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Background:

The Quantum Dimer Model is a lattice model, where each configuration is a covering of the lattice with nearest-neighbour bonds, like in the figure on the left:

enter image description here

Each of the configurations is a basis state in a Hilbert space and they are all mutually orthogonal. A 2x2 patch of spins is called a plaquette. A plaquette is flippable if it contains two parallel bonds (like the green plaquette in the picture; the red one is not flippable).

For each state, one can define two winding numbers, by considering an arbitrary cut through the torus (dashed line in the figure) and counting how many dimers cross the line with a black dot on one side minus the number of dimers the cross the line with a white dot on that same side. In the example above, the winding number in both the horizontal and vertical direction is zero. This decomposes the Hilbert space (on the $L \times L$ torus) into $(L+1)^2$ sectors (each winding number can take values from $-\frac{L}{2}$ to $+\frac{L}{2}$.

The Hamiltonian of the system is

enter image description here

The first term (the potential) gives an energy V to flippable plaquettes, while the latter (kinetic) term flips it. One can check, that the Hamiltonian does not change the winding number, i.e.

$$ \langle b \rvert H \lvert a \rangle = 0$$

for all $a$ and $b$ with different winding numbers. Within each of the sectors though, it is commonly stated that $H$ is ergodic, meaning that for any $c$ and $d$ with the same pair of winding numbers, there exists an $n \in \mathbb{Z}$, such that

$$ \langle c \rvert H^n \lvert d \rangle \neq 0$$

This is stated for example in the original paper:

Any two dimer configurations with the same winding numbers can be obtained from each other by repeated application of the Hamiltonian; no local operator has matrix elements between states of different winding number. The winding numbers therefore label the disconnected sectors of Hilbert space.

One can also find the statement in this (very-well written) review:

For the square lattice, the kinetic term is believed to be ergodic in each topological sector. In this case, there is a unique ground state for each topological sector given by the equal amplitude superposition of all dimer coverings in that sector.

I don't think this statement is true: Consider the two configurations in the middle and on the right hand side of the first figure and convince yourself of the fact that both of them have winding numbers 2 (horizontal) and 2 (vertical). However, there is no flippable plaquette in either of them, meaning the Hamiltonian is identically zero on both! Moreover, I can create many such isolated configurations by swapping around the shaded staircases, in fact, the right configuration was obtained from the middle by swapping the yellow and blue staircases. This way I can create $2^{L/2}$ isolated states (because there are $\frac{L}{2}$ staircases and for each of them I can choose the vertical or horizontal orientation. They won't all have the same winding number, but they are definitely isolated from all other states in the same sector).

The reason for me to ask this question is that the above Hamiltonian, at the point t=V=1, naturally came up in my research. At that point, which is also called the Rokhsar-Kivelson point, there is one ground state for each set of states connected by Hamiltonian in the sense I described above. By the above construction, it is exponentially degenerate.

My question:

People seem to ignore those isolated states and claim that the real degeneracy is $(L+1)^2$. Are those states physically not relevant? Why can we neglect them?

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  • $\begingroup$ I think that, for context, it would be helpful if you could reference a paper (or two) where they make the claim about ergodicity. $\endgroup$ – Ruben Verresen Aug 3 '17 at 12:19
  • $\begingroup$ Are these states actually well-defined on the torus though? $\endgroup$ – Dominic Else Aug 3 '17 at 22:05
  • $\begingroup$ @DominicElse His two colored pictures are on a torus, or am I missing something? $\endgroup$ – Ruben Verresen Aug 3 '17 at 22:18
  • $\begingroup$ @RubenVerrensen I guess you are right. I was wondering if the pictures are well-defined if you try to define them on a larger torus. But now I see that they are (I think the length of the torus has to be even?) provided that you don't worry about matching the colors (which obviously are not actually part of the dimer configuration). $\endgroup$ – Dominic Else Aug 3 '17 at 22:37
  • $\begingroup$ @DominicElse These states are entirely well-defined on any square even-sized torus: Each of the colored stripes defined a diagonal "staircase" if you connect all the vertices within, and it is connected with itself across the boundaries. On each staircase, there are two ways of putting the dimers (on the horizontal or the vertical bonds), which is completely independent of the other staircases. --- If the size is not square, several diagonals are connected along the boundary and thus not diagonal, I guess the number of independent diagonals should be the gcd. $\endgroup$ – Norbert Schuch Aug 4 '17 at 12:15
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Look at this paper: PHYSICAL REVIEW B 69, 064404 2004 It says that most of the states are ergodic EXCEPT the states with no flippable plaquettes

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  • $\begingroup$ I think this barely meets the bar of not being a link-only answer. That being said, it would definitely be a much better answer if it gave more details about the content of the paper and how it applies to the question. $\endgroup$ – David Z Aug 29 '18 at 7:22

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