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I have run a Monte Carlo simulation of the classical Heisenberg model (in the future I am planning to add other interaction terms). I would like to extract information about the property of the system analyzing the fluctuations.

For the magnetic susceptibility, I have used the following,

$$ \chi_{ii} = \beta \langle (\Delta M_i)^2\rangle $$

Is it possible to derive the off-diagonal term of the susceptibility tensor $\chi_{ij}$ from the fluctuations without assumptions on the form of the Hamiltonian?

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I think that the usual derivation of linear response theory should apply. Assume that the effect of a field of magnitude $h$ applied in the $i$ direction ($i$ being $x$ or $y$ or $z$) can be represented as a perturbation term in the hamiltonian $$ H = H_0 - h M_i $$ where $H_0$ is your original hamiltonian, and $M_i$ is the total system magnetization in the $i$ direction, calculated as a sum over all the individual spins. Then the measured value of a possibly different magnetization component $M_j$ in the perturbed ensemble is $$ \langle M_j \rangle_h = \frac{\int M_j\, \exp[-\beta(H_0-h M_i)]}{Q} \approx \frac{\int M_j\, \bigl(1+\beta h M_i\bigr) \exp[-\beta H_0]}{Q} \approx \beta h \langle M_i M_j\rangle_0 $$ where $\langle\ldots\rangle_0$ and $\langle\ldots\rangle_h$ represent averages in the unperturbed and perturbed systems respectively. Here $Q$ is the unperturbed Hamiltonian and I have assumed for simplicity that $\langle M_i\rangle_0=0$ for all $i$.

So, in the above equation, we can identify the proportionality coefficient between $\langle M_j \rangle_h$ and $h$, and this would lead to a formula very closely analogous to the one you are using for diagonal elements of $\chi$: $$ \chi_{ji} = \beta \langle M_i M_j\rangle_0 $$ If the average magnetization in the unperturbed ensemble does not vanish, then $M_i$ should be replaced by $\Delta M_i=M_i-\langle M_i\rangle_0$ and similarly for $M_j$, as you have in your expression. For your classical Heisenberg model, of course, the different magnetization components will be uncorrelated and the off diagonal elements of $\chi$ should be zero. Only if your Hamiltonian has the appropriate reduced symmetry should you get nonvanishing off diagonal components and, in any case, it is expected that $\chi_{ij}=\chi_{ji}$.

It might be advisable to check directly, by also doing perturbed simulations in which small fields $h$ are actually applied in each of the three coordinate directions, separately, and in each case you measure all three components of the average magnetization $\langle M_j \rangle_h$, calculating the relevant three proportionality constants $\chi_{ji}$. It's prudent to try several different values of $h$ and extrapolate to low $h$, to confirm that these perturbed simulations are in the linear regime. Hopefully the results will agree with the fluctuation expression.

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