I was studying applied group theory to condensed matter, specifically representations.

As far as I undersand, we can represent elements of symmetry (rotations, for example) by matrix, being a matrix representation. If these matrices are irreducible, then the representation is irreducible.

Now, consider an Hamiltonian, $H$, which is invariant under the action of a symmetry element, $R$, of a group. Then,

$$[\hat{P}_R, \hat{H}] = 0$$

and, if $\phi_n$ is an eigenstate of $\hat{H}$ with energy $E_n$ then $\hat{P}_R \phi_n$ is also an eigenstate $\hat{H}$ with the same energy.

My doubt

According to these results, what does it mean for $\phi_n$ to transform as an irreducible representation? I cannot understand the meaning 'transforming states as an irreducible representation'. To my (little) understanding, only symmetry elements like $R$ should transform according to a said representations.

  • $\begingroup$ Have you illustrated all these statements with the rotation group you mastered in college? $\endgroup$ Jan 13, 2021 at 16:09
  • $\begingroup$ @CosmasZachos This was explained to me in a rather abstract way. $\endgroup$
    – miniplanck
    Jan 13, 2021 at 16:12
  • 2
    $\begingroup$ A representation of $G$ is a map $D(g)$ which for every $g\in G$ associates one linear operator in some vector space $V$ and which reproduces the group composition law $D(gh)=D(g)D(h)$. Saying that an object transform according to some representation means that the object is an element of the vector space on which the group acts according to the representation. $\endgroup$
    – Gold
    Jan 13, 2021 at 16:13
  • $\begingroup$ I think I got it, thanks @user1620696! $\endgroup$
    – miniplanck
    Jan 13, 2021 at 16:17

1 Answer 1


When we say that an eigenstate $\phi_n$ transforms as an irrep $\rho$ of a group $G$, we mean that it belongs to a subspace of the full Hilbert space which is mapped onto itself under the action of $\rho$ (in the present context, this subspace is an eigenspace for a particular Hamiltonian eigenvalue). That is, $\phi_n$ belongs to a subspace $V$ such that for any $\rho(g)$ for $g\in G$, $$\rho(g)\phi_n \in V$$

  • $\begingroup$ I think I understand now. Thank you very much! $\endgroup$
    – miniplanck
    Jan 13, 2021 at 16:22

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