I heard this statement in one lecture.
Consider a first quantized Hamiltonian $H$ on a single-particle Hilbert space (of finite dimension $N$). If the Hamiltonian possess symmetries that is unitarily represented, then we can bring the Hamiltonian into a block diagonal form with each block $H^{(\lambda)}$ labelled by the irreducible (unitary) representation $\lambda$ of its symmetry group $G_0$. These irreducible blocks do not exhibit the unitary symmetries.
This seems to be an elementary fact that most papers do not give reference for it. Can anyone point out a proof of it?
Note
The lecturer also gave the following exact statement of his claim:
Suppose we have a Hamiltonian $H$ on single-particle Hilbert space (of finite dimension $N$). Assume its group of symmetry is $G_0$. Then the space $\mathcal{V}$ of single-particle Hilbert space, decomposes into a direct sum of vector spaces $\mathcal{V}_\lambda$ associated with the irrep (irreducible representations, labeled by $\lambda$) of $G_0$.
\begin{equation} \mathcal{V} = \oplus_\lambda \mathcal{V}_\lambda \end{equation} Let $m_\lambda$ denotes the multiplicity of $\lambda$th irrep. Denote the dimension of each irrep as $d_\lambda$.
In each vector space $\mathcal{V}_\lambda$, one can choose a (orthogonal) basis of the form: \begin{equation} |v^{(\lambda)}_\alpha\rangle \otimes |w^{(\lambda)}_k\rangle \end{equation} where
- $G_0$ acts only only $|w^{(\lambda)}_k\rangle$, $k=1,\cdots,d_\lambda$,
- $H$ acts only on $|v^{(\lambda)}_\alpha\rangle$, $\alpha=1,\cdots,m_\lambda$.
If you have difficulty understanding @ACuriousMind's answer, please read my comments in that answer for a concrete example.