1
$\begingroup$

Consider an unoriented bipartite quad-graph $\Gamma$ embedded in a a closed surface $\Sigma$ and a state space $\mathcal{H} = \bigotimes_{\text{edges}} \, \Bbb{C}_e^2$ consisting of spins situated on the edges $e$ of $\Gamma$. Equip $\mathcal{H}$ with a family local operators $A_v = \bigotimes_{\text{edges}} a_e$ and $B_p = \bigotimes_{\text{edges}} b_e$ respectively attached to vertices $v$ and plaquettes $p$ (i.e. faces) of the graph defined in the following manner: Each tensor factor $a_e$ (respectively $b_e$) act trivially unless $v \in \partial e$ (respectively $e \in \partial p$) in which case it acts by Pauli-matrices $a_e = \sigma_x$ (respectively $b_e = \sigma_z$). Define a hamiltonian $H$ by summing these local operators, specifically

\begin{equation} H \ = \ \sum_{v} \, A_v \, + \, \sum_{p} \, B_p \end{equation}

Each orientation of $\omega$ of $\Gamma$ can be encoded as spin-configuration $\vec{u}_\omega = \bigotimes_{\text{edges}} \vec{u}_e$ in $\mathcal{H}$ where $\vec{u}_e$ is the standard basis vector $\vec{e}_1$ or $\vec{e}_2$ depending on whether $\omega$ agrees or disagrees with the canonical bipartite orientation (i.e. black points to white). To each Kasteleyn orientation $K$ an eigenstate $[ \vec{u}_K ]$ can be constructed by averaging over all spin-configurations $\vec{u}_\omega$ which are gauge-equivalent to $K$; and such an eigenstate is topological in the sense that its eigenvalue depends only on the Euler characteristic $\chi$ of the surface $\Sigma$.

For each edge $e$ consider replacing each spin-factor $\Bbb{C}^2_e$ by a finite dimensional irreducible representation $V_e$ of $\text{SU}(2)$ and forming the more general state space $\mathcal{H} = \bigotimes_{\text{edges}} \, V_e$.

Question: How should the definition of vertex and plaquette operators $A_v$ and $B_p$ be modified so that the corresponding hamiltonian $H$ possesses topological eigenstates ?

regards, A. Leverkühn

Note: The Kasteleyn condition together with the fact that all the faces of $\Gamma$ are quadrilateral implies that $B_p \, \big[ \vec{u}_K \big] \, = \, - \big[ \vec{u}_K \big]$ and consequently

\begin{equation} \begin{array}{ll} H \, \big[ \vec{u}_K \big] &= \ \Bigg( \# \, \big\{ \text{vertices of $\Gamma$} \big\} - \ \# \, \big\{ \text{faces of $\Gamma$} \big\} \Bigg) \, \big[ \vec{u}_K \big] \\ \\ &= \ \chi\big( \Sigma \big) \, \big[ \vec{u}_K \big] \end{array} \end{equation}

$\endgroup$
5
  • $\begingroup$ What is the role of the SU(2) representation here? (Differently speaking, would and $d$-dimensional complex vector space $\mathbb C^d$ do, or is it relevant that it is equipped with an action of SU(2)?) Also, how would you generalize the orientation <-> state mapping in that case? $\endgroup$ Aug 24, 2016 at 12:20
  • $\begingroup$ On a related note, this looks a lot like Kitaev's Toric Code model (expressed in a different language), for which generalizations to general finite groups $G$ exist (in which case the local vector space has dimension $|G|$). $\endgroup$ Aug 24, 2016 at 12:24
  • $\begingroup$ None --- but eventually I would like the Hamiltonian to be $\text{SU}(2)$-invariant. I have no idea what kind of object ought to generalise the role of these Kasteleyn states. Yes, I'm aware of the $G$-valued version. Thanks :) best, A. Leverkühn $\endgroup$ Aug 24, 2016 at 17:40
  • $\begingroup$ But not even your original Hamiltonian is SU(2) invariant! Why don't you start with looking for a spin-1/2 SU(2) invariant Hamiltonian? --- What is it you don't like about the G-valued version? $\endgroup$ Aug 24, 2016 at 17:59
  • $\begingroup$ Right --- I noticed this right after I replied, sorry. I'm not sure what I want here except to say that I wanted to combine (a variant of) the Kitaev model with $\text{SU}(2)$-spin networks (following Penrose et al.) --- clearly something about the choice of the Hamiltonian needs to bridge the two structures. Thanks for your observations :) best, A. Leverkühn $\endgroup$ Aug 25, 2016 at 1:18

0

Your Answer

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