The AKLT in a spin-1 Heisenberg chain can be realized when we introduce the bi-quadratic exchange interaction in addition to the bi-linear interaction. I would like understand this interaction more physically and where do such high order exchange interaction originate from. For example like a cartoon picture of the electrons and the orbitals of this bi-quadratic interaction?
4
-
2$\begingroup$ I think more context is needed here... $\endgroup$– Kyle KanosCommented Oct 29, 2015 at 11:48
-
$\begingroup$ I mean the Haldane phase in a spin-1 Heisenberg chain can be realized when we introduce the bi-quadratic exchange interaction in addition to the bi-linear interaction. I would like understand this interaction more physically and where do such high order exchange interaction originate from. $\endgroup$– user96952Commented Oct 29, 2015 at 12:26
-
$\begingroup$ So add that (and maybe your current thoughts/understanding as well) to the body of the post, rather than a comment. $\endgroup$– Kyle KanosCommented Oct 29, 2015 at 12:36
-
1$\begingroup$ Comment to the post (v1): Consider spelling out abbreviations. $\endgroup$– Qmechanic ♦Commented Oct 29, 2015 at 14:49
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
You don't need the biquadratic interaction to realize the Haldane phase. The Heisenberg spin-1 chain already does the job. Adding the biquadratic term allows one to have a parent Hamiltonian for the exact AKLT wavefunction. Here is a phase diagram for the bilinear-biquadratic Hamiltonian taken from http://arxiv.org/abs/0806.1839, with the following Hamiltonian
$ H=\sum_i \cos\theta\mathbf{S}_i \cdot\mathbf{S}_{i+1}+\sin\theta (\mathbf{S}_i \cdot\mathbf{S}_{i+1})^2$
[![Phase diagram]](https://i.sstatic.net/OuJWN.png)
answered Oct 29, 2015 at 15:59
-
$\begingroup$ Hello Meng, This partially answers my question. I am aware that introducing the bi-quadratic term gives us an SO(3) symmetric Hamiltonian for which there is an exact MPS representation i.e. the AKLT in this case. I should have been more careful with the terms and used AKLT instead of Haldane in my question. However, what I am really interested is to see something like a cartoon picture of this bi-quadratic interaction in terms of something like the electrons and the orbitals? $\endgroup$ Commented Oct 30, 2015 at 8:03
-
$\begingroup$ @user96952 Just go one more order in the t/U expansion and you will get bi-quadratic interaction. $\endgroup$ Commented Oct 30, 2015 at 19:53
-
$\begingroup$ @Everett You: Thanks, do you have any good reference on this? I am unsure where to look at. $\endgroup$ Commented Nov 2, 2015 at 13:00