{C})\) is Hermitian if \(A = A^\dagger\text{.}\)

Hermitian matrices are also all over the place in quantum mechanics, since observables are generally understood in terms of Hermitian operators on the Hilbert space of states. The eigenvalues of these operators correspond to the possible values of the particular observable variables represented by the operators. This interpretation makes sense because of the following fundamental result in linear algebra:

Another type of complex-valued matrices is important:

Definition 2.3.5. Unitary matrices.

We say that a matrix \(A \in GL(n,\mathbb{C})\) is unitary if \(A A^\dagger = I\text{,}\) that is, \(A^\dagger = A^{-1}\text{.}\)

Unitary matrices preserve the inner product on a complex vector space, just like orthogonal matrices preserve the scalar product on real vector spaces. Given two vectors \(u,v \in \mathbb{C}^n\text{,}\) the standard inner product is defined as \(\langle u,v \rangle= u^\dagger v\text{.}\) Then a unitary transformation \(A\) preserves the inner product, since

\begin{equation*} \langle A u, A v \rangle = u^\dagger A^\dagger A v = u^\dagger v = \langle u, v \rangle. \end{equation*}

This is why unitary operators are important in quantum mechanics: they can be used to do changes of bases while preserving orthogonality between basis vectors according to the inner product. They behave just like rotations but in a complex vector space.

Another nice property of unitary matrices concerns their eigenvalues. While eigenvalues of Hermitian matrices are real, eigenvalues of unitary matrices are generally complex, but of a very particular form: they must lie on the unit circle. In other words, they must have modulus 1. This means that they can be written as \(e^{i \theta}\) for some angle \(\theta \in [0,2 \pi)\text{.}\)

It also turns out that unitary and Hermitian matrices are closely related. Indeed, the theorem about eigenvalues of Hermitian matrices above relies on the fundamental statement that any Hermitian matrix can be diagonalized by an appropriate unitary transformation. That is, for any Hermitian matrix \(H\text{,}\) there exists a unitary matrix \(U\) such that

\begin{equation*} D = U^\dagger H U \end{equation*}

is a diagonal matrix.

Prove that any Hermitian matrix can be diagonalized by an appropriate unitary transformation.

Subsection 2.3.2 The unitarity theorem

Let us now go back to representations.

Definition 2.3.7. Unitary representation.

If all matrices \(T(g)\) of a representation of a group \(G\) are unitary, we say that the representation is unitary.

Unitary representations may seem like very special and constrained. Which raises a question: how common are unitary representations?

Let us consider first one-dimensional representations of a finite group \(G\text{.}\) To any element \(g \in G\text{,}\) a one-dimensional representation \(T\) associates a complex number \(T(g) = r e^{i \theta}\text{.}\) (This is the polar form of a non-zero complex number, where \(r>0\) is the modulus, and \(\theta \in [0,2 \pi) \text{.}\)) Since \(G\) is finite, we know that \(g\) has finite order, and hence there must exist an integer \(n\) such that \(g^n = e\text{.}\) That is, \(T(g^n) = r^n e^{i n \theta} = T(e) = 1\text{.}\) This is only possible if \(r = 1\text{.}\) In other words, what this means is that \(T(g)\) has modulus one, and hence is a one-dimensional unitary matrix, since \(T^\dagger(g) T(g) = r^2 = 1\text{.}\) Therefore, all one-dimensional representations of finite groups are unitary. Note that it was key in the argument that the group is finite.

Does there exist a similar statement for higher-dimensional representations of finite groups? This is the content of the important unitarity theorem, the main result that we prove in this section:

Let \(G\) be a finite group, and \(T: G \to GL(V)\) be an \(n\)-dimensional representation. The proof is constructive: for any such \(T\text{,}\) we will construct an equivalent representation that is explicitly unitary.

To this end, let us first introduce the \(n \times n\) matrix:

\begin{equation*} H = \sum_{g \in G} T^\dagger(g) T(g). \end{equation*}

This matrix is interesting. For instance, it satisfies the following invariance property. For any \(g' \in G\text{,}\)

\begin{align} T^\dagger(g') H T(g') =\amp \sum_{g \in G} T^\dagger (g') T^\dagger(g) T(g) T(g')\notag\\ =\amp \sum_{g \in G} ( T(g) T(g') )^\dagger T(g) T(g') \notag\\ =\amp \sum_{g \in G} T(g g')^\dagger T(g g')\notag\\ =\amp H.\label{equation-invariance-unitarity}\tag{2.3.1} \end{align}

The last line follows because of the Rearrangement theorem 1.4.2, since multiplying the group elements \(g \in G\) by a fixed element \(g'\) only rearranges the terms in the sum. This invariance property will be useful later on.

