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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?

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  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Apr 11, 2022 at 12:25
  • $\begingroup$ @Qmechanic thank you for pointing that out, I completely forgot about it! The [link] (scholar.harvard.edu/files/noahmiller/files/…) from my post leads to course notes by Noah Miller from Harvard University entitled: "Representation Theory And Quantum Mechanics". $\endgroup$ Apr 11, 2022 at 12:32
  • $\begingroup$ An example where this happens is during the calculation of adiabatic electronic states of molecules. You can easily find functions that belong to the same irreducible representation. To find the eigenstates you can then diagonalize the hamiltonian in the substance of functions belonging to the same irrep. Symmetry helps to bring the Hamiltonian into block diagonal form where each sub-block corresponds to an irreducible representation of the point group of your molecule and the dimension of a block is given by the number of functions belonging to the same irreducible representation. $\endgroup$
    – Hans Wurst
    Apr 11, 2022 at 13:46

1 Answer 1

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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.

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  • $\begingroup$ Thank you very much, I think I have a much better idea now! Just one follow up question regarding your last sentence: When I diagonalize $H^J_{nm}$ (assuming it is diagonalizable), the values along the diagonal should be the $n_J$ different energy eigenvalues corresponding to the ($n_J$) different copies of $\mathcal{H}_J$, is that correct? $\endgroup$ Apr 11, 2022 at 16:54
  • $\begingroup$ The $H^J_{ij}$ eigenvalues don't need to all be distinct. Degeneracies might hint at a larger symmetry group though, or they could just be accidental degeneracies. $\endgroup$
    – mike stone
    Apr 11, 2022 at 20:11
  • $\begingroup$ Yes, I assumed this is the case. Thanks for clarifying! $\endgroup$ Apr 12, 2022 at 15:01

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