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Consider the Heisenberg model in zero external magnetic field in the language of Ashcroft & Mermin's book: \begin{equation} \mathcal{H} = -\frac{1}{2}\sum_{\boldsymbol{R}\neq \boldsymbol{R}'}J(\boldsymbol{R}-\boldsymbol{R}')\boldsymbol{S}(\boldsymbol{R})\cdot\boldsymbol{S}(\boldsymbol{R}') \end{equation} On page 715 they introduce the mean field approximation by saying (I'm not including the external field, they do in the book):

Suppose that in the Heisenberg hamiltonian (33.4) we focus our attenton on a particular site $\boldsymbol{R}$ and isolate from $H$ those terms containing $\boldsymbol{S}(\boldsymbol{R})$: \begin{equation} \Delta \mathcal{H} = -\boldsymbol{S}(\boldsymbol{R})\left( \sum_{\boldsymbol{R}\neq \boldsymbol{R}'}J(\boldsymbol{R}-\boldsymbol{R}')\boldsymbol{S}(\boldsymbol{R}') \right) \end{equation} This has the form of the energy of a spin in an effective external field: \begin{equation} \mathbf{H}_{eff} = \frac{1}{g\mu_B}\sum_{\boldsymbol{R}'}J(\boldsymbol{R}-\boldsymbol{R}')\boldsymbol{S}(\boldsymbol{R}') \end{equation}

They then proceed to take the average of the effective field next. What I don't understand is that it seems the interaction energy between any two spins is counted twice by this effective external field. What I would do is to rewrite the original hamiltonian as: \begin{equation} \mathcal{H} = -\sum_{\boldsymbol{R}}\boldsymbol{S}(\boldsymbol{R}) \left( \sum_{\boldsymbol{R}'}\frac{1}{2} J(\boldsymbol{R}-\boldsymbol{R}')\boldsymbol{S}(\boldsymbol{R}') \right) \end{equation} and define the effective field as: \begin{equation} \mathbf{H}_{eff}= \frac{1}{g\mu_B}\sum_{\boldsymbol{R}'}\frac{1}{2} J(\boldsymbol{R}-\boldsymbol{R}')\boldsymbol{S}(\boldsymbol{R}') \end{equation} So that the hamiltonian would be: \begin{equation} \mathcal{H}=-g\mu_B\sum_{\boldsymbol{R}}\boldsymbol{S}(\boldsymbol{R})\cdot \mathbf{H}_{eff} \end{equation} I don't really understand the logic followed by A&M, I think, and I don't understand if the method I would do is wrong or just different.

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  • $\begingroup$ What is the expression given for $\mathcal{H}$ in terms of $\mathbf{H}_{eff}$ given in the book? $\endgroup$
    – Chris Long
    Commented Jul 1, 2021 at 10:42
  • $\begingroup$ @ChrisLong They don't write it. After the approximation, they use the results for the magnetization and susceptibility obtained with free magnetic ions to get the self-consistent equation for $M$, so I guess the hamiltonian they would write would be the same as mine but with their effective field. $\endgroup$ Commented Jul 1, 2021 at 12:12
  • $\begingroup$ I won't answer incase someone else has more insight than me. However, I would agree with you that your Hamiltonian will be the same as theirs except theirs will have an extra factor of a half out front. That said unless $J(0)=0$ then you should add $R'\ne R$ to the summation in $\mathbf{H}_{eff}$ (I think theirs should aslo have this). Additionally, if you are working in mean field theory would then average over $R$ in the Hamilton. $\endgroup$
    – Chris Long
    Commented Jul 1, 2021 at 12:49
  • $\begingroup$ @ChrisLong Yes, the averaging would be the next step. The point is that my effective field is different than theirs and I assume I am missing something, but don't know what. $\endgroup$ Commented Jul 1, 2021 at 14:52

