1
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • $\begingroup$ Have you illustrated all these statements with the rotation group you mastered in college? $\endgroup$
    – Cosmas Zachos
    Jan 13 '21 at 16:09
  • $\begingroup$ @CosmasZachos This was explained to me in a rather abstract way. $\endgroup$
    – miniplanck
    Jan 13 '21 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 '21 at 16:13
  • $\begingroup$ I think I got it, thanks @user1620696! $\endgroup$
    – miniplanck
    Jan 13 '21 at 16:17
1
$\begingroup$

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 $g\in G$, $$\rho(g)\phi_n \in V$$

Improve this answer
$\endgroup$
1
  • $\begingroup$ I think I understand now. Thank you very much! $\endgroup$
    – miniplanck
    Jan 13 '21 at 16:22

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.