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?


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.


This seems to be a strange formulation of the fact that $H$ and $G_0$ commute, so there are "joint eigenstates", in particular, each of the $V_\lambda^{(i)}$ (I'm labelling the irreducible representations by $i = 1,\dots,m_\lambda$ here) can be chosen to be an eigenspace of $H$ with energy $E^{(i)}_\lambda$. So, we pick some abstract vector $\lvert E_\lambda^{(i)}\rangle$ and a basis $\lvert v_{\lambda,j}\rangle,j = 1,\dots,d_\lambda$ of $V_\lambda$, and there's an isomorphism from the vector space spanned by $\lvert E_\lambda^{(i)}\rangle\otimes\lvert v_{\lambda,j}\rangle$ for $j = 1,\dots,d_\lambda$ to $V_\lambda^{(i)}$.

Using the tensor product is a bit of a weird notational choice for this - you can do it as there's the isomorphism I've indicated, but usually you'd just pick a basis of the $V^{(i)}_\lambda$ that are eigenvectors of some generators (those in the Cartan subalgebra if we have a Lie group) of $G_0$, and call the resulting basis $\lvert E_\lambda^{(i)},v_{\lambda,j}\rangle$ for whatever eigenvalues $v_{\lambda,j}$ occur in the $V_\lambda$ representation.

  • $\begingroup$ Sorry, I have rephrased my question so that it looks more intuitive. I actually suppose we do not know exactly its eigenvalues. We hope to utilize its symmetry $G_0$ to the full extent, i.e. bring this Hamiltonian into a simpler block-diagonal form so that we could study these simple blocks (and each block no longer possess those symmetries). In you picture, the Hamiltonian is already diagonalized (we know $E^{(i)}_\lambda$). $\endgroup$
    – taper
    Apr 9 '17 at 12:42
  • $\begingroup$ I have finally understood this by looking at the Spin $\mathrm{SU}(2)$. The simplest example I overlooked before is spin eigenstates, which can be labeled by $|n,m\rangle$, with $n\geq 0$, $m=0,\pm 1,\cdots,\pm n$. Theses states are degenerate for the same $n$ but we have block Hamiltonian labelled by $n$, of size $m\times m$. $\endgroup$
    – taper
    May 3 '17 at 14:54

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