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In my copy of the book, Ashcroft & Mermin write the Hamiltonian as [Eq. (33.4)] $$ \mathcal{H} = -\frac{1}{2}\sum_{\boldsymbol{R}, \boldsymbol{R}'}J(\boldsymbol{R}-\boldsymbol{R}')\boldsymbol{S}(\boldsymbol{R})\cdot\boldsymbol{S}(\boldsymbol{R}'),\quad J(\boldsymbol{R}-\boldsymbol{R}')=J(\boldsymbol{R}'-\boldsymbol{R})\ge0. $$ In the key step of deriving the effective field, they focus on a specific site $\boldsymbol{R}$. It would perhaps be less confusing if they called that site e.g. $\boldsymbol{R}_0$. The first contribution to the effective field acting on $\boldsymbol{R}_0$ is then obtained when $\boldsymbol{R}=\boldsymbol{R}_0\neq \boldsymbol{R}'$, $$ -\frac{1}{2}\boldsymbol{S}(\boldsymbol{R}_0) \cdot \left( \sum_{\boldsymbol{R}'\neq\boldsymbol{R}_0}J(\boldsymbol{R}_0-\boldsymbol{R}')\boldsymbol{S}(\boldsymbol{R}') \right) $$ but if instead $\boldsymbol{R}'=\boldsymbol{R}_0\neq\boldsymbol{R}$, we get $$ -\frac{1}{2}\boldsymbol{S}(\boldsymbol{R}_0) \cdot \left( \sum_{\boldsymbol{R}\neq\boldsymbol{R}_0}J(\boldsymbol{R}-\boldsymbol{R}_0)\boldsymbol{S}(\boldsymbol{R}) \right). $$ Relabel one of the summation indices, use the property on $J$ from (33.4) and sum these contributions and you get the same factor of two as in the book. In principle, there's also a contribution from $\boldsymbol{R}=\boldsymbol{R}'=\boldsymbol{R}_0$ proportional to $J(0)$. I'm not sure if they discard it anywhere, but typically $J(0)$ is considered zero in the Heisenberg Hamiltonian.

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  • $\begingroup$ The problem I have with this is that this way the interaction between $\boldsymbol{S}(\boldsymbol{R}_0)$ and any other site is counted twice in the effective hamiltonian written as $-\mu_B g \sum \boldsymbol{S}(\boldsymbol{R})\boldsymbol{H}_{eff}$ $\endgroup$ Commented Jul 2, 2021 at 18:07
  • $\begingroup$ But that's taken care off by the factor $1/2$ in front, that removes the effect of double counting. In your approach you're effectively only summing over one of the site indices. Maybe it'd be useful for you to explicitly write out the sums for two simple 1D examples with three sites and periodic boundaries: $\mathcal{H}_1=J\sum_{i=0}^2 \mathbf{S}_i\cdot \mathbf{S}_{i+1}$ and $\mathcal{H}_2=\frac{1}{2}\sum_{i,j=0}^2 J_{i,j} \mathbf{S}_i\cdot \mathbf{S}_{i+1}$ where $J_{i,j}=J$ for nearest neighbors and zero otherwise. Unless I made a mistake, the sums and the $H_{eff}$'s. should be the same. $\endgroup$
    – Anyon
    Commented Jul 2, 2021 at 18:40
  • $\begingroup$ Ah, I did make a mistake there. It should say $\mathcal{H}_2=\frac{1}{2} \sum_{i,j=0}^2 J_{i,j} \mathbf{S}_i \cdot \mathbf{S}_j$. Or $\mathcal{H}_2=\frac{1}{2} \sum_{i=0}^2 \sum_{j=0}^2 J_{i,j} \mathbf{S}_i \cdot \mathbf{S}_j$, if separating the summations helps. Also, to be clear, I'm referring to the double counting factor $1/2$ in front of the Hamiltonian. As the 1D example hopefully shows, you can write such sums in different conventions, you just need to be consistent about it. $\endgroup$
    – Anyon
    Commented Jul 2, 2021 at 18:56
  • $\begingroup$ I think the problem isn't the double counting in the original hamiltonian. It seems to me there is an "additional" double counting when constructing the effective hamiltonian as A&M do. I have no problem with equation 33.54 (which is the second eq. in my post) in the sense that I agree the hamiltonian has terms of that type, but it seems to me we double count the interactions if we construct the effective hamiltonian as a sum over $\boldsymbol{R}$ of the terms $\Delta \mathcal{H}$. You can see it with the simpler hamiltonian $\mathcal{H}_2$ (no need to assume nearest neighbor interaction btw) $\endgroup$ Commented Jul 3, 2021 at 10:34
  • $\begingroup$ I think we are $\textit{meant}$ to double count the interactions terms. What we are interested in is constructing a hamiltonian of single independent spins so that each spin feels an effective mean field generated by all other spins. Then we would like to minimize the energy over certain variational eigenstates. If we don't double count, then we construct an hamiltonian in which each spin doesn't feel the complete mean field, but just half of it. Does it make sense to you? $\endgroup$ Commented Jul 3, 2021 at 10:51

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