3
$\begingroup$

I am reading the paper from X. Wen about quantum orders and symmetric spin liquids. It can be found here: https://arxiv.org/abs/cond-mat/0107071

The Hamiltonian he is writing about looks like this: \begin{align} H_{MF} = \sum_{<ij>} \Psi_i^\dagger \, U_{ij} \Psi_j + \text{h.c.} + \text{const.} + \text{Lagrange-multiplier terms} \end{align} where \begin{align} \Psi_i^\dagger := (f_{i\uparrow}^\dagger,\, f_{i\downarrow} ) \end{align} and $U_{ij}$ is some complex $2\times 2$-matrix. A Gauge-transformation $G$ is a transformation of the form $\Psi_i \to G(i)^\dagger\Psi(i)$, where each $G(i) \in SU(2)$. A symmetry-transformation $T$ is of the form $\Psi_i \to \Psi_{T^{-1}(i)}$, where (I guess) $T$ has to be a bijective function from the lattice to the lattice.

He introduces something called projective symmetry group (PSG). Its elements are pairs $(G_T,T)$ of a Gauge-transformation together with a symmetry such that the Hamilton operator is invariant under the transformation $G_T T$. This means $(G_T,T) \in PSG$ has to satisfy \begin{align} \sum_{<ij>} G_TT(\Psi_i)^\dagger\, U_{ij}\, G_TT(\Psi_j)&= \sum_{<T(i)T(j)>} \Psi_i^\dagger\, G_T(T(i))U_{T(i)T(j)} G_T(T(j))^\dagger\, \Psi_j^\dagger \\&=^! \sum_{<ij>} \Psi_i^\dagger \, U_{ij} \Psi_j \end{align} Thus, \begin{align} G(T(i))U_{T(i)T(j)}G(T(j))^\dagger = U_{ij} \end{align} I do understand the definition. And I hope I got it right, how Gauge transformation and symmetries act on the basis. But I don't understand why it is calle a group.

So basically I am looking for the group operation \begin{align} \cdot: PSG \times PSG \to PSG, \qquad (G_T,T) \cdot (G_S, S) = ?! \end{align} It is nowhere mentioned, and I searched a lot. It is probably very easy, but I just can not figure it out. I tried $(G_TG_S,TS)$ and $(G_TG_S, ST)$. Both are of the right form, Gauge and symmetry, but they don't seem to leave the Hamilton invariant. At least I can't see it. I know, that e.g. $G_TTG_SS$ will not change $H_{MF}$, but it is not of the form $(\text{Gauge}, \text{Symmetry})$. So what to do?

$\endgroup$
2
$\begingroup$

Don't bother. I think I found the answer. I will write it down here, cause it is hardly found anywhere. Maybe it helps someone in future.

\begin{align} \cdot : PSG \times PSG \to PSG, \, \qquad (G_T,T)\cdot(G_S,S) := (G_TTG_ST^{-1},TS) \end{align}

This has the required form, since \begin{align} G_TTG_ST^{-1} (\Psi_i) = G_T(i)^\dagger G_S(T(i))^\dagger \, \Psi_i \end{align} is a Gauge transformation. It is two lines to check associativity. And inverse elements are given by \begin{align} (G_T,T)^{-1} = (T^{-1}G_TT,T^{-1}) \end{align} where the neutral element is $(id,id)$. This also makes $IGG$ a subgroup and $SG = PSG / IGG$ as claimed in the paper. So I am quite happy now. :)

$\endgroup$

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.