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Consider an Ising system in an external field, which is different at different sites. The Hamiltonian of the system is given by

$H = -J\sum_{<i,j>}^{}s_i s_j - \sum_{i}^{} h_i s_i$

Here each spin can take two values ($\pm 1$) and $J > 0$ and finite.

I want to write a mean field hamiltonian for this system by taking $s_i = m_i + (s_i - m_i)$, where $m_i$ is the average value of magnetisation at site $i$, and keeping leading order terms in $s_i$.

Substituting the required approximations, I get a hamiltonian of the form

$H_{MF} = -J\sum_{<i,j>}^{}[m_i s_j + m_j s_i - m_i m_j] - \sum_{i}^{} h_i s_i$.

My first question is how to reduce it further to represent a true mean field hamiltonian.

Once $H_{MF}$ is determined, we can write $F = -k_BT \ln Z$, and then by taking derivatives with respect to $h_i$ at $h_i = 0$, we are supposed to get N self-consistent equations for $m_i$. I couldn't figure out how to proceed with this.

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    $\begingroup$ What did you try? This way it looks like "do my work". $\endgroup$ – Norbert Schuch Dec 29 '18 at 17:57
  • $\begingroup$ @Diptanil Roy I am curious: were you given this problem as homework in a class? Or is it part of your research? Or is it just a curiosity? $\endgroup$ – N. Steinle Jan 31 '19 at 0:52
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I could be missing something here, but I think that the average magnetization per site, $m$, should not be indexed, since it is a macroscopic quantity (and an intensive variable). Rather, since the $H$ field is non-uniform it couples with the spins at each site differently, which effects the magnetization via $m = \langle s_i \rangle$.

Since the magnetic field is not uniform (varies from site to site), then the result will not be as clean looking as when the field is uniform. In any case, starting from what you stated,

$$ H_{MF} = -J\sum_{<i,j>}^{}[m s_j + m s_i - m^{2}] - \sum_{i}^{} h_i s_i$$

we need to distribute the sum and the $-J$ across the terms in brackets to obtain

$$ H_{MF} = J(m^{2}Nd) - 2Jmd \sum_{i} s_{i} - \sum_{i} h_i s_i$$

where $N$ is the numher of lattice sites and $d$ is the number of nearest neighbors per site (assuming you're using a rectangular lattice with a wrapping). Some more algebra yields,

$$ H_{MF} = J(m^{2}Nd) - \sum_{i}\Big[ 2Jmd - h_{i} \Big]s_i$$

$$ H_{MF} = J(m^{2}Nd) - \sum_{i} h_{eff,i} s_i$$

where we have our effective magnetic field which varies across the lattice, $ h_{eff,i} = 2Jmd - h_{i} $.

Now, the partition function for each site has a Boltzmann factor for each spin state possible, since each spin state has an energy. But your effective $H$ field is indexed per site, so this seems kind of awkward. We must be careful:

$$ Z =\prod_i \sum_k e^{-\beta H_{MF}} \propto \prod_i \Big( \sum_k e^{\beta (h_{eff,i})} \Big)_i$$

where the product is due to the sites not interacting with each other directly, $i \in \{1,2,3,...N\}$ and $k \in \{1,2\}$ for the two spin states.

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    $\begingroup$ I was under the assumption that it should still be $m_i$ instead of $m$, because the net magnetisation would vary at each site. Shouldn't it? $\endgroup$ – Diptanil Roy Oct 3 '18 at 15:36
  • $\begingroup$ @DiptanilRoy I have edited my answer to reflect your comment - I did not mean to have the $m$ indexed. $\endgroup$ – N. Steinle Oct 4 '18 at 1:20
  • $\begingroup$ @DiptanilRoy What argument do you have that $m$ should have an index? $\endgroup$ – N. Steinle Oct 5 '18 at 17:57
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    $\begingroup$ Why do you claim $m_i=\langle s_i\rangle$ would be site-independent, given the site-dependence of the magnetic field $h_i$? $\endgroup$ – Norbert Schuch Dec 29 '18 at 14:37
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    $\begingroup$ The question is what $\langle\cdot\rangle$ means. Is it an average over sites? No. It is the average value of $s_i$ in the thermal state. As there is no average over $i$ involved, it will depend on $i$ -- unless the Hamiltonian has a symmetry such that the value doesn't depend on $i$ (such as for translational invariance), which is not the case here. --- P.S.: Please use @[username], otherwise I don't get notified. $\endgroup$ – Norbert Schuch Dec 29 '18 at 17:56
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The mean-field equations are just conditions of self-consistency. You assume that all momenta expect the one $s_i$ are in their equilibrium state $\langle s_j \rangle = m_j$ and obtain the effective Hamiltonian for $s_i$. $$ H_{\text{MF}}\left(s_i,\{m_j\}_{j \neq i}\right) = - h_is_i - J \sum_{\langle i,j \rangle} s_i m_j = -h^{\text{eff}}_is_i. $$ You calculate thermodynamic average $m_i = \langle s_i \rangle$ with respect to effective Hamiltonian and demand all the $m_j$ to satisfy the obtained condition. You end up with the following system of equations. $$ m_i = \tanh \left[\beta h^{\text{eff}}_i \right], \quad h^{\text{eff}}_i = h_i + J\sum_{\langle i,j \rangle} m_j. $$


P.S. If you are actually interested in finding a solution, I would recommend to use iterative method. I would expect that when $\beta J$ is small enough you the following scheme would converge. $$ \begin{aligned} m_i^{[0]} &= \tanh \beta h_i, \\ m_i^{[1]} &= \tanh \beta \left(h_i + J\sum_{\langle i,j \rangle} m_i^{[0]}\right) \\ &\dots. \\ m_i^{[k+1]} &= \tanh \beta \left(h_i + J\sum_{\langle i,j \rangle} m_i^{[k]}\right) \end{aligned} $$

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  • $\begingroup$ I want to point out that the question is a poorly formulated. Instead of going along the lines of "I have partially solved the problem, but then I stuck, but I know that you need to take derivate, though I do not know how to proceed" you should clearly state the problem as if it was in a physics text book. $\endgroup$ – David Saykin Oct 7 '19 at 19:58

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