Hubbard–Stratonovich transformation

Remarks

The identity is the Gaussian integral $\int_{\R^N} e^{-\frac12 y^T B y + x^T y} \, dy = (2\pi)^{N/2} (\det B)^{-1/2} e^{\frac12 x^T B^{-1} x}$ with $B = A^{-1}$, which is proved by completing the square. Example: the Curie–Weiss model with $N$ spins $\sigma_i = \pm 1$ and $M = \sum_i \sigma_i$ has partition function $Z = \sum_\sigma e^{\beta M^2/(2N)}$. With $a = \beta / N$, the identity and $\sum_\sigma e^{y M} = (2\cosh y)^N$ give $$Z = \sqrt{\frac{N}{2 \pi \beta}} \int_\R \exp\Big( N \Big[ -\frac{y^2}{2\beta} + \ln(2 \cosh y) \Big] \Big) dy.$$ By Laplace's method the integral is dominated by the maximum, where $y/\beta = \tanh y$; with $y = \beta m$ this is the mean-field equation $m = \tanh(\beta m)$. Non-example: if $A$ is not positive definite, the right side is undefined or the integral diverges; a negative quadratic form $e^{-a x^2/2}$ is instead written with an imaginary coupling, $e^{-a x^2/2} = (2\pi a)^{-1/2} \int_\R e^{-y^2/(2a) + i x y} \, dy$.