0
$\begingroup$

I am reading literature on quantum spin chains and matrix product states, and I notice similar arguments regarding the spin-1 antiferromagnetic Heisenberg model,

$H_{H} = \sum_i S_i \cdot S_{i+1}$,

and of the AKTL model,

$H_{AKLT}=\sum_i \big(\frac{1}{3}+\frac{1}{2}(S_i \cdot S_{i+1}) + \frac{1}{6}(S_i \cdot S_{i+1})^2 \big)$.

The argument is that: (1) The spin-1 Heisenberg model is difficult to solve. (2) The AKLT model is a slightly modified Heisenberg model and hence the two models can be connected by an adiabatic transformation. (3) The AKLT model is gapped with an exactly solvable ground state, the AKLT state which can be represented as an MPS and that it belongs to the Haldane phase. (4) Therefore, the ground state of the Heisenberg model also belongs to the Haldane phase.

I can say that I understand the theory of adiabatic transformations in regards to identification of quantum phases, so statement (4) is no issue to me if statement (2) is indeed true. I also understand the formalism involved on the AKLT model, i.e. it is a sum of projector operator which makes it possible to construct the ground state by arguments.

However, (Q1) I am wondering what makes the spin-1 Heisenberg model difficult to solve. Since when discussed, at least on recent papers, the discussion eventually leads to the AKLT model.

And, (Q2) if statement (2) is really that trivial as often discussed in literature. Because we can define a family of Hamiltonians with three parameters:

$H(p_1,p_2,p_3) = \sum_i \big( p_1 + p_2(S_i \cdot S_{i+1}) + p_3 (S_i \cdot S_{i+1} )^2 \big)$

but there are many ways to define a transformation path from $H_{AKLT}$ to $H_H$, and whether a path is adiabatic is another concern. Hence, simply saying $H_{AKLT}$ and $H_H$ can be connected by an adiabatic path seems to be non-trivial.

I accept that I might be only missing on or misunderstanding key concepts, so if you can point me to them to answer my questions, I would appreciate it.

$\endgroup$
1
$\begingroup$

No, point (2) is not obvious, and indeed there is no proof that this is true, only very strong numerical evidence: The two models can be connected through a path along which the Hamiltonian is gapped (as found through very strong numerical evidence, not analytically), which is generally taken as a definition of being in the same phase (among others, it allows you to connect the two quasi-adiabatically, although also that is highly non-trivial to formalize properly).

Note that the 3-parameter family you give is in fact a 1-parameter family, up to rescaling and a constant offset -- generally known as the "bilinear-biquadratic model" -- along which the interpolation is done.

Regarding your first point, almost all models are hard to solve. It would be surprising if a certain model was easy to solve.

$\endgroup$

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.