1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.