Now notice that \(H\) is Hermitian, since \(H^\dagger = H\text{.}\) Thus we know that it has real eigenvalues, and that is can be diagonalized by a unitary matrix. Hence we can write \(D = U^\dagger H U\text{,}\) where \(D\) is a diagonal matrix with real entries and \(U\) is unitary. Further, we now show that \(D\) has real positive entries. We have:

\begin{gather*} D = U^\dagger H U\\ = \sum_{g \in G} U^\dagger T^\dagger(g) T(g) U\\ = \sum_{g \in G} A^\dagger(g) A(g), \end{gather*}

where we defined \(A(g) = T(g) U\text{.}\) Now consider the \(j\)'th diagonal entry \(D_{jj}\text{.}\) It is given by summing over \(g \in G\) the contributions given by \(A_j^\dagger(g)A_j(g)\text{,}\) where \(A_j(g)\) denotes the \(j\)'th column vector in \(A(g)\text{.}\) Since for each \(g\in G\) and each \(j\text{,}\) \(A_j(g)\) is a non-zero vector, then \(A_j^\dagger(g) A_j(g) > 0\text{,}\) and hence \(D_{jj}>0\text{.}\)

We then define the diagonal matrix \(D^{1/2}\) whose entries are the square roots of the entries of \(D\text{,}\) and \(D^{-1/2}\) as the inverse of \(D^{1/2}\text{.}\) We now form the matrices \(B(g) = D^{1/2} U^\dagger T(g) U D^{-1/2}\text{,}\) and their adjoints \(B^\dagger(g) = D^{-1/2} U^\dagger T^\dagger(g) U D^{1/2}\text{.}\) Why are we constructing these matrices? Note that, since \(U\) is unitary, and hence \(U^\dagger = U^{-1}\text{,}\) the transformation \(B(g) = D^{1/2} U^\dagger T(g) U D^{-1/2}\) is a similarity transformation. Thus the representations furnished by the matrices \(B(g)\) and the \(T(g)\) are equivalent. Our goal is to show that the new representation \(B(g)\) is explicitly unitary, which would prove the theorem, namely that any representation of a finite group is equivalent to a unitary representation.

So let us show that the matrices \(B(g)\) are unitary. We have:

\begin{align*} B^\dagger (g) B(g) =\amp D^{-1/2} U^\dagger T^\dagger(g) U D^{1/2} D^{1/2} U^\dagger T(g) U D^{-1/2}\\ =\amp D^{-1/2} U^\dagger T^\dagger(g) U D U^\dagger T(g) U D^{-1/2}\\ =\amp D^{-1/2} U^\dagger T^\dagger(g) H T(g) U D^{-1/2}\\ =\amp D^{-1/2} U^\dagger H U D^{-1/2}\\ =\amp D^{-1/2} D D^{-1/2}\\ =\amp I, \end{align*}

where we used the invariance property (2.3.1). Thus for any finite-dimensional representation \(T\) of a finite group \(G\text{,}\) we have constructed a new, equivalent unitary representation \(B\text{,}\) given by the set of unitary matrices \(B(g) = D^{1/2} U^\dagger T(g) U D^{-1/2}\text{.}\) We have thus proved that all finite-dimensional representations of finite groups are equivalent to unitary representations.

Remark 2.3.9.

Note here that the requirement of having a finite group \(G\) was crucial in the proof. Otherwise, the expression

\begin{equation*} H = \sum_{g \in G} T^\dagger(g) T(g) \end{equation*}

doesn't even make sense, since the sum would be over an infinite-dimensional set (or a continuous space if the group is continuous).

Remark 2.3.10.

In view of Remark 2.3.9, we may ask: is the unitarity theorem still true for infinite groups, either discrete or continuous? Consider for example the infinite continuous group \((\mathbb{R},+)\text{,}\) and the two-dimensional representation:

\begin{equation*} T(u) = \begin{pmatrix} 1 \amp 0 \\ u \amp 1 \end{pmatrix}, \qquad u \in \mathbb{R} \end{equation*}

that we encountered in Example 2.1.9. We have:

\begin{align*} T^\dagger (u) T(u) =\amp \begin{pmatrix} 1 \amp u \\ 0 \amp 1 \end{pmatrix} \begin{pmatrix} 1 \amp 0 \\ u \amp 1 \end{pmatrix}\\ =\amp \begin{pmatrix} 1 + u^2 \amp u \\ u \amp 1 \end{pmatrix}, \end{align*}

which is not the identity for non-zero \(u \in \mathbb{R}\text{.}\) So this is not a unitary transformation, and one can show that it cannot be brought into a unitary transformation by a similarity transformation. In fact, the same representation restricted to \(u \in \mathbb{Z}\) is also not unitary for the infinite but discrete group \((\mathbb{Z}, +)\text{.}\)

So for what kind of groups, beyond finite groups, is the unitarity theorem true? The general result is that the theorem holds if the group is compact. To define compact groups, we first need to define the concept of topological groups. Topological groups are groups \(G\) that are given the extra structure of a topology on \(G\text{,}\) such that the group's binary operation and function mapping elements to their inverses are continuous with respect to this topology. Then, a compact group is a topological group such that its topology is compact.

In the end, they key point is that because of the existence of this compact topology on \(G\text{,}\) for compact groups we can “replace” the sum over elements of the group by an appropriate integral over the continuous group with respect to some measure (the Haar measure), and the resulting integral then converges. Using this the proof above goes through with minor modifications, and the unitarity theorem holds for compact groups. Examples of compact groups include \(SO(n)\) and \(SU(n)\text{,}\) which appear frequently in physics. Note however that the Lorentz group is not compact.

Note that not only the unitarity theorem holds for compact groups, but Theorem 2.2.4 also applies to compact groups: all irreducible representations of compact groups are finite-dimensional.

login