# How to determine the off-diagonal term of magnetic susceptibility tensor from fluctations?

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?

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.