1
$\begingroup$

I would like to ask the following in the context of symmetry-protected topological phase.

Consider a class of Hamiltonians parametrized by $\{a_1,a_2,...\}$ denoted by $H(a_1,a_2,...)$. Suppose there is a path $p$ in the parameter space, let me denote this as $\{a\}_{\in p}$, such that the Hamiltonian has the symmetry group $G$.

Will these Hamiltonians along the path $p$, should in general obey $[H(\{a\}_{\in p}),M(g;\{a\}_{\in p})]=0$, such that the symmetry operator $M(g;\{a\}_{\in p})$ depends on the parameter space? I.e. if we represent the symmetry operator as a matrix, its elements will vary along the path $p$.

My guess is, yes, but this might only be because I am thinking of the Hamiltonian in matrix form with elements depending on $\{a\}$. For $[H,M]=0$ to be satisfied, it will require (in general) for $M$ to depend on $\{a\}$ so that elements of $H$ and $M$ multiply nicely so that $[H,M]=0$.

I am only beginning to study representation theory. So my follow up question is, can there be a representation of the symmetry operator that need not depend on the parameters space?

Edit: Massive revision to clarify my question. Due to @Max Lein's answer, I narrowed my question such that I remove question about the ground state and then focus on the Hamiltonian.

$\endgroup$

1 Answer 1

3
$\begingroup$

First of all, I don't think your first and second paragraph are compatible with one another: in the first you assume (by your notation) that the group action $M(g)$ only depends on $g \in G$, but not on the parameters $(a_1 , \ldots)$. But the answer is that this an assumption that you need to make whether the group action depends on the parameters.

Secondly, you falsely assume that ground states need to be eigenfunctions of the group action. In case your ground state is degenerate, a ground state need not be an eigenfunction to the symmetry. And if some of your symmetries are implemented antilinearly, then they need not be eigenfunctions at all. For example, if your Hamilton operator commutes with complex conjugation and has a unique ground state, then you can indeed choose a real ground state $\varphi_{\mathrm{g}} = \overline{\varphi_{\mathrm{g}}}$. But $\mathrm{e}^{\mathrm{i} \vartheta} \, \varphi_{\mathrm{g}}$ is also a ground state that is not an “eigenfunction” of complex conjugation. (I put eigenfunction in quotation marks because complex conjugation is only $\mathbb{R}$-linear, but not $\mathbb{C}$-linear.)

$\endgroup$
2
  • $\begingroup$ Thank you for the respond. I had phrased my question poorly and forgot to mention some of what you brought up, like requiring non-degenerate ground state. I will edit the question for clarity as I would like to lead my question more on to the representation theory side. $\endgroup$
    – git-able
    Commented Oct 18, 2018 at 2:13
  • 1
    $\begingroup$ I still think your question is not precise enough, because you do not distinguish clearly enough between the symmetry group $G$ and the symmetry group action $g \mapsto M(g)$. The answer to your question is immediate once you specify what it means for “$H$ to have the symmetry G”. So if you define your group action $g \mapsto M(g)$ to be independent of the parameters $\{ a \}$, then the answer is no. If you assume the group action depends on $\{ a \}$, then yes. Both situations describe different physical situations, so this is not a case of one being wrong and the other being right. $\endgroup$
    – Max Lein
    Commented Oct 18, 2018 at 5:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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