Irreducible representations and Hilbert spaces

Question

I am reading Howard Georgi's book "Lie Algebras in Particle Physics" where he writes the following (chapter 1.14:eigenstates):

"... if some irreducible representation appears only once in the Hilbert space, then the states in that representation must be eigenstates of $H$ (and any other invariant operator)."

The irreducible representation here, as far as I can tell, is meant to be part of a representation $D(g)$ on the full Hilbert space and we assume $H$ to commute with $D(g): [H, D(g)] = 0.$

My question is: what is meant by "appearing only once" in the Hilbert space? Does it mean that, when I write the full representation D(g) as a direct sum of irreps, it appears only once in this direct sum?

To motivate why I think this is the case: in this work explaining Schur's Lemma it is stated that, if the Hamiltonian commutes with $D(g) = \begin{pmatrix}\pi(g) & 0 \\ 0 & \pi(g) \end{pmatrix}$ where $\pi(g)$ is an irrep, then Schur's lemma does not apply but we can say that $H = \begin{pmatrix}A \mathbb{I} & B \mathbb{I} \\C\mathbb{I} & D\mathbb{I} \end{pmatrix}$.

So my questions are: 1) is my assumption correct? and 2) can you point me to an example for the two different cases (an irrep appearing once and more than once) that may potentially clarify my confusion?

Answer 1

If we can decompose
$$ {\mathcal H}=\bigoplus_{{\rm irreps}\, J} {\mathcal H}_J $$ into $\hat H$-invariant irreps of $G$ then Schur's lemma tells us that in each ${\mathcal H}_J$ the hamiltonian $\hat H$ will act as a multiple of the identity operator. In other words every state in ${\mathcal H}_J$ will be an eigenstate of $\hat H$ with a common energy $E_J$.

If an irrep $J$ occurs only once in the decomposition of ${\mathcal H}$ then it is automatically an $H$ invariant subspace and we can find the eigenstates directly by applying projection to vectors in the total Hilbert space ${\mathcal H}$. If the irrep occurs $n_J$ times in the decomposition, then we can project onto the reducible subspace $$ \underbrace{{\mathcal H}_J\oplus {\mathcal H}_J\oplus\cdots {\mathcal H}_J}_{n_J\, {\rm copies}}={\mathcal M}\otimes {\mathcal H}_J. $$ Here ${\mathcal M}$ is an $n_J$ dimensional multiplicity space. The hamiltonian $\hat H$ will act in ${\mathcal M}$ as an $n_J$-by-$n_J$ matrix. In other words, if the vectors $$ |n,i\rangle \equiv |n\rangle \otimes |i \rangle \in {\mathcal M}\otimes {\mathcal H}_J $$ form a basis for ${\mathcal M}\otimes {\mathcal H}_J$, with $n$ labelling which copy of ${\mathcal H}_J$ the vector $|n,i\rangle $ lies in, then the hamiltonian and the symmetry group act as $$ \hat H |n,i\rangle = |m,i\rangle H^J_{mn},\nonumber\\ D(g)|n,i\rangle = |n,j\rangle D^J_{ji}(g), $$ where $D^J_{ji}(g)$ is the representation matrix in representation $J$. Diagonalizing $H^J_{nm}$ provides us with $n_j$ $\hat H$-invariant copies of ${\mathcal H}_J$ and gives us the energy eigenstates.