Trying to understand the ground state of the AKLT model, and having trouble parsing the paper. Here's how it's described in the famous AKLT paper, and the corresponding Hamiltonian:pic.twitter.com/wv9iEBF1k5
Show this thread
-
So in the case of n = 3 spin 1's, represented as 6 spin 1/2s, the ground state would be in P^{otimes 3} * otimes s otimes s otimes *, where P projects on the symmetric subpace of two spin 1/2s, s is the usual spin singlet ^v-v^, and the * states are arbitrary spin 1/2 states.
It seems unclear why the ground state is two-dimensional, since the end spins both have 2 unspecified degrees of freedom.
-
-
No, your intuition is right the ground state space is approximately four dimensional for a long, open-boundary AKLT chain. For a periodic ring there is a unique ground state.
-
Approximately? Is it sometimes not exactly 4?
- 3 more replies
New conversation -
-
-
What you said about constructing product of S=1/2 singlets then projecting on the symmetric sub space of pairs is exactly right. The crucial thing (which I think you already see but for other readers) is that the spins acted on by the projector each come from *different* singlets
-
The upshot about this grouping and projection is that out of every four contiguous spin 1/2’s (every neighboring pair of S=1’s) two of the S=1/2’s must always be oppositely aligned because of the singlet structure. So there’s no way for the S=1 spins to add to 2.
- 1 more reply
New conversation -
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.