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0 votes
1 answer
72 views

Is there a name for a Heisenberg-like model, but instead of the ZZ operator, we have one that favor only spin-up-spin-up configurations?

I understand that the Quantum Heisenberg XXZ model in 1D has the form: $$\hat H = \frac{1}{2} \sum_{j=1}^{N} (J_x \sigma_j^x \sigma_{j+1}^x + J_y \sigma_j^y \sigma_{j+1}^y + J_z \sigma_j^z \sigma_{j+1}...
Kim Dong's user avatar
  • 700
0 votes
1 answer
53 views

How to handle Dzyaloshinkii-Moriya imaginary terms in Heisenberg chain?

The DM interaction has three coordinate-specific terms when splitting it up. Two of these, the DM-x and DM-z terms, are imaginary when we transform them into series of raising and lowering operators. ...
Ravi's user avatar
  • 11
2 votes
1 answer
223 views

Jordan-Wigner Transformations on fermionic system

I've been trying to use Jordan-Wigner Transformations on a given fermionic Hamiltonian. The given Hamiltonian is: $$ \hat{H}= -\sum_{m=1}^{N}(J_z \hat{S}_{m}^{z} \hat{S}_{m+1}^{z} + \frac{J_{\perp}}{2}...
hpSauce_'s user avatar
1 vote
0 answers
204 views

Parity of XYZ model ground state

I am considering the XYZ Hamiltonian (with PBC) $$\widehat{H}_{\mathrm{XYZ}}=\sum_{i=1}^{N} \left(\hat{\sigma}_{i}^{x} \hat{\sigma}_{i+1}^{x}+J_{y}\hat{\sigma}_{i}^{y} \hat{\sigma}_{i+1}^{y}+J_{z}\hat{...
Antonio Anna Mele's user avatar
0 votes
1 answer
241 views

The most general $SU(2)$ invariant spin-$1/2$ Hamiltonian on 5 sites

I have periodic chain of spins $s=1/2$. I want to know what is the most general $SU(2)$ invariant and translation-invariant Hamiltonian. My guess is: $$\sum_i (j_1 S_i \cdot S_{i+1}+j_2 S_i \cdot S_{i+...
Frank's user avatar
  • 1
0 votes
1 answer
58 views

Balance the units of the following hamiltonian

The following image is taken from an article and shows the hamiltonian of a spin chain model. I knew that the dimensional units in an equation must balance. To ensure this, the author took a procedure ...
Bekaso's user avatar
  • 233
1 vote
1 answer
615 views

One-dimensional Ising Model in a three spin chain

I have a system of three aligned spins with $S=\frac{1}{2}$. There are interactions between nearest neighbors, and each spin has a magnetic moment. The Hamiltonian of the system is: $$H=J[S_z(1)S_z(2) ...
Eva Martin's user avatar
0 votes
1 answer
298 views

Hamiltonian for a 1D spin chain [closed]

I am trying to implement the Lanczos algorithm to tridiagonalize the Hamiltonian for a 1D spin chain of length $L$, but I am unable to decipher from my professor's notes (here's a link), what the ...
quanticbolt's user avatar
1 vote
1 answer
2k views

Hamiltonian Matrix for XXZ Model

Given the XXZ model Hamiltonian, $H = -\frac{1}{2}\sum^{N}_{i}(\sigma_{i}^{x}\sigma_{i+1}^{x}+\sigma_{i}^{y}\sigma_{i+1}^{y}+\Delta\sigma_{i}^{z}\sigma_{i+1}^{z})$ The two-site Hamiltonian reads $H ...
user570877's user avatar
1 vote
2 answers
1k views

Block diagonalizing a spin-chain Hamiltonian

$\newcommand{\ket}[1]{\left|#1\right>}$ I am learning about exact diagonalization methods, currently following this explanation. My question is in regards to the part where we utilize the fact ...
Kai's user avatar
  • 3,710
2 votes
0 answers
71 views

Change in ground state after perturbing a hamiltonian

Lets consider a spin $\frac{1}{2}$ chain with $n$ spins and an associated local hamiltonian $H= \sum_i h_{i,i+1}$. We also assume that $\|h_{i,i+1}\|_{\infty} \leq 1$. In this question, we will be ...
anurag anshu's user avatar
30 votes
1 answer
6k views

Detailed derivation and explanation of the AKLT Hamiltonian

I am trying to read the original paper for the AKLT model, Rigorous results on valence-bond ground states in antiferromagnets. I Affleck, T Kennedy, RH Lieb and H Tasaki. Phys. Rev. Lett. 59, 799 (...
Cheng Guo's user avatar
  • 629
4 votes
2 answers
779 views

Undefined amplitudes in the Coordinate Bethe Ansatz for the XXX model?

Rather specific question for someone familiar with the Coordinate Bethe Ansatz... I am considering the Heisenberg XXX-model, consisting of a one-dimensional chain of L sites with a spin-1/2 particle ...
ScroogeMcDuck's user avatar