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My question in the title stems from not completely understanding the last three lines of this answer. I list specific questions at the end of this post.

Setup. Consider a quantum system described over a Hilbert space $\mathcal{H}$ and a symmetry group $G$ of the system. Let $D$ be a representation of $G$ carried by $\mathcal{H}$. Suppose that $D$ is decomposable into a number of irreps where the multiplicity of certain irreps is greater than 1. Let ${H}$ be the Hamiltonian describing the system. Suppose $$ D(g) {H} = {H} D(g), \forall g \in G.$$ $H$ is then an intertwining map for $D$ with itself so by Schur's lemma, $H$'s action on each irrep is proportional to identity.

Multiplicity and Hamiltonian action. Suppose an irrep $J$ occurs $n_J$ times in the decomposition of $\mathcal{H}$. Let $\mathcal{H}_J$ be the subspace associated with $J$. We introduce the $n_J$-dimensional ``multiplicity space'' $\mathcal{M}$ via $$\mathcal{H}_J\oplus \mathcal{H}_J\oplus\cdots \mathcal{H}_J=\mathcal{M}\otimes \mathcal{H}_J $$ where there are $n_J$ copies of $\mathcal{H}_J$.

Here's where I get confused in this answer I linked to above (screenshot because mathjax in the copied text isn't working...):

Part of mike stone's answer.

Specific questions:

  1. What exactly is $H_{mn}^J$? Is it related to a matrix representation of $\hat{H}$ in terms of the basis to $|n, i\rangle \in \mathcal{M}\otimes \mathcal{H}_J$?

  2. What exactly is $D_{ji}^J(g)$?

  3. How does one prove the claims that mike stone makes leading to the last two equations in the answer: $$\hat{H} |n,i\rangle = |m,i\rangle H_{mn}^J, \\ D(g)|n,i\rangle = |n,j\rangle D_{ji}^J(g)? $$ Is the proof related to the argument at the top of page 7 of Noah Miller's notes showing that if $[H, (\pi \oplus \pi)(g)]$ for all $g\in G$,then $H$ must have the following form $$ H = \begin{pmatrix} A \mathbb{I} & B\mathbb{I} \\ C\mathbb{I} & D\mathbb{I} \end{pmatrix}, $$ where $A, B, C$ and $D$ are scalars?

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  • $\begingroup$ It is the same statement as in Miller's notes. Mine is just a a bit more explicit, I think. That that decompostion leads to the operator $\hat H$ commuting with the $\hat D(g)$ group operators is surely obvious (the matrices act on different indices). The converse is also true but takes more effort. My own proof is rather long and uses the group algebra. $\endgroup$
    – mike stone
    Commented May 13 at 12:04

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