3
$\begingroup$

In formula of Hamiltonian for Ising model in one dimension we have $J_{ij}$. usually, we take $J_{ij}$ as a constant. In this way it is called homogeneous Ising model. My question is if we take $J_{ij}$ as a variable depends on sites $i$ and $j$ in which $i$ and $j$ are neighbors, then how its physics differs from homogeneous Ising model? is there any property that we can obtain just from inhomogeneous Ising model while we can not obtain it from homogeneous Ising model?

$\endgroup$
  • 1
    $\begingroup$ Take a look at these : en.wikipedia.org/wiki/Spin_glass#Edwards.E2.80.93Anderson_model $\endgroup$ – biryani Oct 6 '16 at 7:36
  • $\begingroup$ Are you asking about the one-dimensional Ising model with random nearest-neighbor couplings? $\endgroup$ – Norbert Schuch Oct 6 '16 at 9:14
  • $\begingroup$ yes,it is about one-dimensional Ising model with random nearest-neighbor couplings $\endgroup$ – Rose Oct 6 '16 at 10:03
  • $\begingroup$ Then I have the random feeling that the answers below are a bit of overkill ... Nevertheless, I don't see where your question is heading. It would help if either (a) you would ask a clearly defined question, or (b) tell us WHY you are asking about this. $\endgroup$ – Norbert Schuch Oct 6 '16 at 12:21
1
$\begingroup$

If $J_{ij}$ is not a constant, you are introducing disorder in your system, and the physics becomes incredibly more complicated. Basically, the system becomes a spin glass.

There are many popular models which consider the case in which $J_{ij}$ is a variable.

  • In the Edwards-Anderson (EA) model, $J_{ij}$ is taken to be a normally distributed random variable and the interaction is only between nearest neighbors (short-ranged interaction). It was suggested by Boettcher that the EA model has lower critical dimension $d_l\simeq2.5$, which means that below this dimension there is no phase transition (in particular, no phase transition would be present in $d=2$ and $d=1$, while the homogeneous Ising Model has no phase transition in $d=1$ but it has it in $d=2$).
  • In the Sherrington-Kirkpatrick (SK) model, $J_{ij}$ is taken to be a normally distributed random variable, like in the EA model, but each spin interacts with every other spin (long-ranged inteaction). This is equivalent to a mean-field approximation; in a certain sense, it is equivalent to the EA model in infinite dimension. An analytical solution was found by Giorgio Parisi using the replica method.
  • In the bond-diluted Ising model, you simply "turn off" some of the interactions (i.e. you set $J_{ij}=0$) with probability $1-p$. Also in this case, there is no phase transition in $d=1$.
$\endgroup$
  • $\begingroup$ in EA model can we find a point for which we have site-dependent interaction? $\endgroup$ – Rose Oct 6 '16 at 10:09
  • $\begingroup$ @Rose Maybe I don't understand your question. In the EA model the interaction is site-dependent: $J_{ij}$ depends on the site indices $i$ and $j$. $\endgroup$ – valerio Oct 6 '16 at 10:18
  • $\begingroup$ that is right.I mean can we achieve any feature different from homogeneous Ising model with this site-dependence? $\endgroup$ – Rose Oct 6 '16 at 11:19
  • $\begingroup$ @Rose Pretty much everything will be different. Just try to compute the partition function: it is an almost trivial task in the case of the hom. Ising model, but it becomes extremely cumbersome in the case of the EA model (actually, I don't think there is an analytical solution) One thing will be the same, though, as I've written in my answer: there is no phase transition in the 1d EA model, nor in the 1d hom. Ising model. $\endgroup$ – valerio Oct 6 '16 at 11:52
  • $\begingroup$ thanks,and what about their correlation function? $\endgroup$ – Rose Oct 6 '16 at 11:56

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.