Consider the one dimensional Heisenberg Hamiltonian of the form \begin{equation} H = - \sum_{<i,j>} J_{ij} \mathbf{S}_i \cdot \mathbf{S}_j \end{equation} with nearest neighbour interactions. Assume that we have a chain with $N$ spins and that the distance from site to site is $a =1$. Let us assume that $J_{ij}$ is randomly distributed and can only take the integer values $\{-1,1\}$. Furthermore, let us say that the spin at site $i=1$ and at $i = N$ is pointing in the same direction (up).

Now, I would naively think that the ground state of this Hamiltonian would be an alternating chain of spins. That is each spin will either point up or down depending on the coupling $J_{ij}$.

However, in literature on spin glasses people often say that spin glasses arise due to competing ferromagnetic and antiferromagnetic interactions. I can imagine that this competition also exists in the above Hamiltonian, perhaps because we have two boundary conditions.

To put my question(s) bluntly:

  • Can the Hamiltonian above have a spin glass ground state? Or is it simply alternating up and down?
  • If there is a spin glass ground state: is there a way to determine this ground state given that we know all of the exchange couplings $J_{ij}$?
  • $\begingroup$ If each spin can be pointing up or down, than the spins are classical. In this case ground state can be explicitly expressed in terms of $J_{ij}$. After this you can take your definition of spin glass and check if it is fulfilled. No? $\endgroup$ – Gec May 11 '18 at 16:54
  • $\begingroup$ And there are no competing interactions in this Hamiltonian. Each energy term ${\bf S_i S_j}$ can be satiated independently of others. $\endgroup$ – Gec May 11 '18 at 16:59
  • $\begingroup$ Maybe you are right, but I'm not so sure. Consider the case where $N = 4$. If $J_{12} = +1$, $J_{23} = +1$ and $J_{34} = -1$. In this case the boundary conditions at site $i = 1$ and site $i = 4$ leads to that the spin at site 3 (or alternatively at site 2) becomes frustrated (it doesn't know where to point). $\endgroup$ – MOOSE May 11 '18 at 17:07
  • $\begingroup$ Yes, you are right, there can be frustration due to fixed boundary condition. But this effect is weak, ground state degeneration is only equal $N-2$. If you do not fix boundary conditions there will be no degeneration. In systems with strong competition degeneration is exponentially large $\sim e^{Ns_0}$ $\endgroup$ – Gec May 11 '18 at 17:24
  • $\begingroup$ What do you mean by "ground state" in that case? Do you assume an infinite field is applied to the spins at the boundary? $\endgroup$ – Norbert Schuch May 11 '18 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.