In this work, graphene-based systems that are described by mixed spin-3/2 and spin-5/2 are studied using the . A diagram of the structure is shown bellow:

enter image description here

The Hamiltonian used is:

\begin{equation} \tag{1} {\small {H_I} = - J\sum\limits_{\left\langle {i,j} \right\rangle } {{S_i}{\sigma _j}} - {J_\sigma }\sum\limits_{\left\langle {i,j} \right\rangle } {{\sigma _i}{\sigma _j}} - {J_S}\sum\limits_{\left\langle {i,j} \right\rangle } {{S_i}{S_j}} - {K_v}\left( {\sum\limits_i {S_i^2} + \sum\limits_j {\sigma _j^2} } \right). } \end{equation}

Here $\left\langle {i,j} \right\rangle$ refers to the sum over the nearest neighbors pairs, $\left[ i,j \right]$ means sum over the next-nearest neighbors pairs, $J$, $J_\sigma$ and $J_S$ are the exchange interaction constants between sites $\sigma−S$, $\sigma−\sigma$ and $S−S$, respectively (see diagram above), and $K_v$ is the crystal field anisotropy constant. The spins moments can take values $\sigma = ±3/2,±1/2$ and $S=±5/2,±3/2,±1/2$.

How are $J$, $J_\sigma$, $J_S$, $\sigma$ and $S$ chosen?

  • $\begingroup$ It depends what behavior you are interested in. Are you comparing to an experimental system? Another numerical model? $\endgroup$ Mar 23 at 17:55
  • $\begingroup$ You would consider S to be the number of unpaired electron on your magnetic atom, typically. For eg, if you have an Fe atom with a moment of 3 bohr magneton, S would just be 3*(1/2). If you can't eyeball 'S', you can simply infer it from your DFT calculation. All other parameters in your question can be estimated through first-principles, through DFT. I just answered a question regarding the calculation of Js, hopefully that points you in the right direction: mattermodeling.stackexchange.com/questions/1548/… $\endgroup$
    – Xivi76
    Mar 23 at 20:23
  • $\begingroup$ @Xivi76 As this is one of our longest standing unanswered questions, do you think you'd be able to turn this into an answer at some point (if you have time)? $\endgroup$ Apr 18 at 2:59
  • 1
    $\begingroup$ @Xivi76 If you do have something more in mind, I think this question would benefit from a full answer. $\endgroup$
    – Tyberius
    Oct 19 at 20:51
  • $\begingroup$ @taciteloquence why don't you answer it? Since you know what the answer depends on, then you can pick one and answer it based on that! This question has gone unanswered for 16+ months! $\endgroup$ yesterday

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