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Suppose the coupling between two spins is $C_{i,j}<0$, then the classical partition function is given by $$Z=\sum_{\{s_i\}}e^{\sum_{i,j}s_iK_{ij}s_j+h\sum_{i}s_i}$$ where $K_{ij}=-\beta C_{ij}$ and $h=\beta H$ After Hubbard-Stratonovich transformation and some simple manipulations we get $$Z=\frac{1}{\sqrt{\mathrm{det}(4\pi K)}}\int D\phi e^{-\sum_{i,j}\phi_iK_{ij}\phi_j+h\sum_{i}\phi_i+\sum_{i}\mathrm{ln}(\mathrm{cosh}(2\sum_j K_{ij}\phi_j))}$$ where the spin configurations have been summed over.

When trying to apply low $T$ perturbation, $K_{ij}$ is large and positive. Then normally people just assume that strong fluctuations will be suppressed, or more precisely it is assumed that $|\phi_i|<<1$ and that the spatial profile of the field is smooth. Then based on these assumptions, the standard Landau mean-field expansion with a ${\phi}^4$ term can be obtained.

My question is: After the transformation, the variables $\phi_i$ are supposed to be randomly fluctuating numbers that can take arbitrary values. However,the matrix $K$ is obviously not positive definite and as a result some large $\phi$ can contribute significantly to the integration instead of being suppressed. Furthermore, based on this same argument, there seems to be no reason for the fluctuating field to be smooth. Where does my argument go wrong?

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The matrix $K$ doesn't have to be positive-definite, its eigenvalue spectrum just needs to be bounded below. Then we can simply shift K by a constant in order to bring its entire spectrum above zero, so that the new matrix $K'$ is positive definite. Since $s_i^2 = 1$, this simply shifts the Hamiltonian by a constant and rescales the partition function by a constant, which does not affect any observables.

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